If you have a spring mass system that you place in an external medium that can provide some kind of a friction or a drag, in that case if you displace the spring mass system from its equilibrium then the kind of motion that you see, one of the possibilities, is that you will get oscillations that will slowly gap over a period of time. The amplitudes of these oscillations decreases over a period of time. This is happening because in the presence of the external medium, the drag force is causing dissipation of energy of the system. Now the spring mass system is a very common example of a harmonic oscillator.
In ideal situations the harmonic oscillators execute, you probably already know it, simple harmonic oscillations. But when we place a simple harmonic oscillator in the presence of some kind of a dissipating medium, where there is friction or drag involved then you end up getting a damped harmonic oscillator and one of the possibilities of motion is right in front of you so in this video i want to take a look at the nature of the forces of a damped harmonic oscillator and i want to obtain the equations of motion solve them and see what possible motions are there so i'm going to talk about systems without any damping under damping critical damping and over damping. So let us start.
So the spring mass system is not the only example of a harmonic oscillator. You can have other systems like a pendulum executing oscillations with the small angle approximation or maybe something else. But I'm going to use the example of a spring mass system placed in some kind of external medium to represent the damp harmonic oscillator that I am talking about.
So let's suppose that you have some kind of a point mass m. that is attached to some kind of a spring system having a spring constant k and this entire system is placed in a external medium that involves some kind of a drag so whenever an object is moving in this medium it experiences a drag force now the motion of this system mass m is along one dimension so i'm representing this by let's suppose the x-axis now if i displace this mass m by some kind of an amount, let's suppose x from the equilibrium configuration, then in that kind of a case, the forces that are going to act on this mass m are two forces. The first force is, of course, the restoring force due to the spring.
So the restoring force due to the spring that acts on the mass m when it is displaced from the equilibrium is directly proportional to the displacement. and the constant of proportionality is the spring constant. It acts in a direction opposite to the displacement.
Now this is a harmonic oscillator but if I place this setup in an external medium which can cause drag whenever motion is happening, then usually majority of these circumstances the drag force that acts upon the mass m is seen to be directly proportional to the instantaneous velocity of the object. Whatever be the instantaneous velocity, the amount of drag force experienced by the mass due to this medium is directly proportional to the velocity. And let's suppose the constant of proportionality is some kind of a positive constant B, where B can be let's suppose the drag coefficient or something like that.
So in a sense these are the two main forces involved in this kind of a damped harmonic oscillator. One is the restoring force due to the spring system and other is the drag force due to the medium present. Now if I represent all of this into the force equation applying the Newton's second law, it simply leads to the equation that whatever the net forces are acting on the system is going to create some kind of an acceleration and since the acceleration is happening along the x-axis, I can represent this as mx double dot.
What is x double dot? It is nothing but the second order time derivative that is the acceleration along the x-axis. Now what is the net force?
First I said the net force is the restoring force kx acting against the displacement so there's a negative sign here and then there is a drag force bv acting against the motion so there is a negative sign here. So if I write down this equation where I bring all the terms together this simply becomes mx double dot plus now b v what is v v is nothing but x dot the time derivative of x right so this can be written as b x dot plus k of x is equal to 0. if i divide the entire equation by m this simply becomes x double dot plus b upon m x dot plus k upon m x is equal to 0. now to make my calculations easier i'm going to substitute these values with certain terms. So what is b here?
b is nothing but the drag coefficient, the constant of proportionality in the drag force. m is the mass of the object. Let's suppose that b upon m is equal to 2 beta. Beta is some kind of a constant which is equal to b upon 2m.
Why I'm writing 2 beta? Well it will simplify our calculations later on as you will see. And let's suppose k upon m is equal to some constant omega naught square. What is this constant omega naught?
I'm sure you can guess, but we will derive it in the due course. So using these substitutions, let's suppose this is a substitution number one and substitution number two, the equation of motion therefore becomes the second order derivative of x being the displacement plus 2 beta x dot is dx upon dt plus omega naught square x is equal to 0. This is the equation of motion of this kind of a damped harmonic oscillator system. And by solving this, we can get some idea that what sort of motions are possible. Now, if you are familiar with differential equations, then you may know that equations that look like this, where the coefficients 2 beta omega naught, these are all constants that depend upon the system.
equations that look like this have solutions of the form e to the power some parameter alpha t multiplied by some kind of a constant a let's suppose. So by guessing this kind of a solution if I substitute this in let's suppose this is equation number three then what shall I get? Let's try that.
