 Hi, I'm Zor. Welcome to a new Zor education. We continue talking about waves and their energy. And our examples of waves and their energy were too actual. For longitudinal oscillations, we were using a spring. And for transversal, we were using a rope. Now, since we are trying to go just slowly towards properties of light and energy of light, which is transversal oscillations of electromagnetic field, we will concentrate today on the energy of rope as it oscillates. Now, this is the final lecture, which is dedicated to energy of transversal oscillations, mechanical oscillations. And let me just remind you how I have structured this particular topic, energy of waves. Well, the first lecture was actually very simple. It's about longitudinal oscillations of the spring. Now, the second lecture was involved in basically properties of oscillations of rope by itself. And it was a differential equation, which is called wave equation, which delivered the solution to waves on a rope. Now, then I spent some time modeling transversal oscillations with longitudinal, that was the previous lecture. And today I'm going to use this particular model, and I will definitely repeat exactly how it's done, to calculate the energy of transversal oscillations of rope. So, that's the plan. Now, today's lecture is part of the course, Physics for Teens. It's presented on the website Unizor.com, and I do suggest you to basically watch this lecture from the website, because it contains all other lectures in the proper sequence. There are menus, etc., etc. Also, the Unizor.com website contains prerequisite course called Maths for Teens. Maths is mandatory for learning physics, especially calculus, vector, algebra, and something like this. So, I do suggest you to watch this lecture from the website, and you can go to the website. There is a menu. The first level is basically the course, Physics for Teens. Then you go to Waves, and within that part of the course you will find the topic which we are talking about right now. Light and energy and stuff like this. Now, the courses, all the courses are totally free. You don't even have to sign in. There are no advertisements, so it's just pure knowledge for your ultimate enjoyment. Also, the website contains exams, which you can take as many times as you want until you will feel comfortable. Okay, now, back to work. So, first, let me just remind you the oscillations of the rope. So, consider the rope. Well, you can consider it infinite, but if it's not infinite, it's a very, very long one. So, there is one end, which is forcefully oscillating. Forcefully, it means somebody, like myself, I just took the end of the rope and moved it up and down in harmonic oscillations, which can be described as this, where A is obviously an amplitude, well, T is time and omega is angular speed, or angular frequency, sometimes it's called the same thing. So, incidentally, speaking about angular frequency, so we will just be clear. There is a plane frequency called f, which means number of oscillations per second. Now, times 2 pi, it would be an angular, which is basically what is the angle per second, how much in radians we are moving per second. So, that's just another thing. So, we have this particular way how we oscillate the end of the rope. Now, what's the result of this? Well, the result of this is the following. If you will take this as y-axis, this is x-axis, and the rope is stretched along the x-axis, so the rope will start waving at this. Now, I can always have the function called y of x, T, which is basically the vertical level the rope rises at time T at point x. So, basically this is an equation of this particular point. It goes up and down, up and down. And the equation is a times cosine omega T minus kx, where k is some constant. If x is equal to 0, which means it's this point, we will have a equals sine omega T, which is just this one. Now, the further x from the 0, the delay will be basically introduced into the same level of rising this particular point. So, this point, for instance, is oscillating exactly like this one, but with a time delay. Now, let me just change this slightly. Cosine omega T minus k divided by omega x. That's the same thing, right? I just took omega out. Now, in the previous lecture where I was talking about oscillations of a wave and the properties basically of these oscillations, we came out with this very simple thing. So, I have proven there that the speed of propagation of the waves on a rope is related to the parameters k and omega this way. That's part of the previous lecture. If you don't remember it or you didn't really see the previous lecture, I do suggest you to go to the lecture which basically explains it. I think it wasn't exactly previous, it's the one before previous. The previous was about modeling transversal with longitudinal. So, this is very important because if I will use this, I can rewrite it as A times cosine omega minus, sorry, omega T minus x divided by V. Now, in this form, it's very obvious the x divided by V is time. It takes for the wave to come from here to here. This is x, V is a speed, distance divided by speed is time. So, that's the time delay. So, that's why this has much more physical sense. So, in this particular form, this equation, how this particular point is moving up and down, has real physical sense because it oscillates exactly the same as the beginning. The origin of oscillations, but with a time delay. So, let's just keep it in mind. Now, what else can we say? On the other hand, there is a concept of a wavelength and a period. So, the wavelength is lambda, period is let's say tau. And if you divide lambda, which is the wavelength between crest and crest, let's say, or