 Hi, I'm Zor. Welcome to a new Zor education. Today we will continue talking about oscillations where friction is involved. So, this is the second lecture about friction damping, like slowing down basically, the oscillations. The first one, which was called friction damping number one. It was that was the view from the energy consumption. Basically, as the object oscillates, it loses its energy on friction because friction is the force and you have to overcome that force. That means work being done and that's why the total, let's say, potential energy which the object has when you stretch the spring is basically lost as the time goes on. And we have come to certain quantitative characteristics of such a motion. Now, we will continue talking about the friction affecting the oscillations, but now we will try to just do it from the movement as the function of time, basically. So, you will be able to find out exact position, not just the ending points, which we did from the energy consideration, but actually exact position and speed, if you wish, of the object as it oscillates on the spring with a friction. Okay. Now, this lecture is part of the course called Physics for Teens, presented on Unizor.com. This is the course which is based on the prerequisite one. Pre-requisite course is called Math for Teens. Math is absolutely necessary because, for instance, today we will talk about trigonometry and calculus differential equations. So, I do suggest you to watch this lecture and any other lecture from the Unizor.com. You might have found it somewhere else, like on YouTube, for instance, but over there it's just individual lectures. Unizor.com presents a course, basically, which means lectures are organized in a logical sequence. They are dependent. Every subsequent lecture depends on whatever was before. So, it's a course. So, I do suggest that, plus, every lecture has textual notes, which basically can be considered like a textbook, if published separately, if you wish. Also, it has exams in many cases, not in all cases, but in many cases, exams. So, you can self-check yourself. Now, the site is totally free. You don't even have to sign in if you don't want to. No strings attached. No ads. Okay. So, let me start. So, we have a spring with object at the end. This is object. Now, there is a friction here. Now, m is mass, g is free-fall acceleration, mg is weight, basically, and the force of the friction is equal to mu times m times g, where mu is a coefficient of friction. Well, in this case, we're talking about motion. So, it's a kinetic friction. You know, there is a static friction and kinetic friction. Let's just forget about this small thing. Friction has this mu coefficient of friction, and if you multiply it by weight, that actually gives you the force. Force of friction, which acts always against the motion. So, if motion is to the left, the force is directed to the right. If motion is to the right, the force of friction is directed to the left. Okay. Regardless of whether the spring is, let's say, it's straight or it's squeezed, doesn't really matter. Now, the force of the spring, in turn, depends on the position. K is the elasticity of the spring, and x of t is position. So, we're assuming that there is a zero, which is a neutral position of the spring. It's not squeezed, it's not stretched. So, whatever goes to the right of this, this is a stretch, and this is a squeeze. So, to the right, x is positive, and the force is negative, directed towards the neutral position. If the x is negative, which means the spring is squeezed, the force is, again, opposite to this negativity, which means to the positive direction. That's why it's positive. So, if x is negative, force is positive. If x is positive, force is negative. Now, the previous lecture was about energy, and we have found that if you have initial position at times t0 equals to zero, we have position at point A. So, this is stretch by A, and then let it go. Well, it will go to the left in this particular case. It will squeeze after passing the neutral point. It will squeeze the spring, and it will lose A minus 2 lambda with a minus sign, because we are on the left from the neutral. We are losing certain amplitude. So, if this was plus A, this is not minus A, but slightly less on a minus sign, by absolute value. Then A, where lambda is a constant, which was mu, what else? Mg divided by k. I'm not mistaken. All right? Where is my lambda here? Right. Okay. So, that was the previous lecture we came out. Now, I would like to find actually the equation of motion. I would like to find function x of t. Now, there is a difficulty here. Now, the difficulty is that during, let's say, a cycle. A cycle is from the rightmost position to the left, and then back to the rightmost position. Well, this is a cycle, and during that cycle, the force is changing the sign. When the object moves to the left, the force is positive. When the object moves to the right, the second half of the first cycle, the friction force is directed to the left. So, I can't really put it into one formula, like this. This is absolute value, basically. To be exact, this is absolute value of the force. Now, here I have exact formula, because if I access positive, which means it's stretched, force of the spring is negative, directed towards neutral position. If it's negative, it's