Transcription
A foreign kind of motion where you have some kind of resistive process in picture. We did get the equation of motion for such kind of systems. Of this one, I'll just quickly revise whatever we did in the previous class. So we said that you will have an equation of motion for these kind of systems in such a way that if you want to make an educated guess about the solution, you would say that probably the solution for these kind of equations is going to be something like this.
Then we plug in the value of x from here, dx/dt, and d2x/dt2. And then we finally found out that alpha can actually have two values, alpha 1 and alpha 2. And we said that the most general solution would be a solution which will consist of both alphas, alpha 1 as well as alpha 2. And then we said that my solution would be something like this.
This is a general solution that we can think of. A1 and A2 are constants that you can decide from your initial conditions. For example, if you are given at time t is equal to 0, this was your position and this was your velocity, then you can calculate the value of both A1 and A2. These are now two only unknown terms in your solution. Apart from that, everything else is known. You know what is the mass of the object, you would know what is the damping coefficient, you also know omega naught, which is your frequency if there would have been no dissipative forces. Omega naught corresponds to the case in which there were no resistive or dissipative forces present in the system. So this, we said, will be your most general solution for the equation which is this equation. And this equation has both resistive forces as well as your restoring force.
Now we said that if you look at the term within the brackets, there could be three possible scenarios. In the first scenario, we said that you can have (b/2m)^2 would be greater than omega naught^2. Or you can also say that your b/2m is greater than omega naught. Omega naught is a frequency, it can't be a negative number. Therefore, we also can say that rather than taking the square of the term, you can simply take the square root on both sides and you can say that b/2m has to be greater than omega naught. We said that in this case, since b is associated with your dissipative forces, omega naught is basically associated with your restoring forces. So in these kind of cases, your dissipative force dominates your restoring force and hence there will be no oscillations in these kind of systems. And these systems are called as overdamped or deadbeat system.
If you look at the solution here, in this particular case, your solution would be A1. I can actually rather than writing A1, I can take first b/2mt out and then it will be A1 e raised to power plus something t. But this plus here and the term within brackets is smaller than this term over here. So your this term will always be negative. Plus you will have A2 e raised to power minus something t. Effectively, your x would be an exponentially decreasing function. You will have e raised to power minus something into t. So you have an exponentially decaying function with time in this case. So this was one case which we called as overdamped or deadbeat system. And this is what happens in your doorstopper at your homes.
Now if you talk about the second case here, let me erase this portion here.
The second possibility that you can have is (b/2m)^2 is equal to omega naught^2. Or you can say that b/2m is equal to omega naught. In this case, both of the roots alpha 1 and alpha 2 would be equal to each other. If you look at the values of alpha 1 and alpha 2, you would see that both of them would be equal to each other. But nonetheless, still your solution in this case also will be an exponentially decaying function. Let's say your distance becomes 0 plus this term also becomes equal to 0. You will have an exponentially decaying function here and here. So your resultant will be an exponentially decaying function in this case also. And hence there will be no oscillations in the second case also, which is called as critical damping. Again, your x(t) would be an exponentially decaying function and hence there will be no oscillations whatsoever.
Now the third case, which is the most interesting case from our point of view and which we will be talking about in a bit detail, is when your (b/2m)^2 is less than omega naught^2. What would happen in that case? So we are going to talk about now the last case.
In the last case, we are saying that your (b/2m)^2 has to be less than omega naught^2. Or you can say that your b/2m is less than omega naught. Remember we said that b is a damping coefficient, but we never said that it's a unitless number. Hence, the unit of b/2m would be same as would be the unit of omega naught. Then only you can add two terms or subtract two terms when they have the same unit. Right? You cannot add distance into velocity. Hence from here, you can say that both b/2m and omega naught, they have the same unit. B naught is damping coefficient, but that does not mean that it's a unitless number.
Now about this particular third case, I can now write my solution x(t). I can take e raised to power minus b/2mt common out. I'll have A1 e raised to power. Since this is going to be a negative number now, because you are saying that your (b/2m)^2 is less than omega naught^2, I would like to keep the terms within the bracket positive. So I can take minus out here itself. And if you take the square root of minus 1, you will get iota here. And then within the brackets, you will get omega naught^2 minus (b/2m)^2 and this term is further multiplied by time. And then you will have a second term here, which will be A2. Again, in A2 also, within the brackets here, I would like to have the positive term. I can take minus 1 square root out. I'll have iota. And then I can write within the brackets omega naught^2 and this time there will be minus sign also from here, minus (b/2m)^2 multiplied by t here. And this would be your expression for underdamped solution. Let me move this cursor up if I can.
