Today, I'm going to couple a huge number of oscillators.
And that ultimately we'll make that number infinitely large.
And that's the way that you can couple oscillators is through
springs, as we did last time on the air track.
And we can let the cars move like this.
We call that a longitudinal motion.
That means that the motion is along the line of oscillators.
And then, we have ways that we can oscillate things
perpendicular to the direction of the oscillators.
And we call that a transverse motion.
So that goes like this on the oscillators are like this.
And the algebra is identical. That is easier to use a
transverse motion when you do the derivations.
And so, I'll use a transverse motion.
Suppose I have a string and I put on this string [beats?]
masses. I have a tangent,
T. Each mass is M,
and the length between these is L.
And, this is a fixed end. And this and it is also fixed.
It cannot move. And so, this is number one.
This is number two. This is number three.
And then here is the last one, which is number N.
I'm going to have N of these beats on the string.
So, this point here, I can call that zero.
I think of this direction, X, and this is the direction of
Y. And so, at zero there is no
beat. And here, at the position N
plus one, there is no beat, either.
And so, the question, now is, what are the normal
mode solutions to that system? There must be capital N normal
modes. And, if N goes to 10,000,
there must be 10,000 normal modes.
At a particular moment in time, you can imagine,
for instance, that this one is here.
And maybe this one is here. And this one may be here.
And maybe this one is here. The only thing we have to keep
an eye on that this is always zero, and the Y here is also
always zero. That's what we call the
boundary condition. Now, if you make the amplitudes
modest, then it's easy to demonstrate that the tangent
will remain constant in these pieces, modest amplitudes,
and that there is no motion in the X direction either.
At least you can make it negligibly small.
And so, we will only concentrate on the forces in the
Y direction. And so now I'm going to make a
[loner?] version here for particle number P.
So, this is particle P, and here's the location of P
minus one. And here's the location of
particle P plus one. And at a certain moment in
time, let's assume that that particle is there,
that little mass N. And, let's assume that this one
is here. And let's assume that this one
is here. So, this vertical displacement
is YP. This is YP plus one.
And this, then, is YP minus one.
But the strings are just attached like this.
And so, the fact that strings get longer: we will ignore that
because they are small amplitudes.
So the [tangent/tension?] will not change.
We draw this line. We draw this line.
I will call this angle alpha P. And I will call this angle
alpha P minus one. So, the tension is going to be
on that point P on this little mass, M, in this direction.
And there is a tangent in this direction.
And, we will assume that these tangents are,
then, the same for reasons that I mentioned.
So I can write down, now, for that mass for that
location, P. I can write down Newton's
Second Law. And so, I get M,
Y of P, second derivative, D2YDT squared of that position,
P. And, you see I have one
vertical component that drives it down.
That is due to this tangent. And then I have one that drives
it away from equilibrium. And so I'm going to get minus T
times the sine of alpha T minus one.
And I get plus T times the sine of alpha P.
But, of course, I know what the sine of alpha P
minus one is because the sine of alpha P minus one is YP minus YP
minus one divided by L. And so, I can write here Y of P
minus YP minus one divided by L. And here, I have T.
The sine of alpha P is YP plus one minus Y of P divided by L.
I'm going to introduce a shorthand notation.
I'm going to introduce the omega zero squared is T divided
by ML. As time goes on,
you will get more insight as to why we choose that.
At least convince yourself, if you have the time,
that this is, as I mentioned,
one over second squared is the right dimension.
T, by the way, whenever I use this throughout
the lecture, is never a period. It is always tension.
So, I will stay away from the capital T when we're dealing
with a period. So, I can divide M out,
and I can rewrite this to get the M downstairs.
And so we now get Y of P double dot.
And what I'm going to do now, I'm going to take the YP here
and the YP there. And, I bring them to the left.
I have a minus here and I have a minus here.
So, I get plus omega squared times Y of P,
and then I'm going to bring the P minus one in.
And, I'm going to bring the P plus one in.
Notice I have a plus sign here. So, I've got minus omega zero
squared times YP plus one. And here, I have a plus.
Minus-minus is a plus, so when I bring that back to
get a minus. But I already have the minus,
so I get YP minus one. And, that now equals zero.
But that is Newton's Second Law for particle P.
Excuse me?
What is your problem? Yes, thank you very much.
Very good, I appreciate it. This is a two because you have
two P's. You have one here and you have
one here. Thank you very much.
OK, very attentive. So, now we have to do these for
every single object. So, we have capital N
differential equations, one for each particle.
And the only thing that we have to keep in mind now when we
solve it, that Y zero, which refers to that location
zero there must always be zero, and that YN plus one is also
zero. So, if I made a sketch sort of
to warm you up to the idea, and I made a sketch for only
two particles, so this is number one and this
is number two, and this is a fixed end.
And this is a fixed end. Then you can sort of see that
in the lowest normal mode you're going to see something like
this. So, this is omega minus,
and it's going to oscillate like this.
But you can also imagine that in the second mode,
which is the highest one, that number one is up and then
number two is down. So, you get this situation.
And so they oscillate like this.
Just to make you see, I called this little n equals
one. And, I call this little n
equals 2n. They're referring to the mode:
mode one. I will use this n later on,
and this is, then, mode number two.
