 Hi, I'm Zor. Welcome to a new Zor education. I would like to talk about energy of light, well, basically energy of electromagnetic oscillations to be exact. So this lecture, I will start with general energy of waves and particular kinds of waves, waves longitudinal and then transversal. And this lecture is part of the course physics for teens presented on Unisor.com. I suggest you to watch this lecture from the website not only because it's completely free and there are no ads and you don't even have to sign in if you don't want to but you actually have the course, which means there are many lectures logically related divided into parts, parts divided into topics, topics divided into lectures. Every lecture has very detailed notes, basically like a textbook. So you have a lecture and the piece of the textbook basically which relates to this lecture right in front of your eyes. Also, there are exams, which you can take without any restrictions as many times as you want until you will get a perfect score. Okay, back to energy. Now, what's interesting about waves is that they carry energy, but they don't really carry objects. For example, if you take let's say a boat on a surface of the water, if water has waves, the boat will go up and down, up and down. I mean, we are not talking about motors or anything like that. Which means this object will not move anywhere. I mean, not significantly, it will move up and down, but there is no like real movement like automobile on the road or something like this. So the water the water waves carry certain energy because we have to lift it up and down all the time. Sound waves are basically oscillations of the pressure of the air around us and they also carry some energy because they force the eardrums to vibrate and that's how we hear the sound. Light, light also has energy. Well, obviously light somehow goes into our eyes and causes electrical impulses. To the brain. That needs an energy, obviously. Also, if you will direct light, concentrated rays of light on something, it will heat it up. So that's an energy. So all the waves carry energy. All right, so that's some kind of a preamble. Now, today we will talk about very, very simple, the simplest I would say waves, longitudinal waves of the spring. And at the end of the lecture, I will try to give some kind of preview of what will be next lecture and why I have decided to start from the spring. The simple oscillations we have actually learned before when we were talking about mechanical oscillations in the same course, this explosion, that was the beginning of the parts, which I call waves. So we are talking right now about mechanical oscillations, about spring, but we will primarily concentrate on the energy aspect of this. Okay, so let's just imagine a simple spring. So let's say this is the spring in the neutral position. And we are not talking about gravity right now, anything like that. So it's all in space, no gravity in our center. And then we stretch, we stretch the spring. So the difference between this is x. I put x as a function of t, and I'll explain why. But let's just assume this is some distance s, x. Well, first of all, I would like to know how much work I have to do to stretch it from neutral position to neutral plus x. Okay, so what exactly we are doing right now? Well, the springs obey the Hooke's law, right? Remember this? Hooke's law, which says that f of the spring is equal to s function of x is equal to minus kx. k is elasticity coefficient, which is related basically to, it's a characteristic of a spring, what it's made, how it's made, etc. And x is the distance. Obviously this is an approximate law. It's only when the x is not really very large. So relative to the length of the spring, we are stretching it a little. Whatever a little means, obviously it depends on the spring, it depends on our experiment, etc. But let's just take it for granted that the Hooke's law is exact law. It's an abstraction, as everything tells what we are dealing with. Okay, now, so the function depends on the x, right? So the more we stretch, the stronger the spring force which resists us. Spring force is directed against our movement, that's why there is a minus sign. So the spring force is this way and we are stretching. That's our force, this direction, obviously their opposite. So I have to overcome the spring's resistance. Resistance goes towards the neutral position. So if I'm stretching down, so the force of the spring directed upwards towards the neutral position. So I'm stretching away from the neutral position, that's why it's plus here. Now, I would like to know how much work I have to really spend. Well, that's kind of easy. If I'm stretching from x to x plus infinitesimal increment of x, I can consider that during this infinitesimal interval my function f is f of x is not changing. But in theory, if it's not infinitesimal, then the force would grow a little bit. But considering infinitesimal, because I will go into integration of this. Since it's infinitesimal, I can consider that the f of x is exactly kx. And the work is what? The force times distance, now distance is dx, right? And this is my differential of infinitesimal amount of work, which I have to do. Now, if I would like to stretch it to x equals to a, let's