 Hi, I'm Zor. Welcome to Unizor education. I would like to continue talking about potential energy. In this case, in a little bit more complicated case than in the previous lecture. The previous lecture was about the constant force acting on a single object. It was a force of gravity, which we considered to be constant as the object moves. Falls down to the ground. Now in this case, I would like to consider a different scenario. We are talking about the spring now. So the spring, let's say this is a neutral position of the spring. We will obviously attach some kind of an object M to its end. Now the spring can be compressed or it can be stretched, right? Now, first of all, I would recommend you to go to the dynamics section of this course and refresh information about what's there about the springs. What's important is that we have the Hooke's law, which basically tells us that within certain limits, obviously, the force which spring exerts on the object attached to it is proportional to the displacement from the center, from the neutral position. So basically if we have displaced from this position to, let's say, this position we compressed by the length L, then the force would be equal to minus KL. Now, let me explain why minus? Well, it's because my displacement is always in opposite direction to the force. If displacement is to the left, my force is acting to the right. If displacement is to the right, my force would be, if it's stretching, it will, the force will be directed to the back. So it always, the force is always directed towards the neutral position, which means that if L is negative, let's say this is my positive direction of the axis. So if L is negative, then the force will be positive. If L is positive, the force will be negative. And the K is a characteristic of a spring. It's called elasticity coefficient. So it's a specific for every spring, but it doesn't really depend on the object. It's only a spring characteristic. Well, this is the Hooke's law, which is experimental law, obviously, and it's not always true. But within certain limits, let's say, a very small deviation from the neutral, it works fine. So we are assuming this ideal situation, ideal spring, which conforms the Hooke's law. Now, what we will do is we will analyze how the potential and kinetic energy of this particular mass is changing after we have compressed the spring by the length L and let it go. Again, let it go in the same sense as in the previous lecture, when I was talking about raising the object above the ground and let it go. Basically, letting the gravity to really do some work. In this case, we are compressing the spring with the mass attached to it, and then let it go. So we are letting the spring act on its own without our interference. And obviously, there is some work which it will perform, and that's exactly what we are going to do. Now, in that lecture, by the way, in the dynamics part of this course about the Hooke's law, I have also derived a formula for the work which is equal to k L square over 2. So this is the work which I have to really spend against the force of the spring to compress it by the distance L. We will return to this formula a little bit later. It will be actually derived a little bit later in the lecture, and you will see why it's there. But right now, I'm just quoting this as a known formula from another part of the course. It's from the dynamics and about the work. Okay, now, being as it may, now let's just do the formal kind of mathematical approach to this particular problem. So what do we have? First of all, I will introduce a function S of t. This is the position of the object relative to its neutral position. So when we compress the spring, S of t t is time. S of t becomes negative. When it's stretching, it becomes positive. So let's just think about what happens with this spring. So let's say I compressed it by the length L and let it go. Well, first of all, if I compressed it by the length L, it means this particular moment my function at time zero, after I compressed it, that's the time zero, is equal to minus L Now, the initial speed of the object is zero. By the way, this speed is first derivative of my function of the distance. Speed is derivative of the distance. So the first derivative and the point zero is equal to zero. That's what we know about this function S of t. Now, let's just think about qualitatively what would happen. As soon as I let it go, my spring will start pushing the object back to its neutral position, right? Which means that there is a certain force, force which can be calculated based on this particular Hooke's law. So the force would be equal to minus k distance from the neutral position, right? So this is the force as a function of time. Now, my S of t is always negative during the first stage of this process. So my force would be positive. So it goes this way. With a positive force, I will have a positive acceleration, right? Force divided by mass. Now, acceleration, by the way, it happened to be the second derivative of the distance, right? The first derivative is speed. The second derivative is the acceleration. So acceleration is positive. So, which means our object is moving faster and faster and faster. Great. That means that at the very end it has certain kinetic energy, right? Because the speed will increase to a certain degree. What happens next? Well, the spring at this point does not act at all because it's in the neutral position. However, our object has certain speed at this point, right? Now, what happens? Well, it will move forward. It will start stretching the spring. Now, as soon as we start stretching the spring, now the spring will start acting in this case against the motion, right? Because now S of t will be positive as soon as we go beyond this point. So my force will be negative, which means backwards. So it will slow down my object. So, gradually, it will slow down to a point zero. That's the maximum stretch we can achieve in this particular case. After this, the spring will pull it back because there is no inertia anymore. The speed is zero, right? So speed has been decreased to zero and then becomes negative because the force is negative. And it will go back and it will go back to the neutral position and the whole story will repeat again, but in this direction. It will again compress it and then stretch it, compress it and stretch it and that process will continue to infinity. In an ideal situation, obviously. In