 Hi, I'm Zor. Welcome to a new Zor education. I would like to continue talking about forced oscillations. The previous lecture was about one particular case when the frequency of force which acts periodically on the object on the spring has a different angular frequency than the inherent natural frequency of oscillation of the object on the spring. So the object on the spring without external force was described by this differential equation, and the general solution was x of t is equal to a cosine omega 0t plus b sin omega 0t where omega 0 is the square root of k over m. Actually, I can rewrite my equation. If I would divide it by m, I will have x of t plus omega 0 square. x of t is equal to 0. Omega square is k over m. So if I divide everything by m, I will have k over m. That's why. So this is the equation. It's more convenient to use omega 0. And this is a general solution. Now, in case I have an external force, my equation becomes t. So this is the periodic force which acts on the object. And that's exactly what we were doing in the previous lecture. And omega is frequency of the external force. Then this is an equation. And in case omega and omega 0 are different, which was actually the case in the previous lecture, we came up with first one partial solution, and then we added the general solution to an original equation without any force. This is a homogeneous differential equation. This is non-homogeneous. And I explained that in the previous lecture. So what we have to do, we have to find one particular solution to a non-homogeneous equation and add all general solutions to a homogeneous, corresponding homogeneous equation. And that's how we get general solution to a non-homogeneous. Now, in the previous case, when omega and omega 0 were different, we were using one particular solution in the format of cosine with some kind of constant, constant times cosine omega t. We found that constant, and that's how we found the whole solution. And the problem is that solution had omega square, omega 0 square minus omega square in the denominator. So for a case when omega 0 and omega and omega are different, we can do that. In this case, this lecture is dedicated to the case when these two are equal. So I will no longer use index 0. So this is my case. I have only one omega, which is both. The frequency of periodic oscillation of the force and inherent natural frequency of angular, angular frequency of oscillations of the spring on the force, on the object on the spring without the force. So with force and without force, basically. The same omega. The force is acting periodically, and the spring has this inherent natural frequency, also the same. Now, obviously it happens when you are, for instance, you are pushing something in a swing, and you're pushing with exactly the same frequency as the spring itself is moving back and forth. That's exactly the case. In the spring, it's exactly the same thing. So the spring by itself has certain oscillations, and you're pushing with the same frequency. So whenever spring goes into a squeezing mode, you're squeezing even further. Whenever it's trying to stretch, you're stretching even further. And what happens in this case? Well, what happens in this case is so-called resonance, and we will talk about this mathematically. What we have to do is to find a solution, partial solution to this differential equation, and then adding a general solution to a corresponding homogeneous equation, we will have a general solution to a non-homogeneous. All right, so we will rewrite it slightly differently, which is x plus k over m, again, that's omega square. f t is equal to f divided by m cosine omega t. So that's my non-homogeneous solution. All I have to find is partial solution. I will find this partial solution similar to the previous lecture, but in the previous lecture, I had a little simpler form. I had just a cosine, basically. When this is zero, I can have just a cosine, because the second derivative of the cosine is a cosine, so I can always find some kind of coefficient to make it zero. In this case, it's not exactly the same, but you really have to look for in some kind of function which contains sine or cosine, and maybe a little bit more complicated. Now, people were trying to do this, and they found basically that function partial solution of this type, some kind of coefficient times t times sine of omega t would work, and I will explain why it works. It's an intelligent guess. I mean, people were trying sine by itself with some kind of coefficient or a cosine. They got this problem with zero in the denominator, so these functions did not work in this particular case. So next, more complicated than just sine is t times sine. Okay, let's try it, and let's see what happens. Okay, if this is a function, let's take its first derivative, partial solution is equal to q times sine omega t. That's derivative of this times that. Now, derivative of that times this. So, plus q times t times derivative of sine is cosine, but we have to put omega, because I have multiplication interval function, so that's my first derivative. My second derivative of this partial solution is the derivative of this. q times derivative of sine is a cosine. I have to multiply by omega, because it's inner function. Here, again, it's a combination, t and cosine. So, first, we will put q times omega times cosine omega t, and so that's this function derivative of this. This derivative of that would be minus sine, because it's a cosine, so it would be sine. So q times t times omega, omega would be in square and sine of omega t. From cosine, it's minus sine, and internal function omega. You have an omega, now it's omega square. Now, let's substitute it into this equation. So, what happens is, into this equation. So, we will have this, we can