 Hi, I'm Zor. Welcome to Unizor Education. We will continue talking about musical strings. So this is the second lecture, musical strings 2. It's a continuation of the previous one where I have introduced a model of a musical string, a significantly simplified model. And I will talk about the same model and I will try to derive certain differential equations which are basically the same as in the previous lecture, but from a completely different standpoint, from the standpoint of the energy. Well, this lecture is part of the course called Physics 14 presented on Unizor.com. I do suggest you to watch this lecture from this website, because every lecture on the website has textual part, which basically like a textbook. Plus, all the lectures are organized properly in a logical sequence. There are menus and you can go to Unizor.com choose Physics 14's course. This is the waves part of this course. And now we are talking about transverse waves. And this is the second lecture about musical strings over there. So these are hierarchy of the menus which you have to go through. So I do recommend you to take the whole course. There is a prerequisite course called Mass 14's on the same website which is definitely a necessary subject to know before you go to physics. Mass is just a fundamental part of any physics course. Now the whole website is completely free. There are no ads, no financial strings attached, no other strings attached. So just pure knowledge. Enjoy. Now, back to musical strings. First of all, let me remind you very briefly the model which we are talking about and how I have derived certain differential equations in the previous lecture. Very briefly and very fast. I do recommend you to, if you didn't really watch the previous lecture, go into the previous one and watch it entirely. But in any case, the model of musical string which I have introduced is the following. So you have two springs and some kind of a point mass in between. Now the strings are already a little bit under tension. They are stretched a little bit. The same way as musical string, before you are playing this, you have to really turn, for instance, some kind of, you have this mechanism, I don't know how it's called, in a violin or a guitar where you are turning some kind of a bolt or whatever else, which makes the string under tension. So you are stretching the string. So this stretching the strings initial stretching, initial. I kind of model with two identical strings with a mass in between and strings are already stretched a little bit. So let's just assume that initial unstretched length of the combined two strings is L. But now we have stretched it to L plus lowercase L. So right now this is L plus L distance. So they are stretched a little bit. Initial is L capital and increment is lowercase L. So the total length is L plus L and it's stretched. We also assume K or kappa, Greek letter kappa as a coefficient of elasticity of these strings, springs. In this case these are springs. I model string with a spring. And the mass is M, which is point mass. Obviously strings, springs are weightless, assumed. So this is a very simplified model of a real musical string. Musical string is more complex and whether we will learn it or not it doesn't really matter right now because my purpose is to introduce you into complexity of oscillations in this particular case. Alright so now when we pluck the string what does it mean from the models standpoint? Well it means that I will just move this particular mass a little bit up so it goes to this position and the springs are here and they are even more stretched. So they are already stretched here in a horizontal position. But now when I displace my central mass a little bit higher let's call this distance, this is a y axis. So this is y of t. Y of t is the displacement as a function of time. So whenever I did this I stretched it even more. So what I'm trying to do is I'm trying to introduce some kind of differential equation which will help me to resolve this function y of t. And as before I'll just compare the forces which are acting on this mass and connect it with the second derivative acceleration which is basically the Newton's law. So the force is supposed to be equal to mass times acceleration. So acceleration is actually second derivative of t. Now mass we know this is second derivative of t. And now we have to define what is actually f. So f is a force which goes in a vertical direction. So how can I determine? I think I need another. So let's determine the force. Whenever my springs are stretched the force is basically along the springs, right? So what is the force which is acting in this particular direction by one particular spring in this particular case? Well it depends on how much I stretched it, right? I have already stretched it initially by a little bit and now I'm stretching even more. This is hypotenuse, right? So this is l plus l divided by 2, right? If this is l plus l so this is the half. This is my y. So what is the length now? Well it's obviously the square root of l plus l square plus y square. y is a function of t of course, of time. So that's my length. So what's my... Now you know that according to the Hooke's law the force is proportional to the elongation with k or kappa being the coefficient of proportionality, the elasticity coefficient. So if from this length I will subtract the initial length which is l I will get this elongation multiplied by k that will be my force. So the force of one particular spring is k times this square root l plus l divided by 2 square plus y square minus l That's my force. Which is going along this spring. Now that was all part of the previous lecture. Sorry, l divided by 2. This spring was only l divided by 2. Okay. Now if this is the force now I'm interested only in the vertical component of this force, right? Because it has horizontal component and vertical component. Now this spring also will have horizontal component and vertical component and horizontal components will be equal to each other and nullify each other, obviously. So I'm interested only in vertical. So the vertical component of the first force is I have to multiply this force