 Hi, I'm Zor. Welcome to Inizor Education. We continue talking about oscillations and primarily mechanical oscillations right now. Now, before we were talking about basically some aspects of movement itself, as it oscillates. In particular, in a simple case of a spring with an object attached, we came with an equation of displacement of the object of some neutral position. So if this is the neutral position of the spring, then we basically, if we make some effort to move this particular object to this position, stretching, basically, from the neutral position by distance a, and then let it go without any initial velocity. And then this would be an equation which describes the movement where x is displacement of the neutral point. So now we will talk about the different aspect of these movements. We'll talk about energy. So this lecture is about energy of oscillation. Now, this lecture is part of the course called Physics for Teens. It's presented on Unizor.com and I suggest you to watch this lecture from the website rather than from, let's say, YouTube, where you might have found it using some kind of a search engine because the website contains the whole course, which means there is a menu of all lectures and lectures are logically dependent one from another. Like, for instance, today's lecture depends on two previous lectures where I was actually deriving this formula. Plus, there is a very detailed notes for each lecture. It's basically like a textbook. Also, Physics for Teens is a course which has a prerequisite and prerequisite is called mass for Teens. Now, you will see in this particular lecture, for instance, that you do need mass. You need calculus. You need vector algebra. You need a lot of mass. Physics cannot be studied without knowing mass. So today, for instance, I'll use integral. Okay. Oh, by the way, the website is completely free. I mean, there are no advertisements. There are no strings attached. You don't even have to sign in unless you are performing some kind of a supervised studying, but anything is completely free. Okay, so let's talk about energy. When I initially stretched my spring by this distance a, what basically does it mean from the energy standpoint? Well, I performed some work. Now, where is it going? I performed the work, which means it's supposed to be converted into some energy of the spring. So in this position, in a stretched position to its maximum, spring is supposed to have certain potential energy. We know that mechanical energy can be basically characterized as kinetic energy. It's energy of movement. Remember, e kinetic is equal to m v square over 2, where m is a mass and v is a speed. So this is the energy. It's a kinetic energy. Now, potential energy is completely different thing. Potential energy is basically ability to perform some work. Now, what kind of work stretched spring actually can perform? Well, if you put some kind of a weight or whatever else to this particular point, it will push it, which means it will exhort certain force on the object. Object will be accelerated and it will be moving back to the neutral position. Now, I'm assuming that there are no friction, no air, no gravity, just pure spring, ideal spring, weightless, which has only elasticity in it. Okay, so we do have certain potential energy because we have performed work to stretch the spring. Since we performed the work, amount of work we have performed actually is exactly equal to amount of potential energy because you know about energy conservation law. If some work is actually spent, mechanical work I'm talking about, then some energy, mechanical energy, is supposed to be transferred to someone else or something else. Either we are moving something or we are stretching the spring. So in this case, at the very maximum stretch by the by the distance a, my spring is supposed to have certain potential energy, which is equal to amount of work, which I perform. Okay, so what happens next if I just let it go without any initial speed? Well, the spring, which contains certain stretched spring, which contains certain energy is supposed to pull the object back here. Well, from the position where the initial speed is equal to zero, it starts moving, which means it accelerates with certain force, and the force actually depends on how stretched the spring is. You know, the Hooke's law. Hooke's law is that the force is equal to minus kx, where x is the stretched distance or squeezed actually. It's the same thing, whether it's squeezing or stretching. The force is always opposite. If x is positive from here, well, I assume this is neutral position. This is x is equal to zero. So whenever x is positive, the force is negative. It goes back, and if x is negative, then the force will be positive. So it's the Hooke's law, which basically we are considering reflects more or less how the spring behaves. So there is a Hooke's law, so there is always a force which drives this particular object back to the original neutral position. Now the force is always acting while it's moving from here to here, which means the object is accelerating and accelerating. And what happens when it reaches the neutral position? Well, let's talk about this. Potential energy when the spring is in the neutral position should be equal to zero, because spring from the neutral position cannot by itself move left or right, stretching or squeezing. But we do have this object and it reaches certain speed. It has certain mass. So at this particular point, all the energy, all the potential