positive. But in this case, I can't really do it. So, what I'm going to do, I will divide, basically, my first cycle into two halves. One half from this point, from the stretched point towards this, and that would be one differential equation. And then from here back to this, it will be another differential equation. So, that's the idea. Okay. So, I would like also to assume that A is significantly greater than lambda, because if A is smaller, if it's relatively of the same order as lambda, my oscillation will just finish in no time, right? Because every time we are subtracting two lambda from the amplitude. So, let's assume that A is large enough to basically stay on the positive side if we are subtracting two lambda, one half a cycle, two lambda, and another half a cycle, et cetera. All right. So, now, if you remember, if there is no friction, we had a very simple differential equation where we were comparing the Newton force, which is mass times acceleration. Acceleration is second derivative of law of motion. We were actually equating it with the Hooke's law. This is the Hooke's law minus kx of t. Now, this is a differential equation. Now, there are certain initial conditions, and we have assumed that initial condition is stretched by A and no initial speed. That's our initial conditions. And we have basically resolved this to a very simple solution. A cosine omega t where omega is square root of k over m. That was one of the first lectures dedicated to oscillations. Now, with damping, with the friction force, we have a slightly different equation. So, what is our equation in this particular case? Now, it's not only the force of the spring which acts on the object. So, initially, we stretched it. So, the initial motion would be to the left, right? Which means our force is the force of friction directed to the right against the motion. So, this force is towards neutral point. That's why it's negative. As it passes the neutral point, x changes the sign to negative, and the force becomes positive this way. But the friction is always this way, always against the motion. So, this is my friction mu mg, right? Coefficient of friction times the weight. And this is my differential equation which describes the motion. Now, all I have to do is just to solve it with these initial conditions. Now, for convenience instead of 0, I will put here t0 times 0, where time 0 is equal to 0. Why do I do it? Because I will have t1 when object is reaching the end of this squeezing motion. That will be t1. And then back will be t2, etc. So, that's how it's more convenient to have it as an index t0. All right. So, we have this condition. We have this condition. And we have this differential equation. Now, I assume... Now, this is a constant, by the way. Mu is a constant, right? M is mass, and G is free fall acceleration. It does not depend on the motion of the object. This is a constant. Which means my solution to this differential equation should be almost the same as if there is no such a constant. So, I'm guessing... I have an intelligent guess. Let's put it this way. My solution might actually look like this. Where A and B sound coefficients. Now, if you remember, again, I was just writing this for non-friction when there is no this term. It was just A cos omega t. That was solution for x t. So, in our case, I have something similar. And that's why I hope that it will be a solution. Well, it's a good guess, actually. So, let's just try to find out if it's possible. Which means I will take the first derivative, the second derivative, substitute to this, and see if it fits, right? Okay. I can use this. So, my first derivative is equal to cos sign derivative is minus sign. And then there is an inner function, omega times t. So, omega goes out. So, I have a minus A omega sign of omega t. Now, this is a constant. So, it does not appear in the first derivative. Now, the second derivative is, from sign, the derivative is cos sign. And the inner function is omega. So, I will have minus omega square cos sign omega t. Okay. Now, let me change slightly this particular equation. I will divide it by mass m here, m here, and m here. Now, you remember that omega, omega is equal to square root of k over m. That's how it was in the very, very first lecture from there on. That's basically the angular frequency. So, omega square is k over m. So, this equation can be viewed as, with it to the left. So, it's plus, k over m is omega square, x of t minus mu g equals to zero. Okay. So, that's my, the same differential equation, just different, different letters here. Okay. So, instead of second derivative, I can substitute this. And instead of function, I will have this. So, let's see what I will have. Minus a omega square cos sign omega t. That's this one, right? Now, omega square times x of t. This is x of t. So, it's plus a omega square cos sign omega t plus omega square b plus omega square b minus mu g equals to zero. And we see minus and plus. And so, what do I have? I have basically expression for b. b is equal to mu g divided by omega square, which is mu g divided by k times m. And our familiar lambda. Lambda is, if you remember, was introduced in the very first lecture as a critical distance. Critical distance is very important whenever you have a friction because if you are stretching not sufficiently far from the neutral