Suppose I have a sinusoidal wave. And then, if I move this particle down and up, when I move this particle over here, the wave look like that, and then move back, the wave look like this, go upward, look like that, and then down, look like this. So this is one period, right? This is one period, I move down, up, and down. And then as you can see here, the wave, the length of wave propagate, the distance of wave propagate is one wavelength, as you can see over here. That is what I explained, all right? So, move down. Move down, and go up, up, and down. Then, one period of time, the wave is propagating in one wavelengths. That's what I explained using this equation. Suppose, Now we have two waves, so one is right-going wave, and the other one is left-going wave. And I argue that this is complete description of one-dimensional wave. Okay. Let's look at some detail of this wave. How to look at the details of this wave? Using differentiation. If you want the details of physics, we normally use differentiation. So let's take a derivative of this wave with respect to x. And then later on, we will look at the change of this wave with respect to time. Okay, taking a derivative with respect to x of g is simply g', and prime means that I am taking a derivative, With respect to x, or x-ct, it's the same. And h, taking derivative with respect of h is h'. And what about taking the derivative of this wave with respect to time? And that is -cg' because there is -c over there, okay? Energy taking the derivative of h function with respect to time will provide also c and h'. Okay, this is interesting. If I take a derivative again with respect to x on this dh, dy, dx, so what will I get is a g'' and h''. And if I take a derivative with respect to time on g' and this one and this one, what I will get is a c skill '', and c skill h''. Is it big deal? Yes, it is a big deal. That means, the rate of change of wave with respect to x and with respect to time is related again. Okay, let's see some more details, and the dg, dt, as a return over here, look like that, and if you take more derivative, Then we can conclude that the right-going wave with respect to time is related with respect to space like that and the left-going wave also related like that. Therefore, total wave y is related with space and time, is related like this. And this is wave equation, very well-known wave equation. So in other words, the wave (y(x,t)) that has right-going wave and the left-going wave is the solution of wave equation. This is what I want to say to the students today. So the left-going wave and the right-going wave is the result of our thinking. What I mean by our thinking is we postulate all possible waves in one-dimension. Of course, in one-dimension, there is one possible wave is this wave, and one possible wave is this wave. And we express those waves in functional form, that is, gx-ct, and hx+ct, and by using the mathematical operator, we found that those waves follow the wave equation. Okay, in other words, this wave equation, Describes the wave we just talked about y(x,t). So mathematically speaking, what we are going to do is, we solve this wave equation, that is gx-ct and hx+ct. And the type of solution depends on the boundary equation. So this is a wave equation plus boundary condition will determine the possible waves in one-dimension. So, mathematically speaking, we are solving wave equation, provided some boundary condition. And that is not very interesting to all of you students, that is too mathematical. So generally, I can say there is a wave equation. Okay? And if there is a boundary condition, general boundary condition, then we are supposed to solve the wave side that satisfy Wave equation as well as a boundary condition. Okay? That is not very exciting. So let's move on to another very interesting case. Okay, there are typical boundary conditions, like this. And, that this is rigid and boundary condition, and general string mass boundary condition, or if I oscillate the string. With this amplitude, this is a forced boundary condition. What I call this is passive boundary condition. And this is active boundary condition. Depending on what type of boundary condition, the wave that satisfy the wave equation would be different. Okay let's move on to the interesting case. Suppose we have two media. One is rather thin string, and the other one is a thick string. Is a wave and then if this wave meets some discrepancy over here. The only things that allow to be happen is the reflection and transmission. No other physical phenomena is likely possible. So wave is propagating this direction. And then it meets some discontinuity. And it has to be reflected or transmitted. How much is reflected and transmitted? Okay, let's see how this physical, simple physics can be explained. Okay, in this medium. I call medium one, and the possible wave at this medium one would be right-going wave and left-going wave. And left-going wave I denote h1, t plus x over c1, and right-going wave is g1t minus x over c1. So this is complete expression that describes all possible waves that can exist in this medium. And in this medium, there would be only right going wave, so I will denote that as g2. Because g stands for right-going wave. H stands for left-going wave. And this can be written as, like, t, x over c2, c2 is the speed of propagation at the string number two. And, the C 1 is a speed of propagation of string number one. Our objective, immediate objective, is to find out how much transmitted H 1, and how much, sorry, how much reflected H 1. And how much transmitted G2. Okay, The amount of reflection and transmission will be decided by the condition of this continuity. There's nothing else other than condition of this continuity. Okay? Depending on the condition of this continuity, H, amount of H 1, and the G 2, will be decided. What would be the condition of this continuity? One is, as I said before, the two different type of boundary condition