by sine of this angle. Which means y divided by the hypotenuse. So the hypotenuse I know so I will have this which is k times this expression in square brackets times y divided by square root of l plus l 2 plus y square which is equal to if I will divide this by this now this divided by this will be 1 So it's k y 1 minus l divided by 2 divided by square root of l plus l square plus y square Equals k y 1 minus if I will bring this 2 down and put it under the square root it will be 4 and it will cancel this. So it would be l divided by square root of l plus l square plus 4 y square So this is my formula for one particular force. This is absolute value. Let's not talk about direction. Now this force on the second spring is exactly the same vertical component so basically I can double this force and I will get basically the whole force which is acting on my mass in the vertical direction and I can equate it to m times second derivative of y. So double this so m times y square of t is equal to double k y 1 minus l divided by square root of l plus l square minus plus 4 y square of t and I probably have to put minus here because the force is acting if y is positive means I'm going up with my displacement the force goes down. And that's actually a corresponds to Hux law. Hux law is always displacement has to have a minus sign to get to the force. Force is equal to minus k times displacement. So basically that's what we have here. That was the result of the previous lecture and then I was talking about that this differential equation is very difficult to solve. This is also function of t. So what I suggested, I suggested basically to drop y of t as being very small relative to l and then I did certain simplification and I came to basically if this is some kind of a constant then I will have the harmonic equation obviously. Second derivative is equal to some kind of a multiplier by displacement. Normal second order differential equation for spring which gives you harmonic oscillations. But only approximately. So even in this simplified model I do not have clear harmonic oscillation. So forget about the real string obviously it's not harmonic. However it's really very interesting and it's simplified model and it gives you kind of an impression what exactly should be expected. Now I would like to approach this same problem from a completely different perspective. From a perspective of energy. Ok, so now we know the force this is the force, right? This is the force which is applied to our mass when it's moved up or down by y of t. Alright. Now I would like to calculate the potential energy which is accumulated in this mass when I moved it to certain initial displacement let's say A. So y of 0 so an initial displacement is equal to A. What happens? So I moved it up and let it go. Now let it go means the second derivative is equal. I mean the first derivative is equal to 0 that's the speed. Ok, so what happens? Well obviously I have accumulated certain potential energy in the mass. When I let it go the springs are moving it down and my potential energy is lessening but my kinetic energy is increasing. The sum of potential plus kinetic should be constant. The preservation the conservation of energy law, right? So the maximum would be kinetic energy when it goes to horizontal position and then by inertia it will go down and will slow down so the kinetic energy will be less and less but the springs will be stretched when the mass goes down the springs will be stretched again so it will accumulate again potential energy. So potential energy in the beginning goes back to kinetic in the middle and then potential at the bottom and then again back to kinetic potential and they are always exchanging these values between themselves, kinetic and potential energy but always equals to initial potential energy which is basically our work when we were trying to stretch both springs by moving by the distance meaning. So first of all let's calculate how much potential energy we have supplied to the body to this mass between the two springs when we are stretching it. We know the force and the force can be considered as constant around one particular value of y. So if I will multiply it by differential of y by the small increment of the displacement I will have differential of work which is equal to function of time which is this one times dy. That's the work, right? Now we know the force we know the force will be 2k. Well I will use the plus in this particular case because the force of the springs are going down but whenever we are stretching moving up we are working against these forces. So the force which we are applying is exactly equal to this one but with an opposite side. So that's why I put plus plus k y 1 minus l divided by square root of l plus l square plus 4 y square So that's my times dy So this is differential of work This is an infinitesimal amount of work which I have to apply to stretch from y So this is my So this is y and this is y plus dy So on this particular segment I will have to spend as much work. I have to spend as much energy to move from this to this. From y to y plus dy where dy is infinitesimal increment. So that's why I'm using function the force as a function of t in exactly the same way as a function of displacement and then multiply by dy. Now how can I calculate the whole work from here to a? Well I have to integrate dy from y is equal to 0 to y is equal to a Now luckily this integral is actually simple one because well it's minus right? So it's two different integrals. Integral of difference is equal to difference of integrals So this one is just one particular variable y. So it will be integral of y dy Now integral of y is y square divided by 2 That's an indefinite integral. Now in this part again let me just very briefly say it's y divided by some kind of a multiplier. 