Waves and Oscillations4

Foreign, let's start today's class. In this class, we are going to talk about damped oscillations. So far, we have been talking about undamped oscillations, which are your ideal simple harmonic oscillators. There are no realistic kind of scenarios that we face in our day-to-day life. Now, we are moving towards a more realistic case, which will be your damped oscillation. So, you still have oscillations, but these oscillations have some kind of frictional forces which are acting on your object which is moving here. So, what would happen in this case is obvious that your amplitude is going to decrease with time, right? But how your amplitude is going to decrease with time? We would like to have some idea about them. Is your amplitude going to decrease with time as a function of 1 by T? Is it going to decrease as a function of 1 by T square? Or is it going to decrease exponentially? How that decrease is going to happen? We'll get the derivation for that, and these kind of oscillations are again called as damped oscillations.

So, what would happen in this case is that now your energy is going to decrease as a function of time because if your amplitude is decreasing continuously, then your energy is also being dissipated continuously. An example of that kind of oscillations is, for example, you have a vertical spring which is attached here and a mass is attached to that spring, and if it is put inside some kind of liquid, even if it is in air, that is an example of fluid damped oscillation, right? You have a pendulum. If you are doing it not in vacuum, if you are doing it in your lab, that is also an example of damped oscillation.

Now, in the case of damped oscillations, you have some kind of retarding force that is going to act on your object. Now, we are going to make an approximation for that damping force, and we are going to say that this damping force is going to have this kind of expression or this kind of dependence here, where we are saying that your damping force or the retarding force is directly proportional to the negative of velocity. We are saying that more is the velocity, more will be your retarding force, and your retarding force is always going to act in a direction opposite to the velocity of the object. That is why there is a minus sign here. To remove this constant, this sign of proportionality, a constant is introduced, and that constant is B, which is called as your damping coefficient. B can be large, B can be small, depending on the medium in which you are having your damped oscillation.

Now, the question is, is this expression universally true? No, this is not universally true. This expression is most valid when your object is moving in fluids, which basically means that if your object is moving in liquids or gases, in that scenario, this expression is valid. The second assumption is that your object is moving with a velocity which is not very large. Object's velocity is small. That is the second assumption in this case. But this is the best example that we could get for the retarding force in this case.

So, if you take this force also into account in the previous case, when you had an ideal SHM, then you said that your force is equal to minus KX. Now, on top of this restoring force, you have an additional force also. So, now you would say that, just give me a second, someone is very eager to write on the screen. Now, in this expression for force, we have to add another term also, and that another term would be your this retarding force that we have just talked about. So, this would be your expression for force. Now you have a restoring force, but with the restoring force, you also have a retarding force, which is given by this expression over here.

Similar to the previous case of ideal simple harmonic oscillator, we are going to write our acceleration as the second-order derivative of position. That would be equal to minus KX minus B velocity, which is DX by DT. Take everything on one side. This is what you will be getting. You will have M times D2X over DT2 plus B DX over DT plus KX is equal to 0. Divide throughout by M, which is the mass of the object. This is what would be your final expression over here.

Now, these kind of equations, I'm not sure if you are familiar with, but these kind of equations are called as second-order differential equations. And maybe you will read in math, but for the solution of these kind of approximations, you make some kind of educated guess about what could be the possible solution for these kind of equations. And for this particular kind of equations, usually your first guess would be that your solution would be of this form here. It would be some exponential function. I say that this is my solution here, where A is some constant and Alpha is also some constant. Both of these are constants that we don't know as of now. We will determine them later.

Now, if I say that, okay, this is my guess that probably this would be the solution to this above equation here, in that case, if I am saying that this is the solution of this equation, that means this value of X should satisfy this equation, right? So, that I have to take DX by DT and I have to do the second-order derivative of X with respect to time. If you do that, your DX by DT will be A times Alpha e raised to power Alpha T, and your D2X over DT2 will be given by this expression here.

Now, let's go back to this equation and plug in the value of DX by DT and D2X by DT2. Let's see what do we get here. In that case, this is your D2X over DT2, B over M DX by DT is your this expression which is written over here, plus K over M A e raised to power Alpha T is equal to 0. A e raised to the power Alpha T is your X. From this expression, you can take A e raised to power Alpha T common out, right? It's there in all the three terms. If you do that, this is what you will get.

Now, there are two scenarios. Either your this term is equal to 0, or the term within the bracket is zero. If you say that this term is equal to 0, that means your X is permanently equal to 0, which would be a trivial solution, which is not an interesting solution from our point of view. Please, that would mean that the term within the brackets is equal to zero. Excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha 2 has been put here, and you get this expression over here. So, this is going to be now our starting point. I am going to talk about different kinds of damping, damped oscillations, and for each oscillation, this solution is going to be our starting point. I'll start with this value of X. Though I haven't written it, but now I am assuming that you know that this X is now actually a function of time. You know that by default, right? Because you have the time dependence in your solution. So, your X is a function of time here. Yeah, again, this is your solution here.

