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 wiggle
this. This is really a fixed end.
You will see when I hit resonance that my hand is hardly
moving at all because I keep pumping waves in and they keep
coming back at me. And the amplitude will build
up. And then I have to search for
these residences. And when I hit one,
I know I hit one. You will see why I know it.
I feel it in my stomach. I feel it in my tummy.
I feel it in my hands, and my whole body knows I hit
resonance. And you'll see that.
OK, there we go. This is the lowest mode.
This is N equals one. And notice that my hand is
hardly moving at all. So my hand, for all practical
reasons, is really a fixed end. So, you get that solution.
So, I will now try to find the second harmonic,
which would end up, then, as a node in the middle
in addition to nodes at the end. Is that it?
And again, when I hit that resonance, which is a normal
mode, notice that my hand is hardly moving.
And, boy, do I feel it. I really know that I am on
resonance. I can try the third one,
in which case you will see two nodes in the middle,
and two at the end. Is this the third one?
Well, I'm good today, am I not?
You see: very clear. These points stand still.
Normal modes: you're not supposed to do that.
Oh man, you're ruining my demonstration.
So, now I'll try to generate the highest frequency normal
mode that I can. And so, you count the number of
nodes in the middle. And if you find five,
then that's the six harmonic. I think I can do better than
that. But it's not so easy to get a
very high frequency. So, you do nothing because
you'd really ruin it. Yeah, yeah, yeah,
yeah, yeah, yeah, I'm getting there.
I'm getting there.
Oh, no, no, no. You shouldn't have moved,
you see? No, it was not your fault.
I'll try it again. I got another resonance.
I got a resonance. I got a resonance.
Count. Please count.
How many did you see? Five?
I counted 12 nodes in the middle.
What is your name? All right, so that's just the
way that it works but you seem normal mode solutions,
which either result in standing waves.
And the demonstration speaks for itself.
I have here a rubber hose, which we are going to excite in
the fifth harmonic, let's see how that works,
whether we were close. If anyone touches the table,
they are, then, the tangent changes.
And immediately, of course, the resonance
frequency changes because the resonance frequency,
notice, has this V in it. So, the moment you change T,
it's different. But it's not bad.
And so, now I will strobe this one for you because you have no
idea that what is happening that it's going like this.
It's going too fast for you. And so, I make it easy on you.
Could you turn the lights off? Thank you.
So, I'm going to strobe it for you.
And then I can more or less make the rope stand still.
But I can also offset the frequency of my strobe light a
little bit so that you actually see the rope move very slowly.
So, I can purposely give the strobe light a slightly
different frequency. So, that's what I'm doing now.
You see, so you have no doubts anymore about the effect,
no doubts that indeed if the middle goes up,
the one on the side goes down, and vice versa.
What I can also do, to turn this into a work of
art, if I can turn this one off again, yes I can,
I worked with an artist who actually liked these things a
lot. And so, I can also strobe it
twice as often. So, this is actually,
the frequency of the rope is about 9 Hz.
So, my red light was close to 9 Hz.
The green light is 18 Hz, and a little bit off.
So, now you see it twice per oscillation.
Of course, to make it really wonderful, to make it very sexy,
I can do them both simultaneously.
So, now you see the red one doing his own thing and the
green one doing its thing. And what I find interesting,
it may not work for you, but there are moments
occasionally that the red exactly overlaps with the green.
When that happens, I see white light.
Let's see whether we can have a case like that.
Yeah, that was at the bottom. Did you see it?
It's white. It's really amazing.
There it is again at the top. All right.
So, we can have some light again.
Thank you, Marcos.
I can now do the same thing for sound.
These were transverse waves. But there was nothing wrong
with doing the same thing for sound, in which case you would
have a tube which has a certain length, L.
Say it's closed on both sides for now.
And, I can now generate in there pressure waves,
make the air column oscillate, and the air particles in the
lowest mode, all I have to do is what is here in the Y direction
that is a transverse position. I now have to offset the
molecules of the air in the X direction.
So, this is the X direction. And, the direction in which
they move is in the X direction. But, the displacement I will
call psi [SP?], how much they displaced from
equilibrium. So, in the lowest mode,
that means in the fundamental in the first harmonic,
in the middle the motion would be very large like the Y is very
large. And then, it's a little smaller
and a little smaller. And here it would stand still.
