 Hi, I'm Zor. Welcome to a new Zor education. Today's lecture will be also about oscillations and in this particular case, we will talk about oscillations of the rope, just a regular rope. Now, why? We are talking about light, right? Well, here is the reason. I would like to introduce the concept of energy, gradually, energy of light, I mean, and considering the light is oscillation, I started with simple oscillations. Previous lecture was about oscillation of the spring and what are the kinetic and potential energy, etc. Now, the oscillation of a spring are longitudinal because the parts are moving in the direction of propagation of the waves. Light is transversal oscillations of electromagnetic field. So, before going to light, I would like to talk a little bit about simpler transversal oscillations. And oscillations of the rope or oscillation of a string on the violin, they are transversal, which means parts are moving perpendicularly to the propagation of the waves. Now, obviously, whatever we are talking is a model, not the reality. In the reality, oscillation of the rope are very complex and whatever I will talk about as a model of this particular motion is significantly simplified thing. So, in many cases, I will assume certain things which are not exactly what the nature actually provides for us, but our approximation to the nature. So, don't judge whatever I'm saying to, you know, strictly because, again, it's kind of a simplification, primarily for educational purposes. Now, this lecture is part of the course called Physics 14 presented on Unisor.com, and I do suggest you to watch this lecture from this website and use the website's resources. Resources are significant because every lecture contains detailed, textual notes, which basically are like a textbook. I mean, chapter of the textbook for this particular lecture. So, every lecture has corresponding chapter, right next to it on the screen, whenever you go to Unisor.com. The website is totally free. There are no strings attached. There are no advertisements. So, even sign-in is optional, basically. The only purpose for signing-in is to create new educational material, which is the purpose of the teacher, actually, rather than the student. And there are exams which you can take as many times as you want. I do suggest you to take it as many times as it requires to get the perfect score. And now back to oscillation of the rope. So, as I was saying, the rope oscillations represent transversal oscillations, which is much closer to light oscillations of the electromagnetic field than the previous lecture, which was dedicated to spring oscillations. There is a Hooke's law with the spring. There is no exact equivalent whenever you are doing this oscillation of the rope or any transversal oscillation. We really have to devise some kind of a correspondence between the position and the forces to apply them to something like the second Newton's law. And that's exactly what we're going to do. So we will examine the forces which are acting on each element of the rope as it oscillates. And then we will basically use the second Newton's law to create certain equation, the wave equation, which would be the end of this particular lecture. So I will try to examine the movements of every particle, every part of the rope, and I will create certain differential equation which describes this particular movement and obviously suggest a solution to this. And obviously the solution will be sinusoidal oscillations. But you will see basically. All right. So the model which I am using is the following. So you have a rope and let's say it's fixed on very far end, far enough not to take it into consideration. But this end on my side, I will just move up and down in a sinusoidal kind of movement. So the deviation from the middle point, from the neutral position, would be described as something like this. So at moment t is equal to zero, I will have sin of zero zero, so it will be in a neutral position. Now then as t is growing, I will go up to the maximum level of A when sin is equal to one and then move back. Back to the neutral position and to the negative A. So that's how it will go from plus A to minus A in this particular fashion. Again, this is my personal choice how I oscillate this end of the rope. Some other people might do the same, the different thing, but in my case I decided to do it this way. Why? Well, because it's closer representing whatever the nature actually is whenever somebody is doing exactly like this. So it's kind of close to whatever the reality is. Now rope is basically a contiguous curve, right? And it's kind of difficult to deal with this. So obviously I will divide it into really small pieces, connected between themselves. So consider this to be like a necklace, basically. So you have pieces this, this, this, this, this, this, this. So and these are links between these pieces. So this is again, it's a model. Why? Because it's easier to describe whatever the forces are acting and as forces acting an individual piece. Now in more mathematical lingo, obviously that would be differential of the length of the rope. And I will definitely go to there after I will examine the forces which are acting. So one piece of the rope is going up and down and I would like to examine the forces. Okay. So let's start from the very, very beginning. So if this is something like a neutral position and the rope is stretched. And by the way, we're talking about only movements which are caused by my movement of the end of the rope. No gravitation, no other forces involved, etc. Again, simplification. So, so we have these first, second and the third pieces of necklace, which we are talking about, which represents small pieces of rope on a larger scale. So let me just enlarge it even more. So this is one, two and three. Now what happens when I'm lifting this