2kl doesn't matter. This is square root of l plus l square plus 2 square plus 4 y square Now this is actually easy because if you will recall, if you will have this integral b square plus y square. Now if you would like to have its derivative First of all you have to have a derivative from the square root which is 1 over 2 square roots times derivative of internal function Internal function is b square plus y square b square is a constant and y square derivative is 2y So that's exactly the same as we have. y and n square root So all we need is just a couple of multipliers So I don't want to bother you with exact calculations here because I have already done that and I can just give you exactly the result of this. So it's an easy integral The answer is 2k y square divided by 2 minus 1 quarter square root of l plus l square plus 4y square This is indefinite integral and I have to integrate it from 0 to a by y So if I will substitute y I will have 2k a square divided by 2 minus 1 quarter l quarter sorry square root of l plus l square plus 4a square minus Now when I substitute 0 it will be 2k Well this is 0 This is minus so it will be plus here l divided by 4 this is 0 So it will be square root of square root of l square So we can just forget about square root and multiply it by l plus l So this is my answer This is a potential energy from 0 to a grade From this I can just tell the following if this is a potential energy at point y is equal to a what is the potential energy at any point y The potential energy at any point would look exactly the same So I just have to integrate from 0 not to a that particular point y So potential energy at any point y would be equal exactly the same but instead of a I should put y because its calculations are exactly the same so it would be y square over 2 minus l 4 square root of l plus l square plus 4 y square plus 2 k l 4 l plus l So this is potential energy at any point Now if I will add to this Kinetic energy is mass times speed which is first derivative divided by 2 m v square divided by 2 So that's the first derivative So if I will add this one potential energy to this one which is kinetic energy at point y It should be equal to the full energy So potential energy at point y plus kinetic energy and total energy is potential energy which we have just calculated at maximum initial point whenever we are stretching initially our spring And this is a differential equation completely different than the one which we have derived in the very beginning and in the previous lecture and in the beginning of this lecture So we have two different differential equations maybe they have two different results Now this differential equation which is this plus this equals to this is actually the same and the way how you can actually prove to yourself that this is the same is very simple So this is the differential equation basically it's one function is equal to another function then their derivative should be equal right It's two functions are equal their derivatives are equal Okay now if you will take the derivative of this function the derivative of this but this is a constant right this is all a,l,l etc so this is a constant try to see what exactly this expression is So this would be derivative of this would be 0 Now derivative of this if you will take it would be the following Well 2 and 2 will cancel so it's ky square and the derivative by y would be 2ky times the derivative of y right y is a function of time minus 2kl so it's lk divided by 2 right Now derivative of this would be 1 over 2 such squares square roots times derivative of inner function which is 8y and derivative of y is y plus well this is a constant so we don't have anything of this but we do have derivative of this so plus it's m times times y derivative of derivative is second derivative and divided by 2 gives me this basically Now this is equal to 0 because it's all equal to potential energy at initial point a which is a constant so that's equal to 0 so this is differential equation also now if you can cancel y now I might have made some mistakes I don't know but if I did not it should actually be well the same thing as that one and I did check it in the text for this particular lecture I have much more detailed calculations and I had exactly the same equation as this one which is not a surprise because obviously no matter how you approach the same problem you have to have the same result ok so my purpose was that we have approached the same problem which is a simplified musical string model from completely two different positions one was using the Newton's law another was the law of conservation of energy and we came up with well two different differential equations but one can be reduced to another obviously so that's something which probably you should always try if you don't know really a true result which you are trying to achieve if you have a problem and you don't have an answer you should really try to solve this problem from two different approaches and like in this particular case the second Newton's law and law of conservation of energy and you should have the same result and if you do then you are absolutely sure that your answer is correct so that's the kind of simplified analysis of oscillations of musical string again I want to emphasize how complex it is so even in this simplified model the differential equations which we came up with are not really easily solvable I don't even dare to try to solve these equations maybe there is some way I just don't know and I don't want to basically go into this that's something which is not related to the purpose of this course the purpose is to force you to think about different things by modeling them and trying to approach in a simplified way and everything whatever we are doing is basically modeling the nature nature is much more complex than our models alright so basically that's it I do suggest you to read the notes for this lecture and I will try to put some exercises and maybe we'll simple problems solving as a continuation of these two lectures and obviously I will add exams too waves are important and kind of complex it's really difficult part of the physics the real waves I mean these are simplified waves the harmonic oscillations are very simplified thing and in the more advanced courses of physics which you will maybe get in universities you will have more complicated modeling and more complicated results and it's very interesting actually complicated but interesting alright thanks very much and good luck