And here it's a little smaller. Here it's smaller.
Here it's smaller, and here it stands still.
So therefore, in terms of psi,
which is the position of the molecules, this would be a node.
And this would be a node. And this would be an anti-node
in the case of the first harmonic.
And so, I could write down that psi in its end's mode as a
function of X and T is, then, effectively the same as
my Y that I had before, except I have to replace Y,
now, by [xi?]. So, I give it some amplitude,
A of N. Then I get the sine of N pi X
over L, and I get the cosine omega NT through an equation is
completely identical. However, keep in mind that
omega N, which is this value, which is N pi V over L,
that is no different. That is, again,
N pi V over L. But, V is now the speed of
sound. It's nonnegotiable.
V is 300 m per second, and you are stuck with that.
And that has major consequences,
as I will show you Thursday for the design of wind instruments,
whereas with string instruments, you can manipulate
V, and you can change T and mu. You can make the tension and
the strings larger or smaller. You do not have such an option
with wind instruments. The only thing you can play
with is L. So, this is the approximate
speed of sound. More often than not do we
express the standing wave in the case of sound in terms of
overpressure. So, we don't look at the
displacement of the air molecules, which I did here,
which is psi. But we want to know where the
pressure is larger than 1 atmosphere, and where it is
lower than 1 atmosphere. It is the same idea,
but it is very often done. Now, keep in mind that if these
particles flow in this direction and pile up here,
that the pressure here becomes higher and the pressure here
becomes lower because the particles move away from it.
So that means where you have a node in psi, you always have an
anti-node in pressure. So these other locations by the
pressure becomes high. And the overpressure over and
above ambient is zero here. There was nothing that prevents
the pressure from building up. These particles are free to
move. And so, if you write it down in
terms of P, which is now overpressure,
it is not the total pressure. But it is what is over and
above one atmosphere. Then you get some amplitude in
pressure. I give you just a P of N.
And now, I get the cosine of N pi X over L, and then I get the
cosine of omega NT. And you understand now why you
get the cosine because the anti-nodes are now here at the
ends, which is the consequence of the boundary condition.
If I drew a curve as I made a plot of the pressure,
I will do it here if it doesn't become too cluttered.
So, here is now zero, and here is L.
If I put here the pressure, then in the first harmonic,
the fundamental, here is a pressure node in the
middle where there was an anti-node from psi that is now a
pressure node in the middle. And then the pressure will
change in this way. So, this is now N equals one.
So, the pressure here at one moment in time is high.
Then it's zero here, and then it is low here.
It is below ambient pressure, and that of course is changing
with time, with the cosine omega T term.
And, again, you see that lambda one is, of course,
2L. There is no difference there
with the string. I can demonstrate this to you
in a very, very nice way. I have here a two which is
closed at both ends. So here it is.
But at one end I have a piston. And so I can change the length,
L. You will see in a minute why I
like to change that length, L.
We are going to drive the inside with a microphone.
Here's the microphone. And we're going to do that with
a frequency. We really wanted to do a 403
Hz, at 8.03 Hz has a rather low wavelength.
I wanted to have it shorter. So I do it three times 8.03.
So, the frequency in [UNINTELLIGIBLE],
which is 24.09 Hz. And so, the wavelength,
lambda, which is V divided by the frequency.
That's correct. The wavelength,
lambda, equals V divided by the frequency is then very roughly
14 cm. So, that gives you an idea
about this much. In here, we have a microphone
that we can move in and out, and a microphone measures
pressure. It has a membrane.
And so, it's sensitive for pressure.
And this microphone is connected to a loudspeaker so
you can hear it. It'll also show to you,
the signal the microphone. It'll show to you as we
recorded it on an oscilloscope. If I put the microphone at a
pressure node, you hear no sound.
And if I move it to an anti-node, you will hear sound.
I will create in here pressure nodes, nodes,
nodes, nodes,
which are always half a wavelength apart.
Think about why that is, if you hear the nodes are half
a wavelength apart. The same will be true for
sound. I am then going to put this
microphone somewhere at a node that we all agree you hear
almost no sound. And we see no signal from the
oscilloscope. And then, I'm going to move it
back and search for one, two, three, four nodes.