This expression here, in this particular case, which will be your underdamped system. Now you are saying that your restoring force is stronger than your dissipative force. And this will be the only case in which you will have oscillations about the equilibrium position. In the case of overdamped and critically damped system, there are no oscillations. Your position simply decays exponentially as a function of time. Which basically means that if you have a pendulum and it will just stop there. There will be no further movement and hence there are no oscillations here.
Now if I want to know that at time t is equal to 0, your pendulum was here. In the case of overdamped and critically damped, at let's say time 2 seconds, where was my pendulum? That's when with these kind of solutions, they come into picture. You can predict at what amount of time, even though your displacement is an exponentially decaying function, but you can still make the predictions about the location of pendulum. In this case, at any point of time, even though there are no oscillations, critically damped and overdamped made whole oscillations where there will be oscillations about equilibrium position. And in this case, underdamped case, even though there are oscillations, but these oscillations are not going to stay for an infinite amount of time. Which basically means that again it will come and let's say in this direction it went up to distance equilibrium position, opposite direction, a distance, a less distance because there are some dissipative forces acting on it again. So maybe it will stop somewhere here itself.
This is a function exponentially and eventually oscillations and finally position. But that happens only in underdamped system.
There is some mathematics here where you can reshuffle these terms here and there, and eventually where you say that this is equal to omega t plus phi. Your x(t) would be given by this expression. Now this omega is not the omega for your undamped oscillation. This omega is basically equal to. Okay, uh, so this omega would be now equal to the omega that I was talking about. Will be equal to the square root of omega naught^2 minus (b/2m)^2. And this frequency is called as your damped frequency.
I lost that page. No, I can't get that page back. Anyway, let's forget about that. It is multiplied by t. So we have defined this term as your omega d or simply omega. Many times it is written with d to denote that this corresponds to a damped frequency. But since we are using omega naught now for undamped case, so you can simply call it omega also. If you look at the numbers here, your omega is always going to be less than omega naught. That would be the first observation from your relation between damped frequency and undamped frequency. That your damped frequency would be always less than undamped frequency. The second observation you can make here is that in the case of underdamped oscillation, even though you are saying your amplitude is going to decay with time, but your frequency does not have any time dependence. Look at the numbers here. Omega naught is constant because omega naught is sqrt(k/m). B is a constant, m is the mass of the spring, that is also constant. So your damped frequency, even though this frequency is less than your undamped frequency, but it is still a constant number. Which basically means that your time period is also constant.
So we talked about these three systems when you have your (b/2m)^2 greater than omega naught^2, in which case you are saying that your damping term is larger than your restoring force.
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And let's say hypothetically three centimeters. What would happen is because there are damping or resistive forces present, let's say air is present in this case. So let's say up to 2.7 centimeters, which would be your this case here. What would happen? It will come back to its equilibrium position. So again, it would go less than 2.7 centimeters. That's it. Probably it will go to 2.4 centimeters. Again, it will come to equilibrium. And let's say this time it will go up to 1.8 oscillations, right? So we are saying that in this case, you have this relation for critically damped, which is given by the red line here. Resistive forces are less than critically damped. Together equilibrium positions.
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Right? So from here to here. So if you look at the amplitude, it looks like amplitude as compared to from here to here, between these two points here, and between these two points here. So years of time period because.
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At time t tends to infinity, ideally, because you are talking about an exponentially decaying function.
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Okay, um, it feels as if as if overdamped position. But as I said before, solutions for the three cases, finally find out, then you will see that in the case of critically damped, your solution comes out in such a way that your system is going to come to its equilibrium position the fastest.
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Right? So you can associate them with dissipative and restoring forces, but you cannot say that they actually represent dissipative and restoring forces. Okay, thank you. Thank you.
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Zero at equilibrium position, right? Now that's what I'm saying, right? Yeah. Happy. So, excuse me.
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Sorry, man. Okay, I thought you were asking a question. So let's revise what I just said that when you have b/2m less than omega naught, that would correspond to your underdamped system. Depending on how much is your b, you will have oscillations here. And we said that is cosine of omega t plus phi. And this omega is your damped frequency, which is given by this expression here. That is going to represent your amplitude. So if you look at the amplitude here, you are saying that basically your amplitude is going to decrease exponentially as a function of time. That means here, cosine terms of oscillations. So you still have oscillations in underdamped system. You still have an associated frequency. You still have a time period also for underdamped oscillation. And your frequency as well as time period, they are constant in underdamped oscillation. Even though your amplitude is continuously reducing as a function of time, but.