position, then the force of the spring will not overcome the force of the friction and the object will not oscillate at all. So, you need to stretch it greater than lambda to start oscillations. So, that's what lambda was. And we're still having exactly the same thing. So, it's a very useful constant. It combines in itself the friction itself, which is characterized by mu. It combines the weight of the object and the strength of the spring, elasticity of the spring. So, the stronger the spring, the smaller will be this dead distance when the object cannot oscillate. The weaker spring, well, obviously, the dead interval will be greater. The more heavy, the heavier we have, the more friction basically we have. Again, the more friction we have, the dead interval will be wider. So, this is a very nice kind of characteristic of this particular system. So, we have B. Now, what is A? Well, we can very easily find A if we start talking about initial conditions. Now, first of all, let's just use this one. At t is equal to zero, we should have A. Now, if t is equal to zero, cosine is equal to one of zero, right? Cosine of zero is one. So, we have A plus B. So, A plus B, this is initial position A. We know B, B is lambda. So, A is A minus lambda from here, right? So, we have our equation of motion. So, our equation is x of t is equal to A, which is A minus lambda, cosine omega t plus B, which is lambda. Now, what's important is, this is our equation during the first half of the first cycle. Okay, found it. Now, how does it look? Well, it's a sinusoidal type, but its amplitude is less than A. It starts with A, but then it goes all the way down to the end point and we will calculate what exactly this end point is. So, it's basically a smaller amplitude. So, how can we find out what is the opposite point when you're squeezing the spring all the way to the left? Well, you're squeezing all the way to the left and it stops at that point, right? It's squeezing the spring and the friction as well and it stops. Now, when does it stop? When it stops when its speed is equal to zero, right? Now, speed is the first derivative. So, when is this equal to zero? Well, when sine is equal to zero, right? So, sine is equal to zero when the angle is equal to what, pi times n, where n is any integer number. Recall your trigonometry, which means t is equal to pi over omega times n, where n is any integer number. If n is equal to zero, that's time t zero and we have the initial, when the spring is stretched initially, we have speed is equal to zero. Then, what is the next value of t when the spring is also, when the object is not moving? Well, that's when this thing with n is equal to one, right? So, we have omega t is equal to pi, right? n is equal to one. So, t is equal to pi over omega. Omega t is equal to pi. Cosine of pi is equal to minus one. And what do we do here now? Well, if we will substitute this t to this equation of motion, we will have cosine of pi minus one. So, it would be x of t one. That's the t one is a point where n is equal to one, right? So, t one is equal to pi over omega. t one is equal to pi over omega. And position is equal to, this is minus one. So, it's minus a minus lambda minus a minus lambda plus lambda, which is equal to minus a minus two lambda. Am I right? Yes. Plus minus a plus lambda and plus lambda. Two lambda, yes. This is exactly the same result which we have obtained from the energy consideration during the previous lecture. We were talking about how energy is spent during the friction. And we have found that the left most position when the spring is squeezed is at this particular point. Now, I prefer to do it this way rather than two lambda minus a, because it actually shows that it's on a negative position. We assume that a is large enough to accommodate subtracting lambda a few times, right? So, that's a negative position, which means the spring is squeezed. And initial was a on the stretch side. On the squeeze side, it's negative, but by absolute value is less than a by two lambda. So, half a cycle eats up two lambda from the, excuse me, from the amplitude. And that's again exactly as we have found in previous lecture from the energy consideration. Okay, that's it. That's the end of the first half a cycle. Now, let's start the second half cycle. Now, we have a different direction of motion, which means our friction is directed differently. Friction is directed negatively because the object on the second half of the first cycle would move from left to right on our picture. So, what is the equation of motion? Well, it's exactly the same thing here, except here and now we have force is negative because it goes against the motion, which means to the left, which means against the positive direction of x. Now, this simplified equation would look like this. My initial conditions would be different. Now, we are talking about T1 and this would be minus A minus two lambda. Right? Because that's the end point as it reached the leftmost part. And the speed is still zero at the very, at the leftmost point. We are looking for the same kind of solution here. So, what would be my solution in this particular case? So, x and x and the derivative and the second derivative are the same. But in this case, we will have this plus mg. Right? That's the difference. From which we found that k. Now, if we