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To avoid possible
misunderstandings,

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my lectures start at 9:30
Eastern Standard Time,

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which is different from lobby
seven time, which you may have

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noticed.
The clock in lobby seven is

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seven minutes slow.
Today, we're going to cover

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coupled oscillators,
which is a big part in 8.03,

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though we leave out damping in
order to avoid a major

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complication.
Imagine that I have a pendulum

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length L mass M,
and I have another pendulum,

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also mass M length L.
And I connect a spring between

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them, as you see there,
spring constant K.

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Imagine now at time T equals
zero that I give this object,

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which I call object number one,
and this is object number two,

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give it a certain position.
I give it a certain velocity,

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so I have four choices.
And, I let the system go.

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What you're going to see is
something extremely chaotic,

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and our task today is to
predict what the position of

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this one is at any moment in
time, and what the position of

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that one is at any moment in
time.

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And, to show you how chaotic
that motion is,

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I will just show you this.
So, I take this one and just

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displace it from equilibrium.
I displace this one from

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equilibrium, and at time T
equals zero, I will give the

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one, my left hand,
just a certain velocity.

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And, if you now look at the
position of the individual

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object, it would seem nearly
impossible to come with an

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analytic solution,
which tells you what these

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motions are.
You'll see that amplitudes

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build up.
Of certain ones,

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amplitude goes down.
This one is hardly moving at

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all now.
Now it's picking up again.

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And so, our task today is to
work on that.

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What is by no means obvious
that I will show that to you,

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that any motion no matter how
you start it off is going to be

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the superposition of two normal
[mode?] solutions.

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Anyway you start it can always
be written as the superposition

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of two normal mode solutions.
What is a normal mode?

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A normal mode is,
in this case,

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that both objects have exactly
the same frequency that is

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fundamental to normal mode and
that they are either in phase

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with each other or out of phase
with each other,

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nothing in between because
there's no damping.

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So, it's either in phase or
it's out of phase.

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That is the normal mode.
In other words,

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if I call one of those
frequencies omega minus,

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minus means it is the lowest
frequency.

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There are two frequencies in
the system because there are two

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objects.
Omega minus,

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I call that the lowest
frequency.

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Then it would mean,
if they are in phase with each

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other, that they come to a halt
at the same moment in time.

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That means in phase and in the
same direction.

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So, they come to a halt at the
same moment in time in the same

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direction, and at the same
frequency.

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If, then, I have another
frequency which is higher there

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are two normal modes because we
have two objects.

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If we have three objects there
are three normal modes.

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If we go to the higher normal
mode, the higher frequency,

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they have the same frequency
that there are 180° out of

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phase.
So, when one comes to a halt

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here at the other one comes to a
halt there.

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That's what it means 180° out
of phase.

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I can excite,
and I will excite,

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this system into its normal
modes only.

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I can excite this normal mode
alone and this normal mode alone

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if I choose the correct initial
conditions.

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So, for any randomly chosen
initial condition,

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the motion of each object can
be written as a linear

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combination of these two modes.
If you take my word for that

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for now, but of course I will
demonstrate to you and I will

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prove that to you,
it would mean,

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then, that [SOUND OFF/THEN ON]
function of time can be written

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as having some kind of an
amplitude, X zero,

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I give it a minus sign.
Of course, it's related to that

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normal mode frequency times the
cosine of omega minus T plus

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some [phi?] minus,
plus some other amplitude,

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which I call X zero plus,
which is going to be related to

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this frequency times the cosine
of omega plus T plus phi plus.

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Now, let's look at this.
Let's try to see through this,

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what this means.
It means that if I know my

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initial conditions,
that I can determine this

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amplitude.
I can determine this phase.

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I can determine this amplitude,
and I can determine this phase.

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There have to be four
adjustable constants because I

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have the choice between the
positions at T zero,

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and the velocity.
So, there must be four.

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What you see is that omega
minus and omega plus,

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which are these normal mode
frequencies, are independent of

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the initial condition.
So, now I'm going to write down

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the position in time for object
number two.

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And I know that in this mode,
it must have the same

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frequency.
So there must be here a cosine

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minus T plus the same,
exactly the same phase because

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I told you, normal mode means
same frequency in phase,

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or same frequency out of phase.
And, it is the next one which

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is going to be out of phase.
But, you will see that shortly.

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So, this term must become omega
plus T.

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For now, I will say plus phi,
plus what you're going to see

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how the out of phase comes in
very shortly.

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Notice this omega minus must be
that omega minus.

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Otherwise it wouldn't be a
normal mode.

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This omega plus must be this
omega plus.

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Otherwise it wouldn't be a
normal mode.

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These phi's are the same to
allow them to be in phase.