between trough and trough. And tau is the time it takes for the wave to go from one end to another of one particular wavelength. That's also speed of propagation, right? So, we have another formula. Now, what is the time, what is the period? How period is related to frequency, for example? Well, obviously, tau, it's the time it takes for one wave to go. And frequency means how many times per second? So, obviously, they are inverse to each other. This is the time it takes for one wave. Let's say one wave to cover the distance of one wavelength is, let's say, five seconds. Then, what's the frequency? Well, we have to divide it. It will be one-fifths of the wave per second, right? Which means that I can put it here, lambda times f, right? I will continue this, okay? And f can be done from here. So, that's omega divided by 2 pi. All these simple calculations, I did not, like, memorize it. I logically tried to come up with all these based on the concept, on the physical concept. So, this is a speed. If this is a speed, then if you divide wavelengths by the time it takes to cover it, that would be the speed, right? The only thing which I do remember from the previous lectures is this one, because there is a little bit more, a little longer, maybe, proof that this is true. Everything else is pure logic. From here, we can come up with what? If this is equal to this, then k is equal to 2 pi divided by lambda, right? I have lambda here, sorry. And instead of f, I put omega divided by 2 pi. So, k would be equal to 2 pi divided by lambda, right? This is equal to this. So, let's just keep it in mind, and that's all, basically, which is needed right now. Now, I will wipe it out, and then we will go to properties. Now, let me go back to the model which we have built, how to model my transverse oscillations with a longitudinal. So, if you remember, we have decided to have a tiny spring from each point so that the neutral position of the spring is exactly on the neutral position of the rope, okay? Now, the previous lecture was dedicated to how we built this model in such a way that the oscillations of the spring, of the springs, all these tiny springs, each one is carrying an infinitesimal piece of the rope with its mass, obviously. So, how can we arrange so these springs would be exactly modeling the waves on the rope? Well, we have basically decided that we have to establish these springs in such a way that they have the same amplitude of oscillations, and this is simple, because if you will stretch the spring by A, it will oscillate from plus A to minus A, okay? So, the amplitude is satisfied. Now, we know the mass which is attached to each particular tiny spring. The mass is, it's a differential of mass and infinitesimal, which is mass density per unit of lengths times differential of the x, okay? This is x. So, we know the mass, we know the amplitude. Now, we have to create these springs in such a way that their oscillations will be exactly the same as oscillations of the rope. So, the rope has omega as an angular speed, so we have to really come up with the spring which has exactly the same frequency of oscillation. Now, frequency of oscillation depends on the spring, and again, if you don't remember it, you have to go back to the corresponding lecture. So, we have the formula, and the square is equal to the coefficient of elasticity divided by mass. Now, mass is dm, it's this one, so this is the mass at the top. Now, obviously, oscillations of the spring depend on properties of elasticity of the spring and what kind of a, how massive the object at the top of the spring in this particular case, right? So, it's equal to ke divided by mu dx. So, we have this. So, we have to basically come up with coefficient of elasticity from this formula, k is equal to mu omega square dx. So, omega square, omega we know because this is the rope. So, we know what exactly kind of oscillations we are forcing the rope to move because this is the end of the rope which somebody is forcefully moving up and down with certain angular frequency. So, we just will take the springs in such a way that their coefficient of elasticity is this one. So, everything seems to be like doable. Okay, so after we have gotten all this, so we will do next the following. We will stretch all springs up to the level a to whatever amplitude we need, but we will release them not at the same time but with certain time delay. So, we release first from the left and move releasing mechanism to the right. For instance, there is some kind of plank which I put here which prevents them to go down and then they will moving the plank out with the speed of propagation of the wave on the rope. If I will do that, then my springs will start oscillating exactly in sync with oscillations of corresponding rope as if all these individual pieces are attached to the rope, they would do exactly the same thing. And that's how my model is working. Okay, so everything seems to be done properly. The only thing is we have to do this releasing this thing with proper speed of propagation. And again, speed of propagation we know basically, right? So, that's all part of this view. So, what exactly do we need? Spring have the same release delay. Okay, now after we release them, we will have the following equation for each spring. Now, this is the deviation of the top of each spring. Well, each it means on the distance x from the beginning. A very tiny spring where we have attached a tiny piece of rope. Not connected obviously to other pieces of rope. So, now the rope doesn't exist. We have broken it down into small pieces, but infinitesimal pieces actually. And this is for each infinitesimal piece which is on the distance x and y is