say. What should I do? I should integrate it from 0 to a kx dx, which is equal to 1 half k a square. Now, if you are not familiar with integration, how I get this? Well, all I can say right now is go to Unisor.com. There is a mass for teams course, and there is a very large pot calculus. So, again, I suggested from the very beginning of this course of physics that you have to know calculus, you have to know the vector algebra. All these are presented in the mass for teams course or anywhere else wherever you want. Maybe you just know it. So, I take it for granted that this is the formula. Alright, fine. Now, this is amount of work, which I have to spend stretching the spring. Now, where is it going? Well, obviously, now, when the spring is stretched to the distance a from the neutral, it has certain potential energy. So, this is actually potential energy as a function of this amount of stretch, which I have actually done. Well, obviously, it depends on the spring, but I'm not really putting any kind of indices here because it's kind of understandable. We're talking about only this particular spring. Alright, fine. So, we have basically found what is the potential energy if I'm stretching it to the point by distance a from the neutral. Now, a can be any. I can, with the same basically token, I can say that p of x is equal to one-half kx square. So, x can be any. It can be a, it can be anything. So, for any lengths I'm stretching, I can use any letter here. It doesn't matter. Letter is not important. What's important is it's the same here and here, right? So, obviously, usually kind of x is used as a variable. So, yes, potential energy depends on how much we stretch it. We stretch it by a, the potential energy is this. We stretch it by x, potential energy is basically this. Fine, let's talk about kinetic energy. Now, what happens if I just let go the spring after I have stretched it? Well, it will, force of the spring will have no resistance because I'm not really involved any. Okay, so it will move if there is an object here of mass m. This force will start moving this object upwards, towards the neutral position. And since force is always working, and before I cross the neutral position, force is always in one direction, which means there is some kind of acceleration in that direction, right? Okay, so the speed would be increasing and increasing since the acceleration is always positive towards that particular direction. The speed will increase, mass is constant. So, kinetic energy. Kinetic energy is one half and the square, right? Remember this. It's a mechanical energy. So, the speed V will be increasing, absolutely, and that's why we will have the increase of the kinetic energy. When we are crossing the neutral position, my potential energy would be zero because x is distance from the neutral. But my kinetic energy will be substantial. Now, in theory, if the conservation of energy is working, the kinetic energy at the point of crossing my neutral position should be equal to my initial potential energy, right? When the speed was equal to zero, when I stretched it, at that moment I let it go. Alright, so that's why it will cross the neutral position and go further. Now, it will be squeezing the spring. Now, when we squeeze it, spring will again resist the squeeze. The same Hooke's law is working for both stretching and squeezing, within certain interval, obviously. And so, my force will be directed towards, my spring force will be directed towards neutral position. So, it will slow down, slow down, slow down until it will stop. At that point, kinetic will be equal to zero and my potential should be equal to exactly the beginning if the conservation law is working. So, what I would like to know is the following. Since I know the potential energy at any point, and if x is function of time, whatever, so I can put p of x of t, so it will be x of t here too, right? That's the same thing. I will derive the formula of kinetic energy also dependent on time. And I will see if the conservation law is working. Okay, fine. Now, how can I determine my kinetic energy, my speed, etc.? I have to know this x of t. I have to know how distance is changing with time. So, what do I know? I know the Hooke's law. So, I know the force which acts on the mass, this one. And I also know the second Newton's law. Second Newton's law says m times acceleration is equal to force, right? So, I will use fs here. And so, this is minus kx of t. x is a function of t. It moves by itself only using the force which is provided by the Hooke's law. Now, what is a? a is the second derivative of x. So, it's a d2x of t d2 square is equal to minus kx of t. Now, when we were talking about mechanical oscillations, I had exactly the same formula. Also, in the mass for Tien's course, I was using something like this to explain how to solve differential equations. So, I'm not talking about how I solve it. I just move in and give you the answer. Now, this is a second order differential equation, second order because it's a second derivative. So, I need two initial conditions because there are many functions which satisfy this. But I need only one function and to make it only one, I need two initial conditions. One initial condition is my initial stretch at point t equal