So if I substitute this guess of a solution into the equation number three d upon so d2 upon dt2 a to the power e alpha t where alpha is some kind of a parameter whose value i am interested in finding out plus 2 beta d upon dt a e to the power alpha t plus omega naught square a e to the power alpha t again is equal to 0 then a is a constant it comes out i'll get alpha square e to the power alpha t here i will get in the second term beta multiplied by a alpha e to the power alpha t plus omega naught square a e to the power alpha t is equal to zero. Now if I take the common terms out, so what are the common terms here? Well a and then e to the power alpha t is a common. So if I take this out in the bracket I am left with alpha square plus 2 beta alpha plus omega naught square is equal to zero. Now a here is a constant and e to the power alpha t is not necessarily equal to zero.
Which brings us to the the condition that the potential solution will come from alpha square plus 2 beta alpha plus omega naught square is equal to 0. So what is this? If you take a look this is just some kind of a quadratic equation and I am sure that you know how to find out solutions of quadratic equations right. So the solution of the quadratic equation alpha is nothing but it comes from the quadratic formula.
minus b which is 2 beta here plus minus root over b square which is 2 beta square which is 4 beta square now i think it's going to become clearer why i took 2 beta in the substitution earlier so 4 beta square minus 4 ac where a is 1 and c is omega naught square this is 4 omega naught square and in the denominator i have 2 into 1 just 2 so this becomes alpha is equal to so 2 to cancel you have minus beta plus minus if I take the 4 out I'll get 2 which gets cancelled and I'm left with beta square minus omega naught square. So the alpha parameter that I was talking about in the solution that I assumed have two possibilities right. So therefore from what I can do is I can say that the solution of the equation number 3 can be written as x t is equal to some constant a1 e to the power alpha 1 t plus some constant a to e to the power alpha 2 t where alpha 1 is equal to minus beta plus let's suppose root over beta square minus omega naught square and alpha 2 is minus beta minus root over beta square minus omega naught square so if i substitute these terms this simply becomes xt is equal to some constant a1 e to the power minus beta plus root over beta square minus omega naught square t plus a2 e to the power minus beta minus root over beta square minus omega naught square t. The exponential to the power minus beta t term I can take out as a common term.
So this simply becomes x t is equal to e to the power minus beta t and in bracket I have a1 e to the power root over beta square minus omega naught square t plus a2 e to the power root over there's a minus sign here beta square minus omega naught square t so this is the nature of the general solution of the equation of motion that i was talking about so whatever equation of motion we obtain for the case of the damped harmonic oscillator this one basically equation number three its general solution is this point number four now before we talk about what kind of motion this point number four reveals to us, there are certain situations possible here. So there are four possible scenarios that can arise out of this solution. The first scenario is if beta is equal to zero. Now, if you remember, beta is nothing but b upon m, b upon 2m rather, where b is the drag coefficient.
and m is the mass. So in a sense beta has something to do with the nature of the drag that is happening. It is basically giving us an idea about how strong the drag force is.
This is actually known as the damping parameter. It gives us an idea about how much damping is going to happen because it not only depends on the drag force but it also has a inverse proportionality with the mass. So if there is no damping, if we have an ideal setup, where there is no drag force, no dissipation, no friction involved. In that kind of situation, we have an ideal harmonic oscillator.
So the nature of the solutions are kind of obvious. I'm going to derive it, of course, but you'll end up getting simple harmonic oscillations. Now, the next situation arises when, if you look carefully, there is a relationship between beta and omega naught.
So let's suppose Beta square is less than omega naught square in that kind of a situation. Let me give you a hint. I'm going to derive these results later on. You will get oscillatory motion, but it will correspond to not SHM, but it will correspond to what is known as underdamping when the amplitude dies out over a period of time.
OK, and the third situation corresponds to when beta square is equal to omega naught square. This is the situation where the oscillations will actually die out. You will not even get a single oscillation.
