 Hi, I'm Zor. Welcome to a new Zor education. We continue talking about oscillations, and a few lectures starting from this one will be dedicated to oscillations with some other external or internal forces acting to, well, basically prevent indefinite oscillations. Before we were completely ignoring all these external forces and just assumed that we have something like a spring with an object going back and forth, etc., without any friction, without any resistance, and we have derived certain equations of the motion. Now, we will do exactly the same in case of certain forces acting on this object, and this particular lecture, which is called friction damping 1, is about how the friction affects the oscillations. Now, this is number one, and there is a number two that will be the next lecture, also about friction, and this one will be more oriented towards how energy is actually spent. The next lecture will be really about equations of motions with differential equations and stuff like this. So, that's the plan. Okay, now this lecture is part of the course called Physics for Teens, presented on unizor.com. I suggest you to watch the lecture, this one and all other lectures from unizor.com, because it presents the course, which means there is a logical connection between the lectures. For instance, in this particular lecture I'm using the results of the previous one, so it's a course, so take the course and you'll be fine. Now, there is a prerequisite course, the Physics for Teens, it's called Mass for Teens, on the same website, and I do suggest you to familiarize yourself with all the concepts presented there, well, unless you know them, which is fine, because physics uses the mass always, I mean, mass is the foundation of the physics. Okay, so, first of all, let me just explain this particular name. Well, you know what friction is, so I will not stop on this, what's damping? Damping is something which affects the oscillation, which slows them down, which prevents them to indefinitely continue without any change. Like if you have ideal spring, and it's the vacuum in space, there are no forces around, it will basically oscillate indefinitely. Now, we are talking about something down to Earth, and in particular, the example which we will use is, if this is a spring and some kind of an object here, and we assume that there is a horizontal surface and there is some kind of a weight, which is mg, where m is the mass and g is acceleration of the free-falling acceleration on the surface of the Earth, and there is certain friction, and there is a coefficient of friction, mu. This mu multiplied by the weight gives exactly the force of the friction, which always goes against the direction of the movement. Well, if there is no movement, there is no friction, but if there is a movement left or right, then the force of friction acts when the body moves to the right, the force acts to the left, and if the body moves to the left, the friction force directed to the right, so it's always resisting. So, damping is gradual reduction of the intensity of the movement, in this case intensity of oscillations. That's what damping actually is, damping effect, and in this particular case, we will talk about damping as a result of the friction. So, if you just have this spring, you stretch it and let it go, well, it will start oscillating, but the amplitude will be lower and lower, less and less, until it will stop, basically. Now, we will examine exactly in two lectures how this mechanism of damping works in case we are talking about friction. So, that's the plan. Now, other examples of damping, which is not like a friction, for instance, what is the whole thing, the oscillation is in some medium, let's say, in water. Water has certain viscosity, and there is a resistance, obviously, of the water, and obviously, this oscillation will slow down, slow down, and will stop eventually. So, there are different reasons, but maybe if it's in the vacuum and in space, there might be certain internal resistance to indefinite movement. For instance, if spring is not ideal spring, and there is some kind of a inner resistance between the molecules, and gradually the energy is spent, basically, to move these molecules around, then again, the oscillations will slow down. So, there are different examples of how damping actually can occur, and in practical life, there is no oscillation without damping. So, we always have to, if we want to continue our oscillations indefinitely, we have to feed it up with energy. For instance, put the battery into the watch, or wind up the watch, something like this. Okay, now, so we were talking about friction as a damping force. Now, I would like to talk, in this lecture, I would like to talk only about energy. Now, the previous lecture was about potential energy of the spring stretched or squeezed by certain distance. But in this particular case, I will just complicate the issue adding the friction. Now, if we are talking about energy, and there is some external forces involved, and we would like to somehow make some equations, what we have to do is, we have to examine basically the work, which this particular force of friction is doing. So, obviously, the work is equal to the force of friction times distance, whatever the distance is. Now, in this particular case, I would like you to notice a very important detail, that the force, which is basically mu times force of friction, which is mu times mass times acceleration, free fall, it's constant. So, no matter how stretched or how squeezed our spring is, the force will still be the same. No matter how fast or how slow the object at the end of the spring would move, the force of friction will be the same. There is one little detail, there is a distinction between the friction during the motion, which is sometimes called kinetic friction, and the friction, which basically is when the object stays still at rest, which is called static friction. Static friction is usually a little bigger than the kinetic friction, but we will ignore this difference just for simplicity. Okay, fine. So, this is the work. So, whenever our object is moving, we have to overcome the