And, there are only two modes because there are only two
particles. So, let us now proceed with the
equation that we have. And it let us write down for
this system the two differential equations.
And so, look at this one. We're going to substitute for
P, first, number one, which is this one.
That's one differential equation.
And then, we're going to put in for P number two.
We get a second differential equation.
So, if you're ready, then we're going to get Y1
double dot plus two omega zero squared times Y1 minus omega
zero squared. And then, we get Y2.
That's this particle. And then, we get plus YP minus
one, which is plus Y zero, which happens to be zero,
by the way because Y zero here is zero.
So now, we go to [D?], and this is zero.
Now we go to the second particle.
And so we get Y2 double dot plus two omega zero squared Y2
minus omega zero squared. And, now we get first P plus
one, which is number three, which is Y3,
which happens to be zero because Y3 is this point.
And that's zero in this specific case with only two
objects. And then we get,
here, plus YP minus one. And, that is Y1 plus one.
That is Y1. And, that is zero.
So, these two coupled oscillators will have to be
solved. And in the normal mode
situation, we are clearly going to put this in as our trial
function, cosine omega T. They must oscillate with the
same frequency, omega, otherwise we wouldn't be
dealing with normal modes. And so, these are our trial
functions that we're going to put in these equations.
And we will put them in, number one.
We are going to put them in number two.
And then we will put them in particle number P.
And then, we can all the way go to N.
And if N is 10,000, we have to write down 10,000
differential equations on the blackboard.
That will take the rest of the hour.
So, I'm going to do number one here.
So I get a Y1 here, which is A1.
The second derivative always gives me a minus omega squared.
So, you get minus omega squared times A1.
I ditch the cosine omega T because each term will have a
cosine omega T. Then I get plus two omega zero
squared times A1, and then I get minus omega zero
squared times A2, right, because now I get a two,
plus A zero, which happens to be zero.
But I just put it there; you will see shortly why I want
to keep it there. I go to particle number two.
I get minus omega squared times A2.
It becomes a little boring. Two omega zero squared times A2
minus omega zero squared. Now, I have a Y3 here.
So, I get an A3, which happens to be zero.
But that's not so relevant right now.
And then I get plus A1. And, the reason why I started
off with one and two: that now you can see how we can
put it in the P's particle. So, the P's particle now is
going to be rather easy. Maybe I should do that in
color, minus omega squared AP plus two omega zero squared AP
minus omega zero squared. And, now we are going to get AP
plus one plus AP minus one. And that equals zero.
Also, this equals zero. And also this equals zero.
And so, now you see the differential equation for
particle number P. And so, you can go on now to
particle capital N, and now you have to solve N
differential equations. That's a zoo.
That's a terrible thing. Now, we will take a shortcut,
which is not very rigid. But, it really with save a lot
of math. And, that is we will use our
intuition, something that we know sort of from experience,
if you have a lot of beats on here, fixed here and fixed
there, and you ask yourself what's going to happen in the
lowest possible mode, then you just know that you get
something like this. It goes like this,
and like this, and like this.
And, you know that on the second normal mode,
the one that follows that has a higher frequency,
you expect that the side goes up.
This goes down. And then it will oscillate like
this. So, we use that experience,
which is not very rigid in order to decide on our trial
function. This would be mode one,
and this would be mode two. And so, not going to put in as
a trial solution A for particle P, which is in mode N as in
Nancy. This is the mode.
I want a sinusoid in there that is always zero here and zero
there. And it can have an amplitude,
of course, which I can freely choose.
So, this C of N is the amplitude of this sinusoid.
This is C1, and this value is going to be C2.
Each one can have its own amplitude.
And then I get the sine of PN pi over N plus one.
So, let's look together at this equation so that we have a full
understanding what we are trying to put in there.
Notice that P equals zero that the sine is always zero.
That's obvious because we wanted that because this point,
the zero point, is not moving.
Notice, also, that if you put in P equals N
plus one, that AN plus one is also always zero.
It's [put in?] P N plus one. The sine of a multiple times pi
is always zero because N is now our mode.
It's going to be one, two, three, etc.
These are the modes that we are looking for.
For instance, if we take N equals one --
Let's take N equals one. So, we have particles and they
are all in mode one. So then we get that AP in mode
one, we have a C1, which is the amplitude of that
sinusoid, and then you would have the sine of N pi divided by
N plus one. And you can,
indeed, convince yourself that that is exactly this sinusoid
with an amplitude C1. And, you can also convince
yourself that P zero, A zero, and particle N plus one
is, again, zero, of course.
And if you take N equals two, excuse me?
Yeah, boy, yeah, yeah, yeah, yeah,
the N is a P, right?
Thank you very much. Is that what you were saying?
Yeah, thank you very much. Today is not my day.
Yeah, there is a P. So, for particle number,
when P is zero you see this goes to zero.
And, when P is N plus one, this again goes to zero.
So, you see here that AP2 is now C2 times the sine of 2P pi
divided by N plus one. And, that exactly is this
curve. Notice that if you put in P
equals zero, you get a zero. If you put in P is capital N
plus one, you get a zero. But you will also find,
now, a zero precisely in the middle when P is capital N plus
one divided by two. And so, that is the consequence
of the introduction of this function.