its vertical deviation. Well, it will have exactly the same cosine of omega t minus x divided by v. So, this is the time delay where we know what v actually is. We were already talking about this. Now, from another, from the classical point, it's omega t minus kx where k is equal to 2pi divided by lambda. I just did it before, right? So, we can look at the equation in this way or in this way. Now, this seems to be simpler since we already know what k is. Written in such a way, k doesn't really present physical sense. But written in this way, that's obvious that this is a time delay. But that's the same thing basically, right? Okay, so we know the equation. We know how each spring is basically moving, right? So, this is an equation for each spring. Which is located at x coordinate here and it depends on the time how it moves. What it means is that we can actually talk about energy of the spring and basically equate it to energy of the rope. Why? Because each infinitesimal piece of the rope, if it's on the rope, it does some movement. But if it's on the spring, it does exactly the same movement. If the movements are the same, then the energy which this particular piece of the rope, infinitesimal piece of the rope, would be the same. And that's basically my purpose of converting relatively complex movement of the rope with all these tension things, etc. It's kind of difficult. We are considering each point as attached to the spring, which has exactly the same parameters of oscillations as the rope. We can do it that way. And on the spring it's much easier. We have already covered this in all the details. So, that's exactly what I'm going to do right now. So, now let's just consider what exactly we want to do. Now, what does it mean we want the energy of the rope? The rope is long. Infinite may be rope, right? So, what is the energy? It's infinite. Which means we have to really somehow have a different task. So, the different task which physicists consider to be reasonable in this case is what is the amount of energy each wave, the piece of the rope of the length equal to wavelength from crest to crest. Because as the waves are moving, this difference in X coordinate between one crest and another remains the same because the waves are just moving synchronously. All the waves are going without changing the distance between them. So, the distance between them, which is a wavelength, that's basically the way how we are measuring the energy of the rope. So, we need amount of energy which is contained in all the rope pieces of the lengths equal to wavelengths. So, considering that we know everything about the rope, sorry, about the springs, so let's just find out the energy of one particular spring and then we will integrate it by X, let's say from zero to lambda, which is wavelengths, or from 10 lambda to 11 lambda. It doesn't really matter because it's all, because it's periodic. Since it's periodic, I just can cut any piece of X which has lengths of lambda and integrate all the energy of all the springs there. Some simple, right? So, let's talk about what's the first. First is potential energy. Potential energy of a piece of the rope attached to a string on the distance X at time t. So, how can we do that? Well, here it is. When we were talking about the spring, we were talking about expression for potential energy which is equal to this, K e times mass times, sorry, times Y square, right? So, if spring has coefficient of elasticity and we have stretched it, now Y is the direction of stretch. In previous lecture, we were using horizontal stretch, so it was X here. But now, X is a different coefficient. Stretching is towards Y. So, Y is a stretch. The more we stretch, the more potential energy spring has. And it depends only on the amount of stretching and the coefficient of elasticity. So, this is our equation for end mass. Sorry, yes, end mass. We have to end the mass. Now, this is the mass obviously at the end of this. Now, it can be converted into one-half. Now, instead of K e, we have just derived what exactly this is. It's omega square. Now, dm is mu dx, right? And Y square is A square times cosine square. So, my equation, Y of Xg is equal to A cosine omega t minus Kx. So, cosine square of omega t minus Kx, where K is equal to 2 pi over lambda, right? So, we basically know everything about potential energy of one particular spring, which carries this mass and stretched by this. Because this is how the spring is moving, well, the top of the spring is moving, right? So, that's the amount of stretch. A square times cosine square, this is the square of the stretch. And omega square, and it's here. It's K e equals to omega square dm. Yes, that's the proper way. So, we have this. This is amount of potential energy, which is carried by one particular spring. It's based on the formula which we have derived in one of the previous lectures, which are dedicated to the spring. Now, all we have to do is we have to integrate it. Let's say from 0 to lambda, u of x of t, which is 1 half omega square mu A square integral from 0 to lambda, cosine square omega t minus Kx and dx. Now, this is a trivial integral. Now, in the notes for this lecture, and every lecture has notes, it's like a textbook, basically. I do it a little bit more detail how to get this integral. It's really very simple because cosine square is converted into cosine of a double angle. So, cosine of double, it will be 1 plus cosine of double angle divided by 2, something like this. And if it's a plain cosine, not the square, then it's a trivial integral, integral from the cosine will be a special sign, whatever. And I will just give you the result. What's interesting is that the result does not depend