So no oscillations and this is known as critical damping and the final possible scenario is when beta square is greater than omega naught square. In this kind of a setup you will not get any oscillations. So no oscillations. However the system is going to take a little bit longer amount of time to come from a displaced configuration to its equilibrium configuration.
This is known as over damping. Now, these possible scenarios can be written by a comparison of the beta term here and the omega naught term. But I can also write them down in terms of the terms B, K and M. OK, so depending upon which textbook you're reading, sometimes somewhere is written in terms of the comparison between beta and omega naught. In others, it can be written as a comparison between B and let's suppose K and M.
Now, what is beta? Beta is equal to small b upon 2m. What is omega naught square? it is equal to k upon m.
So let's suppose in the third setup, beta square is equal to b square upon 4m square, which is equal to omega naught square, which is equal to k upon m. So this simply means that b square is equal to 4k m. That's it.
So beta square is equal to omega naught square corresponds to the drag coefficient b square is equal to 4k. Km where K is the spring constant and M is the mass. Similarly, I can show that the second condition reflects B square is less than 4Km, the first condition is of course where b is equal to zero and the last condition is where b square is greater than 4Km.
K being the spring constant, m being the mass and b being the drag coefficient while beta being the damping parameter. So under these four possible scenarios, the general solution will reveal to us different kinds of motion. So let us explore these different kinds of motion of the oscillating system that we have taken here for these possible scenarios. So let's first look at the possible scenario number one where beta is equal to zero. So if beta is equal to zero then equation number four simply becomes let me write it down here it simply becomes x t is equal to e to the power minus zero which is just one and you will have a1 here you e to the power root over minus omega naught square so that involves an imaginary number so e to the power i omega naught t right and the second term involves plus a to e to the power minus again beta is zero so root over minus omega naught square becomes i omega naught t now this is something actually we can simplify it by considering the fact that e to the power i theta can be written as cos theta plus i sine theta right if i use this this simply becomes a1 cos omega naught t plus i a1 sine omega naught t plus a2 cos omega naught t plus i a2 sine omega naught t if i combine the cos and the sine terms separately i end up getting a1 plus a2 cos omega naught t plus i a1 here should be a negative sign of course a1 minus a2 sine omega naught t now what is a1 plus a2 is just some constant that depends upon initial configuration a1 minus a2 is the same so i can write this as let's suppose a c cos omega naught to a c some real constant and let's suppose d here is some complex constant sine omega naught t.
Now of course we are interested in the real solution but even if I am interested in a real solution I can write the expression in this particular manner. What is this? It is actually a sinusoidal variation.
I can in fact show that the real terms can be simply written as some other constant a let's suppose cos omega naught t plus some constant phi. I can show that both these two expressions are actually equivalent to each other. This a result xt that we have obtained for the scenario where beta is equal to zero that means there is no damping is the obvious result of a sinusoidal variation which is a simple harmonic oscillations so when you have an ideal harmonic oscillator setup no damping no dissipation of energy happening you end up getting a sinusoidal variation so you will have some kind of a function that looks like this So here I have the solution for the oscillating system when it is displaced from its mean or equilibrium by some kind of an amount, let's suppose A. Okay, so the y-axis here represents the displacement with respect to time x and this is the time axis.
So this is a sinusoidal variation. All right, this is the result that you get. So when there is no damping, Then the obvious result is simple harmonic oscillations.
Now, this was obvious enough. This was quite obvious. What about the second case scenario?
So let us explore the second case scenario where beta square is less than omega naught square. So in the second case scenario, when beta square is less than omega naught square, or as I said, this is equivalent to b square being less than 4km. This represents a situation where I can rewrite the general solution point number four in a manner where I'm going to make an assumption here. I'm going to assume that let's suppose omega naught square minus beta square can be written as a further term let's suppose omega one square.
Then I can substitute these terms here. What will I get if I substitute these terms with omega one? I will get minus omega 1 instead of beta square minus omega naught square. So let us write down the solution then.
So for this kind of a scenario we will get x t of course the first term is e to the power minus beta t. The further terms involve a1 e to the power as I said beta square minus omega naught square is equal to minus omega 1 square. So from the minus term because there is a root over sign here I'll get an imaginary i omega 1 t.
a2 e to the power minus i omega 1 t. Just take a look here. I'll just get minus and then i omega 1 t. So the general solution for this kind of a scenario is this. Now before I look at the variation of this kind of a function, I can simplify it further by making some assumptions.