And, of course, the ratios of the individual
P's for a particular mode state, for instance,
mode number one, when the system is oscillating
like this, the ratios of the amplitudes of the individual P's
is then given by this sign. P1 has its own amplitude.
P2 has its own amplitude. P3 has its own amplitude.
So, P1 will be here. P2 will be there.
P3 will be there. P4 will be there,
and so on. And so, then the amplitude
first goes up. And then the amplitude goes
down again. But now comes the question,
what is omega N? What is the frequency omega at
these normal modes N equals one, N equals two,
N equals three, N equals four,
and N cannot go all the way down, then, to capital N.
And for that, we have to return to this
equation. And I'm going to write it now
slightly differently. I'm going to take the [ace?] to
one side. So, I'm going to write down
here AP plus one plus AP minus one, and divide that by AP just
rearranging. And then you will find that it
is minus omega squared plus two omega zero squared divided by
omega squared. Take a look at this,
and convince yourself that this equation is identical to this
one. If you look in French,
French will take you from here one step further,
which is pure trigonometry. And I decided not to go that
route. But you can use the values for
AP that we have defined, namely this one.
And so, you can now massage the trigonometry.
And, you can find that this ratio is very simple.
It's twice the cosine. So, this whole thing is twice
the cosine of N pi over N plus one.
That is correct. N pi over N plus one,
so, there's no physics there. It's purely a matter of
trigonometry. We can now put an N here,
Nancy, because we know now that we are going to get solutions as
a function of the mode number N. And then, with a little bit
more trigonometry, and you really want to check up
on French there, which is page 141,
he then comes up with the normal mode frequencies,
which equals our goal. So, I will give you the result,
but it really is implicit, already in here.
You will get two omega zero. The result is by no means
intuitive times the sign of N pi divided by two times N plus one.
And, of course, I'm going to look through that
result with you. It now looks very opaque.
So, this, then, is the solution to omega N.
This is the solution. I will take one color to show
you how these link. This is the solution for A.
If you know the mode, and you know which particle it
is, and you have specified the amplitude of the sinusoids.
Then this tells you each particle what the amplitude is.
If you know the mode, N, then you know that this is
going to be the frequency. And so, you can write down now
that Y, which is the displacement as a function of
time, A is amplitude. Y is displacement.
Particle P in mode N, and now we can put in the
amplitude that we know. That is the PN amplitude,
APN. And now it's going to oscillate
with cosine omega N times T. And, of course,
you can always add a phase angle depending upon at T equals
zero what the particle is doing. And so, this one gives you the
amplitude. This one gives you the
frequency. And this, then,
is the time dependence of the displacement of particle number
P in mode N.
What I want to do now is to take a specific example which I
also will try to demonstrate, which will give you tremendous
insight. You can actually do it on the
left here because these are all very opaque.
But when you see an example worked out, and you see actually
how it oscillates, then it comes to life.
I'm going to have five beats on a string: one,
two, three, four, five, fixed here,
fixed here, one, two, three, four,
five. The tangent is T.
The mass of each one is N, and the separation between them
is L. And so, N is five.
So, keep in mind that N plus one is six.
The reason why I write that down: because you're going to
need the N plus one. And then, omega zero is the
square root of T divided by ML. So, I'm interested in knowing
what the frequency is in the lowest possible mode,
which is going to resemble something like this.
And so, that frequency, omega one, is then two omega
zero times the sine, N equals one,
capital N plus one is six, 180° divided by 12 is 15°.
So, there's the sine of 15°. I write it now in degrees
because I have a better feeling for degrees than I have for
radians. And that is 0.51.
I now go to omega two. And, I get the same thing,
except I get 30°. So, I get 0.5.
Sorry, I get exactly omega one, omega zero, right,
because the sine 30° is one half.
And, that eats up this one. And, I get omega three.
And, now I get the sine of 45°. And, that is approximately
1.41, not exactly, but as the square root of two,
so that is about 1.41 times omega zero.
Then, I go to omega four. And so, I get 60°.
And that can only be approximated again by about 1.73
omega zero. And then I have omega five
which is the last one. So I get this sign of 75°.
And that, then, becomes 3.73,
no, 1.93 times omega zero. What I want to concentrate on,
because that's part of the demonstration,
is not so much on the meaning of omega zero.
It's just some arbitrary thing that I have called omega zero.
But, I want to concentrate with you on the ratios of the higher
frequencies to the lowest one. And so, I call the lowest one
omega one. I simply call this omega one.
If that one is omega one, then the next one is 1.93 omega
one, again, not exactly but approximately.
It is this one divided by 0.51, the ratio of the frequencies.
And if I take this one and divide it by that one,
then I get 2.73 omega one. Then I take the next one.
I get 3.35. And the last one,
then, is 3.73. So, the bottom line is that the
ratio of these frequencies are not at all very nice numbers,
as you may have expected. But the ratios are quite
bizarre: 1.93 times higher than the lowest one,
2.73 times higher, 3.35 times higher,
and 3.73. And so, the general solution to
that system is, then, the linear superposition
of all these normal modes. That's the general solution.