friction, which means we have to spend energy to overcome this and spend this amount of extra energy wherever we go, left or right. We have to spend this amount of energy where G is the distance we are moving. Okay, so, now, let's assume that we have a system of coordinates with only one dimension, which is X, and there is a neutral position, neutral position of the spring, which is X is equal to zero, which means whenever my object is stretching, its position is positive, whenever my object is squeezed, its position is negative. And this is just the kind of a wall to which the spring is attached, and this is the floor where the object is lying upon. We are obviously assuming that this is a point mass object. So, okay. Now, let's go back to the previous lecture, and the result of this is that if my spring is stretched or squeezed by some distance D, I'll use the same distance here and here, but it's different. This is just the distance covered by the object, and this is the distance we have stretched the object from the neutral point, positive or negative. Then the potential energy, which depends on the position, is equal to KD square over 2. That's the result of the previous lecture. So, D in this particular case is the distance we have stretched or squeezed. So, if this is the new position and this is D, this is the neutral and this is the stretched position of the object, then this object has potential energy equal to this one. That's the result of the previous lecture. The formula was derived clearly without any problems. So, I'll use it, and if you just don't know, I suggest you to go to the previous lecture again through the Unisor.com. You can go and you'll basically be familiar with this particular result. So, this is the potential energy. So, what happens if we have stretched our spring by some initial distance? So, we have a moment at moment T0 equals to 0. My position in this moment is equal to A. So, let's say we are talking about stretching by A. So, let's use A in this particular case. Well, then, obviously our spring has certain potential energy, which is U. A is equal to KA square over 2, right? That's the potential energy which our spring possesses. Now, if I will stretch it with my hand and let it go so there is no initial speed, I'm not pushing it back and forth, left or right, which means that my first derivative at this moment is equal to 0. So, only the spring now at that moment is pulling the object back to the neutral position. Not so fast. There is a friction. So, what's the friction? Friction is a constant force. Now, what's the force spring pulls the object? Well, that's the Hooke's law, right? The Hooke's law that the force on the spring is equal to minus K, I'll use X. So, if X is positive, direction of the force is negative. Positive means I have stretched it, so it goes back to the original direction. If my initial position is a squeeze, which means my X is negative, the force will be directed forward. So, negative and minus it will be plus. So, this is the Hooke's law. Now, I have to be a little bit more precise here because both are functions of time. If my X is changing as time is going, my spring is oscillating, the force is also changing. So, what's interesting is this force is constant because it depends on the weight and the coefficient of friction. This force is changing. So, let's see what happens in initial moment T0. Well, my force of spring at moment T0 is equal to minus KA, right? I would like to make one very simple comment. Hooke's law is symmetrical whether we are squeezing or we are stretching the spring. The distance is the most important part. So, if my spring is stretched by certain distance or squeezed by certain distance, the absolute value of the force is exactly the same. We're talking about ideal spring, obviously. So, now, which means that I can actually do whatever I'm doing with A positive. So, I assume that at initial time I'm stretching the spring. If it's the opposite way, it would be exactly the same kind of logic. So, let's just concentrate on A positive. In this case, my spring would pull back to the neutral position with this force. So, we don't forget the spring, the object's friction, right? Friction is mg, constant, by the way. This is depending on time, but we are talking about initial moment of time. This is constant. Now, friction is always directed against the movement. So, this guy is pulling back to the original position. This guy resists, which means it's directed that way, and we have to overcome. So, only if the absolute value of this force is greater than the absolute value of this force, the movement starts. Otherwise, there is no movement. The object would be at rest. So, the absolute value of this, we are assuming A is positive, as I was saying, is Ka. And Ka should be greater than new mg for movement to start. So, if the friction is very big and we are stretching just a little, it will not be sufficient. The pulling force of the spring back to the neutral will not be sufficient to overcome the friction. So, the object will just stay. I have stretched it a little bit, but it will not move back. It will just stay there, because the force of the spring to basically contract will not be sufficient to overcome the friction. So, this is a very important inequality. So, my force should be sufficient, force of the spring should be sufficient. And for this, my stretch should be greater than mu mg divided by Ka. I call this lambda. Now, I suggest basically to call... I didn't really see it in some textbooks or internet. I suggest this, at least for the purpose of this lecture, to call a critical distance. So, unless we stretch greater than the critical distance, the object will not move. It will just stay. What happens if we will stretch it greater than critical distance? We will investigate right now. But if it stretches less than the critical distance, it's not moving. Absolutely the same if we squeeze it. So, there is this lambda on this side and lambda on this side. This is minus lambda, right? Because this is zero. This is x is equal to zero. This is minus lambda. So, these two points are very important. Unless we stretch beyond that