字幕表 動画を再生する 英語字幕をプリント Today's a big day for the physics department and for MIT. Professor Frank Wilczek was sharing the Nobel Prize in physics for his seminal work when he was a graduate student on quantum chromo-dynamics. We are now ready to tackle the normal modes in continuous mediums. And I will do that starting first with a string which is fixed at both ends. One way you can do that is you can go back to the results of last lecture when we have capital N bits, and you can make N go to infinity. And the shapes that you get for your continuous media for the strings when they're fixed at both ends are sinusoidal motions would be the first harmonic, the lowest frequency, and then you get high harmonics with higher frequencies. The question, now, is one of those normal mode frequencies for those continued media? And I will derive these today. I will derive these normal mode frequencies using a different approach than the N go to infinity route. And, I find the same results, of course. So, imagine that I have a string here and the tension is T, and the mass per unit length is mu. And, I wiggle one side, and I generate in there a wave. So, I moved the [NOISE OBSCURES] angular frequency, omega, and I do that with an amplitude, A. I will then generate waves whereby this is what we call, in physics, the wavelength. That is the distance that the disturbance travels in one oscillation time. If I call this a positive direction, this wave will propagate with beat V in that direction. And we know what V is. V is the square root of T divided by mu. We derived that last time. And so, Y as a function of X and T, Y being in this direction, X being in this direction, is that an amplitude, A, times the sine, or if you wish the cosine, that's fine with me, times 2 pi divided by lambda times X minus VT. Let's first make T equals zero. Then you see what you have here, the sine 2 pi over lambda times X, if you take X zero, then you find zero. And, if you make X larger by an amount, lambda, you get zero again. So, that's why lambda is called the wavelength. The parent repeats over a distance, lambda. Clearly, this must be a solution to the wave equation because any single valued function I told you which is a function of X minus VT, satisfies the wave equation. And so, this one does too. Now, lambda is V, which is this V times the period of one oscillation, for which I will write a P today, not a T, because I don't want to confuse you. And, that P is obviously 2 pi divided by omega. This is my omega with which I am shaking. And so, omega, which is, then, my frequency, is 2 pi divided by lambda. I will introduce for 2 pi divide by lambda a new symbol, which I call K which is called the wave number. It has dimensions, meters minus one. This is done in most books. It is unfortunate that French calls K1 over lambda. It's not his fault because in the old days it was always done that way. I will always follow the convention of Beckafee and Barrett [SP?], and I will call K always 2 pi over lambda. If I use that new K, then I can rewrite this in a form that you see very often. But there's nothing wrong with that form. That would now be A times the sine of KX minus omega T. Notice the 2 pi lambda over lambda becomes K. So, that is KX. But, omega is always K times V. And so, this form is nice in the sense that if you have these two numbers, you know immediately what the frequency is. That's nonnegotiable. And you know immediately what the wavelength is. It's 2 pi divided by that number. And, you even know that the ratio of those two, omega divided by K is UV, which is that V. So, it is a nice way of writing things down. Now, I'm going to not only generate a wave on this side that moves in this direction. I'm going to generate one that goes in this direction. And then I want to see what they do together. And so, this is now the wave that goes in the positive X direction. And now, I'm going to have another one: Y2 XT, same amplitude, same frequency. Therefore, same wavelength, but now it's going not in this direction. But it's going in this direction. Notice there is a plus here, but there is a minus there. And so, I want to know, now, what some of these two is because one is going like this. Another one is going like this. So, I want to know what the superposition of those two waves do. And so, Y, which is then the total displacement is Y1 plus Y2. And so, I get the sine of alpha plus the sine of beta. That is twice. So that becomes 2A times half the sum, the sine of half the sum. Half the sum becomes KX times the cosine of half the difference. Half the difference would be minus omega T. But minus and plus for cosine, so I will just write down cosign omega T. And this, now, is a very unusual wave. All the spatial information is here. So, this is all the spatial information. And, all the time information is here. And so, whenever this sine is zero, if the X is such that the sine is zero, just tough luck. Then, that location, X, never moves. The cosine omega T could never make it move. It always stands still. We call those nodes, and I will show you, of course, some examples. This has a name, a very nice name. It's called a standing wave, whereas those that I have there we call traveling waves. Now I'm going to make the string, fixed at this end, and fixed at this end. I may shake it a little bit here. You'll see a demonstration of that. And now, at this X equals zero, and at this X equals L, I now have boundary conditions that I have to meet. Y cannot be anything but zero there because it's fixed. And so, I now demand that X equals zero and at X equals L that Y must become zero. And so, the only way that that can be done is that you only allow certain values for K to exist. And, those values for K, which I will now give an index, N, as in Nancy which is going to be the normal mode, N equals