energy, which used to be when the object was in this position, will be converted to kinetic energy at this particular moment. What happens next? Well, there is an inertia, and the object will start moving further, squeezing the spring. What happens in this case? Well, the spring will slow it down, right? Because the force is always acting opposite to direction. If x is negative, then the force will be positive, so the force will prevent movement. What does it mean if it's already moving with certain speed? Well, it will slow it down up to a zero point of speed zero. When speed is equal to zero, it means that the spring is squeezed to the position where its potential energy should be equal to the same kinetic energy at this point or potential energy at this point. So energy, which we have supplied to a spring, stretching it, is converted into potential energy in this. Then the potential energy is diminishing, but the kinetic energy speed of the object is increasing. By this point, all energy is kinetic, potential is zero. Then the kinetic energy starts diminishing because the speed is diminishing, but potential is increasing. But the sum, potential and kinetic energy at any time, because of energy conservation law, must be exactly the same constant and equal to the initial amount of energy at the stretch by the distance a. And that's the most important part about how the energy is converting. My work, when I stretch, is converted into potential. Then potential is decreasing, kinetic is increasing. Some of them is constant. At this point, everything is kinetic because potential is zero. And then the potential is increasing, kinetic decreasing, up until the maximum squeezed point. And then the potential energy at this particular point is equal to potential energy at this point. And then it goes back. So as soon as the spring is completely squeezed to the point where potential energy is equal to the initial, then it will start straightening again, back to stretching again, back to neutral, and through the neutral back to this initial stretched position. Because again, energy is preserved, the constant. And that's how it will move back and forth, back and forth, according to this law, where x is the displacement of the neutral position. Great. This is preamble. Now let's talk about mass. First of all, what kind of potential energy I will have at this particular point? Because everything depends on it. Because now from this point, we can calculate all other things, kinetic energy at this point or at any other point. So we do need initial amount of energy at this point. How can we calculate it? Well, here comes the mass. Let's consider that we have already stretched the object to this position x. And now we are trying to move it by infinitesimal dx into position x plus dx. It's infinitesimal. dx is differential of the distance and it's infinitesimal, which means from x to x plus dx, we can assume that the force is not really changing. The force we have to apply to stretch this spring. And the force depends on the x. So the force at point x is equal to this one. Right? So what we have to do right now is we have to add the force which I have spent from x to dx to another force, which I spent from x plus dx to x plus 2 dx, etc. So we are dividing our interval from 0 to a into infinite number of infinitesimal small dx entries in intervals. On each one, we spend certain amount of work. It's differential of work, which is equal to f of x, which is this times dx. Right? Work is equal to force times distance. And we are assuming that from x to x plus dx, the force is constant. And it's equal to f of x. Distance is dx. So this is differential of work. What do we have to do now? We have to integrate it from 0 to a, which is equal to integral from 0 to a minus kx dx. Well, minus actually doesn't really make sense, because it doesn't matter which direction force actually is going as a vector. In this particular case, we are interested in absolute value of this force, because actually this is a vector equation. But now let's forget about the vectors, because x is directed along the same line as the force f. So they are collinear. So let's forget about this sign, and the integral of this will be kx squared divided by 2 using formula of Newton-Leibniz. We have to substitute instead of x. Now, I draw this because the derivative from x square is 2x, since divided by 2 will be x. And k is just a linear multiplier. And that's equal to kA squared divided by 2. So what we know is that at point A, my potential energy u of A is equal to kA squared over 2. This is energy at this particular point of stretching. So now let's talk about how the object moves. It moves from this point to this point, and then to squeezing the spring to this point. So that's how this potential energy is converted into this kinetic energy at this point, because at neutral point our energy is potential energy is equal to zero. So everything is kinetic. So if everything is kinetic, then v of zero, well, I will use v0. This is a speed at moment of crossing the neutral point. So if this is v0, then mv0 squared over 2, which is kinetic energy, should be equal to maximum potential energy. From this v0 is equal to square root of k over m. Well, that's with absolute value. So absolute value, because when it moves here, the vector of speed is going this way. When it moves here, vector of speed is going that way. But the absolute value of speed, which we are talking about right now, is this. By the way, if you remember, square root of k