the real spring, there are some losses of energy because of the friction and stuff like this. We are not talking about this. So this is our analysis of the situation. So in the beginning, we do have certain potential energy because the force will start moving an object towards a neutral position. Then, when we don't have any more potential energy, because there is no spring force exhausted in this particular case, now, but we do have kinetic energy. Now kinetic energy will move the object to that place to the end of this stretching thing. At the end, we don't have any more kinetic energy. It will be only potential because if left alone, which we wish we did, the force of the spring will start pulling it back. So it will convert, again, back to kinetic energy at this point and potentially be zero, then goes to a compressed mode and it will, again, it will lose the kinetic energy and the potential is growing. So again, my point is to calculate potential energy and kinetic energy at any given time and calculate the full mechanical energy, which is their sum, and basically show that it's really a constant. That's my plan. So let's do it. Okay, so we don't need this anymore. This is an obvious thing. So we have assumed that we have compressed our spring by the length L, initial length L. That's why f of 0 is equal to L and this is this. Okay, now, what can we say right now? Well, this is basically gives us a differential equation, right? Minus f of t is minus k s of t divided by m is equal to s second derivative of t. What is this? Well, back to calculus. This is a differential equation and these differential equations I did consider in the mass 14th course, which I might recommend you basically to refresh if you want to. But, however, I will just borrow the solution to this differential equation wholesale and the solution is basically as follows. Now, if we have an equation, something like x of t is equal to minus k divided by m x of t, right? Something like this we have. So x is a function instead of s. Now, there is a general solution. x of t is equal to general means it depends on the constants. You remember the differential equations are equations which usually, if we don't have any initial conditions, give you the whole set of solutions. So this is the set of solutions. Now, it's c 1 times cosine square root of k over m t plus c 2 sine of square root of k over m t. This is a general solution to this equation. If you will have a second derivative of this, you will get this, no matter how c 1 and c 2 are. Now, what are c 1 and c 2? Well, that's what we will do. Since s of 0 is equal to 0, now if we will put 0 here, this will be 0, this will be 1, and we will have only c 1. So c 1 is equal to minus l. Okay, fine. Now, the second condition. If we will make the first derivative of this and substitute 0 at t, what we will have? It will be it will be minus minus c 1 square root of k over m sine of square root of k over m t plus c 2 square root of k over m cosine of square root of k over m t. That's my first derivative. And if I will substitute 0, this will be 0, sine of 0 will be 0, that goes. Now, this 0 will be 0, so cosine of 0 is 1, and I will have c 2 square root of k over m is equal to 0, which means c 2 is equal to 0. That gives us basically the expression for s of t. So, let me just write it down. So, based on this differential equation, and this initial condition, I can write that s of t is equal to minus l cosine of square root of k over m t. So, I know my function of distance from the neutral depending on time. Now, at time t is equal to 0, I will have cosine of 0, which is 1 minus l. So, this is my initial compression mode. And then it starts basically oscillating. That's how cosine is oscillating. So, from minus l, it will go to plus l. And basically what we have here is these oscillations will have a period of what? 2 pi divided by this, or 2 pi multiplied by m over k. I inverted it, right? Regular cosine has a period of 2 pi, but if you will multiply it by some factor, this factor goes into denominator. That's why it's reversed here. So, I have a period. So, with this period, my mass at the edge of the spring will oscillate back and forth, from minus l relative to neutral to plus l, from minus l to plus l, etc. Okay, fine. Now, we are ready to calculate the potential and kinetic energies of our mass. So, we don't need this anymore. So, I have a little bit more real estate. Okay. So, we know S of t, that's the distance from the neutral position. Now, we can obviously have the V of t. That's the speed. This is the first derivative, right? Derivative of cosine sine minus sine, with minus it will be plus, so it will be l. Then this will be inner coefficient and sine of square root of k over m t. So, that's my speed. Okay. So, I have my distance as the function of time and speed as the function of time. Okay, fine. So, next thing is, let's define the potential energy of the mass in its compressed mode. So, as soon as we compress to distance to the position minus l, what is the potential energy at this particular point? Well, as I was saying, just let it go and see what happens, right? Well, the force of the spring will perform work. This work, which the force of the spring will perform, is actually a potential energy which my object has in this particular position, right? So, let's see what is this potential energy. Now, the work is usually its force times the distance. Now, the force is variable. Now, in this particular case, what's easier is to use this expression for the force, not as a function of time, which I can also have, because s is function of time, but as a function of t, of s, of a distance. So, in this particular case, my work is different on different segment of the way of the movement of this edge of the spring, right? But in a small area from s to s plus ds, my differential of the work is f of s times ds. So, the force during which infinitesimal increment of the distance times the distance, that would be infinitesimal increment of the work. And if I would like to have a complete work, I will have to integrate it from point minus l. That's the maximum compression to zero, back to zero, back to neutral position. So, from this compressed position to a neutral position, and that would give me my work, which is performed by the spring, right? Which is equal to integral of minus l to zero. f of s is minus ks ds. That's a simple thing. Integral of s, indefinite integral is s square