I always mentioned. One is passive boundary condition and the other one is active boundary condition. In other words, under this continuity the velocity has to be continuous. Otherwise two of the strings would break down. And also the force acting in this direction has to be singing. Because at the discontinue there's no mass. String one and string two as a discontinue, there is no mass. Therefore, the force in median one string one and string two has to be same. So, I have two boundary condition, that means I have two equation. And here, I have H 1, and G 2 is two of those, provided that G 1 is given. Then I can solve it. And if we do know these two waves satisfy the wave equation. We don't have to worry about it. And our objective is to find out H one and g two compare with g one. In other words, how much wave is reflected with respect to g one, how much wave is transmitted with respect to g1, that we call reflection corrections and transmission correction. And those are reflection corrections and transmission correction all depends on the characteristics of this medium that is string number 1. And the characteristics of this media is during number two. And this is very vial, important approach that allows us to understand transmission and the reflection let's see how to do it. Okay, as I mentioned before. The velocity of a string at x equals zero minus, zero minus means that I used the coordinate. Actually core general over here. General minus is adjusted before x equals zero. That means it's string number 1. The string number. The velocity of string number 1, and the velocity of string number 2 has to be set. And the force at string number 1 plus the force at string number 2 will add to the mass between String number 1 and string number 2 that is 0, because there is no mass. Mass times according to the Newton's second law that is a result on the force acting on 0 mass on the between medium 1 and medium 2 or if you like string number 1 and string number 2. Has to be 0. This is boundary condition and we have solution, therefore we can solve it. If you see my book, then you will find out how to solve it and let's see and a result, is very simple. H 1 is the amount of, amount of waves reflect. And g2 is amount of wave transmitted, and g1 is instant wave. The amount of reflected wave compared with the instant wave is totally depends on this strange letter Z and the Z is the characteristic [INAUDIBLE] service string. Which I did not explain in detail. Okay. Characteristic impedance of string is related with density of the string per unit lance, L, and the propagation speed in medium one, LC. This is very interesting. If two strings are same, then z one is equal to z two. Then this is zero. Therefore, there is no reflected wave. Okay, if z2 is very small compared with the z1. For example, z1 is 1 billion and z2 is 10, then this is 1. In this case, H 1 over Z 1 is one. That means the total reflection, right? What about this case? This expresses how much wave is transmitted. When z1 is very small compared to z2 and this zero, there will be no transmission. If z1 is very large compared with z2, then there is did I say z1 is very large? Right? In this case, z2 is twice of z1, that is not possible anyway, because transmitted away, they are twice bigger then. And you have to very careful about this. This is only talking about the amplitude of the transmission. Okay, but the energy and power of a transmission has to be the velocity of medium gt and the velocity of a g1 if you calculate the velocity of g2 and the velocity of g1. Then this would be two z two, z one plus z two, therefore, actually, when z one is much much larger than z two, the amount of energy transmitted to the medium two Is middle ratio. Okay anyway, I will go some details in the next lecture. Today's lecture let's see the generally what's going to happen. I guess using this graph. Okay when Z1 is the same as Z1 over z3 is zero. What was the physical situation of this? The rigid wall condition right, there is a string. And there is a rigid wall. And then Z2 is very large then our physics says there will be total reflection. So wave is then to wave reflective no transmission. Increasing reflection. Some small transmission. But increasing z one over z two, transmission is increasing. When z one equals z two, that means string one and string two are identical. There would be total transmission, no reflection. When Z1 over Z2 increase there is a reflection as well as a transmission. But interestingly the phase of this wave and this wave, this is reversed, right? Looking over there z one over z two is getting large and large. It's a very interesting phenomena occurs. So, that means depending on the duration between z1 and z2, in other words characteristic impedance of a string. The amount of reflected wave and transmit wave This is [INAUDIBLE] and also the phase is changing. All right as you can see over here. So in this lecture what I emphasizing that. The characteristic impedance of a string determines all the reflected and transmitted characteristics of a wave. And I didn't talk about precisely what is the characteristic impedance of a string yet. And I will talk about character. What is characteristic impedance of a string in the next lecture. So let me summarize today's lecture, okay? Let me summarize today's lecture. First, what I emphasized is oscillation in time produce oscillation in space. In other words, oscillation in time is not independent with oscillation in space. This relation relate how the oscillation in time is related with oscillation in space. That is, simply, this. K equal to omega over c. [SOUND] This is what you have to remember in this lecture Okay. And then the other one. I introduced today is the amount of reflected wave and amount of transmitted wave is completely controlled by characteristic impedance of a string, z1 and z2. Okay, we will talk about the details of those characteristic impedance next lecture.