And then, I know that this distance is going to be two
lambda. And I have, therefore,
measured lambda. And, the amazing thing is,
which actually surprised me, that you can do that to an
accuracy of about one million. That is so enormously clear
when you are at a node, if you move your mike by 1 mm
you can really see that you're no longer at the node.
So we'll know lambda probably to 1 mm accuracy.
Once we know that, the speed of sound is lambda
times F. We know what F is.
That's known to one part in, we couldn't be off by more than
1 Hz. So, that's a very well known.
And so, we now have a measurement of the speed of
sound, a high degree of accuracy.
We measure the speed of sound. And so, we catch three birds
with one stone. First of all,
I can show you the nodes in anti-nodes.
So you see how they build up there.
And then, at the same time we can measure the speed of sound.
The only reason why we make this end movable is that as I
move it and bring the mike first at a node, I can make sure that
the length is just at resonance. And so, that's just my
beginning to make sure that the node is as close to zero as it
possibly can be. And so, if we now get the image
up there, and then I think we are going to get the light
situation like this, and going to turn on the sound,
here is the sound. I can turn up the volume
seeking hear it better. Is this connected,
Marcus? Oh yeah, of course,
thank you. I didn't connect the microphone
yet. It has a switch.
OK, let me first use the, oh, this is awful.
This 2409 Hz, by the way.
I'll turn it down a little. Is that better?
OK, let's first move it around. Let's first bring to a node.
Boy, there's nothing left anymore, right?
And now, I'm going to change L. OK, I cannot do much better.
So, what L is, is not very important.
I set L so that I think the system is at resonance in a node
that you can calculate if you know how many nodes there are.
So, I'm going to search for, well here, you see the
anti-nodes, by the way. You see that?
I'm moving it in now. I'm moving it further that way.
And here, you get to another node.
You see that? You see that wonderful,
and I go again to an anti-node. And here comes another node.
And this note I'm going to measure the position.
I have a ruler in there, and the ruler gives me the
position of the mike. And this one is 52.8.
So, 52.8 cm is this position here somewhere.
And, I can put a mark here for you, not that it will help you
very much because my accuracy is 1 mm.
And, this is a very crude way. So, now I'm going to pull it
back. You're going to see an
anti-node. You can hear it,
a little bit more sound. And you're going to see a node.
That's number one. Here's node number two,
node number three, node number four.
And I'm going to look; and I'm going to look now at
the reading. And I read 24.3.
So, I read 24.3 cm. So, I subtract them.
That is five, 28.5 cm.
And this is two lambda. And I know this really to,
I would say, a millimeter.
It's certainly no worse than two millimeters.
And, there is no uncertainty for sure in the F,
if you can give me the light back.
So, I can now calculate what the velocity of sound is.
So, that is 28.5. I divide that by two.
It gives me 14.25 cm. This was just a rough number
that I gave you. And now I multiplied that by
the frequency 2409. And then I find 343.
So, V is 343 I would say plus or minus maybe 2 m per second.
So, we have measured the speed of sound to a high degree of
accuracy. And, of course,
at the same time you have seen this wonderful resonance normal
mode behavior where you see the nodes and the anti-nodes,
in this case, of pressure.
Earlier you saw them [NOISE OBSCURES].
Now you have seen them in terms of longitudinal motion.
There is energy in a traveling wave.
If I have here a traveling wave, which is moving in this
direction, has tension T, mu is the mass per unit length,
then I can write down Y as a function of X and T.
I can give it a certain amplitude, A,
times the sine. And I can write this now in
many different ways. But let's write it down in this
form: P times X minus VT. And, we know this V,
that V, let's call it V squared is T divided by mu.
So, this is a traveling wave that goes into the plus X
direction. Is any matter moving in the
plus X direction? No.
But is anything moving? Yes.
It is moving in the Y direction.
So, since these particles have mass, and they are moving in
this direction, there is kinetic energy due to
the motion in this direction, not due to any mass that moves
along. No mass moves in the direction
of here. Suppose I carve out here the
section, DX. So this is the direction of X.
And, I take a small section, DX.