to zero is equal to a, right? That's how I will... Now, my initial speed is equal to zero because what I said was that I just let it go. I did not push it on anything. I just let it go and then the spring itself started oscillating without me. So, these are two initial conditions with this equation and I will just write down the result. X of t is equal to a times cosine omega t, where omega is equal to k over m square root. So, this is a solution. Okay. Now, if this is x of t, then v, the speed, it's the first derivative. It's x of t. First derivative is equal to, derivative of cosine is minus sine, so it's minus a and omega and sine of omega t. My second derivative which is acceleration minus a omega square cosine omega t. So, if I will multiply it by m, now what is omega square? Omega square is equal to k over m, right? From here. So, omega square would be k square divided by m square. I saw k divided by m. That's omega square. So, if I will multiply it by m, m a, I will have minus a k and cosine, right? Let me just release some space here. So, knowing the speed v of t, I can calculate my kinetic energy as a function of t of time. So, it's one half m v square. One half m v square, which is what? a square, omega square, sine square omega t equals one square m. So, what is omega square? Omega square is k over m, and I multiply. So, I have only k, a square, sine square omega t. Okay. Now, let's recall what is my potential energy. Potential energy. Now, it's function of x. x is a function of t, so I can put function of t is equal to one half k x square. Now, x square is this. So, it's a square, cosine square omega t. Now, what will be k plus p? Total energy. Total energy. k of t plus p of t. Hope you remember trigonometry. One half k a square, k a square times sine square, sine square plus cosine square, which is one. So, this is the formula for a total energy. As you see, it's a constant, and that's exactly what we had initially when x was equal to a, my potential energy was exactly this. Initially, when I stretched it to a, my potential energy, which I gave because I was working on it against the spring, I gave it potential energy. And that's exactly the sum, potential energy as the time goes, diminishing while we are moving towards neutral. And kinetic energy is increasing, sine is increasing, until the point when I cross the neutral position, then my potential will be increasing, kinetic energy will be decreasing. So, and the sum would always be the constant. And by the way, this sum depends only on the spring, not on a mass. So, the total energy of any mass on the spring, well, within the limits of our experiment, when the Hooke's law is working, etc., etc. The total energy depends only on how much work I actually did by stretching. So, we don't have any gravitation, so it's just the mass, which means when I stretch this mass doesn't really play any role. Only the spring resists my movement, and that's the amount of work I do, and that's the amount of potential energy I gave to the spring. And then this potential energy is decreasing, increasing, kinetic energy is vice versa, increasing, decreasing, but the sum is always this. This is obviously the confirmation of the law of conservation of energy. Okay. Now, this is basically all I wanted to talk today. I just wanted to talk about this energy of a spring and how the spring is vibrating, oscillating. Now, let me just give you an idea of why I did that. It's very simple. The next thing which I will do in the next lecture, I will talk about vibration or waves of a rope. So, if you have a rope vibrating up and down one end, the waves will go. Now, how I will model it, I model it this way. I consider that there is a spring here. This is a neutral position of spring. Now, this is a stretched position of the spring, and this is the squeezed position of the spring. So, with every piece of rope, I consider I will have a spring. Now, initially, all the springs are the same and they are in neutral position, but as soon as I make a move with the end of the rope, my first spring will stretch and if they are connected by the ends, all these springs, it will not only stretch itself, but it will also pull the next one, the next pull the next one, etc., etc., and that's how waves will go. So, that's why I consider a rope as a combination of the springs, like these, and that's why I started with analyzing the spring, and then we will gradually move to a more complicated case. So, this is just a longitudinal oscillations, which means that the oscillations are only in the direction of the movement itself. Now, this is a combination. Each piece is oscillating longitudinal, but the waves are transversal. So, the direction of the propagation of the waves and the direction of the movement of the end of the spring are perpendicular to each other. In this case, the direction of the waves and the direction of the movement of, let's say, the end of the spring is the same. In this case, they are perpendicular. So, this is a more complicated picture. That will be the next lecture. Okay, that's it. I suggested to read the notes for this lecture. So, after you have listened to the lecture, it's very, I think, beneficial if you read it. And basically, that's it. Thank you very much and good luck.