 Hi, I'm Zor. Welcome to Unisor Education. We continue talking about waves, and in this particular case we continue talking about transversal waves. This lecture is part of the course called Physics for Teens, presented on Unisor.com. There is a prerequisite course, which is mass for teens on the same website. And I do suggest you to watch the lecture as the part of the course. And that means you have to go to Unisor.com and follow the menu. If you found this lecture on YouTube or somewhere else, just by itself, I mean obviously it has its own benefits. But don't forget, I'm using always information in previous lectures in subsequent ones. So there is certain logical sequence. And then you obviously have the benefits of having exercises, having exams if you are on the website. So I do suggest you to use the website for basically education for taking the whole course. By the way, the site is completely free. There are no advertisements, no strings attached. Even your login is not mandatory. You can do it just for purposes of exams and there is certain educational process, actually, which you can get involved. Anyway, so we'll talk about transversal waves. Now, the title of this lecture is Musical Strings 1. And there will be probably musical strings, too. Well, it implies we are talking about strings as part of the musical instrument, like guitar or violin. Now, that's actually a very complex thing to study in full details. So I will try to simplify this as much as possible. I will build certain simple model and then I will investigate how the simple model behaves. So before, the only example of transversal waves in the previous lecture was when you have a rope, let's say, and you're waving it up and down and it actually propagates the waves. So the waves are propagating one way, but every little piece of the rope was actually moving perpendicularly, or almost perpendicularly. They talk about this almost. In this case, I would like to talk about a string used in the musical instrument, like violin. And I think I will try to basically exemplify the same thing, how these transversal oscillations are happening and I will try to mathematically approach what exactly is happening with every piece of this string. Now, again, as I was talking, it's too complex to basically research this thing in all the details. So I will try to do it step by step and in today's lecture I will exemplify the oscillation of a string on a very, very simple model. So what exactly is the model that I talked about? Well, a string is something which is actually fixed at two ends with certain tension. You remember in violin or guitar we have certain things we have to turn to make the tension on that string and depending on the tension, the tone of the sound is changing. The more tense we are making the string, the higher the tone, which means more frequently it oscillates. But let's just have it mathematical. So this is the string and if we will pluck it somewhere in the middle, which means we will just make this shape out of it and let it go, it will start oscillating between these two points, obviously, but it will start oscillating from this to this, etc. So that's basically how the sound is produced. And as I said it's kind of complex to research right now in full details. So let's just make a very, very simple model which resembles this particular type of oscillations and we will try to mathematically research how this would work. And even there actually we will have problems. So what exactly is the model we are talking about? Let's consider again two ends, but instead of a uniform string I will have two weightless springs with certain mess in the middle. And the length of the whole thing is the length of the string, of the musical instrument. So whatever it is, the guitar for instance has something like this, almost a meter maybe, or less, 60 centimeters. So let's imagine that we have this type of thing, this type of arrangement. So there is a mass m and each particular string, let's assume these are identical strings and each of them in lengths is equal to half of that distance. So it has certain coefficient of elasticity, k or kappa. And now we have to really think about this tension thing which we are doing in the very beginning with any string, the musical instrument. Now what does it mean? Well it means that these strings, these springs, I was kind of mixing string and spring. So these two springs are supposed to be tense already at certain tension. What does it mean? It means it should be a little stretch. So the neutral length of each spring is smaller and we have just stretched it a little bit and this one. So it's already at certain stretch. So let's assume that each string has its own lengths and there is a certain stretched additional length which we have stretched it into. So this is modeling our string on a musical instrument which is already stretched a little bit. So let's just think about there is something which is a length of the string or two springs in a neutral state. And then there is a certain additional length which we have stretched it to to create this tension. Which means that each of these springs has the lengths L plus L divided by 2, L plus L divided by 2. So this is my model which models an oscillation of a string stretched between two ends. So this is my initial model. Now let's think about what happens if we will pluck this particular construction similarly to plucking a string on a musical instrument. That means we just have to take this mass, shift it a little bit up and let it go. And it will obviously oscillate on these two springs between two fixed ends. And our task right now is to mathematically approach how it happens, what exactly is happening in this case. Well, let's think about it. As soon as you will move this mass, point mass a little bit higher, you increase the lengths of these strings a little bit