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• 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