Third case, when your b/2m is greater than omega naught, which will be an overdamped system. A public system for displays release oscillate. Again, it's not going to oscillate. It will simply return to its equilibrium position as an exponentially decaying function of time. Now, if you keep on increasing the damping, the time period that your object needs to come back to its equivalent position, that also keeps on increasing. Equilibrium position. So these are three possible scenarios that we did talk about at damped simple harmonic oscillator. And now we were talking about mechanical simple harmonic oscillator. We have already talked about these three curves here. The blue curve represents underdamped motion, red represents your critically damped, and black represents your overdamped system. If you are given three, let's say three different fluids, exactly the same mass and the same spring, if somebody asks in which case, if you pull the spring down, it will come back to its equilibrium position fastest, that will be your critically damped system. In the critically damped system, your system is going to come back to its equilibrium position fastest.
Now we were talking about underdamped oscillation. We said that this first term here represents your amplitude. The second term represents your phase, which is your cosine factor over here. And this represents your amplitude, right? Now A here, if you plug in the value of b is equal to 0. If your b is equal, b is equal to 0. Is equal to dv, that is equal to 0. Again, this solution should reduce to your ideal simple harmonic oscillator.
Which is what we derived for an undamped mechanical simple harmonic oscillator, right? Another point here is that A, here we said this is your amplitude. Here, but A represents the amplitude.
Which would have been your amplitude if there were no resistive or dissipative forces present in the system. System that would have been given by A. Now there are couple of terms defined which is called as your amplitude relaxation time and your energy relaxation time.
Now I want to define a time in which your this amplitude reduces to 1/e of its initial value. That means I would like this amplitude to be A/e. What would be the value of time for that? Can you tell me that? Let's unmute some of the people.
2m/b. Right? That would be your time in which your amplitude would reduce to 1/e of its initial value. And that is called as your amplitude relaxation time. It is defined, sorry for the spelling mistake here. Amplitude relaxation time is defined as the time in which your amplitude decays to 1/e of its initial value. Similarly, there is another term which is called as energy relaxation time. And energy relaxation time, similar to amplitude relaxation time, is defined as a time in which your energy reduces to 1/e of its initial value.
We derived the expression for energy and we said that energy is equal to 1/2 k a^2. Well, K was your spring constant and A was your amplitude. So you can say that your energy is directly proportional to amplitude squared. There is another way to do here also. Like in the case of undamped oscillation, we can calculate the potential energy, 1/2 k x^2, and we can calculate the kinetic energy also. That is, this whole term is your amplitude, right? Amplitude is maximum displacement and that will be given by this expression over here. So if we are saying that your energy is directly proportional to its amplitude squared, in that case, I can say that your energy expression, that would be some constant. If I take care of this term, I'll have e raised to power minus b/m into t. Well, E naught now represents your energy in the case of undamped oscillation.
Factor multiply. Okay, actually now, if I want to. The people who are interested, please go ahead and do this calculation here and see that do you actually get an expression which has e raised to power minus b/m over t in this case also. Please do that. Similar to amplitude relaxation time, energy relaxation time is defined as the time in which energy decays to 1/e of its initial value. Now, if I want to calculate that value of time in which your energy becomes 1/e of its initial value, I would like my E to become E naught over e. And that time in that case would be given by m/b in this case. So this would be called as your energy relaxation time. We have two terms, one for amplitude and one for energy. I think there were couple of doubts.
Excuse me. Yes, oscillations absorbed there. And one did yet amplitude or energy? Yes, ma'am. Okay, okay. Your energy is proportional to amplitude. Okay, okay.
Um, good morning, ma'am. You said critically damped, we will get to the equilibrium position the fastest as compared to all the three cases. An actual example here in which you can see that if you solve the equations here, because we haven't gone into the exact solutions so critically damped and overdamped. So if you solve those equations here, um, let me show you the I have one figure in which it has been actually solved for the three cases. Just give me a second.
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Can you see on my screen? Yes, ma'am. Okay, so here an actual scenario has been taken where your equation has been, for example, underdamped. So if you look at the exact solution here, you're happy. So it looks like initially overdamped, fastly. All right. But after a certain amount of time, you look here. So after the critically damped, here, over time, you can actually solve this equation if you know a bit of programming. Just plug in the value of k, 30, m, 20, and b key value. Always second-order differential equation. Now you can try solving that position as such, but permanently is.
And I'll start in the next class.