more. So you're increasing the tension. But now let's think about how the forces are distributed. If you will lift this point mass a little bit up, this is one spring. This is another spring. Now the springs will push this way because we are stretching the spring. So this is stretched and this is stretched. Well, it was already stretched here. Now we're stretching even more. Now at initial position these two stretched strings also have certain forces this way and this way. But these two forces are along the same direction and they neutralize each other. So that's why if you don't move this point mass from its initial position, nothing happens, it does not move. But as soon as you are shifting it a little bit up, then these two forces start acting and they are not opposite to each other. There is an angle here, right? So if you will take the horizontal components and vertical components. So these are horizontal. This is vertical component of each force and this is horizontal component. So horizontal components are neutralizing each other but the vertical components will push it down. So this is the force which pushes back to initial position. And then it goes by inertia even further down and that's how it oscillates. Okay, so let's just calculate what exactly is happening in this particular case. We had the first initial length of each strength L plus L divided by 2. Now we will introduce a variable, the deviation along the vertical line. So the deviation from the initial position would be y coordinate y as a function of time. So at time t equal to 0, we have initially shifted our point mass, our central mass by certain lengths and let it go. So y of 0 is equal to, well I put it a, certain initial distance we started plucking. And initial speed is equal to 0 because we didn't really push anything, we just let it go. So this is our initial position. So this is our main variable y which characterizes how far from the central position our point mass is at any moment time. So now we can basically calculate what kind of forces are we talking about. So we have stretched this spring to a hypotenuse. We have one calculus which is equal to y of t. Another calculus is initial stretched lengths of this spring. So what will be the length of the spring after we have shifted the point mass up? It will be obviously square root of L plus L divided by 2 square plus y square of t. This is the length of the spring after our mass is positioned at any point y of t as a function of t. So this is the length now and as we were saying this is already stretched length and what's the neutral length? Length of this spring in the neutral position, well that's L divided by 2. We have already stretched it here, but initially the length in the neutral not squeezed length of this spring was L divided by 2. So the difference between these two is the increment of the length relatively to its neutral length. Now we can apply the Hooke's law. Now the Hooke's law is telling that the force which this force along the length of the spring. So this force is equal to increment of the length multiplied by certain coefficient of elasticity. So this is the force which goes along this and the same thing along that. Now as I was saying we can actually represent this force as a sum of vertical and horizontal component. Horizontal components are canceling each other but the vertical component is acting exactly along the movement and that's what we need to calculate. So what is this piece? Well it's the force times sine of this angle. Now sine of this angle is proportional to ratio between this and this. Hattitus divided by hypotenuse. Obviously these two triangles are similar to each other so if we will multiply f by sine of this it will be exactly the same thing as multiplying by the ratio between this and this. Now the ratio of this and this is y of t divided by the length of the whole thing which is by the way this one. And this is my vertical force from one spring. Now we have two springs and each of them has exactly the same vertical component. So I have to really multiply it by two and if I would like to really know what is this vertical component is I should really have this expression whatever this expression but with a minus sign. Y minus sign you see if y is positive then the force goes down which means opposite to growing of the y component. If y component is negative considering this is zero positive negative. If y is negative then the force must be positive right? So everything else is positive in this case. So basically that's why we have this minus as typical in any kind of Hooke's law application. So we have this as the force which is acting it's a combined force of two springs. Now what is my differential equation? Well that's equal to m times acceleration the second Newton's law second derivative of y. And here is our differential equation right? Okay so let me just represent this differential equation in a slightly different format. So first of all I will divide by m so I will have y of t. Now if this is equal to this with a minus sign I would put everything into the right side of this equation and have it equal to zero. So that would be zero equals. So y now m I will put here as a denominator so that would be what? It would be 2k divided by m. Then this okay y of t also would be here. And then I will have this divided by this you see this is the same. So if I would divide this to this there would be 1 minus. Now this divided by this so it would be l divided by 2 divided by square root of l plus l square divided by 2 plus y square of t. So this is my differential equation. Well let me tell you it looks ugly and I don't really know how to solve this differential equation. But my first purpose was well first of all any differential equation can be solved numerically or whatever else we want to. But analytically it's sometimes difficult this is difficult