Then the kinetic energy, DE kinetic, tiny little bit of
energy is one half times the mass, DM times the velocity in
the Y direction squared. That is 8.01,
right? Kinetic energy is one half MV
squared. Never confuse this V with that
V. That is the velocity of
propagation. This is the velocity of the
string in this direction. So, I can write this down.
DM is obviously mu times DX. If I have a length DX here,
and I have mu kg per meter, so I get here one half mu times
DX, and for this I write down DY DT squared is the velocity in
the Y direction. And I use partial derivatives
because I do it at a given location for X.
So, what is DY DT? Well, that's easy.
That's a piece of cake. There is my function.
So, I get an A. Then I get [NOISE OBSCURES]
minus V. So, I get a minus and a K and a
V. And then, the sine becomes a
cosine. And so, I get K times X minus
VT. I should have put brackets
around there, too, but that's
self-explanatory. And now I want to know what the
square of this is. So, I'm going to square this.
Now I can calculate what the total energy is,
kinetic energy, in one wavelength.
All I have to do is integrate, now, from zero to lambda to get
the total energy in one wavelength.
So, I'm going to do that. I'm going to write down here
now E kinetic. Follow me closely.
I have a half. I have a mu.
And then, I get the DY DT squared.
So, I get an A squared. I get a K squared.
I get a V squared, and then I get the integral
from zero to lambda of this function: cosine KX minus VT.
And then I have my DX, which is this DX.
That's what the integral is in the direction of X.
Yeah? Yeah, thank you very much.
Extra course credit. What is this integral?
Maybe you don't remember, but I've done this often enough
that I do know, the integral of cosine squared
of that function is lambda divided by two.
And I leave you with that. That's a very easy exercise.
And I can also take the V squared out and write for that T
over mu. So, [in kinetic?] now,
and this is in one wavelength. This is my shorthand notation.
So all in calculating for you now is how much is in one
wavelength. So you're going to get,
now, one half mu to get an A squared.
The K squared I can write down as 4 pi squared divided by
lambda squared. For the V squared,
I can write down T divided by mu.
And then, I have my lambda over two.
Well, this mu kills this mu. This two kills this four and
this two. And, one lambda kills one
lambda here. So, now I have the final result
that E kinetic in one wavelength equals A squared times pi
squared times T divided by lambda.
Now, if you look at this and you ask me, is that obvious?
I would say, not to me.
There is a T. There is a lambda.
My goodness, and I can also cocktail the
whole thing. I can get the T out and get a V
back in again. I will admit that I don't have
a very good feeling for this function except for one.
I do know that always the energy in the wave is always
proportional to the amplitude squared.
You're going to see that when we do electromagnetic waves in
8.03. It's always proportional to A
squared. So, that's the only one for
which I may not have a feeling. But I know it's got to be
there. And the rest I will leave you
with that to see whether perhaps you can talk yourself into
understanding why you see the symbols where they are.
Now, clearly, there is also potential energy
because it takes energy to make that straight line into a curve.
And, that means work that you have to do.
You have to squeeze to stretch it.
There is a tension, and you have to stretch that.
And so, you have to work to just get the shape.
And that is potential energy. And the potential energy per
wavelength, which I want you to do on your own.
It's worked out nicely in French.
I will not do it today, happens to be exactly the same
as the kinetic energy, which is by no means obvious.
So, the potential energy per wavelength is the same as the
kinetic energy per wavelength. And so, what that means,
then, is that the total energy in a traveling wave is twice
this: kinetic energy plus potential energy.
That is the total energy in a traveling wave.
I can now make you see in a nice way an energy balance to
compare traveling waves with a standing wave.
If I have a standing wave, and the standing wave is like
this, and I look at this picture at the moment that the wave
stands still, that VY is zero,
so it goes like this. That's a standing wave.
You've seen that. I do it at this moment,
and there's no kinetic energy. There is only potential energy.
A little later, there is only kinetic energy,
and there is no potential energy.
And although later, there is only potential energy
and no kinetic energy. If I pick that moment that it
stands still, then I know that that's the
total energy because it's only potential but it is the total.
In other words, for a standing wave,
if this has an amplitude, A, it must be the same
potential energy as you have in the traveling wave because it's
simply due to the fact of the shape.
It has nothing to do anymore with motion.
So, for a standing wave, the total must be that number,
whereas for a traveling wave, you have kinetic energy plus
potential energy. So, you have twice as much.