Now, I'm going to talk about three different cases here. First case, in the case when your this expression B upon 2M square is greater than Omega naught square, right? So, your square root term in both of the cases will be positive terms. This will be also a positive term, and this will also be positive, right? This is a positive term because remember you are adding, you have e raised to power minus B upon 2M T, and then you are adding into something which is less than B upon 2M. Right? If you take the square root of B upon 2M whole square minus Omega naught square, this is always going to be less than B upon 2M. And that's why I'm saying that your this term would be some e raised to power minus expression multiplied by T. Similarly, for your second term also, you have minus your minus here. This is also going to be some exponential function and exponentially decreasing or decaying function. So, if your B upon 2M whole square is greater than Omega naught square, that means remember B we defined as damping coefficient. So, this basically means that your resistive forces or the resistive term dominates your stiffness parameter which is associated with Omega naught. And you will have an over-damped or deadbeat system. This is called as your over-damped system and also called as deadbeat system.

Now, the second case. The second possibility is that if both of these terms are equal to each other. In that case, what would happen is that your solution would be something A1 e raised to power minus B upon 2M T plus A2 e raised to power minus B upon 2M T. Usually, I wouldn't go into the details of that scenario. If your roots are equal, in this case, we are saying that both of your roots are equal. In that case, the solution usually you take is of the form A1 plus A2 T multiplied by E raised to power minus P upon 2M T. So, but I'm not going to go into the details of that. But excuse me, ma'am. [Music] You can hear me now, right? Okay. So, now if you are saying that your this term is equal to 0, which basically means that you are a simple quadratic term is equal to zero, and we all very well are aware of the solution of these kind of equations, right? Our quadratic term equation solution is easily found out. Another thing is that when you will do probably this kind of second-order differential equations in math, you will realize that these equations that you finally get, this quadratic equation, this is called as the characteristic equation, and this is a quadratic equation. If my this term is equal to 0, you can easily get the roots of this equation, right? If you have a simple quadratic equation of this form VX plus C is equal to 0, you say that your X would be equal to minus B, which is this term over here, plus minus b square of this, minus 4 times AC, 4 times A is 1, C is K over M, divided by 2A is 1 here.

So, now this is going to be a solution. You have a plus and minus sign here, right? Another thing that has been done over here, just this 2 has been taken care of in the numerator itself. We'll see that why we have done that. You'll see that very shortly. So, I'll get something minus B over 2M plus minus. If I take 4 common out from here, square root of 4 would be 2. 2 and 2 would cancel each other out, and here you will have B upon 2M whole square minus K over M. The reason we did take 4 out here so that you can get this expression here. K over M, remember from your ideal simple harmonic oscillator, K over M was equal to your Omega square, right? That's how we defined our Omega square. So, that's why we have taken 4 common out from here. If you further go, I can say that there are two roots now. One root will have the plus sign here, and the other root is going to have the minus sign in between. These will be two roots. But I said that the solution of this equation was of this form. Now, of course, either of your Alphas, whether it's Alpha one or Alpha two, I have defined Alpha one where we have retained the plus sign here, and I have defined Alpha minus as Alpha two where you have written the minus sign here, right? So, because these two terms are separate from each other, now you have the solution. Either one of these is going to be a solution. If you say that this term is a solution of your equation that we started with, the second-order differential equation that we started it, which is your this equation D2X over DT2, so the solution of this equation is going to be X is equal to A e raised to the power alpha 1 T. That is going to be the solution. Similarly, you are this another term. And now since we are talking about two different alpha 1 and alpha 2, it's better to define the constants also separately here. So, your this term is also going to be a solution. Both of these terms will be a solution of this equation. You can go back and plug in the value of alpha 1 and alpha 2 here, and you will see that they will satisfy the situation. Both of these terms are solutions. But if you talk about the most general solution, if you have two roots for any equation, then there is something which is called as the most general solution, and that would be a linear combination of both the solutions, which basically means that you can simply add these two terms. Remember in the case of simple harmonic oscillator, ideal simple harmonic oscillator also, we said that your solution can be a sine function, it can be a cosine function, or it could be a linear combination of both functions. We can take any of those as a solution for your ideal simple harmonic case. Similarly, here your first term, I am repeating that again, your first term is a solution, your second term is also a solution. But if you talk about the most general solution, that would be a linear combination of both the terms, and that is why we are saying that this would be now my solution here for that second-order differential equation that we obtained in the case of damped oscillation, where your alpha 1 and alpha 2 are defined here. I'm stressing on alpha 1, alpha 2 time and again because there is some physical relevance attached with these two terms here.

Now, plug in the value of alpha 1 and alpha 2 here. You have alpha 1 multiplied by T. So, you will have e raised to power minus B T upon 2M plus square root of whole of this term. K over M has been replaced by Omega naught square here. I have defined Omega naught as your frequency for undamped oscillation.

I would like to relate my frequency for damped oscillation with this frequency. If I can find some kind of relationship between my frequency for a damped oscillator with an undamped oscillator, that would be a very good expression. So, that is why we have put the value of K over M as Omega naught square, and this is further multiplied by T here. Similarly, the value of Alpha