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These phi's are the same,
even though they are out of

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phase.
But you will see very shortly

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that I will get a minus sign
here, which will take care of

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the 180°.


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I can, with this system,
excite the lowest mode,

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so that it only oscillates in
that lowest mode.

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So, this term is not there,
and that this term is not

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there.
I can do that.

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I can always oscillates
something in one of the normal

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modes.
But, if I set it off at random,

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then it is going to be a
superposition.

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Even though it is early for
you, can you show me using your

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hands and your legs,
whatever you want to,

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when I set this off at just the
right initial conditions,

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what will that lowest normal
mode look like?

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How will these pendulums
oscillate?

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Very good, very good.
So you have a good instinct.

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Very good.
Let's do that.

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So, let's set them off in the
same direction and let them go.

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Those are ideal initial
conditions for this mode.

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And you will see that it is a
normal mode, namely,

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they have the same frequency,
and they are in phase with each

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other.
They come to a halt at the same

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moment in time.
So, what you see already is

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that this value here must,
also, be X zero minus.

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That's a must.
You have just seen it.

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In this case,
what it means is that you might

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as well remove the spring.
The spring is not doing

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anything.
The spring never gets any

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longer, never gets any shorter.
What do you think the pendulums

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will do when I excite the other
mode, which has a higher

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frequency?
Very good, can I see some other

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hands or heads or legs?
Good.

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That's this,
symmetry.

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There is a symmetry in the
system.

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And that's why you people are
so smart and see immediately the

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connection.
So, let me make here a drawing

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of the omega minus,
and then the amplitudes are the

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same.
And here, I make a drawing of

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the omega plus.
Higher frequency,

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and the amplitudes are now the
same, but 180° out of phase.

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So, it's easy to excite that.
There we go.

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Notice they oscillate with the
same frequency,

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and they're coming to a halt at
the same moment in time.

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But, if one is here,
the other is there.

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And so, therefore,
this one here must be minus X

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zero plus.
That minus sign,

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now, gives me the 180°.
So I can still keep that phi

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plus there.


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It is utterly trivial,
without any work,

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to calculate,
or tell you what omega minus

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is, that just the frequency of a
single pendulum,

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so that's easy.
So, omega minus is simply the

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square root of G over L.
There's no spring,

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right?
Each one is doing its own

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thing, and the spring is never
pushing or pulling.

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But we should be able to
calculate omega plus.

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So let's do that.
And I will only make a drawing

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of one of those pendulums.
I don't need them both.

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So, here's one of those
pendulums.

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And let this separation
distance be X from equilibrium.

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And, I call this angle theta.
And so, here is that spring.

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And, the other one is on the
other side.

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Let's pull in all the forces
that we can think of.

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So, there's gravity,
MG.

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There is tension,
T.

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But there's more.
What else is there?

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Spring.
How much longer is the spring

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than wants to be?
2X, exactly.

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In other words,
there is a force here.

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I call that the spring force,
FS, which is bringing it also

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back to equilibrium.
And, that one is 2KX.

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Now, the tangent is very close
to MG.

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We have discussed that several
times.

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I'm going to introduce a
shorthand notation.

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Omega zero squared is G divided
by L, and omega S squared,

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making reference to the spring
is K over M.

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And now I have to set up the
differential equation.

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I have to apply Newton's Second
Law.

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And to remind you,
the magnitude of the spring

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force is 2KX.
So, Newton's Second Law,

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MX double dot equals,
now, there are two forces that

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I have to take into account:
first, the spring force,

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which is minus 2KX.
And then I need the horizontal

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component of the
[tension/tangent?].

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Remember, that's only one that
is important here,

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which is T sine theta.
But, T is MG,

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and sine theta is X divided by
L if L is the length of the

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pendulum.
So, we get minus T,

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which is MG times the sine of
theta, which is X divided by L.

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And so, this is the
differential equation.

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I'm going to divide M out,
and I'm going to bring

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everything to one side.
So, I'm going to get X double

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dot plus two omega S squared
times X plus omega zero squared

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times X equals zero.
And our task now is to solve

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00:14:02,824 --> 00:14:07,564
this differential equation.
And that, of course,

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00:14:07,564 --> 00:14:12,910
you can do in three seconds
because you recognize this

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00:14:12,910 --> 00:14:17,449
differential equation.
It is X double dot plus

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something times X.
And so, the new frequency,

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which I will call omega plus,
that new frequency omega plus

207
00:14:27,737 --> 00:14:35,000
is the square root of two omega
S squared plus omega squared.

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00:14:35,000 --> 00:14:40,444
And, what you see is something
that I already anticipated,

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which was consistent with your
intuition.

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It is larger than omega zero
because this is the effect of

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the spring.
And so, omega zero is the one.