to solve analytically I mean I don't know how in any case. I mean maybe if you will just you know spend some time and maybe there will be certain ways of solving this but that's not my purpose. My purpose is not to go into mathematical difficulties but just to explain actually what physically happening and have some mathematical proof that this is exactly what's happening. So how can I approach it? Well physicists are in cases like this don't hesitate to simplify mathematical formulas assuming that certain variables are too small too big or whatever else something else. Now if you remember even the Hooke's law is always commented okay this is the nice law but only if we are deviating from the neutral position by a little bit. So if you will significantly deviate from the neutral position it would be much more complex because there are much more complex things involved in this thing. But somewhere around the neutral position these oscillations do obey the Hooke's law. And that's the same thing with every other physical law it's always to a certain degree of precision. So I would do exactly the same. Now let's think about it. What is YFT? YFT is vertical displacement of this center point which basically models the displacement of this string from its initial horizontal position up and down. And this displacement is usually very small. I mean just think about you have a guitar it's a relatively long string right and you are plucking it just a little bit. So the distance by which you are deviating the position of the center point of this string from its initial position is really small. Now relatively to total lengths even divided by two YFT is always small so this A is very small. So what I would do I would just say that you know what relatively to the whole thing this contributes very very small amount to the denominator. Let's get rid of it. Let's just assume that this is zero. Well I know it's wrong but it's not that wrong. I mean approximately it will still give me certain mathematical result which resembles the real physical situation. And again that's how physicists are doing in many many different cases. As a mathematician I do have my reservations about this but as somebody who really wants to know how the physical process is actually doing with a certain precision I can really do it safely. And let's see what happens. Now I have the completely different situation. Now if this is not here square root and this is a square so it would be just L plus L divided by 2 right. And since this is divided by 2 and this is divided by 2 two can actually cancel each other. So the whole thing would be L divided by L plus L this. Now 1 minus L plus 1 minus would be equal to common denominator L plus L minus L so it would be L divided by L plus L. So the whole square expression in square brackets would be equal to this. So my equation would be Y of t this is plus plus 2 K over M L over L plus L Y of t. Alright now let's think about what this is. Well if you remember the previous lectures we always had equation of this type. This is the equation of harmonic oscillations. We had it for the first lecture about waves and this is basically the angular speed or angular frequency. And the result of this is something in case Y of 0 is equal to A and Y of 0 derivative. Speed is equal to 0 in this case it's something like Y of t is equal to A times cosine negative t. So we did have this equation and we did have this solution with these initial conditions. If you don't remember just go back couple of lectures it's all derived over there. And we have exactly the same situation here. So this is my omega square. So from this follows that vertical displacement of my point mass in between the two springs actually is making harmonic oscillations with the frequency which is determined this where K is coefficient of elasticity and mass of that central point mass. L is initial length of a string which we are modeling without initial stretching and tension. And L is basically the length by which we are trying to stretch this particular string. Now what's interesting here is that the greater the L the greater will be frequency. And the greater frequency means higher tone of the string. More frequently it oscillates the higher will be the sound the musical instrument makes. So this is basically a mathematical explanation why it happens. The more tense you have made this particular string the higher will be the tone it produces. So again we have harmonic oscillation of this central point. Now granted this model of two springs instead of a real string like we have in the musical instrument is a simplification. But it can be expanded and I would like to say that there is a very strong similarity between oscillation of this on two springs and oscillation of a musical string. So this similarity actually brings us to an opinion at least that the oscillations of the real string of guitar or violin really resembles at least each point. It resembles the harmonic oscillation to a certain precision obviously. In reality it's significantly more complex and obviously it's more complex because there is also the resonator which also produces certain secondary waves. So it's not just the string which produces this particular sound it's also kind of reflected through this resonator etc. So much more complex nevertheless in the beginning of that thing like in a very very basic type of understanding what's happening. This is probably not a bad model and this model gives us relatively precise description of oscillations which are coming which are occurring in this case. And we also have a very nice explanation that the tension is very much related to the tone of the sound which musical instrument makes. The more lengths we have really stretched our string the more frequently it oscillates and higher the sound is. Well that's it for today. Thank you very much and good luck.