thing up? So I lift this thing up into this position. What happens next? Obviously this is a link. So this thing would pull this thing up. Something like here. And this thing, as it's moving up, will move up, move up, move up the next bead on this necklace. All right? So what happens is if I will connect this to this. Here's my first simplification. Obviously this is longer than this. And this is longer than this. I ignore it because everything would be eventually infinitely small. And the difference would be infinitely small of a higher order. So we can do it. All right, but what's really important is let's just consider these three beads. I'm particularly interested in this one. And I'm going to devise some kind of an equation for movement of this point. So I need two neighbors, basically. Now next assumption. All assumptions and not exactly the same as it's in the nature, but anyway it's a reasonable assumption. Assumption is that there is a tension obviously between these links. And I assume the tension is exactly the same everywhere. Well, yes. I mean, if you have just one long rope and you have it under tension, the tension at every piece is exactly the same. So every piece is not moved. It experiences the force to the right and the force to the left, which is basically a tension. Exactly the same. If you will put two some kind of gadgets which measure the pressure. We will put it here and here between this point. Now these two gadgets will actually show exactly the same result everywhere on the rope. So I assume again here that there is something like a tension of the rope. And basically this tension of the rope is the source of the movement because this is pulling this one. And that's basically what tension is. This pulls this and that's what tension is. Now how tension is working? Well, it's working in both directions. So the tension pulls a basically down and b, well not down, towards b. And the tension of this particular link is pulling b towards a. That's what tension actually is. Same thing here. b is pulled towards c by the tension of this link and c is pulled to the b. So b basically if this is a straight line, b would experience the same force to the left as the force to the right. And it will stay in place. But in this case we have a different situation because this bead went up first and then it pulls up this. And only then it pulls this one. So there is a difference between these angles. If this is alpha and this is beta, alpha is greater than beta. And that difference actually is the source of the movement for point b. That's very, very important to understand. The angles are not exactly the same and that's why b moves up. Well, why? Now it's very easy to understand because there is a force here and force here. This is tension and this is tension. Well, obviously there is a force here, which is also tension and force here, which is also tension. And these two forces are equal to each other. However, these two are equal in absolute value, but they're not acting against each other completely. Because what we are interested in is vertical movement. And the vertical movement in this case is t times sine alpha. And in this case it's t sine beta. Right? So if this is alpha and this is alpha, this is t. So this is t times sine alpha. Same thing here. If this is t and this is alpha, then this is t times sine beta. And the difference between these two gives vertical movement. Now, to tell you the truth, there is a difference in horizontal movement as well. Because it's not really going up, it's going a little bit to the left because this link has a fixed link. However, again, when we are going to infinitesimally small distances, this would be t times cosine and this would be t times cosine. And the cosine around zero would be very, very close to one. Sine is different. You see, sine around zero is this type of a curve, right? Cosine around zero is this. And difference between cosine and one is a difference of the second order. Difference between sine and zero is the difference of the first order because they are proportional to the distance. And here we have like a tangent. So that's why we can ignore difference between cosines when the angle goes to zero. But the difference between sines we cannot ignore. Alright, so what is the difference between sines? Again, I will do something which mathematicians would probably have to go more rigorously, I would say, but it's definitely valid assumption. And again, it's all about infinitesimals of the higher order. Now, that difference between sines is very much difference between cosines, between tangents. Why? Because tangent is sine over cosine. And the cosine as we just talked about, they are very, very close to one and we can ignore them. So the difference between these is infinitesimal of the higher order. So that's why it's a valid approximation. Now, what is t times tangent alpha? What is tangent of this angle if this is a curve? And this is a curve, actually. What is the tangent of the tangent of tangential line made angle with horizontal? Well, that's actually the first derivative, if you remember. The first derivative of a function is tangent of this angle, tangent between the tangential line and horizontal. Now, if these kind of calculus things are not really familiar to you, you just have to stop right now. Because whatever will follow will be a lot of different calculus things and you really have to know calculus. Either you can go to this website and take the course mass proteins. There is a course preceding and kind of prerequisite for physics proteins. So I'm just using this as a given that the tangent of alpha is basically the first derivative of this curve. Now, we're talking about curve, obviously, because all these pieces will be infinitesimally closed to each other. And that's a curve. Okay, now, since we are talking about