on time. Why? Because we are integrating on the period. And when you integrate on the period, the dependency on time basically disappears because no matter how you integrate, whatever goes up in one place goes down in another place and they nullify each other. So, that's basically the result of this. And there is only dependency on all these parameters and, obviously, lambda. And the result would be... And the result would be, I'll put it here, one-quarter mega-square mu a-square lambda. So, there is no dependency on t on a time. And, obviously, there is no dependency on x because we are integrating by x. So, this is amount of potential energy in a spring wherever it is, but not in a single spring. We have all the springs which amount to one wavelength. That's why it doesn't depend on time or x. Great. How about kinetic energy? Well, kinetic energy... Again, let's start from a single spring. What is it? One-half mass, which is gm in this case, times square of the speed, right? Since we know this, what is the square of the speed? Well, that's derivative by time, square. Now, derivative by time, it's very simple to take. That would be a sign. And then we will just have to take... Now, dm is obviously mu times dx. And then we will have to do exactly the same thing. We have to integrate it. And again, I put some more words into a textual description here. I'll just give you the answer again. To integrate sine square would be really simple. It's exactly the same as to integrate cosine square. And what's remarkable is that the result is exactly the same. Why? Because sine and cosine are very much alike. They're just shifting one against another. And if you take the whole wavelength, then you will have exactly the same result. And the result is one-quarter, omega square mu. Great. Sum them up. Total energy is... Now, I shouldn't really put it as u of x now and t, because it doesn't depend. I should put really u lambda and k lambda. And a lambda is equal to... Which is energy concentrated in one wavelength of a rope. So one-quarter and one-quarter and one-half. All right. So that's the total amount of energy. Now, let me just graphically represent the energy. Let's say you have... Now, let's just think about it. It means that our point went all the way up, and then it will go down, right? So this is... For this particular x, the kinetic energy of this particular piece, infinitesimal piece, the kinetic energy will be zero at this point. Why? Because the speed first goes up, then goes down, which means in the middle, on the top. It's supposed to be equal to zero, right? Since it changes the direction. Which means my kinetic energy would be zero at this point. Now, at this point, it goes all the way down and continues to go down. So this will be the fastest piece. So this would be, therefore, my maximum of my kinetic energy. Again, this would be kinetic energy at the minimum, at zero. Because speed is zero. And here, again, it would be on the maximum. So this would be my kinetic energy. Okay? Now, what about potential energy? Well, potential energy should be opposite to kinetic, because their sum is supposed to be the same. They still have the conservation of energy. So whenever the point is at the very maximum, the corresponding string is at its top, which means it has the greatest deviation. So my potential energy would be at maximum here. And at this point, spring is back in the neutral position. So this potential energy would be zero. So if I do the potential energy, it would be like this. Whenever kinetic is at minimum, potential at maximum. Whenever kinetic is at maximum, potential is at minimum. So that's the graphical representation of both. But their sum is constant. Why? It's very simple. Potential energy depends on cosine square of something, if you remember, right? Genetic energy depends on square of sine of the same thing. And if you add them together, cosine square plus sine square is equal to one. So that's why the energy, total energy is constant, does not depend on anything. I mean, it depends on properties of springs, etc. It does not depend on the time or coordinate. And that is almost everything. Okay. Back to energy. So energy of one particular wave, as I said, was equal to one-half omega square mu A square lambda. Okay, that's energy. Now let's think about what is power? Power is amount of energy or work which is exerted in the unit of time. So what would be... Now this is the period. Well, this is the wavelength, right? What if we will divide wavelengths by the time it takes for the wave to cover the distance? Which means we will divide it by period, by tau. What is this? Well, this is speed. So speed would result in the power. This is average power, obviously. Average power of one wavelength, would be equal to one-half omega square mu A square and time and speed of wave propagation. So the more frequent oscillations are or the faster these waves are delivering, the greater average power they are delivering, all these waves. Now these are all mechanical waves. We did not talk about light waves yet. Next slide. Okay, now I do recommend you to read the text provided for this lecture, the notes. It's like a text book, basically. And well, you have to really think about all these little formulas. It's interesting what happens with energy, how it's delivered, etc. Another thing is, which I would like actually you to pay attention to, do not try to memorize all these formulas. Try to find physical sense, whatever I was doing in the very beginning of this lecture. I was trying to try to find the physical sense of all these coefficients, kx in the formula for harmonic oscillations. Okay, that's it for today. Thank you very much and good luck.