point, which depends on the coefficient of friction, weight and how strong the spring is. Because this is the elasticity coefficient of the spring. So, how strong the spring and how strong the friction. That's what basically it's all about. So, these two points on both sides of the neutral, one is for stretching and another is for squeezing. This key is an interval, which is basically a dead interval between them. So, if we stretch not further than these two points, stretch or squeeze. Stretch not further than this and squeeze not further than that. The spring will not contract back to or stretch back whatever to original position. Okay. So, this is a critical distance and these are two critical points. All right, fine. We have found this. Now let's assume and again now we're talking more about stretching. We are assuming that we are stretching greater. So, our A initial stretch is greater than lambda, than the critical distance. So, it's further from the neutral position than the critical point, stretch critical point of stretching. This is critical point of squeezing, this is critical point of stretching. Okay. So, let's assume we are further. So, then the movement starts. Well, the movement starts but let's just think about it. It will start back to this position, this direction. Now, let me ask you this question. Will it reach the point symmetrical to this one as if there is no external forces as the real oscillation? Real oscillation without any kind of a damping will be from A to minus A. If you remember it's a cosine function with the amplitude A. In this case, my question is, will it reach the point minus A? Well, answer is no because the point minus A, the potential and energy if D is equal to A or D is equal to minus A, potential energy is the same. So, potential energy of stretched by A is exactly the same thing as potential energy as squeezed by A. And it's impossible because from here to here we have to overcome the constant force of friction. So, we have to spend some energy. So, it will definitely not reach that point minus A. Because otherwise it will be the breaking of the energy conservation law, right? Now, what energy conservation law says? Well, let's consider it will stop somewhere at point B. And point B actually, depending on where is A, point B can be before or can be after the neutral point as it moves to the left. But it will be somewhere. Okay. So, what do we have at point B? At point B where the object stops, its potential energy would be this. And it should be smaller than this one, right? By what exactly part it should be smaller? Well, by the amount of work we have to spend on friction to overcome the force of friction. Because the force of friction as our object moves here, now force of friction moves directed there. So, we have to really exhort certain force to overcome it and spend some energy. What's the energy? Well, the energy is distance and force, right? The work is equal to force of friction times distance. What's the distance? Well, if this point is A and... Okay, let me give another little picture. So, this is our object. This is spring. So, this is x is equal to zero. This is lambda. This is minus lambda. This is A. And let's say B is somewhere here. In this case, B is negative. But B can be positive as well. It all depends on the difference between potential energy and the work which needs to be done to overcome the friction. So, the distance is A minus B, obviously. I'm talking about absolute value of distance, right? So, the distance is A minus B. So, my loss in potential energy, which is this, should be equal to work, which I have to spend overcoming the friction on the distance A minus B. So, it's F friction times A minus B. It's M... It's mu. Mg times A minus B. Right? Now, this is kA squared, half minus kB squared. So, it's k over 2, A squared minus B squared. That's what it is, right? Difference between them. Or, I hope you all know by heart that A squared minus B squared is A minus B times A plus B. If you don't believe me, you can just multiply and you will see. So, that's what we have. This is equal to this. From this, we basically cancel A minus B. And we put k back to here. So, what do we have? We have A plus B over 2. A plus B over 2. k goes there and A minus B cancels. Is equal to mu Mg divided by k, right? So, k goes here. A minus B cancels. A plus B over 2 stays. And we have mu Mg divided by k, which equals to lambda. We just introduce this critical distance. Same thing. It's the same value which defines the minimal distance we have to stretch to start movement, right? Remember, f spring by absolute value, which is kx, whatever, should be greater than absolute value of this, which is equal to Mg mu Mg. From which k greater than mu Mg divided by k, and we have introduced this new variable lambda. So, this is the same critical distance. Well, that's why I think it's important to have a name for this particular constant. This is a constant of the system, the whole system, which includes spring characterized by elasticity and the object which is characterized by its mass and the force of gravitation and the coefficient of friction. So, it's all together. It characterizes the whole system. Every system has this particular constant, lambda, which defines its characteristics, well, related to oscillation, of course. There are some other characteristics, but related to oscillation is basically one particular constant concentrates the whole system's property. That's very nice, I think. So, this is a plus b over 2 equals lambda. This is a very important equation. Now, what is a plus b over 2? That's the midpoint between a and b. So, it looks like this particular critical point of stretching, if we stretch beyond it, then the coordinate where object stops when it moves to the left should be symmetrical. So, in this particular case, b is not here, b is supposed to be on the same distance, this should be b. So, this and this should be equal. Lambda is a midpoint between a and b. That's what it is. And it's very, very important. So, whenever you have a situation of this type and the friction actually is involved, you always calculate where exactly are critical