This will be the lecture number 2. Today, we will continue to talk about wave propagation using one dimensional wave. And the lecture 1 and the lecture 2 essentially covers what is written in my text Chapter 1. Okay. On the last lecture, we learned about dispersion relation. [SOUND] That essentially describes how the oscillation in space, 2 pi is wave number that is K is wave number that is 2 pi over lambda, so and also omega is a frequency in radian so if it express the omega in a frequency in hertz, then 2 pi over lambda is related with 2 pi f over c, therefore we got the lambda which is wave length in space is related with the frequency like this. So this is a very exciting and meaningful and very representative equation that relates with the space and time oscillation. And I will show how it is actually related by demonstrating a simple experiment and I need some volunteer from our students. So, you volunteer because you think you're most handsome boy? [LAUGH] >> [LAUGH] >> Okay, hm? >> [INAUDIBLE] >> I will. >> Roll up. >> Okay, let me go up, so everybody can see this? Okay that's good. Now, okay what you can see now is the frequency as well as wave lengths or frequency which is the oscillation in time. And you will see the how the oscillation in time is related with the oscillation in space, okay? I will oscillate this with a very low frequency, I think that this is about five hertz, okay? And you will see this big wave lengths. And then if I oscillated this with a rather small period of oscillation. In other words, larger frequency. Then you will see more rapid fluctuation in space. So, when I oscillate this with higher higher frequency, then you will see more rapid oscillation in space. So this demonstration essentially exhibits how this dispersion relation is really happened in space. Okay. And then in the last lecture we talked about the wave equation. In two different ways. One is I argue that in all the possible. One dimensional way all the possible one dimensional way is composed by two components. One is right going way. [SOUND] This is right going because as time goes on, the function, g, will move in the right direction. And also another possible wave, is the left going wave, and then we show that there is two wave, actually satisfy the. One dimensional wave equation, by differentiating twice of this, the wave that I argue that this is the only possibly wave that this can exist in one dimension. And then, [SOUND] we didn't go detail but it is written in the text when we consider the. Forces acting on small section of a string, string is the mechanical component that can only transmit tension. Okay? Beam is the mechanical element that can transmit bending. So because the string is the element that can only transmit string not allowing bending transmission so if we push the string at one point. There must be a discontinuity in a slope. Okay, say this is the tension acting on a string. And say this is the infinitesimal lengths of string. Then, using Taylor expansion, we can say the tension over here will be TL and dTL ds ds, okay? As you can see here there is a force unbalanced because of these two different tensions. This string will move following the famous Newton's second law, okay? Newton's second law says, simply, the mass of this. Small section of string will move with a certain acceleration. So if I write what I said, using the mathematical expression, then I can say there will be some mass times x rotation, in this direction. That is, low L. Multiply ds, so this is the total mass of this element. And y double dot, y is the coordinate in this direction and x is coordinate in that direction. And I can write there is another acceleration component in x direction. Therefore, the business we have to do is equate one balance force in this part with mass times acceleration over there. So we can do that because as you can here, there is a component in x direction and then there's a component in y direction. The component in x direction of a string, the force acting on this small segment of element in x direction has to be equal to low lds. X double dot, x double dot is acceleration in x direction. And then I could say the force component in y direction has to be equal to the [INAUDIBLE] l multiply dsy double dot in y direction. And the result will give us exactly same wave equation that is of course, d squared y. The x squared equal to one over c squared, d squared y dt squared. Assuming that the motion in x direction is very small compared with the motion in the y direction. That makes sense because the string, as we demonstrated just before, is the element that is moving up and down in this direction. Of course, there is a very small amount of motion in x direction, but that is small. And then we will show how it is small later on by looking at some details that is derive in the test. And also you can approximate this ds is of course, square root of the dx squared and the dy squared. And the square root of the dx squared plus dy squared can be written as, if I take out the dx, then dx square root of one plus of dy dx squared. And the dy dx is the slope of the string. And then we assume that the variation of a strings elevation with respect to x is small. Therefore, dy dx square is much, much smaller than dy dx. This is what we call linearization. So by doing well known linearization, We can obtain this wave equation. So as one of the students asked in the last lecture, this wave equation is obtained by assuming general one dimensional h wave. And then looking at the detail of the behavior of this general solution, we obtain the wave equation. And that the other one is we look at how the small element of string will behave in space and time. By applying Newton's Second Law upon the small segment of element, we obtain or gain wave equation. What does it mean? First, this solution is complete. Okay, second, depending on the boundary condition. The shape of the solution will be determine. So mathematically speaking, the business we have to find out all the solution is to solve this one dimensional wave equation. Upon or given [INAUDIBLE]. I raise a very well known mathematical problem and if you remember the [INAUDIBLE] problem, you can invite some. Some also [INAUDIBLE] a central solution and then you can estimate the weight of our solution that satisfy the boundary condition. And I will not go into detail about that, because that is too mathematical. And I'm sure everybody in this case would remember [LAUGH] those other [INAUDIBLE] program at least. Or you can do it, I mean right away. But at least if you remember which group you have to look at to understand the boundary value problem, then you can do it. Okay, and the last one, what we did in the last lecture was okay, if there is two string, a thin string and a thick string. Experiencing tension, okay? The possible wave in this section, say this is the string number one. Because this is string number one, I said this wave is g. The wave of g, I always use g for right going wave. Because this is at a sting number one, I use a subscript of one over there and x minus ct. And because there is a discontinuity, okay? So something must have happened. That is there will be some reflected away at string number one. And I use h for the reflected away propagating in relative x direction, h1 (x + ct). And then here something will propagate in this string number two. Only in positive x direction. So I call g because it's right going wave and then I'll use a two because this is string number two, x minus ct. Somebody will ask whether this c is the same as this c. Okay. And the speed of propagation would even go through the detail or it depend on the string. Characteristic string. So for generality, let's put this a C one and C two. Okay. c1t and c2t. And the business we have to work on at the moment is determine how much each one will get with respect to g1. And how much hg2 we will get with respect g1? In other words, we want to know the ratio between g1 and h1 which we call reflection coefficient Physically that makes sense. And we also want to know how much transmitting that's the ratio between g1 and the g2 that's what we want to have. So, we need to know r and towe provided that there is right going wave, left going wave and there is a right going wave. To know these two unknowns, we absolutely need at least, not at least, absolutely need to have two equations. Okay. If you have three equations then it's over determined. If you have one equation that we call the [INAUDIBLE] So those two equations has to come from this continuity condition. So one is the wave, I mean the displacement or velocity at this point, must be continuous. That is one equation. And the force acting on this element, and that element, has to be far low. Newton's Second Law. So, two equations. One is the kinematic continuity equation. And the other one is force balance equation will provide us the solution of R and tau. And the result show that, in the last lecture, this is essentially controlled by this strange interesting Z. 1 and Z2. In other words how much reflected totally depends on new concept, specific impedance of the medium stream one and stream two. And I will go into details about this today. And how much is transmitted depends on again Characteristic and cadence of string. So, immediate conclusion is that If there is two different medium. In this case, string number one and string number two. How much reflected, how much transmitted, totally depends on Z one and Z two. And what it Z one, and Z two? That is our question. So let's go back to our PowerPoint or our acoustics text. What is this? What is the Z1 and Z2? Okay?