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00:14:53,626 --> 00:14:58,976
Here, an omega plus now is the
square root of two omega S

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squared plus omega zero squared.
And, omega S squared is defined

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this way.


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00:15:07,000 --> 00:15:13,000


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So, if I now turn to my general
solution, if for now you accept

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the fact that that is the
general solution,

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the superposition of the normal
modes, then the key point is

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that independent of your initial
conditions is omega minus.

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That was the square root of G
over L.

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00:15:37,875 --> 00:15:40,750
I never put in any initial
condition.

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Independent of the initial
condition is omega plus.

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That's this one.
I never put in any initial

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conditions.
Independent of the initial

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conditions is this ratio,
which is plus one.

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And, independent of the initial
conditions is this ratio.

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That is minus one.
In other words,

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00:16:02,718 --> 00:16:06,507
if I didn't tell you at the
initial conditions,

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00:16:06,507 --> 00:16:10,214
which I haven't,
I can predict that this ratio

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is plus one.
And I can predict that this

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ratio is minus one.
If I make this three,

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then this is minus three.
If I make this five,

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00:16:20,429 --> 00:16:24,300
then this is minus five.
The ratio is minus one.

234
00:16:24,300 --> 00:16:28,501
So, the ratios are independent
of initial condition,

235
00:16:28,501 --> 00:16:32,208
and the frequencies are
independent of initial

236
00:16:32,208 --> 00:16:35,091
condition.
If I tell you the initial

237
00:16:35,091 --> 00:16:39,375
conditions, then of course you
can also calculate the

238
00:16:39,375 --> 00:16:44,566
individual amplitudes.
Suppose, now,

239
00:16:44,566 --> 00:16:51,437
I start this system off in some
way at time T equals zero,

240
00:16:51,437 --> 00:16:56,861
which I can choose.
And it used the following.

241
00:16:56,861 --> 00:17:00,357
At T equals zero,
I make X1 C,

242
00:17:00,357 --> 00:17:05,540
just some number C that we can
choose, 3 cm,

243
00:17:05,540 --> 00:17:11,862
whatever you want to choose.
But, I make V1 zero.

244
00:17:11,862 --> 00:17:14,562
So, as I release it:
no speed.

245
00:17:14,562 --> 00:17:18,846
And, suppose I make X2 zero and
I make V2 zero.

246
00:17:18,846 --> 00:17:24,433
So, if you don't know what that
means, of course I'm going to

247
00:17:24,433 --> 00:17:28,437
demonstrate it.
It means that this one here,

248
00:17:28,437 --> 00:17:33,000
the other one I offset,
that's all I do.

249
00:17:33,000 --> 00:17:36,483
And I let this one go,
pull my hands off,

250
00:17:36,483 --> 00:17:40,488
and then I want to know what's
going to happen,

251
00:17:40,488 --> 00:17:43,710
right?
Because I release this one with

252
00:17:43,710 --> 00:17:47,193
zero speed.
That's what you see in there.

253
00:17:47,193 --> 00:17:51,983
There is one of an infinite
number of initial conditions

254
00:17:51,983 --> 00:17:55,640
that you may choose.
So, this now has to be

255
00:17:55,640 --> 00:17:58,775
substituted into my general
solution.

256
00:17:58,775 --> 00:18:04,000
And so, you have to take the
derivative of X1.

257
00:18:04,000 --> 00:18:07,989
So, you have to do X1 dot and
put that equal to zero.

258
00:18:07,989 --> 00:18:10,751
I leave that in your competent
hands.

259
00:18:10,751 --> 00:18:14,818
Then you have to take X2,
and you have to take X2 dot,

260
00:18:14,818 --> 00:18:17,810
and you have to make that equal
to zero.

261
00:18:17,810 --> 00:18:21,033
And I leave that into your
competent hands.

262
00:18:21,033 --> 00:18:23,565
That's easy.
This and you'll find,

263
00:18:23,565 --> 00:18:27,631
then, that in this particular
case for this particular

264
00:18:27,631 --> 00:18:30,930
example, phi minus equals zero,
and phi two,

265
00:18:30,930 --> 00:18:36,035
and phi plus equal zero.
It will not take you more than

266
00:18:36,035 --> 00:18:39,834
a few minutes to find that.
I didn't want waste your time

267
00:18:39,834 --> 00:18:43,090
on taking a derivative of such a
simple function.

268
00:18:43,090 --> 00:18:47,025
So, if we take this right now,
then I will substitute these

269
00:18:47,025 --> 00:18:50,417
results in the equation.
So, you did the hard work.

270
00:18:50,417 --> 00:18:53,605
You did the velocities.
I will do the positions.

271
00:18:53,605 --> 00:18:57,133
So, I will substitute in that
equation T equals zero.