Now, I can convince you. And that was my plan earlier
today that if I have one traveling wave with amplitude
one half, A, and we have another traveling wave again with
amplitude one half, A, I know that I'm going to
make a standing wave with amplitude A, right,
because one half A and one half A, and we add them up,
you got twice this. You've got these [two A's?].
But, there must be a certain amount of energy in that wave
that comes in with one half A and a certain amount of energy
in this one. And when you add those energies
up, you must exactly get the energy in a standing wave
because no energy was lost. So, therefore,
I make the following statement now, which is testable,
that to traveling waves each with amplitude one half A,
I'm going to compare them with a standing wave with amplitude
A. There must be the same energy
in a standing wave with amplitude A as there is in two
traveling waves with amplitude one half A.
Did we follow that? Was that too difficult?
Because, we know that we can make that wave the standing
wave. OK, we know that the total
energy in a traveling wave is this.
But, we have two of them. So, in the traveling wave,
in two traveling waves, we have two.
Then we get that two there. That's nonnegotiable.
Then we get pi squared. Then we get T,
and we divided by lambda. But the amplitude A that we
have there is now half A, yeah?
Is that too difficult? Give it an amplitude half A.
So now, I get one quarter times A squared.
So, this is the energy into traveling waves.
Do we agree? What is the energy in one
standing wave with an amplitude A?
Well, we have the answer here. In a standing wave,
all the energy must be A, pi squared, T divided by
lambda. Look at it.
They are the same. These two, and these two,
and these four eat each other up.
Just a second. So, you see the consistency,
I'll give you a chance, that two traveling waves making
up one standing wave in the exercise I did earlier that,
indeed, energy is conserved. The energy in the two traveling
waves with half the amplitude gives you, then,
a standing wave with A. Yes?
A squared, thank you very much. Always A squared.
Isn't that what I said? Amplitude is always A squared.
Thank you very much.
If I generate traveling waves, I, Walter Lewin,
start shaking strings or some other instrument is making
traveling waves, then there is an energy flow in
the direction of propagation because these wavelengths keep
moving, these waves, and so there's an energy flow.
There's no mass flowing. Oh, there's energy flowing.
The mass is only doing this. And so, therefore,
if I generate a certain amount of energy, then I need power to
do that. Namely, power is energy per
unit time. And so, I can now very easily
calculate for you the power to generate a standing wave.
That is utterly trivial because the energy in the standing wave
is 2A squared, pi squared T divided by lambda.
And all I have to do now is to divide it by the time that it
takes me to generate one oscillation.
So, that's the period of the oscillation.
And the period of the oscillation, I try not to put
too many P's in there. And I hate to call it T.
So, the period of the oscillation is the same as one
over the frequency. And, one over the frequency is
V divided by lambda. So therefore,
the power is simply the energy that I'm generating.
And, V divided by lambda is the same, is one over the period of
one oscillation, energy divided by period,
the time that it takes to make one oscillation is [per?]
definition power. And so, then you get the result
here. You get a lambda squared here,
and you get a V there.
This is a traveling wave. It's good that you asked that.
I'm generating a traveling wave.
I have to work for every wavelength that I produce.
This is the amount of work that I have to do.
Do we agree? There is energy in the
potential energy because I have to give it that shape,
and there is kinetic energy because it moves.
If I generate this amount of energy, what is your name?
James, if I generate this amount of energy and I divide
that by the time that it takes for me to generate that energy,
I get the power, the average power that I have
to put in. It's really average power.
And so, I have to divide this by the period of my oscillation,
the time that it takes to make one oscillation.
And that is lambda divided by V.
That is the period and so I had to multiply this by V divided by
lambda. So this is really what we call
frequency. And so, that you get the power.
And so you can move energy from one point to another without any
math moving. All right, this is an ideal
moment for a break and for a fatherly talk.
I've had some very interesting e-mail exchanges with quite a
few of you. And as a result of that,
I have decided to lower the 10% of the mini-quizzes to 5% and to
raise the homework from 10% to 15%.
I think that will make probably most of you happier,
though not all. Some of them thought it was
fine, but I think most of you will be happier.
And I will make that change this afternoon on the website.
So you'll see that reflected when you look tomorrow morning.