over m is omega here. That was in one of the previous lectures. All right, so this is what angular frequency or something like this is called. Okay, so we have found the maximum speed. Now why is it maximum? Because all the potential energy converted into kinetic. Then the speed is reduced by the resistance of the spring, right? And it goes from this value, from maximum value, to zero at point of maximum squeeze. And all potential energy becomes again this one, where a is initial stretch. Okay, now let's talk about not just this moment when it crosses the neutral position, but at any distance. Let's say I'm interested at distance x is equal to d. What is potential energy and what is kinetic energy? Well, potential energy, you know, would be u of d. Exactly the same way as I calculated potential energy at point x is equal to a. I can put exactly the same calculation. It's just integral would be instead of zero to a, it would be zero to d. Which means it would be k d square over 2. So this is potential energy at any point d. Now v d would be speed. So m v d square over 2 would be kinetic energy plus u of d should be equal to full energy of the system, right? u of a. Because all this energy when we are somewhere in the middle here is partially converted to kinetic energy and whatever is left is potential energy. So this is a general for any d, right? Now from which we would, we can find out the speed at this particular moment when we are at this particular position where our object is at position d of the neutral point, right? So what is this? m v d square over 2 is equal to u of a which is k a square minus minus u d k d square over 2 from which v d absolute value is equal to k over m square root and then square root of a square minus d square, right? So this is speed of the object at any position which is at the distance d from the neutral point. So speed actually depends on the position, right? If d is equal to a maximum stretch we will have zero here. Obviously because we are saying that we just let it go without any initial speed after we stretched it. If d is equal to zero we will have the same formula as before. v zero would be square root of k over m and then square root of a square which is a, right? That's what we have just derived before. And if d is equal to minus a it will be once more it would be zero. So it goes to the minus a position when you're squeezing the spring. When the spring is actually squeezed by the object which by inertia continues moving here and spends its kinetic energy to convert into potential energy of the spring. And just as an interesting curiosity what if you will graph the speed as it depends on the distance of the neutral position. What kind of graph this function is? Well let's call it y is equal to some kind of a coefficient let's say alpha square root of a square minus x square. I'm using x and y because it's more kind of typical for functions, right? Basically it's d instead of y and x instead of d. Now what is this? Well y square is equal to alpha square times a square minus x square, right? What is this? This is basically some kind of a circle depending on alpha. If alpha is equal to 1 let's say it's pure circle because y square is equal to a square minus x square. And this is the circle because x square plus y square is equal to a square which is x square plus y square that's the hypotenuse and that's the distance from the zero. So it describes a circle. Well in this particular case maybe only a quarter of a circle because we are talking about the positive as x is changing from zero to a. But then if it's from minus a to zero it would be another quarter of a circle, etc. So basically it's a circle if we have some kind of alpha it would be squeezed the scale of the y would be different because it would be y over alpha square plus x square so it's basically a circle squeezed by some kind of a coefficient alpha. But in any case it's just some kind of an interesting detail of what exactly this dependency represents. And in this case for instance this is the beginning point where x is equal to a, right? d is equal to a this is g and the speed is equal to this is speed b. So in this particular case we have speed is equal to zero but the distance is a. As the distance diminishing speed is increasing then again when the speed is has reached its maximum at point of distance equal to zero at this point then if object goes further to the left squeezing then the speed is diminishing but the potential energy would be increasing because the distance of the center of the neutral point would be increasing to the negative side by absolute value. All right so basically that's it that's how energy is supplied and then transformed from potential to kinetic and then from kinetic to potential and then back and the circle is repeating again and again and again and that's how the equation in this particular case describes the movement and this thing or this thing rather describes the speed at any point where object is located from minus a to plus a on the spring of the central point or of the neutral point. That's it so thank you very much I recommend you to read the notes for this lecture it's basically the same thing but just expressed as a textbook and it it better lies in your in your mind actually after you read it with your own eyes. So we have like two different avenues your visual and audio as as you're watching the video and then you will have some kind of a text show impression of the same thing. All right that's it thanks and good luck!