will put a mega, now a mega, what is a mega square? Mega square plus mg minus, so it would be minus here, right? Minus here, minus here and minus lambda. Okay, minus lambda. Now, if we will substitute into this formula T1 and T1 is what? Pi over omega, so we will have cosine will be minus. So, it will be minus A plus B is equal to, my initial position is minus A minus two lambda. Right? Now, B is lambda, so what's the A? Let's start again. Minus A plus lambda equals two lambda minus A. So, A goes here. So, we will have lambda minus two lambda plus A is equal to A. Wait, wait, wait, wait. Now, B is minus lambda, I'm sorry. B is minus lambda, so it's minus lambda. So, this is minus lambda, minus lambda. So, A is equal to A minus three lambda. Okay. So, my equation of motion would be on the second half would be different, would be A minus three lambda and this would be minus lambda. Now, the T is changing from T1 to T2. Now, T1 is equal to this, T2 is equal to, whenever we put two instead of n, so it's two pi over omega. So, during the times from T0 to T2 from pi over omega to two pi over omega, this is the equation. So, equation is different. That's what's very important. Equation on the left is one, equation on the right is another. So, the graph, if you will put the graph, you will just divide it. This is zero, which is T0 and this is T1, this is T2, this is T3, et cetera. This is initial position, this is left most, this is right most, this is left most, this is right most. But what's important is the difference between these is always pi over omega. So, it's a constant difference but the amplitude will be smaller. It's reducing amplitude by two lambda every times. So, we have found the second equation, okay, and what's the most, what's the right most point? It's where T is equal to two pi over two. Two pi over omega, right? The T2. Now, if it's two pi over omega, then cosine is equal to cosine omega times T would be cosine of two pi, which is the same as zero, which is one. So, I will have A minus three lambda minus lambda. So, my most right most point would be A minus four lambda. And that's exactly the same result as we have from the energy consideration. We are losing two lambda in amplitude on every half cycle. We have lost two lambda on the way back, way to the left, and we are losing another two lambda on the way right. So, squeezing minus two lambda, stretching another two lambda. So, now it's minus four lambda. And as you understand, it will continue doing exactly the same thing. So, every time amplitude is diminishing by two lambda. So, I assumed in the very beginning that A is big enough so we have a few cycles basically until it's finished. But every time you are subtracting two lambda and eventually you will get situation A minus whatever number of lambdas, it will be within the dead zone and the object will stop oscillating. So, it's very important to understand that if there is a friction, constant friction, obviously, we are talking about a finite time before it completely stops. In the next lecture when I will talk about instead of friction, we will talk about viscosity. Like, for instance, the whole oscillation is under water, let's say, and the water prevents it. There the slightly different laws of physics are actually acting and we will talk about this. And over there, oscillations will not stop. They will infinitely continue but will be smaller and smaller and smaller. Here is that stop with a friction. So, what's the somewhere of the whole thing? First of all, I probably shouldn't really call it period because the period means it's repeating exact motion. So, it's not really a period but it's a cycle. So, the cycles between rightmost and then the next rightmost position, which is shorter and shorter and shorter. So, the cycles had exactly the same timing from the time perspective. So, after each pi over omega, you have another maximum deviation from the neutral. At 0, at pi over omega, at 2 pi over omega, at 3 pi over omega, 4 pi over omega, etc. That's how it goes. So, the timing of leftmost and rightmost position are the same interval. But amplitude will be shorter and shorter and shorter. So, the graph would look like this. And as I was just talking about, the equation of this part is not exactly the same as the equation of this part. Here, we have the first one, which was a minus lambda cosine plus lambda. And here, we have a minus 3 lambda and minus lambda. So, it's a different equation. But they are glued together here. And so, the x of, this is T1. So, the x of T1 in the first and the second are exactly the same. And so, it's for T2, for T3, etc., etc. And even the first derivative would be the same if you can check it, if you wish to. Because the first derivative is 0 will be always at these points. So, we have these two pieces of this graph glued together relatively smoothly. Because they are at the same value and at the same first derivative, which means the tangential line is horizontal. That's the first derivative. And that basically completes all the research of how the friction, constant friction, affects the oscillations of ideal object and ideal spring. Okay, I suggest you to read the notes for this lecture. They are relatively well organized, like a textbook basically, as I said. Other than that, that's it. You get it. Thank you very much and good luck.