I will also be more careful that the people here in the
front row don't have five minutes, and the people in the
back maybe only one minute. And so, therefore I'll make
sure each one of you from now on has the same amount of time.
And so what I'm asking you now is not to start until everyone
has, and then I will blow the whistle, and then I will give
you five minutes. So, will you start handing this
out? [SOUND OFF/THEN ON] OK.
I'm now going to change the boundary conditions,
which is something that you will need very much for your
problem set. Here is a string,
mu one, V one, and here is a string,
mu two, V two. And obviously they're the same
[tangent/tension?], one string.
And so, V1 is the square root of T divided by mu one,
and V two is the square root of T divided by mu two.
I will call this point here, for simplicity,
X equals zero. That is the junction.
And this is the Y direction, and this is the X direction.
And I'm going to have a wave coming in from the left,
which I called the incident wave.
I will give that an I moving in this direction.
And then, there will be a reflection which I call R.
Something may come back so that goes into this direction both in
medium one. This is medium one and this is
medium two. And then something is being
transmitted into the second medium.
I will give that subscript TR. And, that is going into medium
two. And that's what I want to
evaluate now. Now, I have boundary conditions
at X equals zero. I told you earlier.
All of 8.03 hangs together: the wave equation and boundary
conditions. And that's true.
The boundary condition is that at X equals zero,
Y one must be Y two unless the string breaks.
So, the Y right here must be the same as the Y right there.
Otherwise there would be a break of the string.
But not only that, DY1, DX, must also be DY2 DX.
If that were not the case, there would be a kink in the
rope at the junction, and the kink would mean that
there is a tension here and that there is a tension here which
would give a net force down. But the junction itself has
zero mass. And so, that would give an
infinite acceleration. So you can never have a kink in
the string, not even when it is connected like this with another
string. So, it must be something that
is always fluid here, which could be like that,
which could be like that. It cannot be like this.
And so that's, then, the condition that the
derivative, the spatial derivative from the left must be
the same as on the right side. So, let us start with an
incident wave that comes from the left.
And so, Y1, YI, I, incident,
has some amplitude A of I. That is the amplitude of the
incident one times the sine. And, I'm going to write this
now. I can write that in many
different ways. But I'm going to write this,
now, as omega 1T minus K1X. That's clearly a traveling wave
that moves in the plus direction because the signs are different.
And now I have a reflected wave, amplitude A reflected
times the sine, obviously the same frequency,
but now I get plus K1 times X. Notice the difference in sign.
This one is going this way. This one is coming back.
The amplitudes are different, but the K's are the same
because the wavelength in medium one is the same.
The K is 2 pi over lambda, and that wavelength is not
going to change for those ways as it is for this one.
And so, now I get to transmit it one, this amplitude
transmitted times the sine of the same omega because if this
junction shakes up and down with frequency of omega,
[NOISE OBSCURES] given. You can't change the omega.
You can say omega is different from the right side than it is
from the left side. In other words,
omega one is V1 times K1. But, that is also V2 times K2.
It's the same omega. And so, I get here minus K2
times X. So, this goes into the second
medium. The minus sign indicates that
it's going in this direction, but the K2 is different because
the speed is different because V2 is different from V1.
And so, with the same frequency I get a different value for K.
So, now if I go to my boundary conditions, Y1 equals Y2 that
you see here, then I get there that A of Y,
A of I, so, X equals zero, at any moment in time I must
meet that condition. So, A of I plus A of R must be
ATR. If not, then the string would
break. I can now take the derivative
of my function, DY1 DX.
And so, when I do that, I get the first one is going to
be an A of I. And then the derivative against
the X gives me a minus K1 here. And then, I get a cosine of
omega 1T because X1 zero, remember, so I get a cosine
omega 1T, but each term will have a cosine omega 1T.
So, I will dump all the cosine omega 1T's.
So, I go to my next one, which gives me now plus A of R
times K1. And that now,
so this is the left side, this is my DY1 DX,
and my DY2 DX in my second medium is now this one gives me
a minus K2 times A times minute times minus K2,
and I dump the cosine. So, this is the second
equation. And you could solve these two
equations easily, except you have three
questions, two equations, sorry, with three unknowns.
You have AI, AR, and A transmitted.