You give them very modest amplitudes.
And, you can choose the amplitude of each one of them.
That's effective, like, say you were choosing C1,
C2, C3, C4, C5. And, you also can give them
initial velocities, if you want to.
So, at T equals zero, they do not all have to stand
still. You have that choice too,
of course. So, you can change the relative
phase between the five different modes.
What I will do is I want to make life simple.
I will generate all five normal modes for you.
And I will start them off all at zero speed when I show you
the simulation. And, the first one that I'm
going to show you, then, is number one.
And then, I'm going to show you number two.
I want you to appreciate that if I showed you the
superposition of one and two, so I let it oscillate in this
mode and in this mode, I start off at a certain
position of these particles. And they start to oscillate in
this mode in that mode, that the shape that I have will
never ever become the same as it was at time zero.
And, why is that? So, you just let it oscillate.
You can wait 100 billion years, and you'll never see the same
shape. If I start with a certain
shape, and it never ever, ever comes back to that shape,
why is that? Yeah, that's right,
that the sine of 15° is the killer.
That is not the ratio of two integers.
Therefore, you'll never get it back to the same position.
Maybe approximately, but you never get it back.
This demonstration is going to be a cocktail between very
low-tech and very high-tech. And, I will start with myself,
which is very low-tech. And that is this.
I was sitting in my office and I said to myself,
gee, what will I see? So, I took a pencil and a just
sketched very roughly the sinusoid right here.
And, I know according to these solutions, if you accept them,
that these beats, these particles must lie on
that sinusoid. And C1 is then the choice of
that amplitude of that sinusoid. In the second mode,
you pick another value for the amplitude, say,
C2. And then, the beats have to lie
on that sinusoid, and so on.
What you will see, however, is that these beats
are connected with straight wires.
So, you will not see those nice arcs.
What you will see is, of course, this.
The red lines are the actual strings.
And so, for instance, if you go to the second mode,
we call that a second harmonic, if you like,
then notice that this point here and this point here never
reached the amplitude, C2.
The sinusoid does, and the C2 in that equation
does, but those points will never reach that because their
location is such that they never make it to that point,
here. This [won't?] stand still,
then. That was intuitive.
So, we have that here. And if you go to higher
frequencies, particularly the very highest one,
neighboring beats always are out of phase with each other.
You see that: up, down, up,
down, up. And again, this little particle
will never reach the amplitude C5.
This one does. This one does not.
This one does not.
But this one does. And so, as I'm going to show
you this simulation, we will keep this going because
it will be great to anticipate what we may be seeing.
So, the first thing that I'm going to show you is one
complete oscillation in the normal mode number one,
which I have set to be 15 seconds with the help of
[UNINTELLIGIBLE], who has guided me greatly in
this demonstration. So, let me first of all give us
the right light conditions. And, now I will start the last
15 seconds. I have given amplitude a two,
which is very large. And, of course,
that's unrealistic, these high amplitudes.
But I want you to see the relative position of these
particles.
There we go. Now, you will see,
they will make one complete oscillation.
And that, of course, is no surprise.
If I clicked only once, it will stop now.
And that will be. If I click twice it will start
again. No, thank goodness only did
once. Now, I'm going to show you the
second one. And I give it the same
amplitude. So, C2, I give it two.
And, I want you to count how many oscillations it makes
before comes to a stop. It will, again,
be exactly 15 seconds. And so, we will have to agree
that the number of oscillations that it makes is now 1.9.
It misses the two. It will knock it back to the
two complete oscillations. And so, you're going to look at
this mode. So these two particles will
never reach the value, two.
Two is marked here. And it will go down,
up, down, and it stops very just short of two oscillations.
So we'll do that now. So, I make amplitude of the
first one zero. And now we get the amplitude of
number two. And, I make that two.
And, there we go. Now, count.
That's one oscillation. And it will stop just short of
two. You can't tell that,
of course. Why?
Because you don't have that resolution.
But it stopped just short of two.
Now we're going to this one, again, give it amplitude of
two. And now we're going to count.
And you will definitely be able to see that it just misses 2.75
because 2.75 is something that you can eyeball.
So, we're going to number three now.
Number three already has two interesting points which don't
move at all. This point won't move and that
point won't move. They will reach the plus two
and the plus two. Zero, and we get a two.
And there we go. That's one.
That's two. That's two and a half.
And that's 2.73. You see, it's just short of
2.75. So, that is this number.
Now we're going to number four. Number four,
this one will stand still. And others do not have the
maximum amplitude of two.
So, again, count. There we go.
One, two, three, and it stops there.
It stops at 3.35. So now, the last one before
we're going to cocktail them, there is this one.
So, again, this point will be plus two.
So this one will never reach that.
This one will reach the plus two.
OK.
One, see how it doesn't make it to two?
It doesn't get that high. Two, three, three and a half,
and it stops. Just under the 3.5,
it stops. Sorry, just over the 3.5,
it stops short. This would have been 3.75.
It's just under there. Now I will cocktail for you two
and four. If I cocktail two and four,
then this point would stand still, and this,
of course is going to be starting up.