272
00:18:57,133 --> 00:19:01,000
I know already that the phi's
are not there.

273
00:19:01,000 --> 00:19:08,303
And so, then I get that C at
time T equals zero is going to

274
00:19:08,303 --> 00:19:13,214
be X zero minus T zero.
So, this is one.

275
00:19:13,214 --> 00:19:18,251
T zero, so this is one,
plus X zero plus.

276
00:19:18,251 --> 00:19:25,555
That's this initial condition.
Then I go through the second

277
00:19:25,555 --> 00:19:31,474
initial condition that is zero,
is X zero minus,

278
00:19:31,474 --> 00:19:37,508
minus X zero plus.
There is a minus sign here.

279
00:19:37,508 --> 00:19:41,225
And so, what do I find I have
solved now?

280
00:19:41,225 --> 00:19:45,871
The general solution,
I will find that X zero minus

281
00:19:45,871 --> 00:19:49,681
is one half C,
and F zero plus is also one

282
00:19:49,681 --> 00:19:56,000
half C, which of course should
not come as a surprise to you.

283
00:19:56,000 --> 00:20:02,000



284
00:20:02,000 --> 00:20:06,398
So, let me write down,
now, the general solution that

285
00:20:06,398 --> 00:20:10,036
we have for this specific
initial condition.

286
00:20:10,036 --> 00:20:13,589
So, we know everything.
We know X zero one.

287
00:20:13,589 --> 00:20:16,803
We know X zero one minus,
X zero minus,

288
00:20:16,803 --> 00:20:20,271
not one X zero minus.
We know X zero plus.

289
00:20:20,271 --> 00:20:23,063
We know phi's.
We know everything.

290
00:20:23,063 --> 00:20:26,024
So, I'm going to write it down
here.

291
00:20:26,024 --> 00:20:30,000
So, X1 is going to be one half
C.

292
00:20:30,000 --> 00:20:38,480
Remember, we found it at one
half C times the cosine of omega

293
00:20:38,480 --> 00:20:46,113
T equals phi zero plus one half
C times, this is minus,

294
00:20:46,113 --> 00:20:53,462
by the way, cosine omega minus
T cosine omega plus T.

295
00:20:53,462 --> 00:21:02,226
That is X1, and X2 is one half
C times the cosine omega minus T

296
00:21:02,226 --> 00:21:10,000
minus one half C times the
cosine omega plus T.

297
00:21:10,000 --> 00:21:13,503
Take a deep breath,
substitute in there T equals

298
00:21:13,503 --> 00:21:17,753
zero, and you see immediately
that X is C and substitute T

299
00:21:17,753 --> 00:21:20,884
equals zero into seconds,
and then you see,

300
00:21:20,884 --> 00:21:24,835
indeed, that X2 is zero:
no surprise because that's my

301
00:21:24,835 --> 00:21:28,264
initial condition.
Now, I remember from my high

302
00:21:28,264 --> 00:21:32,663
school days that the cosine
alpha plus the cosine of beta is

303
00:21:32,663 --> 00:21:37,136
twice the cosine of half the sum
times the cosine of half the

304
00:21:37,136 --> 00:21:41,993
difference.
So, I can write down X1 as

305
00:21:41,993 --> 00:21:48,901
twice the cosine of half the sum
times the cosine of half the

306
00:21:48,901 --> 00:21:53,736
difference.
So, that two that I get eats up

307
00:21:53,736 --> 00:21:58,917
these one halves.
So, I get C times the cosine

308
00:21:58,917 --> 00:22:05,710
omega minus plus omega plus
divided by two times T times the

309
00:22:05,710 --> 00:22:11,812
cosine of omega minus,
minus omega plus divided by two

310
00:22:11,812 --> 00:22:16,478
times T.
I've just rewritten it in a

311
00:22:16,478 --> 00:22:20,608
different form.
And, X2 as a function of time,

312
00:22:20,608 --> 00:22:25,657
I now have the cosine of alpha
minus the cosine of beta.

313
00:22:25,657 --> 00:22:30,521
That is twice the sign,
half the sum times the sign of

314
00:22:30,521 --> 00:22:36,709
half the difference.
So, now I get here the sign of

315
00:22:36,709 --> 00:22:44,003
omega minus plus omega plus
divided by two times T times the

316
00:22:44,003 --> 00:22:50,309
sign of omega minus,
minus omega plus divided by two

317
00:22:50,309 --> 00:22:54,018
times T.
Notice when T is zero,

318
00:22:54,018 --> 00:23:01,189
that the sign of this one is
zero, consistent with what our

319
00:23:01,189 --> 00:23:07,000
initial conditions,
X2 zero, remember?