And of course you cannot solve two equations with three
unknowns, but what you can find, that is the only goal I have,
is the ratio of the amplitudes of what comes in over what comes
back. That's my whole goal.
If you calculate for this one K1 over K2, and you substitute
that in here, then you can replace the case
by the ratios of V's. You see that?
So you take K1 divided by K2, and you get rid of your K's.
So, when you do that, you get A of I minus A of R
times V2 equals A transmitted times V1.
So, that is simply combining these two.
And so, now all you have to do is calculate the ratios of those
amplitudes for this equation, one equation with three
unknowns, and the second equation with three unknowns.
And, what comes out of that, this is also worked out in
French, what comes out of that is that the reflected amplitude
over the incident amplitude is V2 minus V1 divided by V1 plus
V2, and A transmitted divided by the amplitude or the incident
one is 2V2 divided by V1 minus V2.
Sometimes I call this with shorthand notation,
R, reflectivity. It's the ratio of amplitudes.
And, sometimes I call this shorthand notation TR,
which is transmitidity (sic), the ratio of the amplitude that
penetrates the second medium divided by the one that came in.
So, that is just my shorthand notation.
We can now do some interesting examples, and you will begin to
understand what is happening here.
Take an example, for instance,
that mu two is infinitely large.
That's a wall. So, in other words,
that second string is a wall. If that second string as a
wall, you better believe it that that point cannot move.
The incoming wave couldn't possibly move the wall.
So this is the same as what we earlier called,
it's a fixed end. In other words,
V2 is a zero, right, because if mu two is
infinity [NOISE OBSCURES] zero. So, now, I'm going to go to
these equations, and I'd say,
what is now R? V2 is zero.
Well, if V2 is zero, I get minus one.
Hey, hey, minus one, that means when a mountain
rolls in, what comes back? A valley.
And when a valley rolls in, what comes back?
A mountain. And we have seen that.
We demonstrated that last time, a fixed end.
And what do you think TR is going to be?
What do you think it's going to be?
Zero. Nothing goes through.
And, look at it. If V2 goes to zero,
ah, TR is zero. So, aren't we happy?
Take now an example that V1 is smaller than V2.
So, in other words, mu one is larger than mu two.
When we have a case like that, notice that R is always larger
than zero. If V1 is smaller than V2,
this is a speed now. This is always larger than
zero. What that means,
just a second, I'll give you a chance.
What that means is that a mountain comes back as a
mountain. Now, the amplitude of the
mountain will be changed, but it comes back as a mountain
because that's the plus sign. And notice that the
transmitivity (sic) is always larger than zero.
It never gets a minus sign. Of course it cannot get a minus
sign. That you can't explain to your
kid brother because if a mountain comes in here,
that point is going to move up no matter what.
And if this point is going to move up, what goes into the
second medium as a mountain. And if this point goes down
because it is a valley, then what goes into the second
medium is a valley. So, it is completely illogical
that here you never can get a minus sign.
But here you can. That was a question.
Think you very much, my goodness.
Yes, V1 plus V2, thank you very much.
Thank you very much. So now, we take an example
which is absolutely thrilling, V1 equals V2.
That means mu one equals mu two.
That's another way of saying there is no junction.
[NOISE OBSCURES] anything will reflect.
Of course not. So I predict without even
looking at the result that R is zero, and that the transmitivity
is going to be plus one. Let's check that.
When V1 is V2, this is zero.
And when V1 is V2, you just corrected me at the
right time, by the way. When V1 is V2,
you see that this goes to plus one.
With a minus sign it would have gone to infinity.
I would have caught it. But you got it earlier.
OK, so now let's do a specific example whereby we give some
numbers because that's what I plan to demonstrate.
So, the example that I have in mind is that V1 is 2V2.
So, V1 is 2V2. In other words,
mu one is one half mu two. If I put that in that equation,
I find that R equals minus one third.
And, I find that the transmitivity is plus two
thirds. It's just a matter of sticking
those numbers in. And that means,
then, the following, that if this is my incident
wave, so this is now the incident one,
and the incident one, say, has an amplitude one,
and is moving in this direction, that the returning
one has an amplitude which is three times smaller.
And it is flipped over but it has the same length because it
has the same speed. So suppose the returning one
were here. Then you would see the
returning one with only one third of its amplitude.
