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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