320
00:23:07,000 --> 00:23:11,205
It's all there.
Initial conditions are all

321
00:23:11,205 --> 00:23:14,487
there.
Initial conditions are all

322
00:23:14,487 --> 00:23:18,487
there, just written in a
different form.

323
00:23:18,487 --> 00:23:24,538
Now, imagine now in your mind
that these two frequencies are

324
00:23:24,538 --> 00:23:29,461
not too far apart.
Then, these equations smell of

325
00:23:29,461 --> 00:23:31,741
what?
Beats.

326
00:23:31,741 --> 00:23:35,574
See, this here is the fast
term.

327
00:23:35,574 --> 00:23:41,880
This one and that one,
and this one is the slow one.

328
00:23:41,880 --> 00:23:48,556
And so, if these two are close,
then what you'll see is

329
00:23:48,556 --> 00:23:53,996
something quite remarkable.
At T equals zero,

330
00:23:53,996 --> 00:24:00,301
this one stands still.
And, the cosine term is going

331
00:24:00,301 --> 00:24:06,973
to be one because T is one.
This one is going to oscillate

332
00:24:06,973 --> 00:24:10,723
happily with this frequency.
But, this cosine term is very

333
00:24:10,723 --> 00:24:13,881
gradually going to zero.
And, as that cosine term

334
00:24:13,881 --> 00:24:17,434
gradually goes to zero,
this one will stop oscillating.

335
00:24:17,434 --> 00:24:19,802
But, this sine term becomes
plus one.

336
00:24:19,802 --> 00:24:22,763
And so, the other one will
start to oscillate.

337
00:24:22,763 --> 00:24:26,644
And then, a little later in
time the cosine term will become

338
00:24:26,644 --> 00:24:30,000
minus one.
So, it starts to oscillate.

339
00:24:30,000 --> 00:24:33,753
But when that happens,
the sine term is zero again.

340
00:24:33,753 --> 00:24:36,831
So, it stops.
So, you see a beautiful beat

341
00:24:36,831 --> 00:24:39,758
phenomenon.
The first one will gradually

342
00:24:39,758 --> 00:24:43,211
come to a halt,
and the other one will pick up.

343
00:24:43,211 --> 00:24:46,439
And then, the other one will
come to a halt,

344
00:24:46,439 --> 00:24:49,217
and then transfers in [a way?]
energy.

345
00:24:49,217 --> 00:24:51,919
It is, of course,
consistent with the

346
00:24:51,919 --> 00:24:54,396
conservation of mechanical
energy.

347
00:24:54,396 --> 00:24:58,000
And, I want to demonstrate
that.

348
00:24:58,000 --> 00:25:03,000



349
00:25:03,000 --> 00:25:07,464
So, we have this year.
All I have to do is offset X1

350
00:25:07,464 --> 00:25:10,791
over a distance,
C, that we can choose,

351
00:25:10,791 --> 00:25:14,468
and then we release this one at
zero speed.

352
00:25:14,468 --> 00:25:19,545
And then we'll just watch it.
And then, you should see that

353
00:25:19,545 --> 00:25:23,309
strange beat phenomenon.
You ready for this?

354
00:25:23,309 --> 00:25:26,023
This will have to hold in
place.

355
00:25:26,023 --> 00:25:29,000
Three, two, one,
zero.

356
00:25:29,000 --> 00:25:34,000



357
00:25:34,000 --> 00:25:37,343
Look, this one is standing
still.

358
00:25:37,343 --> 00:25:41,000
Look, this one is standing
still.

359
00:25:41,000 --> 00:25:46,000



360
00:25:46,000 --> 00:25:49,534
That's beating.
And you see that the energy is

361
00:25:49,534 --> 00:25:52,125
transferred from one to the
other.

362
00:25:52,125 --> 00:25:56,680
And that is a beat phenomenon
that follows immediately from

363
00:25:56,680 --> 00:25:59,272
this.
What would happen if I moved

364
00:25:59,272 --> 00:26:05,490
the spring up?
Suppose I moved the spring here

365
00:26:05,490 --> 00:26:13,023
higher, say, halfway.
Anyone without looking too much

366
00:26:13,023 --> 00:26:19,542
at the blackboard,
sort of use your intuition.

367
00:26:19,542 --> 00:26:25,771
What would happen?
Would the same phenomenon

368
00:26:25,771 --> 00:26:27,509
happen?
Yeah?

369
00:26:27,509 --> 00:26:32,000
Yeah?
The spring what?

370
00:26:32,000 --> 00:26:34,823
Yes, so would the same
phenomenon happen?

371
00:26:34,823 --> 00:26:38,000
But what would happen with the
beat period?

372
00:26:38,000 --> 00:26:44,000



373
00:26:44,000 --> 00:26:46,206
Now you may look at the
blackboard.