So, that's about here. So, you would see something
like this. And it would be moving in this
direction. And this amplitude is one
third. And now, what goes into that
media number two is a pulse with two thirds the amplitude.
So, this is the two thirds mark.
So, the amplitude is only two thirds.
But, since the speed is lower, because notice,
speed is lower. Therefore it shrinks.
And so, not only will amplitude be two thirds of the incident
one, but this length will only be half.
And so, you will see this. And this is now two thirds.
And this is moving with velocity V2.
This is moving with velocity V1.
And this is moving with velocity [UNINTELLIGIBLE].
So, that's the meaning of R and TR.
There was a question here. Mu one, V1 is 2V2.
Then, mu one is one half mu two.
Is that one? You have a good point.
Let's just remove it. OK now?
Thank you. So, this now I want to
demonstrate. And the way we are going to
demonstrate it is we're going to use this machine.
And this machine allows me, this has a medium here whereby
the speed is twice as high as the speed here.
I'm going to generate here a mountain.
And the mountain will go into that medium.
And the first thing I want you to notice is that as it enters
this medium where the speed is lower, the speed is higher here
than there, I want you to see two things: that,
first of all, the mountain go through as a
mountain. Whatever goes through must
always have the same polarity. If the mountain comes in,
a mountain goes through. So look at the two things.
The mountain remains a mountain, and it will shrink.
And then, later, we'll overlook the reflection.
So, here, we have to meet it completely dark,
don't we, for this? And I think that's more
romantic. Yet, that's what they like,
yeah. OK, so I'm going to generate
here a mountain as fast as I can.
The speed here is larger than the speed over there.
And you will see that the pulse shrinks.
But the mountain remains a mountain.
Are you ready for that? You see that the mountain
remains a mountain? Can you see that it shrinks?
I need some light because I think I broke something.
No problem. Fixed.
OK, now we want to do something else.
Now I want you to see that when I drive in a mountain that a
mountain goes through, but the valley comes back.
And, the valley that comes back has the same length as the
incident one. That is, the minus one third,
remember? So, the valley is very shallow,
but the mountain will come back as a valley when it hits this
point. So, we have already seen that
the mountain goes through with a mountain.
You have already seen that the pulse gets shorter.
And now you are going to see that when it hits this point,
the reflection will be minus one third.
So, the mountain will come back as a valley.
Are you ready for this? And there it comes.
Could you see it, that it came back as a valley?
Shall I try again? You see, the speed is very
high. That's the problem.
It's very hard to see. I'll try once more.
Yeah, yeah, I could see that the mountain came back as a
valley effect. What would happen if I drive a
mountain in from this end? What would come back,
a mountain or a valley? So, if I do it from that end,
the mountain returns as a valley.
If I do it from this end, the mountain must return as a
mountain. Shall we take a look at that?
And it comes back as a mountain.
I'll do it once more.
I think I broke that one, too.
Yeah, boy, I only have one problem.
And that problem is going to be your problem.
And that is the following. I would like to explore now the
possibility that mu two is going to be zero.
That means V2 has become infinitely high.
If mu two is zero, there is nothing here.
It means there is tension on the rope.
But there is nothing here. It's empty.
We call that an open end. So, imagine in your head that
something is holding it tight. The end can move freely.
That's no problem. But, there is no mass there.
Remember, it was like the mass-less string,
and the rod. So, now, in this case,
which is really an open end, there's nothing in medium two,
I can now go to my equation and ask what R is.
And, I will be very pleased. When mu two is zero,
when V2 goes to infinity, when V2 goes to infinity,
there's this plus one. Mountain comes back as a
mountain. All of the predicted that last
time an open end. A mountain comes back as a
mountain. Physics works.
What do you think is going to be the transmitivity?
Zero, yeah, that's what you think.
Put in there V2 equals infinity.
That takes you by surprise, plus two.
Holy smokes! What is going on here?
If this were true, we would have solved the energy
crisis of the world because a pulse comes in and the whole
pulse comes back. Everything that went in comes
back. Mountain comes back as a
mountain, but there is something in addition that goes into
nothing. I want you to think about this,
and you may not be able to sleep tonight.
And if you can't, it is not my fault.
It's the fault of physics. See you next time.