 Hi, I'm Zor. Welcome to Indizor Education. So we are continuing talking about oscillation as a kind of movement. Now, the previous lecture was about oscillation of the pendulum, which is basically just going like this. This lecture is about the spring, which is stretching and then contracting again. Slightly different way of movement, but again, it's kind of an oscillation and there are obviously proper equations which the movement should satisfy and we are going to come up with these equations right now. Now, this lecture is part of the course called Physics for Teens presented on Unizor.com website. I suggest you to watch this lecture from this website. There is a prerequisite for the entire course of physics. It's the course of mathematics on this website. It's called Maths for Teens. And the website contains some other courses like US Law for Teens, for instance, etc. Now, the site is completely free and there are no advertisements, so I do recommend you to watch it from the site because every course has certain logical segments of educational material presented. It's not just you just found one particular topic which you are interested in, somewhere on YouTube where all these lectures are actually stored and just watch this lecture. All the lectures are interrelated. It's the course, basically. Now, the Maths for Teens is a completed course. Physics for Teens is being done as we speak. Right now, we are in the Mechanics and we are talking about springs and spring oscillation. All right, now, back to business. Let's consider, we have the following situation. Here is our spring and this is the spring in its neutral position. Neutral means that it can stretch a little or it can contract a little and this is the neutral level. Then we have a little weight here. Obviously, a point object which has certain mass, M. Now, what happens with the spring if we just hang this object? Well, it will stretch a little bit, right? Under the weight of this object and we are talking about the weight, so there is something like a G, which is a free fall acceleration. So, the weight is M times G. This is the weight, M times G. All right. Now, everything is not moving. It just stretched a little bit and that's it. Now, what happens if I will take this particular object and pull it down a little bit and let it go? Well, obviously it will start oscillating. Up and down, up and down. And if we are talking about ideal spring, which doesn't have internal friction and air resistance we ignore, etc., etc., it will just continue oscillating indefinitely. So, this is kind of an ideal situation and we would like to come up with the law of the motion in this particular case. All right. So, first of all, we are dealing with the spring and we have to know how spring actually reacts when you are stretching or contracting it. Well, there is so-called Hooke's law and the Hooke's law tells that if there is a spring at certain neutral position, then the force, this spring, exhorts, if you are stretching it, is proportional to the distance you are stretching. So, if you have a spring like this and you are stretching it like this, now if this is the neutral state of the spring and this is a stretched one, so the difference between these two, this difference, is the difference I am stretching from the neutral position. And the force which spring in this case, it resists the stretching, it's trying to contract again. So, this force is equal to minus k x, where x is the distance I am stretching. k is some kind of a constant. It's the elasticity of this particular spring. Obviously, different springs have different elasticity. So, in this particular case, I assume that k is given. So, we know what kind of a spring this is and what it's coefficient of elasticity. Now, I have to warn you, however, that this is a very, very approximate law, this Hooke's law. It works more or less precise if these stretching are really small. I mean, if you are stretching the spring significantly more than its original length in a neutral position, then the Hooke's law will not be as precise. So, under certain circumstances, within certain relatively small degree of stretching, this is working. And we are assuming that that's exactly the case right now, because that's the only thing which we know about this spring. If it's a more significant stretching, etc., we don't have any laws to govern derivation of our equations, right? So, we are assuming that this is the case and the Hooke's law is is working. Now, another thing which should be given. So, we have mass, we have acceleration of the free falling, we have the elasticity of the spring, and then I will probably have to say what exactly is the amount of stretching. So, if this is the neutral position and I have stretched, let's say, to this level, so this is initial stretching of this object on the distance L from the neutral position of the spring without any load, right? So, L is also given, that's initial stretching. And then I let it go. Now, obviously, what we have to do right now is position of this particular object at time t would be a function of the time. So, x is the distance from the neutral position to current position of the object. And that's the function which we have to