374
00:26:46,206 --> 00:26:49,061
Do you think these have a
certain separation?

375
00:26:49,061 --> 00:26:51,591
Right, you make that separation
smaller.

376
00:26:51,591 --> 00:26:54,251
You make the effect of the
spring smaller.

377
00:26:54,251 --> 00:26:57,950
So, the omega minus becomes
even closer to the omega plus.

378
00:26:57,950 --> 00:27:01,000
And so, the big period will
increase.

379
00:27:01,000 --> 00:27:05,827
It will take longer for one to
stop.

380
00:27:05,827 --> 00:27:12,034
Let's try that.
So, are we going to move these

381
00:27:12,034 --> 00:27:17,000
up?
I'll put it roughly halfway.

382
00:27:17,000 --> 00:27:22,000



383
00:27:22,000 --> 00:27:26,153
OK, are we ready for this?
And so we're doing it again.

384
00:27:26,153 --> 00:27:30,692
And then you'll see the same
phenomenon, except it will take

385
00:27:30,692 --> 00:27:33,923
longer for the first one to come
to a halt.

386
00:27:33,923 --> 00:27:37,000
So, I have decreased the
coupling.

387
00:27:37,000 --> 00:27:48,000



388
00:27:48,000 --> 00:27:54,117
It's still swinging happily.
Now it's beginning to change

389
00:27:54,117 --> 00:27:58,159
its mind.
You see, and now it standing

390
00:27:58,159 --> 00:28:01,000
still.
It took longer.

391
00:28:01,000 --> 00:28:04,874
Do we agree?
Now, there is something

392
00:28:04,874 --> 00:28:09,745
mind-boggling,
something that you really want

393
00:28:09,745 --> 00:28:13,619
to see.
Suppose I bring this all the

394
00:28:13,619 --> 00:28:18,712
way to the top.
That's interesting because then

395
00:28:18,712 --> 00:28:24,579
omega minus is exactly the same
as omega plus because,

396
00:28:24,579 --> 00:28:31,000
look, this term goes away.
What now will happen?

397
00:28:31,000 --> 00:28:33,390
It's almost cutting out the
spring.

398
00:28:33,390 --> 00:28:37,468
There is no spring anymore.
What now will happen if I start

399
00:28:37,468 --> 00:28:39,226
one here?
Three, two, one,

400
00:28:39,226 --> 00:28:40,000
zero.


401
00:28:40,000 --> 00:28:45,000


402
00:28:45,000 --> 00:28:47,005
What do you think?
Yeah?

403
00:28:47,005 --> 00:28:49,098
Ah, brilliant.
Brilliant.

404
00:28:49,098 --> 00:28:52,151
One will swing,
and the other won't.

405
00:28:52,151 --> 00:28:55,901
Isn't that shocking?
You have two pendulums,

406
00:28:55,901 --> 00:29:00,000
and they're not even connected
anymore.

407
00:29:00,000 --> 00:29:05,510
And you start this one
swinging, and the other one will

408
00:29:05,510 --> 00:29:09,591
never start to swing.
Isn't that amazing?

409
00:29:09,591 --> 00:29:14,591
For $30,000 tuition,
you learn something fantastic

410
00:29:14,591 --> 00:29:19,183
except that the big period is
infinitely long.

411
00:29:19,183 --> 00:29:24,591
So, with a bit of patience,
so let's demonstrate that.

412
00:29:24,591 --> 00:29:30,000
So let's demonstrate this.
So, there we go.

413
00:29:30,000 --> 00:29:32,691
Unbelievable.
Physics works.

414
00:29:32,691 --> 00:29:38,373
And look at this equation.
When omega minus is omega plus,

415
00:29:38,373 --> 00:29:43,657
this term is always zero.
So, you see X2 remains zero.

416
00:29:43,657 --> 00:29:46,747
Ha, ha, ha, that's what you
see.

417
00:29:46,747 --> 00:29:52,928
And, when omega minus is omega
plus, this cosine term is always

418
00:29:52,928 --> 00:29:56,517
plus one.
Ha, ha, ha, that's what you

419
00:29:56,517 --> 00:29:59,806
see.
Isn't it amazing the power of

420
00:29:59,806 --> 00:30:02,000
physics?


421
00:30:02,000 --> 00:30:06,000


422
00:30:06,000 --> 00:30:11,314
So, it is truly remarkable that
we can describe for any initial

423
00:30:11,314 --> 00:30:16,714
condition the motion in terms of
the linear superposition of the

424
00:30:16,714 --> 00:30:20,742
two normal modes.
And so, what originally looked

425
00:30:20,742 --> 00:30:25,885
like an impossibility when I
started the first 30 sections of

426
00:30:25,885 --> 00:30:28,714
my lecture with the
demonstration,

427
00:30:28,714 --> 00:30:33,857
when I started them off in a
random way, it looked so chaotic

428
00:30:33,857 --> 00:30:39,085
that it was almost unimaginable
that we would be able to short

429
00:30:39,085 --> 00:30:43,628
that out and be able to predict
the motion of each one

430
00:30:43,628 --> 00:30:48,000
individually as a function of
time.