curve, we will talk about coordinates. This is x, this is y, and this curve will be y of x. However, I will take another parameter, y, because this thing is moving. So it's not like a fixed function. It's a function which lives in time and it's changing. Now, we are talking about right now derivative, derivative only by the distance by x. So it's called partial derivative. And again, that's part of the calculus which you have to know, partial derivative. So basically what I'm talking about is that this is a difference between partial derivatives of this curve. On the left, let's put it a little l here. Minus on the right. Or if you wish, I can put it as partial derivative at point a and point c, because again, they are infinitely close to each other. Okay, great. Now, what is the difference between two values of the function very close to each other? Again, let's go back to calculus. The difference between, let's use another letter, between f of x plus delta x minus f of x. Now, with delta x being very, very small is about the same as first derivative times delta x. And whenever we are talking about infinitesimal delta x, that would be, instead of delta, we will have d. That's differential. And here we will have differential as well. So that's assuming this differential is infinitesimal. So I can actually replace difference between these two functions with derivative of this function again by dx. Now, what's the derivative of the first derivative? It's the second derivative. So it will be d2 y of x t to dx squared. That's the second derivative by x times dx. So this is the force which acts in upper direction. That's my fb. So it's a difference between vertical part of this t and vertical part of this t, which are just approximated in some way. And that's the second partial derivative. Tension being assumed being constant. Right? Okay, so we know the force. Right? Great. If you know the force, we can go to Sir Isaac Newton and use his second law. Now, the second law is mass times acceleration. Acceleration in the vertical direction we are talking about only. So what is the mass? Well, we are talking about infinitesimally small piece of the rope right now. So let's forget about necklace and beads. Now we are talking about piece of the rope. Necklace was basically like to explain maybe it's better. So obviously this is a rope and infinitesimally small piece has infinitesimally small mass. Right? So instead of m, I should really put dm. Now what is acceleration? Well, if I have a function y, what is acceleration? Acceleration is the second derivative by time. So it's d2 y of xt by time d2. Now what is mass? Okay, mass is very much related to the lengths. Now gx is the lengths infinitesimally small. dm is a mass of this. If I have a certain rope of certain lengths, let's say l and its mass is m, then m divided by l is density of the mass, right? Per unit of lengths. So obviously dm is equal to mu times dx. So this is density of mass. So if I will, and this is the constant for a rope, we just have the total mass and you divide it by lengths and that would be mass per unit of lengths or density of mass. So mu b density and this is related. So I can replace this with times dx. And this must be equal to each other. Basically that's it. Let me wipe out this and I will write this. So what we have is g times second derivative of our function by x is equal to mu times second derivative of our function by time. This is the wave equation basically. T is supposed to be given, the tension, because that's how this particular rope is initially positioned and tensed. Mu is again a constant and well, we are now talking about differential equation, which means we need some initial conditions. Okay, initial conditions determine, because there are many different solutions to these differential equations. And again, I can suggest the differential, the solution to this differential equation in the following form. I will put y of x comma t is equal to sine of omega t minus kx. Why did I suggest this? Well obviously because I know that this would be a solution. Now, what I suggest you to do and I would like basically to save some time, just take the first derivative by x, the second derivative by x, then the first derivative by t, and the second derivative by t, and basically compare. Now I will do it very quickly right now, but you can read about this in the notes of this lecture. So I'll do it as fast as I can just for time sake. So the first derivative by x would be, it would be, so it's dy x t by dx is equal to, okay, a times cosine of this ink times minus k. Now the second derivative by x is equal to minus a sine x k, first derivative cosine of omega, my second derivative is minus a sine omega t minus kx times omega square. Something like this right? And as you see, my second derivative here and second derivative here, they would be very much like this one if t divided by mu, let's put it this way, t divided by mu. So if my t divided by mu is equal to everything corresponds, basically this is k square, I can just replace this with k square, omega square divided by k square. So if I have this, if my omega and k are chosen in this function in such a way that they are proportional, then it will be a solution. So that's most important thing right now. So I have come up with many different solutions for many different a, basically a can be any positive number, real positive number. And omega and k again can be in a very wide range as long as their ratio is the same as ratio between t and mu. Since t and mu are given, so we know this ratio, so we know this ratio, so we can basically, for instance, the ratio is two. Well, I can choose this one, let's say whatever, three, and that would be, square would be nine. So to get the ratio of two, I need to have k square equals to four and a half, so square root four and a half. So there