determine, all right? Now, first of all, I told you that we have stretched initially at moment t is equal to 0 to the length L from the neutral position. That means that we have an initial condition this. Then I told you that we just released this particular object without pushing it in any direction, which means that my first derivative, my speed, is equal to 0. Now, we will come up with some differential equation for x of t. These are initial conditions, and we will apply these conditions to some constants, which will be part of the general solution of the differential equation, and we'll see what happens. All right? All right, now. So, let's think about what kind of forces are acting at any moment t on the object of mass m if it's on the distance x of t from the neutral position. Well, there are two forces. One force is the weight, and that's always the same, m times g. And another force is into the opposite direction, f, which is equal to minus k x of t. Why? Because x of t is the stretching of the of the spring. This is position where we are stretching the spring. Minus because it's directed to the opposite direction relative to the weight. So, we are assuming this down is a positive direction of the x-axis, right? That's what x is. And the k is a known coefficient of elasticity of this spring. Now, the combination of these two forces is basically the resultant force, which gives our object certain acceleration, right? So, if w minus f or f minus w, it all depends on their their relative position, that would be well, actually, I should say plus because the minus is already here. That should be the resultant force, and it should be equal to m acceleration, and acceleration is a second derivative of x of t. So, x of t is the distance, the first derivative is speed, and the second derivative is acceleration. All right, so basically this is a differential equation which we have. Now, let's put whatever we know about this. So, w, the weight is mg. Now, the force of the spring, the elasticity is minus k x of t, and it's equal to m x second derivative of t. This is differential equation, and it's a linear differential equation. In the mass for t-ins, I touched these, and I explained how to solve these relatively simple differential equations. So right now, I'll just write down the solution to this particular differential equation, and then we will talk about what exactly are the constants it depends upon. Now, the solution to this is something related to trigonometry. If you remember, the derivative of a sine is a cosine, derivative of a cosine is minus sine, so that's how we are relating to second derivative and the and the original function. Now, this is a little complication, but everything is taken care of. Now, my general solution to this particular differential equation is the following. It's one constant times cosine of t-con times square root of k over m plus another constant sine of t times square root of k over m, and then I need a free member plus mg divided by k. This is the general solution to this equation. All right, now what we can do right now is we can differentiate these ones and twice, compare with this, and it will satisfy this particular equation. So, just out of curiosity, let's do it relatively quickly. The first derivative is minus c1 from the cosine is a sine, and then there is an internal function sine of t square root of k over m. Now, plus c2 again square root of k over m, the derivative from a sine is a cosine, t times square root of k over m. Now, this is the constant, so the first derivative is 0, and my second derivative is minus retains. Now, this square root and this square root will be k over m cosine of t square root of k over m, and this will be also minus c2 no, not a square root anymore, because it's a square root times square root, so it's playing k over sine of t square root of k over m. Now, what happens if let me make this a little bit simpler, I'll divide it by m and see what happens now. So, the second derivative is this one. Now, if I multiply it by k over m, if I multiply this xt by minus k over m, I will have this minus k over m cosine, which is this one, and then minus from the, it will be k over m sine, which will be this one. So, if I will multiply my x of t, this one, by this k over m, this will go to this, this will go to this, and what happens with this guy? If I will multiply it by k over m, I will have only g left, g left. So, this is exactly this multiplied by k over m, by minus k over m plus g, right? Okay, so this is a solution. So, we have the expression, but it's depending on two different, it depends on two different three constants. How can I get these constants by initial conditions? So, this is the initial position. We stretch it by l. So, if I will put 0 into this function, now, this would be 0. This would be 0, so cosine of 0 is 1. So, I will have c1 plus mg over k equals l. So, c1 is equal to l minus mg divided by k. We've got that. Now, now from this, I have to take the first derivative. Now, the first derivative, this will give me sine of something with some coefficient. So, in 0 it will be 0. This will be cosine. So, it will be c2 times square root times cosine of 0, which is 0. This will not be participating