431
00:30:48,000 --> 00:30:52,505
I want you to appreciate that
we have two coupled oscillators

432
00:30:52,505 --> 00:30:55,209
here.
That gives you to normal modes.

433
00:30:55,209 --> 00:30:59,490
If you have three coupled
oscillators, we will get back to

434
00:30:59,490 --> 00:31:02,418
that, then there are three
normal modes.

435
00:31:02,418 --> 00:31:05,798
And if you have for,
then you have four normal

436
00:31:05,798 --> 00:31:09,243
modes.
Now, clearly what we need is a

437
00:31:09,243 --> 00:31:11,337
general recipe.
In this case,

438
00:31:11,337 --> 00:31:15,750
the problem was a beautifully
symmetric we could do all mega

439
00:31:15,750 --> 00:31:19,565
minus just a split second.
Omega plus: you could see

440
00:31:19,565 --> 00:31:23,229
exactly the motion.
So, in no time [you get?] that

441
00:31:23,229 --> 00:31:27,493
the ratio of the amplitude was
plus one and was minus one.

442
00:31:27,493 --> 00:31:30,783
But let me tell you,
if I break the symmetry,

443
00:31:30,783 --> 00:31:34,997
oh man, all hell breaks loose.
For instance,

444
00:31:34,997 --> 00:31:39,752
if you change the ratio of the
masses, you make one longer than

445
00:31:39,752 --> 00:31:43,893
the other, you have three beats
connected with springs.

446
00:31:43,893 --> 00:31:47,958
If you break the symmetry,
it is very, very hard work.

447
00:31:47,958 --> 00:31:50,949
And, for that,
we need a general recipe.

448
00:31:50,949 --> 00:31:54,401
And, I'm going to give you the
general recipe.

449
00:31:54,401 --> 00:31:58,619
And then, I'm going to apply
that general recipe to this

450
00:31:58,619 --> 00:32:03,636
case.
So, everything that you have

451
00:32:03,636 --> 00:32:09,610
seen will then come out after a
lot of algebra,

452
00:32:09,610 --> 00:32:15,844
but at least you know that the
recipe is working.

453
00:32:15,844 --> 00:32:22,207
So, the first thing that you do
that's number one,

454
00:32:22,207 --> 00:32:30,000
you give each object a
displacement from equilibrium.

455
00:32:30,000 --> 00:32:35,126
Even though you are free to
choose the direction,

456
00:32:35,126 --> 00:32:41,427
I always, matter of discipline,
set them all off in the same

457
00:32:41,427 --> 00:32:44,417
direction.
That's not a must.

458
00:32:44,417 --> 00:32:48,689
But, that reduces the chance of
mistakes.

459
00:32:48,689 --> 00:32:52,213
But, you give them a
displacement,

460
00:32:52,213 --> 00:32:56,699
and I always do them in the
same direction.

461
00:32:56,699 --> 00:33:01,184
That's number one.
Once you have done that,

462
00:33:01,184 --> 00:33:05,776
you write down Newton's Second
Law for each,

463
00:33:05,776 --> 00:33:11,116
which means if you have two
objects, you have three

464
00:33:11,116 --> 00:33:16,000
unknowns.
Well, that is three unknowns.

465
00:33:16,000 --> 00:33:23,000



466
00:33:23,000 --> 00:33:28,288
So, you want to find one of the
normal mode frequencies.

467
00:33:28,288 --> 00:33:33,576
You want to find this one,
and independently you want to

468
00:33:33,576 --> 00:33:37,711
find that one,
which has to come out of that

469
00:33:37,711 --> 00:33:42,326
recipe, what are in principle
the three unknowns.

470
00:33:42,326 --> 00:33:48,192
This ratio plus one holds here,
but that doesn't hold in other

471
00:33:48,192 --> 00:33:51,653
cases.
And so, in principle you have,

472
00:33:51,653 --> 00:33:55,788
as an uncertainty,
this amplitude and omega.

473
00:33:55,788 --> 00:34:03,000
So, you always end up with two
equations with three unknowns.

474
00:34:03,000 --> 00:34:08,760
So if I see two objects,
then you have three unknowns.

475
00:34:08,760 --> 00:34:14,956
And you have two equations.
So, that looks like a problem.