You see a1, what is it? It's just a constant that depends upon the initial conditions, right? So let me suppose that this is equal to some other constant a e to the power let's suppose i some other constant theta and a2 is equal to half of a let's suppose e to the power minus i theta if i make these assumptions then the form of this solution becomes e to the power minus beta t and then you will have half of a this will be common and then you will have e to the power i omega 1 t and then here you will have e to the power i theta plus e to the power minus i omega 1 t and you will have e to the power minus i theta. Okay, so let me write this in a simplified fashion. So, a e to the power minus beta t half, all right, and then you have e to the power i omega 1 t plus theta plus e to the power minus i omega 1 t plus theta, all right.
So if you have this kind of an expression, you can make use of the identity. Let's suppose you have e to the power ix plus e to the power minus ix divided by 2. Then some of you may already know that this is what? This is cosine of x, exponential to the power minus x plus exponential to the power minus ix divided by 2 is cosine of x.
So if I make that substitution here, it simply becomes e to the power minus x plus ix. e to the power minus beta t into cosine of omega 1 t plus theta. This is xt. So the displacement of the harmonic oscillator when this second case scenario is true, the parameters are such that beta square is less than omega naught squared is this particular solution. Now what is the mathematical meaning of the particular solution?
Because it's a product of two functions you see. A, of course, it depends upon the initial conditions. But what about, let's suppose this term.
This term is nothing but some kind of an exponential decay, right? But what about the cosine term? Cosine term is an oscillatory term, right? So what you end up getting is you get this oscillatory term, which is further modulated by an exponential decay term. you get an oscillatory term which is further modulated by an exponential decay term.
The graphical representation looks quite interesting so let me show you. Okay so if I displace the system by some kind of an amount at time t is equal to zero, let's suppose I displace it by a, y-axis here simply represents the displacement x with respect to time and I do not provide it with any kind of initial velocity. In that kind of a situation, These are the oscillations you get.
You see the cosine term is creating these oscillations. The system is executing oscillations, but the exponential decay is causing the rapid decay of the amplitude over a period of time. These amplitudes are decaying or decreasing over a period of time in an exponentially decaying manner. Now in this kind of a setup, since I gave the initial conditions of some displacement, but no velocity, it automatically turns out if you try to calculate it, theta comes out to be zero here. So the oscillations are actually, this oscillation is nothing but cos omega 1 t, while the modulation that is happening, this modulation is a e to the power minus beta t and here this is of course in the negative axis minus a e to the power minus beta t.
So you have these oscillations that are being modulated by an exponential d. So when you displace the entire system from its equilibrium, when damping is there, then the system starts oscillating. But after some time, its amplitude slowly decreases and it exponentially decays and it slowly, slowly, slowly decreases.
And with time, the exponential decay term, the system comes to rest. This is the classic scenario of what is known as for the condition beta square less than omega naught square under damping. It is called under damping because you still have oscillations.
Okay. It is the damping is not so high that oscillations have vanished. The oscillations are there, but the oscillations have exponentially decreasing amplitude relationship with time. So whenever the beta square term is less than omega naught square term or the b squared term is less than 4 km, you end up getting a system that executes oscillations that die out over a period of time.
Now it is interesting to note here that the oscillations are happening with a frequency omega 1. Now if you compare this with the ideal harmonic oscillator, where the frequency of oscillations was omega naught, what was omega naught? Omega naught was nothing but if you remember point number two, omega naught was just root over k upon m, right? omega naught was root over k upon m which is just the spring constant k upon the mass square root this was a frequency of oscillations the angular frequency of oscillations for the ideal harmonic oscillator but for the damped harmonic oscillator in the case of under damping the frequency of oscillations is omega 1 where omega 1 is what omega 1 is omega naught square minus beta square so omega 1 is actually root over omega naught square minus beta square which simply means that omega 1 is actually less than omega naught.