are many different solutions. What exactly I have to satisfy to choose which one exactly, that's what the initial conditions are serving. When differential equations are involved, you have some kind of general solution, and I'm not really pretending that this is general solution, there are other solutions, I just suggest this one. But to choose which one exactly is a matter of sufficient number of initial conditions. Now what initial condition we can know about this particular thing? Well, as I was saying, I was moving up and down the end. If you remember from the very beginning, the distance I'm moving as a function of t was a times sin omega t. Now, it's not an accident that I'm using omega here and omega here, obviously. So I have to choose in this solution the same a as this a and the same w as this w. And the only thing which I have to really think about is how to choose k knowing that k is equal to what t divided by mu omega square square root, right, from this formula. So all you need to do is to choose k which satisfies this, if this is my initial condition, and that would be fine because that x is equal to zero, x is equal to zero, that's the beginning of the rope, the one which I'm actually going with up and down. So if x is equal to zero, this function is equal exactly this, and that's my initial condition. Okay, so we've got the wave equation and we've got this particular condition on omega and k. I would like to spend some time to talk about what is omega over k. It's not such a simple thing, actually, it has a physical meaning of this. And you know what meaning is? It's the speed of propagation of the wave. That's something which was, well, unexpected for myself to tell you the truth. But the derivation is really very, very easy. Now think about this way. If this is the wave, okay, now let's talk about crest. Let's say the crest of the wave was at x. Now let's say that we are moving a little bit. The wave is moving. So let's consider that the next wave is something like this. It's moving. You say, okay. So this is x plus delta x. So during the moment, during the time period of delta t, the crest actually moved from position x to position x plus delta x. Now, if this is the maximum, it means that the function sign of omega t minus kx is equal to 1, right? That's what it means. Now, when we have moved, it means the sign of omega t plus delta t minus kx plus delta x is also equal to 1, right? Which means they're equal among themselves. Now, what follows is the difference between them is equal to 0. And now, considering they are very, very close to each other, the difference between signs, I will basically approximate with the difference between arguments. Because you know that sine x divided by x, as x goes to 0, it goes to 1. So they're very, very close to each other. Again, the difference is infinitesimal at higher order, which means that difference between them omega t plus delta t minus kx plus delta x minus omega t minus kx goes to 0, as delta x, as delta t goes to 0. That's what it means. Well, if you will open the parenthesis, wt will cancel with this one, wt cancel with this one. So omega delta t minus k delta x goes to 0. Or delta x divided by delta t would be equal to, not equal, go to omega divided by k, right? These are approximately equal. So delta x divided by delta t. Now, and what is delta x divided by delta t? That's the lengths by which my crest has moved during the time delta t. So as delta t goes to 0, that would be speed, distance divided by time. So the speed of the wave is exactly this. So what's interesting is that for the solution, this is equal to v square. And our wave equation can be rewritten as v square times d2 y of x t divided by dx2 is equal to d2 yx. Sometimes it's written this way, sometimes it's written this way, which is the same thing. Also another little nuance. Sometimes the solution which is suggested is this one. Sometimes suggested solution is slightly different kx minus omega t. It's just a different form because the difference is just a sign. Now, what's the difference between these two? Well, at time t is equal to zero, this is positive. I mean close to zero, we go up since this is positive. If times goes from zero, increasing from zero, this would be negative where x is equal to zero in the beginning, which means my initial movement would be down. So it doesn't really matter whether I move this way or this way. Same thing. But solution would be, you know, it looks a little bit different, but everything else is exactly the same. Actually, this is probably more prevalent in the textbooks, which I have seen. What else? Yes, this is very important thing. You see, whenever you are, but I would like actually to talk about this formula. This is equal to this square. So I think from the physical intuitive standpoint, the more tense the rope is, the faster the wave will propagate, right? I think it's kind of natural. Same thing. The heavier the rope is, the slower the speed of propagation of the wave is. So that's just physical sense of this. But that's very interesting and important formula. Okay, basically that's it. I wanted to talk about, I did not touch the energy yet because for energy we need something like this wave equation. We understand how the movements, etc. So next lecture would be about the energy of the rope, of the movements of the rope. Again, not yet about light. Light is much more difficult and kind of advanced because light is electromagnetic field oscillations. It's not as physical, not as tangible as, let's say, the rope. So that's why I started with rope. Anyway, that's it for today. Thank you very much. I suggest you to read the notes for this lecture. It has basically all these calculations and you can just take a look at them yourself, maybe do it yourself. That would be nice. That's it. Thank you very much and good luck.