just write it down, q omega cosine omega t plus q omega cosine omega t minus q t omega square sine omega t plus omega square times function x. x is this. So, it's q t sine of omega t, and it's equal to this. Okay. Now, lo and behold, this is the same as this, plus and minus. So, what do we have now? This is the same expression, it's just double. So, it's 2 q omega cosine omega t on the left, and on the right I have f0 and cosine omega t. Now, that's supposed to be equal for any t, because it's the equivalent of two functions, which means that these coefficients must be the same. Cosine and cosine, that's fine. So, from here we have q is equal to f0 divided by 2 m omega. So, that's a solution. So, instead of q, I can put f0 2 m omega, and this is my partial solution. I have partial solution to a non-homogeneous differential equation, this one, and now all I have to do is to add a general solution to a homogeneous corresponding, and we know what that is. So, my general solution to this equation is partial solution, which I have just found, plus general solution to a corresponding homogeneous equation. That's it. That's a general solution to our equation in case of, well, let's call it as it is, in case of resonance, in case of the frequency of the periodic oscillation of the function of the external force is equal to inherent frequency of oscillation of the spring with an object without the force. So, that's a general solution. Now, again, as you remember from the previous lecture, I modified this expression by having this type of conversion, a over d is equal to cosine phi and b over d is equal to sine phi. That's very easy to prove that this is exactly what it is. There are d and phi, which actually can be replaced in this, which means it will be this part plus d, and then that would be a divided by square root, so it's a cosine phi. This would be a sine phi, so it would be a cosine of the difference. Again, I actually explained in detail in the previous lecture, so I don't want to repeat the same thing. So this is my general solution. d and phi are two constants, basically. So for any d and phi, this is a solution to our differential equation. Now, depending on the initial conditions, we can find d and phi. So let's just substitute some concrete numbers and we'll see what happens. Let's say that our initial position of the object is at the neutral position and it's not stretched and not squeezed, the spring, and no initial spring, no initial speed. And then we start basically pushing with certain frequency, omega, with certain force. So what happens? Well, let's just substitute... Oh, by the way, this is t. Okay, let's just substitute these two and see what happens with d and phi. And we will find the equation which describes oscillation with this periodic force in case our object initial is in neutral position and no initial speed. Okay, x of 0. So this is x. Why did I put x of p? It's x of t. Okay, initial position is 0, so I substitute 0 here and 0 here and that would be 0, right? So if 0 here is 0 and this would be d times cosine of phi... Well, minus phi and phi, cosine is even function. So that's equal to 0. Now, the first derivative. Okay, so what's the first derivative of this? Okay, it's f0 to m omega t times cosine omega t times omega plus f0 to m omega times this sine of t. So that's some of this and this would be minus sine of omega t minus phi. Okay, we substitute 0 for t. Now, this would be 0 because t is equal to 0. This would be 0 because sine of 0 is 0. So I have minus d omega, etc. So it's d times omega times sine of... This is 0, so it's phi is equal to 0. So we have two things, this and this. Well, omega not equal to 0. Now, sine of phi and cosine of phi cannot simultaneously be equal to 0. Now, we have d times cosine is equal to 0 and d times sine is equal to 0. So d must be equal to 0 because if d is not equal to 0, it means that sine and cosine both are 0, which is impossible because sine squared plus cosine squared is equal to 1, right? So d is equal to 0, which means this is 0 and the only thing which describes our oscillation in this particular case is this function. So for this very simple case, when my initial position is 0 and they don't have any initial push to the object, my motion of the object is described by this equation. So what is this? Well, don't forget this T. T is very important, which means as the T coming, as the T going, as the time is increasing, this without the T would be just a plain sinusoidal. But multiplied by T, it would be first a smaller one but then bigger one and bigger one and bigger one and bigger one. So it will be inscribed into the angle, basically, with greater and greater oscillations. Well, in theory to infinity, but obviously this is not practical because the spring has certain limits. But in any case, within certain time frame, within certain movements where the elasticity of the spring is really described by the Hooke's Law, which is not very far from the neutral position, this would be a function which describes the oscillations. And precisely because there is this multiplier T, that's what makes it resonance. That's what makes the whole thing go with a greater and greater amplitude. And that's exactly what we're using when we are entertaining our children on a swing, when we are pushing a little bit, a little bit more, a little bit more, but we are pushing in sync, synchronously with the oscillations of the swing itself. Well, basically that's it. I do suggest you to read the notes for this lecture. So you have to go to Unizor, go to Physics for Teal, the part which we are talking about right now is called Waves, and within the waves there is mechanical oscillations as the topic, and this is the last lecture in that topic. Well, that's it. Thank you very much and good luck.