And, this will start up because I give them both the same
amplitude. I will not go to plus two now,
but I will make it plus one. Otherwise, you get two large
values and it becomes a little bit unrealistic in what you see
there. So, I'm going to give number
two a one. That's amplitude,
and I'm going to give number five zero, and number four also
a one. And now, already you're going
to begin to see that the motion that you're going to see becomes
already sort of a little bit chaotic, a little erratic.
So, it's a superposition now of two normal modes.
This one, which I start off at a one here, and this one which I
start off as a one here, and at T equals zero are
released in both with zero speed.
So too I have the one in there? Yes I do.
There we go. This motion is already not so
predictable. But it's still sort of
symmetric for obvious reasons because this one stands still.
And now I want to do is start them all five.
Now, if we start all five, the start will be very
asymmetric because, look.
Particle number one: positive because I set them all
off positive. Positive, positive,
positive, positive, positive.
So it will start up very high. Now, look at the number five.
Positive, negative, positive, negative,
positive. So, number five will start very
low, and number one will start very high.
And then, when it starts to oscillate, it will take more
than the age of the universe to come back to that same shape.
But it is extremely erratic. You and I, no one can really
relate any more to what's going on.
And it is even possible to imagine that the motion in a way
is very simple, namely, a superposition of
five, very well behaving normal mode solutions.
It is a linear superposition of five very simple normal mode
solutions. But the net result is total
utter chaos, at least that's the way it appears to us.
But, it can be dissected into five very simple modes.
So, these were transverse motions.
And the same idea holds for longitudinal motions.
So, you can have five beats with six springs and then the
oscillation is in this direction.
We call that a longitudinal oscillation.
In this case, the displacement is
perpendicular to the oscillators.
We call that transverse. But the algebra,
as you can imagine, is identical,
except that the displacement are, then, not in this direction
for the longitudinal one, but in this direction.
We will shortly answer the domain of waves to make you see
the idea. The big difference would be
transverse waves and longitudinal waves,
sound. This is a pressure wave.
My direction to you, the air, the pressure wave is
doing this. So, the air is oscillating in
the same direction that it moves.
That is a longitudinal wave. But, this is a nice moment.
Will break for five minutes and will start exactly 5 minutes
from now. [SOUND OFF/THEN ON]
All right. Thank you very much for the
performance. That was prearranged,
by the way. So, we are now ready to make
the step to continue this medium whereby N goes to infinity.
Well, you can argue that it goes to as many items as we can
line up on a string. It's close to infinity.
And, it should not come as a surprise, of course,
that now you are going to get that the entire string,
which is now continuous mass, so you no longer have
individual beats that the entire string is now going to oscillate
as a sinusoid in its lowest mode.
So this is N equals one. And then it's going to
oscillate like this. For N equals two,
N equals two. And N equals three,
going up like this, and this goes like this.
It should not come as a surprise.
I will not pursue that today. We will get back to that later.
What I want to mention, though, what is interesting is
that the ratio of these normal mode frequencies will now be
one, two, three, four, five, and so on.
So, now you get that second mode is twice the frequency of
the first, which is what we didn't have there..
That's the big difference between number of N,
which is finite, and an infinity number of these
oscillators. What I want to pursue today,
will get back to this in the future.
What I want to do today is to generate a disturbance in a
medium, which has an infinite number of coupled oscillators,
which is a string to generate in there.
So, I take a string and I wiggle the end.
And then, I want to evaluate with you what's going to happen.
So, for this, I need some assistance from
someone. Nicole, would you mind?
Hold those firmly in your hand. Now, most of you may think that
this is a spring with a P as in Peter.
But, no, it is a string with a T as in Tom.
You will see that. I'm going to use this as a
string. I'm going to put tension on it,
T, which is what we needed, also for the N coupled
oscillators, and the amount of mass that we have we express
that normally in terms of the mass per unit length.
Remember, in the other case, we had little M divided by L.
Well, we call that now [gnu/mu?].
So, that's how much mass per unit length we have.
And what I want to do now is just shake my hand.
And then you tell me what you see.
Ready? Here we go.
Are you ready, Nicole?
What did you see?
Just some in what you see right after I do this.
What did you see right after I did this?
The disturbance moved. That's number one that we have
to understand. Why does it move?
Now, look what happens at Nicole's side.
I generate a pose which is like this.
I would call that a mountain for now.
And, only look at the moment that the mountain reaches her in
something comes back to me. And then, stop looking because
things begin to wander back and forth.
And tell me what comes back at me.
I'm going to send a mountain to Nicole.
What came back at me? A valley.
Now I'm going to send a valley to Nicole.
What do you think is coming back?
Very good. I don't know why it is.
It's very hard to generate a valley.
Let me do a mountain again. This is a mountain.
That comes back as a valley. I'll try a valley.
OK, I'll try to do a valley. So I go down and up.
Yeah, that was a good one. Well, because if you it worked.
Thank you very much. You did a great job.
So, now we have to understand two things, and that is,
why does it propagate? And why does a mountain come
back as a valley, and why does a valley come back
as a mountain? Continuous medium:
infinite number of coupled oscillators.
I start here with a piece of that rope.
That's called this position, X.
And, I call this position X plus delta X.