So the ideal harmonic oscillator has a angular frequency of oscillations omega naught and the damped oscillator has a frequency which is less than that of the ideal harmonic oscillator setup. Now the reason that these amplitudes are decreasing over a period of time is that the system is losing energy because of the friction involved or rather you can say because of the drag involved due to its interaction with the medium. The system is losing energy.
The system is dissipating energy. And the nature of this dissipation is also exponential because you can calculate the energy from let's suppose the amplitude square. If you do an energy analysis, you will find out that whatever the energy was there in the setup at the beginning, let's suppose the energy is E0, then whatever the energy is there after a certain interval of time t, if we do a calculation, you can see that the This has a relationship of exponential to the power minus twice beta t. So the graphical representation is just an exponential decay. For what?
For the energy. So energy, if it was initially E0, it slowly decreases with an exponential curve as the system loses energy due to its motion. Now let us come back to the scenario number 3, where beta square is actually equal to omega naught square. what is going to happen in scenario number three now in this setup where beta square is actually equal to omega naught square if you go back to the general solution then the terms will probably involve uh you know beta square and omega naught will cancel each other out and we'll end up getting e to the zero now that will probably not give me the right answer so what i'm going to do is i'm going to solve the equation of motion again the point number three for this setup so let us use point number three here in scenario number three. So the point number three is just the equation of motion that is d2x upon dt2 plus 2 beta dx upon dt plus omega naught square was there but since omega naught square is equal to beta square here I can write beta square x is equal to 0. This can be further simplified by writing d2 by dt2 plus 2 beta plus beta square x is equal to 0. Now what I can do with the term in the bracket which involves this operator kind of an expression d2 upon dt2, I can simplify it into two individual operators.
So I can write it as d upon dt plus beta multiplied by d upon dt plus beta. If you take a close look, these two expressions operated one after another is actually overall this expression. Just take a look at it. It's actually the same. So this equation can be written in this manner.
Now if I make an assumption that d upon dt plus beta acting on x is equal to let's suppose y in that kind of a situation I will simply get an equation that is d upon dt plus beta y is equal to 0, right? Because this is nothing but y. So let me solve this first.
So this is dy upon dt is equal to minus beta y, which can be written as dy upon y is equal to minus beta dt. If I integrate both sides, I will end up get ln. y is equal to minus beta t plus some kind of a constant.
Let's suppose c and left hand side, if I just write y, right hand side, I'll have to take an exponential. So exponential is minus beta t plus c. The c can be further simplified by adding a constant here. Let's suppose I add a constant here.
Now what is y? y is this. This is y. y is equal to this expression. So if I write this expression here, I simply end up getting d upon dt plus beta x is equal to a e to the power minus beta t right this is the y that i assumed so what is the left hand side this is nothing but dx upon dt plus beta x is equal to a e to the power minus beta t right now if i take the exponential term to the left hand side what do i get dx upon dt e to the power beta t plus beta x e to the power beta t is equal to a Now, if you take a look at this particular left hand side, I can write this left hand side as the time derivative of x e to the power beta t, right?
So if I differentiate this by time, then these are the two terms that I will actually get if you take a look at it. So therefore, I can substitute this entire term in this manner. So this is equal to a.
Now, if I do an integration with respect to time on both ends sides, this simply gives us X e to the power beta t is equal to t plus some kind of a constant. Finally, we get X, which is a solution in the scenario number three is equal to a t plus b e to the power minus beta t. This is the solution that we end up getting. For our case of what we had beta square is equal to Omega naught square So for the case of beta square is equal to Omega naught square This is what we end up getting now if you take a look at it first of all this is one function that you have here, which is like some kind of a straight line and There is this further function here, which is the exponential decay And of course an exponential decay is going to overcome the first function. So what you actually end up getting As time tends to infinity after a long period of time, the displacement of the system actually tends towards zero.
That means the configuration tends towards the equilibrium. There are no oscillations here. If you notice, no oscillations here. The system simply moves towards rest, towards the equilibrium without oscillating in any manner.
So this is what's known as critical damping. It's called critical damping because this lies at what is the boundary between over damping and under damping. You see in under damping you have oscillations, in over damping you do not have oscillations.
So in between you have critical damping where there are no oscillations but the system comes to rest in the quickest manner possible. It takes the least amount of time for the system to come towards rest. So having said that if you look at the let's suppose a graphical representation of this function.