in the first derivative, because it's a constant and the first derivative is also 0. So, I will have c2 times something is equal to is equal to 0. So, c2 will be equal to 0. So, here is my final formula. x of t equals l minus mg divided by k cosine of t times square root of k over m plus mg divided by k. So, that's the formula and it gives basically the behavior of the position of the object at time t as a function of time t. So, what can I say about this? First of all, it's the oscillations are trigonometric function. So, which means it's really going up and down, up and down to infinity. That's one thing, one observation. Now, another observation is what happens if l is equal to mg divided by k? If l is equal to mg divided by k, what does it mean actually? Well, it means that lk equals to mg. Now, let's think about what is l times k? l is my initial stretch of the spring relative to the neutral position. Times k gives me the force with which spring tries to pull back. mg is the weight. So, it looks like on my spring I have stretched the spring exactly to the position which is producing the force directed up exactly equal to the force directed down. So, the force of the spring is completely balanced the force of the gravitation, the weight. And obviously, this thing will not move at all. So, if it's already stretched a little bit by this particular length, l, which is equal to mg divided by k from the neutral position, then this is an equilibrium point. So, my object will not move up and down. Only if I will stretch it a little bit from this equilibrium point, then it will start moving up and down. So, otherwise, if l is equal to this expression, this gives you zero and we will have a constant position of my object, it will not move at all. Now, if l is less than mg divided by k, that means that I didn't really stretch it, I really pulled it a little bit up. So, it will start from the upper point and then it will go down, up, down, down, etc. So, that's all kind of simple thing. So, basically, that's it. So, we have derived this equation. Now, I would like you to recall the previous lecture where we were talking about pendulum. Now, in case of pendulum, if you remember, we had a differential equation which cannot be solved in simple algebraic formulas and we had to really consider that the angle which we are tilting this pendulum is really small, in which case the sign of something can be replaced with that something, just because the sign of an angle and angle itself in regions are very close to each other whenever the angle is very small. Sign of the angle and angle are very close if the angle is small. So, we made this particular simplification and then we were able to come up with another equation also depending on trigonometric function, so-called harmonic movement. Harmonic means it basically behaves like sine or cosine. And here also. Now, does it mean that in this particular case we really didn't need any simplification? We did need that simplification in case of the pendulum and in this case we really didn't resort to it. Well, not so fast. The universe is infinitely complicated, let's put it this way. We have simplified already by using the Hooke's law. I was telling you that the force of the spring is proportional to its stretching or contracting, only in case these stretching or contracting are very, very small. So, that's where we have made this simplification. And unfortunately it's also there and the real life obviously is much more complex, etc. And as in many other cases, physicists decided to simplify their lives and instead of a very, I don't know, really very difficult kind of research, they decided, okay, we will make some nice formula which we will just base our calculations upon, etc., etc., but you have to know the margin where you can move within these calculations. So, the formula for Hooke's law or the formula for any of these equations or the pendulum equations, well, they are okay within certain boundaries. As soon as you are out of these boundaries, don't rely on these formulas anymore. So, for instance, the formula for the period of pendulum which I gave in the previous lecture, in this case, we also can say what's the period of this particular spring-related oscillation. Well, simple, it's 2 pi divided by this coefficient, or I will multiply by its reciprocity. So, this is my period of oscillation in this particular case. So, it's mass, the greater the mass is, the slower will be the oscillations. The stronger the spring is, which means this coefficient is greater, then the faster it will oscillate. So, that's what this period actually means. So, the stronger the spring, the faster it oscillates, and the bigger the mass, which we are hooking to this end, the slower it will be. Okay, that's it. I do suggest you to go to unisor.com to this course and read the textual part of this lecture. This and any other lecture has textual parts. And then there are exams for every topic. For instance, for this topic of pendulum, well, this is the pendulum and the spring, actually. Combine them together. There will be exam as well, and I do suggest you to take it just for your own satisfaction, challenge yourself. That's it. Thank you very much and good luck.