I call this Y. I call this angle theta plus
delta theta. And, I call this angle theta.
We have a tension, T, on the line,
and mu is the mass per unit length.
So, you tell me what the mass per one meter is,
and I know what mu is. If the length,
the mass per unit length. Well, if our displacements are
not absurdly high, then we can make the same
assumption that we made with the beated (sic) string that the
tension is the same on both sides.
It's an approximation. But, for modest amplitudes,
it's a very reasonable approximation.
So, we have a T here. And, we have a T there.
And they are, then, to a decent to a
reasonable approximation, the same.
Just like with the beats for modest amplitudes,
we don't have to worry about motion in the X direction.
The only thing that matters is the motion in the Y direction.
So, I will concentrate exclusively on the motion in
this direction, which drives it back to
equilibrium. And so, F of Y on this segment
is, then, minus T sine theta because this component is down,
minus T sine theta plus T sine theta plus delta theta because
this component in the Y direction is driving away from
equilibrium. But for small angles,
and we have to have small angles, otherwise our
assumptions are wrong, these are not the same.
For small angles, the sine of theta is the same
as theta in radians. And so, this becomes a theta.
This becomes theta plus delta theta, and so this thing becomes
T delta theta. That's an approximation for
small angles. Now, I will apply Newton's
Second Law. The amount of mass that is in
here is DM. And, I will calculate shortly
what DM is. It's a little bit of mass.
We're going to make DX go to zero, infinitesimally small
amount of mass. And so that mass times Y double
dot must now be this force that we just calculated.
So, it must be T delta theta. But, what is DM?
Well, we know that the length of the string is delta X.
So, DM must be delta X times mu because mu is the amount of mass
per unit length. And, if my length is delta X,
then DM is mu delta X. So, I can write this now as
delta X times mu times Y double dot equals T times delta theta.
We are getting there. Now, since we are in a limiting
case, we are going to make delta X zero.
The tangent of theta, so that's becoming,
then, this direction, the tangent of theta is DY DX,
right? That is DY DX.
And the reason why I use partial derivatives is that I
think of it as the time not changing.
Any moment in time, this is DY DX.
That's the only justification for the partial derivatives.
I take the derivatives on the side and on this side in X.
So, the left side, I take D tangent theta DX,
and I do it on the right side. Now, the derivatives of the
tangent of theta of the function is one over the cosine squared
of theta. That can't take you more than
20 seconds to confirm that. You can do that in many
different ways. So, this is the derivative of
the function itself. And then, of course,
I have to multiply it by D theta DX because I take the
whole function derivative in DX. And so, here I get,
then, D2Y DX squared. But, for small angle
approximation, cosine squared of theta is one.
And so, I'm going to substitute now this result into my
differential equation. I read this as delta theta,
which is here. And I read this in my mind as
delta X, which is here. Now, mathematicians would
probably never do that, but physicists have no problem
with that. So I'm going to write now,
here, mu times delta X. And here, I write D2Y.
DT squared. I use partial derivatives
because I'm not changing X. That's the justification for
the partials. And now, I get T.
And now, this delta theta, I'm going to write [whether?]
this times delta X. So, now you get delta X times
D2Y DX squared. And, now I'm doing something
that mathematicians would never do.
I'm going to divide out delta X.
Don't tell your 18.0 whatever people that I did that.
So, now what you have is that mu divided by T times D2Y DT
squared, constant value of X is now D2Y DX squared.
And, believe it or not, this is a big moment in our
life. You have here with differential
equations of Y, which is a function of X and T
whereby here you take the double derivative in time.
And here you take the double derivative in space in location.
What is a possible solution to this differential equation?
You can just see it by looking at it.
You immediately see what the solution must be.
Any function, any single valued function,
you can come up with any one. I don't care with which one,
any single valued function of X, plus or minus a constant
times T will satisfy this differential equation.
Just look at it. You can see immediately that it
works. Take the second derivative in
time. You get a C square out,
and you get the second derivative of the function.
Take the second derivative in X; you would only get the second
derivative of the function, and that's all.
So, all it requires is that C is the square root of T divided
by mu. Then, I'll bet you a month's
salary that any single valued function will satisfy this
differential equation. What is the dimension of that
C? What is the dimension of that
C, meters per second? It's a velocity because if I
have apples here, I must also have apples there.
And so, this can only be an apple if C has the dimension of
a velocity. So, therefore,
you might as well write this as plus or minus VT.
And, you might as well write V for here, a velocity.
And we might as well change now this differential equation in a
way more uniform way, which is what I'm going to do
now, which is one over V squared times D2Y DT squared equals D2Y
DX squared. And, this equation is what is
generally called the [rate?] equation.
It will be with us until the end of the course until death do
us part. It is really a big moment
because you're going to see this equation many times for many
different systems. But now you have seen it being
derived for this very specific case.
Let's now evaluate the meaning of that V.
Well, if I have, here, X, and here,
Y, and I picture [us?] a function.
It could be a sine. It could be a cosine.
I pick one that is way more imaginative.
I pick this one. [WHISTLES] That's my function.
It has to be single value, though.
You have to be careful. It must be single valued.
You cannot go back. That's my function.
And so, that's my function, F, at time T equals zero.