one, two, three, four, is now going to be N pi divided by L. Clearly, if X is zero. Oh sorry, we're here. If X equals zero, then surely we are here. Then the whole thing is always zero. That's not an issue. But you see now, if X equals L, then you get N times pi. And so, again, the sine becomes zero. So, I've met this boundary condition that this point cannot move. So, lambda of N, which is 2 pi divided by K of N, that's the way that I define K of N, then becomes 2L divided by N. And, being that one, two, three, four, five, omega N, which is always K of N times V, therefore becomes N pi V over L, and the frequency in hertz, if you prefer that, is 2 pi smaller would become NV over 2L. And so, what are you going to see when you plot dysfunction, this standing wave function? Well, it depends on the N number. Let's take N equals one. And so, the length is L. If you plot that sinusoid for N equals one. It looks like this. And then, the cosine term will make you do this. That's the task of this cosine term. The sine term is just this thing, which would then have an amplitude. That could be different, of course, for the different modes. But it would be this amplitude that you see here. Recall this one, the fundamental. It's the lowest normal mode, but we also call it the first harmonic. I will do both. I will sometimes refer to this as a fundamental because I'm more used to that. But I will also refer to it as the first harmonic. Lambda one is 2L. That's just staring you in the face. In order to make a full wave out of this, you have to add this. So, lambda one is 2L. And, of course, when you look here, lambda one, 2L divided by one. That's exactly what we have there. So now we go to N equals two, which is called the second harmonic. So, let me write this down. So, this is called the fundamental, which is also called the first harmonic. And this is then called the second harmonic whereby the point in the middle stands still. And then, the cosine term will change shape like this. It is unfortunate that there are books that call this the fundamental and is the first harmonic. That's enormously confusing because now you had to start from zero which is the fundamental, and then N equals one is the second harmonic. I will never do that. This is always for the first harmonic. This is always the second harmonic. And so, for N equals two, you see immediately that lambda two is simply L. And that's exactly what you see here. And then, you can put in the third harmonic and I will demonstrate this very shortly. So here you have an omega one, and here you have an omega two. And, they follow this pattern. Omega two is exactly twice omega one because when you make N equals one, you have omega one. And when you make it two, you get twice that much. So, now you see that the ratios, omega one, omega two, omega three, relate as one, to three, to four, to five. And so, you can also write down here then that omega N is N times omega one, and therefore that F of N, which is the frequency in hertz is also N times F1. So, it's very easy to think in terms of the series of these harmonics. They're also sometimes called overtones. When we talk music next Thursday, I will often use the word overtone. So, if the lowest frequency in this mode were 100 Hz, then the second harmonic would be 200 Hz. The third harmonic would be 300 Hz, and so on. Now, when I'm driving this string at one end, I will have a wave going in and have a wave coming back. That is exactly recipe that I need for a standing wave. I get reflection here, and so I have one wave going in, and I have one wave coming back. And, if I drive these at these discrete frequencies which are set by the boundary conditions, then the system will react very strongly. We'll go into resonance. You build up a huge amplitude because you keep feeding in waves. And, they keep coming back. So the whole thing starts building up. And, that is the idea of resonance. And, resonance and normal modes are one and the same. So, that's the normal mode of these strings fixed at both ends. We also refer to them, sometimes, as natural frequencies. These points here have a name. They are called nodes. So, this is a node and this is a node. And these points here, also here and here, the ones that have the largest amplitude are called anti-nodes. In Dutch, we call them tummies, very strange. We call the [barker?], barker is this. So, the boundary condition leads to discrete values of the resonance frequencies. And, if this were quantum mechanics, we would call these [igan?] solutions and igan states. So, if you want to write down, now, the situation for your Nth mode, in its most general form, then you would get Y as a function of X and T in the Nth mode would have its own amplitude, A of N, whatever that is. You can pick that for different values of N. And then, you have here the sine of N pi times X divided by L. And here, you have the cosine of omega NT. That, then, meets the boundary conditions for two fixed ends. And, any linear superposition, any combinations of various values Nancy N, and various values of A will satisfy the wave equation. So, this string can simultaneously oscillate in a whole series of these normal modes. And when we do music next Thursday you'll see that. I will demonstrate that to you. I want to show you now is that if I take a string, I need, again, even though it has spring-like qualities we will treat it as a string. Then I will show you that if I drive this at the end at the proper frequencies, that I can generate the resonances. And, I can make you see the normal modes. So, who is willing to assist me this time? You were dying last time also, right? But Nicole won the battle then. So, hold it in your hand firmly. Just walk back and do nothing. Just walk back, walk back. We need a little tension on there. OK, that is fine. So, I'm now going to wigg