Okay. If you look at the one of the strings of a guitar, and it look like for example, it has a small connecting point and then it has a string, go on and on, and then there is a hole, and then there is some stop over here. So the string over here will vibrate. How to understand the vibration of this string? How? As we used before, we would like to look at the vibration of the string in terms of mathematical point of view or theoretical point of view. Then, based on this mathematical interpretation of the vibration of a string, we'll investigate, we'll see the physical meaning of the vibration, and then use this physical understanding where we'll try to find the good way to apply the string vibration in practical application. Those are the three element we have been using in vibration study. How to understand the string vibration theoretically? Basically, there are two ways: One is look at the vibration of a string in terms of energy or in terms of global vibration. In other words, I am looking at the vibration of a string far away, or that we call the integral approach. But most common and conventional way to have the mathematical and theoretical understanding of vibration, is to look at the vibration of a string upon very small element of a string. That is called differential approach. By looking at the small tiny element of a string, this will show us, I mean essential characteristics of vibration. So suppose I look at a very small infinitesimal element of this string, then want to know how it moves, how it vibrates. Then we need to apply Newton's second law, as we did before for single degree of freedom system or two degree of freedom system. So we have to find out all the forces acting on the string. The string can carry the tension. For example, for the guitar vibration, we are controlling tension. Let's say this is a very small element, so call it ds. Of course, ds has a dx and dy. If I denote that this is tension T, then over here, the tension will be T plus dt. So dt is a small variation of tension, using Taylor's expansion. Then, Newton's second law said this unbalanced force will cause the movement of this small section, ds, either this direction or this direction. Let's call this is d squared x, dt squared, and this is d squared y dt squared. I have to put the mass component over here assuming that mass per unit length is rho L, and then the mass will be rho L multiplied by ds. There should be a rho L multiplied by ds too. If I write again more clearly, this expression of Newton's second law applying to the string vibration, and here is dS, and I have a tension T, and I have a tension. But it will vary because of this small infinitesimal distance. Then I can say T plus dt, dS and dS. But as you can anticipate, by carefully looking at the vibration of a string, the vibration in this direction will be very small compared with the vibration in this direction. Therefore. I would like to say this can be written as T plus dT, ds dx dy. But dx certainly have a more contribution on the variation compared with the dy. So I would say dT dx, and dS can be approximately by dx, or more rigorously you can say, ds is a square root dx squared plus dy squared, and they can be written as dx square root one plus dy dx squared. If you expand this, then it can be approximately dx if dy dx is small compared with one. So that has to be equal to. I should use equal to. Again, using the same token as I used before for this, I would say the mass will be Rho L multiplied by dx ad then acceleration in this direction will be d squared y, dt squared. That's the result of applying Newton's second law to the infinitesimal element of a string. Now, let's get the equation, governing equation, that certainly as the word say, governs the vibration of a string. Now, let's see how the Newton's second law can be expressed in y direction and say, I want to measure the displacement of a string of y with respect to X-axis and because it vibrates. The y displacement of y would varies or changes with respect to time. So in Y direction, I can see that the tension in Y direction has to be T and then, if I say the angle over here is a Theta, then the force in this direction will be T sine Theta, and T sine Theta is approximately same as dy, dx, and that has to be minus direction because I set up the coordinate. Plus in this direction. Then, what would be this term look like? So I have the force T plus dt, dx, and dx and looking at it in positive direction. So I put a plus over here. And then the angle over here would be expressed as dy until you see this expansion, dy, dx plus d squared y, dx squared dx simply meaning that the incremental increment increase of the angle would be this one. I mean, the change of this with respect to x therefore that is dx squared as the length increased by dx. That has to be balanced by equal Rho L dx multiplied by d squared y, dt squared. It looks complicated but, physically meaning that the unbalanced force in the Y direction has to be expressed as mass times acceleration and apply Newton's second law or upon the small element of a string hoping that this will reveal us the physical model to understand the vibration of a string. Now. That's complicated but, let's look at some of the term. This term, minus T, dy, dx will be canceled out T multiplied dy, dx and then we have T plus d squared y, dx squared dx. So let's write down T d squared y, dx squared dx, and the other one is dt dx dx multiplied dy, dx. So I have another term dt, dx, dy, dx, and a dx, and I have another term dt, dx, dx and dx squared y, dx squared dx. So I have dx squared. As you remember, dx is small. The squared of the small stuff will be much much smaller than the small stuff of dx. So I neglect it. Then that has to be equal to Rho L d squared y, dt squared dx. Those dx are common. Therefore, we got, we are approaching to the final train station. T, d squared y, dx squared, plus dt, dx, dy. Sorry, dy, dx. That has to be equal to Rho L. d squared y sorry, dt squared. I can write this one more compact form like this d squared Ty, dx squared, right? Is it right? I don't know. May be right or may be wrong. That has to be equal to Rho L. d squared y, dt square. Okay. Another linearization or approximation is waiting for us. When you look at the vibrational of this guitar, small string, the tension variation along this excellence would be very small. In other words, this term would be very small compared with that term. So, I can write finally d square y, the dx square as to be equal to Rho L over T, d square y, dt square. That is a fantastic equation. We start with a more complicated expression that apply Newton's second law, but we end up with a very compact, looks very simple, governing equation. So, we have governing equation of a string that looks T is tension, d square y, dx square is equal Rho L, d square y, dt square. That looks very interesting. This is the acceleration multiplied by mass. So this is the inertia term. The string vibration of guitar, for example, inertia has to be balanced by d square y, dx. What is d squared y dx squared? That is the string of guitar's curvature. Multiply by the tension. So, that is interesting. First, if I increase the tension, this term will increase, therefore the inertia term has to increase. If I increase the mass, meaning that I move on the vibration of guitar from thin one to the thicker one increasing Rho L, then the inertia term also increase for the given tension multiply d square y, dx square, because that is given. Increase in this will decrease of that, therefore decreasing frequency, therefore decreasing pitch. Now, what about how to express the excitation? Let's look it again. We have a string like that. That has to be balanced by inertia, Rho L, dx, d square y, dt square. We can see that there is a tension and tension. What about I have a force acting on this string? That could be general. Then if I say that is the force per unit length f of x dx. Then that term has to be somewhere over here. But looking at because we assumed that the force directed in this direction that has to be minus f(x), and dx is a common one. Therefore, the governing equation that employs, that expresses excitation force would look like that. T d square y, dx square, minus Rho L,d square y, dt square has to be f(x). This is force per unit length. If I excite a guitar using this, then this force per unit length can be changed as Delta, the position located X_0 multiply by I mean this kind of Delta function. How to solve this case. That's very challenging questions. Let me show again the vibration of a guitar. You can hear very tiny difference of a sound is radiating, I mean the vibration of a string depending on where I excite it. Another one you can observe is the vibration style. Vibration of this string can be superposition of many different vibration. This kind of vibration and this kind of vibration and so on and so on. But also you can argue that, what is the boundary condition over here and over here and over there? Again, let's use the simplest case. Then I could say if I understand the vibration of a string, the fixed boundary condition, then certainly, understanding based on this observation will get us to understand the vibration of string for general boundary. That can be expressed by mass times mass, and the string and the damper over here. So, using the same argument as I used before, let's study this one for us to understand this, and then understand the general vibration of string when it is excited by general force. Mathematically, it is called inhomogeneous partial differential equation.