Let us take for V always a positive number for simplicity.
I'm going to call it even speed.
Speed is always positive, right?
And, I want to know now, if I look a little later in
time when there is a minus sign here, what the function looks
like. So, at T equals zero,
I gave it to you. What would it look like a
little bit later in time if there is a minus sign there?
Any suggestions? The function has shifted?
In what direction? Use your hands.
Who thinks it's in this direction?
Very good. It's in this direction.
So, you'll see a little later in time, you will see it here.
And, what is it doing? It is moving with speed V in
that direction. Now, we are going to evaluate
the plus sign. What will happen?
We now look at the function a little later in time.
A little later in time, it has moved in this direction.
And it's moving [UNINTELLIGIBLE] in this
direction. So, now, you can look through
the meaning of this equation. You now understand why,
when I wiggled here, why the string had no choice.
It must propagate that function that I generated.
And it must propagate that with the speed, square root of T
divided by mu. We derived the speed of
propagation for that string. Mu is the mass per unit length.
T is the tension. If I asked you,
is it obvious that the higher tension gives you a higher
speed, is it completely obvious to me?
Sort of, not quite. But I accept that.
Is it obvious if I made mu large, that I make it a very
thick, very heavy per meter, that propagation speed is
lower? Yeah, maybe.
Now that I know the answer I would say, yeah,
that's quite obvious. But it's not so trivial.
So, in any case, we have derived two things.
We have derived that there is such a thing as speed.
But we even have derived the speed itself,
the square root of T over mu. So, if we had done the
experiment again with a higher tension, then the [pulse?] would
have moved faster. But now, there is something
else that we had to explain. Why on Earth is a mountain
coming back as a valley? And why is a valley coming back
as a mountain? And that, now,
is the result of boundary conditions.
Some people who have lectured 8.03 make a very simple
statement. They say 8.03 is only about two
things: this equation and boundary conditions.
And all the rest follows. It's quite accurate.
So, we have here the string that Nicole and I were holding.
And, here is the answer. That's where Nicole was.
I hope I spelled that correctly.
And we know that that end must stay fixed, cannot move.
I'll put the line a little lower.
I'll put it here. This is the end.
And, my pulse came in. This is the pulse.
And, let us evaluate the moment in time that this part of the
pulse reaches Nicole. Are you ready for that?
So, this part is here.
And the part that, yeah, maybe it's in heaven,
is here. I have to make these a little
steeper to make it look alive, make it a little steeper.
And this, yeah, who knows what happened with
that? But, Nicole knew very well that
this point cannot move. Therefore, she very sneakily,
without telling you and me generated a pulse that came back
to me, which made sure that at all moments in time this point
stood still. So, at this very moment in
time, she must've generated the pulse, which had this
displacement, so that this part is exactly
the same as this. And so, [UNINTELLIGIBLE] stands
still. But she must have done that at
every moment in time. She must have done that when
this part arrived, when this part arrived,
when that part arrived.
So, that means he must've generated a pulse on her side
that is a valley that now looks like this.
So, this part is here. And at this moment in time,
all she has to do is generate this pulse.
And so, the net result is that if you took a photograph of this
string at this moment in time, you would see something very
bizarre. It is the sum of this with
this. And, you try to draw what that
looks like. For one thing,
this point will be here. That's for sure.
And then, whatever you see here, well, you try to add the
two up. And this thing is moving in my
direction with peak V because she is generating a valley.
And so, the consequence of the boundary condition is since this
point is fixed, a mountain must come back as a
valley, and the valley must come back as a mountain.
And given a little bit of time, when this point here has passed
Nicole completely, then there is,
of course, a very nice, healthy post on the way back to
me, which is mirrored, now, this way.
The mountain is a valley, but it also has mirrored this
way. That's why I made the pulse
purposely asymmetric. So, that is what is happening.
So, now I want to do this experiment again with Nicole,
of course, because she knows how to do it.
And you're going to look at this with completely different
eyes. Your eyes were closed when we
did it the first time. You are blind.
Let's face it. But now, you've seen the light.
This is a big moment in your life because you now know,
first of all, why it propagates.
And now when it arrives there, you know that the mountain
becomes a valley. So, I'm just going to do
exactly the same thing, only to allow you to look at it
now through different eyes. And that's what education is
all about regardless of whether it's physics or whether it is
art. Without education,
you cannot appreciate it. Now you can.
Watch it. You ready for this,
Nicole? You see, it moves.
It has no choice. And the mountain comes back as
a valley. I will do that once more very
clear. You deserve an A for this
course. That's clear.
Oh, what, you don't want it? I'll make it a B.
[LAUGHTER]
I can change the boundary conditions.
I don't have to keep this point fixed.
And, I can do that in the following way.
Here's my string. I have here a metal rod.
We put oil and grease on it so it's completely frictionless in
this direction. And, we mount here a mass-less
ring, mass-less. But the tension,
of course, is there. And, mass per unit length is
mu. None of that changes.
But here is a mass-less ring, and this is a rod with zero
friction. Those are very different
boundary conditions. This point can now move up and
down. And, it will.
However, the shape of that string right here is now very
special. At all moments in time,
what will the shape of this string be when we photograph it
no matter when you photograph it?