points and then you say the following. If you stretch by certain distance, then the object will start moving and it will end its moving at the point symmetrical to the start at the critical point lambda, which is basically a fixed constant for the system, which is this one. Absolutely the same thing is about squeezing. If you squeeze it beyond minus lambda, it will again start moving and symmetrical point relative to minus lambda would be the result of movement. So, if a is here, for instance, then object will... So, well, it cannot be so. It's too much, right? If a is here, for instance, so we are squeezing up to this point and then let it go. Then it will stop on a symmetrical point here, all right? So, that's very, very important consideration, which is based only on the energy considerations, because the law of conservation of energy is supposed to be observed. Now, let's just examine a few cases, what exactly happens depending on how much we stretch, okay? So, here is our object and let's say this is minus lambda, this is zero and this is plus lambda. And these are critical points where lambda is a critical value mu mg divided by k. So, my first note about this is that if we are stretching less than lambda, and I'm talking only about stretching, squeeze will be exactly the same. So, if I'm stretching less than by lambda, we saw that there is no movement. So, let's assume we're stretching beyond lambda, okay? So, a is greater than lambda. Let's just assume that a is less than lambda but less than 2 lambda. Now, again, a plus b over 2 is equal to lambda. So, b is symmetrical to a relatively to lambda. So, 2 lambda is here. So, we are talking about a being somewhere between lambda and 2 lambda. Well, if that's the case, then symmetrical point as you see would be here, right? So, if my a is between lambda and 2 lambda, my symmetrical point b, you can actually calculate it from here. b is equal to 2 lambda minus a, right? From here. Or minus a minus 2 lambda. That's kind of a more convenient thing. Okay, so what happens with b? b will be between 0 and lambda, right? So, the object will start moving. It will move beyond the critical point to the same distance. This is a. So, this is the distance a minus lambda. And we have to subtract from lambda this. That's how we will get 2 lambda minus a, right? 2 lambda minus a. 2 lambda minus a. Now, but this is a dead zone, remember? From minus lambda to lambda, if we are stretching or squeezing our spring, there will be no movement because the force of the spring when we reach this point will be insufficient to overcome the friction force. So, it will move from here to here and stop. Okay, let's increase it by another piece. Let's say this is 3 lambda. So, from 2 lambda a, 3 lambda. So, if my a is here, then I'm stretching by lambda and another lambda and a piece. It will be symmetrical relative to this one. So, it will be somewhere here. This will be between minus lambda and 0. It will be minus, minus lambda, b and 0. Still dead zone. So, if we will stretch it not beyond the 3 lambda, it will start moving, I mean above 2 lambda but less than 3 lambda. It will start moving but it will still stop in the dead zone because the dead zone is from minus lambda to lambda. Okay, let's go further. I have something dropping every time. Okay, let's go to 4 lambda and put our a here. So, from 3 lambda a to 4 lambda. Now, symmetrical relative to lambda would be what? Minus 2 lambda b lambda, which means here. Minus 2 lambda. This would be b. Now, this is already moving zone. We are squeezing beyond the critical point, which means it will start moving back. Where? To a point symmetrical to a critical point of the squeezing. So, it would be somewhere here. That would be c point. So, b plus c over 2, the midpoint should be minus lambda. Which means c is equal to minus 2 lambda minus b. And we know what b is. So, it's minus 4 lambda plus a, or a minus 4 lambda if you wish. So, what happens is, if we stretch sufficiently above the 3 lambda, it will start moving this way and then this way and stops. And we are losing 4 lambda amplitude on this. So, we started with a somewhere here, but then we will stop somewhere here. So, we are losing 2 lambda amplitude. Now, in this particular case, it happens that the object will be again after the second move. So, the first move, let's say it's the first half of cycle. And then another, to a different direction, it would move back into the dead zone and stop. Now, if I will move a beyond 4 lambda, it will stop a little further in the dead zone on the positive side, but still stop. And only if I will start significantly further, like a is equal to 6 lambda or something like this, it will go this way and then this way lose 2 lambda on the cycle. Full cycle is back and forth. And then, if it will be beyond the dead zone, it will start again oscillating. So, it will start oscillating again and again and every time it will lose 4 lambda on amplitude. 2 lambda this way and 2 lambda this way. And that's very important. This is a damping effect of the friction. So, the friction results in this particular case is we are losing 2 lambda on each half a cycle. Now, that's basically the end of this lecture. The next lecture, I will go into examining what exactly the equation of motion, how exactly this happens and what is the function which basically describes the movement of this thing. But what's important is that every time our half a cycle is losing 2 lambda and eventually whatever the A is, after finite number of back and forth movements, we will end up in a dead zone because every time we are subtracting some constant 2 lambda from the initial amplitude. And it will stop in a dead zone and that will be the end of the motion. Well, that's it for today. That's the view from the energy perspective on the oscillation with constant friction. And again, next lecture would be about equation of motion like a function of time t and we will come up with some properties and then we'll probably have exactly the same result from these analytical approaches that we will lose 2 lambda on each half a cycle. Alright, until the next time, thank you very much and good luck.