Then again, here I give the system some kind of an initial displacement. All right. Then with time, whatever the angular or rather the displacement from the equilibrium is simply comes towards this equilibrium and goes to rest. This corresponds to critical damping.
All right. Now, let's explore the last possible scenario, which is scenario number four, where beta square is greater than omega naught square. So when beta square is greater than omega naught square, it also corresponds to b square being greater than 4k m as I already demonstrated. Under this situation, what happens? Well, first of all, let me make another substitution.
There is omega 2 is equal to, let's suppose, or rather omega 2 square is, let's suppose, beta square minus omega naught square. If I make this substitution, let's go back to the general solution I was talking about in point number 4. So here I can substitute a beta square minus omega naught square with omega 2. If I substitute beta square minus omega naught square with omega 2, then Quite simply put, the solution is going to look like x t is equal to e to the power minus beta t and then you have a1 e to the power omega 2 t plus a2 e to the power omega 2 t. So combining bracket as well as the exponential outside it, I'll get a1 e to the power minus beta minus omega 2 t plus a to e to the power minus beta here should be a minus sign here plus omega 2 t so the solution for this kind of a situation is this particular expression now if you take a look at this this is just the sum of two exponentially decaying functions this is an exponential decay this is an exponential decay although the decays are happening at different rates So this is decaying with b minus omega 2 in its power. So if you take a look at it, of course, this is going to decay faster, right?
The second term b plus omega 2 is greater than b minus omega 2. So the second term is going to decay faster. So the second term is going to approach zero quickly. So it is the first term that is going to dominate.
So quite simply put, over a period of time, as t tends to infinity, the term x is basically going to be approximated by a1 e to the power minus beta minus omega 2 t. Clearly there are no oscillations happening and it is more or less an exponential decay that is happening. This situation is known as over damping. Now it's called over damping because it is a similar to the critical damping situation. It's just that it takes a little bit of longer amount of time compared to the critical damping.
So the mathematical or the graphical representation would be when I displace the system by some amount, the system without oscillating comes towards its equilibrium configuration over a period of time. So the critical damping is different from overdamping that in the sense that in critical damping the amount of time taken is the minimum possible. In overdamping it can take a large amount of time.
So let us do a quick revision of all four possible scenarios. with the graphs. So here I have plotted all four graphs together. Alright, so if we do a final analysis in the harmonic oscillator setup which was placed in the presence of an external medium when there was no damping then that correspond to the let's suppose the drag coefficient being equal to zero or beta being equal to zero it let's. to what we can see here is simple harmonic motion.
You see this simple harmonic oscillations are happening over a period of time. So when there is no damping, the harmonic oscillator provides simple harmonic oscillations. However, now that damping is happening in the red line, this corresponds to under damping case. So the under damping case happens when the beta square is less than omega naught square or b square is less than 4k m. As you can see in the under damping case oscillations are happening but the amplitudes are decaying in an exponential fashion.
Okay the amplitudes are decaying in an exponential fashion and these oscillations are happening with respect to some frequency omega 1 which is less than omega naught. All right this is the case where damping is there. but it's not that high so the oscillations are still happening and it dies out over a period of time. However, what if damping is large enough?
If damping is large enough we reach a particular situation of what is known as critical damping. So in this kind of a case where critical damping corresponds to beta square being equal to omega naught square or b square being equal to 4 km, this is a situation where there is a rapid decrease of the displacement but there is no oscillations. there are no oscillations.
In the case of critical damping, the system takes the minimum amount of time to come from its displaced configuration to its equilibrium configuration. For further situations like this blue line here where β² is greater than ω0² or b² is greater than 4 km, these correspond to what is known as over damping. So in the cases of over damping, the system slowly comes to rest without oscillations but it takes its own time.
So a harmonic oscillator when placed in external medium that can provide some kind of dissipation of its energy, that can provide some kind of a resistance, can show, can demonstrate these possible scenarios. Without damping you have sinusoidal oscillations. With slight damping you get oscillations where the amplitudes decrease exponentially.
With further damping the system comes to rest. With further very high over damping the system comes to rest but very very slowly. So these are the different interesting possibilities All right, so I'll see you in the next video.
That's it for today. Thank you very much.