You can photograph it before the poses there,
after the poses there, at any moment in time.
What will this point look like? It comes in at 90°.
DY DX, if this is Y, this is X, at that location DY
DX must be zero. If it were not zero,
this would be zero. This would be zero.
This is zero. This is zero.
That is whole zero.
If it weren't zero, if it was this,
then there would be a force on this ring because the tension
would be in this direction. But the ring has no mass.
And so, the acceleration of the ring would be infinitely high,
which we don't allow. So, therefore,
in the extreme case that you can go to this situation,
you will now see something very different.
You will see that the string, at all moments in time,
will have to be like this.
If now I sent in a mountain, what you think will come back?
The mountain comes back. Mountain goes in,
mountain comes back. Who thinks it a mountain goes
in, a mountain comes back? Very good.
And that's the consequence of the fact that it is open now
because the only reason why a mountain goes in and a valley
came back, there was only one reason.
The end could not move. Now the end can move,
and I will demonstrate it to you.
And now, the mountain will come back as a mountain.
We referred to this in physics as a closed end,
and we refer to this as an open end.
And when you have an open end, and this is the pulse that
comes in, say it has amplitude, A, then what comes back at some
point in time is again a mountain going in this direction
with speed V. This comes in with speed V.
We call that the incident pulse in this.
We call that the reflected pulse.
This has amplitude, A.
And this has very interesting consequences,
namely, at the moment that this point here reaches that
mass-less ring. The mass-less ring must go up
by an amount, 2A because it generates,
that ring generates this pulse. And so, the ring generates this
pulse. But this was also there.
And remember, you have to add the two
together. Like, this one was added to
this. That gives me zero.
Now you have to add this A to that A.
And so, what you'll see is that if [E?] was your ring,
it will go up to 2A. So, it will make a huge
excursion, goes twice as high as the incoming one,
and then it will go back to zero.
And then the mountain rolls back.
And needless to say that we, of course, would like to
demonstrate that. Now, to make a rod,
which is nearly frictionless, it's difficult,
but we can use a lot of oil, and a lot of grease,
and a lot of soap. So, that was not our major
hang-up. But when we looked at the
Amazon.com, and we wanted to buy a mass-less ring,
Marcos and I really tried, but it didn't work.
We couldn't buy a mass-less ring.
And so, therefore, it is not so easy to
demonstrate this in a way that I have there.
So, we will demonstrate it to you in another way.
And that is with this instrument.
I will first explain it. This is not a string.
These are rods, all the same length.
They're connected here with some metal.
And so, you can move these. And then, the pulse that you
generate will propagate. So, they are coupled.
I don't know how many there are.
Do you know how many there are? OK, let's say I count 40.
Then it's 40 coupled oscillators.
And now, I have the option with this machine that I can hold
this one fixed, which is then a closed end.
But I can also let this one open, and then it's an open end.
So, if I hope this one close, and I sent in a mountain here,
then a valley will come back. But if I keep it open,
and I send in a mountain, then a mountain will come back.
And you should be able to see that the end gets a huge
amplitude at the moment that it reaches the maximum.
And, so that is what is on our plate now.
And, we will make it extremely romantic for you,
believe me. We're going to do this very
romantic way.
I told you.
So here, I have a clip here. So I will first lock this in
place so that this N cannot move.
That's what I will do first. And, from this side,
I will then generate a mountain.
The speed at which it propagates is actually quite
decent, not as fast as it was with a string.
And so, I want you to see that first of all it propagates and
then comes back as a valley. So, the end here is now fixed.
It's a fixed end. Ready?
Mountain, and now it's a valley.
You see it? OK, now it's always a pain
because the system is a very high Q system.
So, it doesn't want to damp out.
I can try to send in. Yeah.
I know exactly what you're thinking.
We are aware of this. If they try to com calm it
down, it makes it worse sometimes.
I will now try to generate the valley.
It is a little harder. I don't know why this,
why it's a little harder. I have to talk to my
psychiatrist about it. It's easier to go up and down
than to go down and up. I don't know why that is.
So, I'll go down and up, make a valley,
and then when it comes back, it's a mountain.
There it goes. And it comes back as a
mountain. But you see it?
Did you? If you didn't,
just say so. We can do it once more but I
don't think we have to. Now comes the big thing.
Now I'm going to make this, and freely moving.
So now, it's an open end. And, I will generate a
mountain, now. And I want you to not only
appreciate that it comes back as a mountain.
But above all, the end will have twice the
amplitude at one moment in time when the top of the mountain
reaches that end. And then, of course,
it will go back to zero. And the regular mountain will
grow back to me. So, if you're ready for this,
there goes the mountain. Sweep high, and it comes back.
[LAUGHTER] What is so funny about that?
Did you see that huge amplitude?
I'll do it once more because I don't want you to forget that.
Let's give it just ten seconds. Oh boy, it's like an ocean.
I will do it once more. I'm not going to try a valley
because that's what the problem comes in.
I will simply go up, and then let's look again at
the end and see whether we can see that double amplitude of the
mountain. Mountain: whoa,
biggie, oh, man. Whoa, OK, have a good weekend.