over two. So, it's minus ks square over two from minus l to zero. If you substitute zero for s, it would be zero. And then it would be minus. This is the formula of Newton, Newton Leibniz formula for integration. So, it would be minus and minus. So, it would be plus and l should be equal to s should be equal to l. It would be k l square over two. Because that's actually the same, obviously, work which we initially spent compressing. So, whatever energy we spent or whatever work we performed by compressing the spring, the spring will actually perform returning back to a neutral position. So, that's why the potential energy at the very end of this compression at point minus l is equal to the same work which we have spent. This is another kind of manifestation of conservation of energy. Energy and work, they're all together. They're all kind of the same thing. So, this is my potential energy at the very, very edge. Now, fine. Now, let's see what would be my kinetic energy in two opposite ends on the most compressed mode. Well, that's where kinetic energy is equal to zero because the speed is equal to zero. But let's calculate it at the neutral position when it has the maximum speed. Okay, I have the period, right? So, my period is from most compressed position to neutral to a stretch position back to neutral and back to a compressed position. So, that's my t. That's my period. So, from the compressed position to neutral, that's t over four divided by four, right? One, two, one, two. So, this is just one quarter of the whole period. So, I have to calculate my kinetic energy as my kinetic energy at moment t divided by four. That's the kinetic energy at the neutral position, right? Now, that's m v of t divided by four divided by square, right? So, I know v m divided by two v square l square k m sine square of square root of k over m t. Now, t is this. So, it's not just any t, it's t over four. So, it's k l square over two sine square of square root of k over m. Now, t over four is pi over two m over, sorry, m over k, right? k m m k. So, I have pi over two, sine of pi over two is one and it's equal to k l square over two. The same thing as potential energy at the moment of stretching. So, whatever the potential energy at the moment of, sorry, not stretching, the compression, whatever the moment, whatever the potential energy at the moment of compression, maximum compression is, is the same as kinetic energy at the moment of passing through a neutral point when the speed is maximum, after that speed actually goes down. So, in this case, they are equal and you might suspect that in between these two points, between the most compressed and the neutral position, the sum of these two remains constant, right? And that's what actually makes the same thing as before with, for instance, gravity, when the potential energy, as we are saying, is converted into kinetic energy. Potential goes down, kinetic goes up, but their sum is always the same. That's why we are saying that one is converted into another. So, let's just do this final calculation and find out what exactly is a potential function as a function, potential energy is a function of time, kinetic energy is the function of time, we will add them together and we will see that at any time it's the same and should be exactly the same as this, right? So, that's my last little thing. Okay. Now, we know that if we compress by the length L, the potential energy is equal to work, which, which is performed to compress it, the same thing with stretching. The potential energy, if we stretch by the length L, will be exactly the same thing. So, we can definitely say that my potential energy as a function of time is equal to K s square of t divided by 2, where s is a displacement of the neutral point. Now, my kinetic energy as the function of time is equal to m v square of time divided by 2. And since we know s of t and L of t and V of t, we can calculate them both at any given time, right? So, at any given time, the potential energy is equal to K L square over 2 cosine square of, right? And my kinetic energy is equal to m 2 v square, which is L square K over m sine square of K over m t, which is equal to, let's just cancel m and we will have K L square over 2 sine square root of this. And now, it becomes obvious that as the time grows from 0, my cosine is decreasing from 1 down. So, that makes the whole thing decreasing. Now, this is a sine. It's a sine. So, as t is increasing, sine is increasing. So, this is increasing. So, my potential energy as the time goes by from 0 to 1 quarter of the period to a neutral point, my potential energy is decreasing. My kinetic energy is increasing. What about their sum? Well, if you will sum this and this, you will see that K L square over 2 cosine square plus sine square, which is 1. So, we will have this constant, which proves that it's a valid point, actually, to say that potential energy of the compressed spring is gradually converging into kinetic energy at the very neutral point where it's the maximum. And then, if we continue this, these are sine and cosine. Just believe me, if we will continue, we will get exactly the same story. The kinetic energy at the neutral point will actually cause the spring to stretch, and at the very end, the kinetic energy will be equal to 0 because the speed will be equal to 0, but the potential energy will be exactly equal to exactly the same thing. It will stretch to the same exactly distance it was compressed before, and that's how it will oscillate from minus L to plus L, from minus L to plus L, and the kinetic and potential energy together will always be equal to the same constant, but they are this amount of energies distributed among potential and kinetic in this particular fashion, which basically proves, again, that full mechanical energy in these simple kind of situations is preserved. The conservation of full mechanical energy is basically proven here. Okay, that's it. I suggest you to read the textual part for this lecture. It's available on unidisor.com. I think it would be a very good idea if you will try to calculate, basically, this potential energy as a function of time and kinetic energy as a function of time just by yourself. Do all these calculations without looking at my notes to this lecture or without listening again to a lecture. Try to do it yourself. That's a great exercise and you will have this answer, I hope. All right, so that's it. Thank you very much and good luck.