476
00:34:14,956 --> 00:34:19,195
Number three:
you're going to put in the

477
00:34:19,195 --> 00:34:25,500
condition for normal modes,
which means that X1 is C1 times

478
00:34:25,500 --> 00:34:29,847
cosine omega T.
But, X2 can be some other

479
00:34:29,847 --> 00:34:35,826
amplitude, which we had to
calculate also times the same

480
00:34:35,826 --> 00:34:40,038
omega T.
I don't have to put the phase

481
00:34:40,038 --> 00:34:43,763
in there because it's either in
phase or out of phase,

482
00:34:43,763 --> 00:34:47,067
and out of phase means that you
get minus signs.

483
00:34:47,067 --> 00:34:51,073
And so, you see immediately
that if you substitute this in

484
00:34:51,073 --> 00:34:54,658
Newton's Second Law,
that you get for two equations,

485
00:34:54,658 --> 00:34:56,696
three unknowns.
C1 is unknown,

486
00:34:56,696 --> 00:35:00,000
C2 is unknown,
and omega is unknown.

487
00:35:00,000 --> 00:35:04,447
So, you substitute this in
here, and then comes the

488
00:35:04,447 --> 00:35:07,828
problem.
How are you going to deal with

489
00:35:07,828 --> 00:35:10,763
two equations with three
unknowns?

490
00:35:10,763 --> 00:35:14,411
Well, think about what I told
you earlier.

491
00:35:14,411 --> 00:35:19,837
If you don't know the initial
conditions, you can never find X

492
00:35:19,837 --> 00:35:23,039
zero minus, so you can never
find C1.

493
00:35:23,039 --> 00:35:29,000
But the ratio is independent of
the initial conditions.

494
00:35:29,000 --> 00:35:34,663
So that means the ratio C1 over
C2 cannot be dependent on the

495
00:35:34,663 --> 00:35:37,873
initial conditions.
In other words,

496
00:35:37,873 --> 00:35:43,348
you can always solve these
equations in terms of C1 divided

497
00:35:43,348 --> 00:35:47,690
by C2 and omega.
And, I will show you that that

498
00:35:47,690 --> 00:35:50,805
works.
So, you follow this recipe,

499
00:35:50,805 --> 00:35:54,581
and we will do that religiously
together.

500
00:35:54,581 --> 00:35:57,884
We will find,
then, in this case two

501
00:35:57,884 --> 00:36:03,633
variables for omega
We will find an omega minus,

502
00:36:03,633 --> 00:36:07,583
which has an associated value
of C over C2.

503
00:36:07,583 --> 00:36:12,191
I'll put a minus there.
And, we will find an omega

504
00:36:12,191 --> 00:36:16,705
plus, which has its own ratio
associated with it.

505
00:36:16,705 --> 00:36:20,655
In the case of our simple
symmetric system,

506
00:36:20,655 --> 00:36:25,357
this would be plus one and this
would be minus one.

507
00:36:25,357 --> 00:36:31,000
But that is not the case when
you break symmetry.

508
00:36:31,000 --> 00:36:34,382
So, my task is,
now, to apply this recipe in

509
00:36:34,382 --> 00:36:37,134
its most general form to the
system.

510
00:36:37,134 --> 00:36:39,730
It will be 16 minutes of
grinding.

511
00:36:39,730 --> 00:36:44,449
We will go through each step,
and out comes something that we

512
00:36:44,449 --> 00:36:47,674
already know.
But at least you'll see that

513
00:36:47,674 --> 00:36:52,078
I've made no assumption which I
did there about symmetry.

514
00:36:52,078 --> 00:36:56,325
Therefore, this is an ideal
moment for you to have your

515
00:36:56,325 --> 00:37:00,966
five-minute break so you can
take a deep breath to get ready

516
00:37:00,966 --> 00:37:06,000
for this 16 minute marathon.
[SOUND OFF/THEN ON]

517
00:37:06,000 --> 00:37:09,831
All right: general recipe.
Someone asked me during the

518
00:37:09,831 --> 00:37:13,807
intermission whether this omega
is the same as that one.

519
00:37:13,807 --> 00:37:16,843
Yes, of course,
otherwise it wouldn't be in

520
00:37:16,843 --> 00:37:19,807
normal mode.
We're going to search for the

521
00:37:19,807 --> 00:37:22,481
normal modes.
And, in the normal mode,

522
00:37:22,481 --> 00:37:25,879
the frequencies must be the
same of all objects,

523
00:37:25,879 --> 00:37:30,000
whether you have five or six or
two or three.

524
00:37:30,000 --> 00:37:33,894
But the ratios are different.
That's a different issue.

525
00:37:33,894 --> 00:37:36,923
However, they are in phase or
out of phase.

526
00:37:36,923 --> 00:37:41,322
So the ratios can be negative,
positive, but you never had any

527
00:37:41,322 --> 00:37:45,504
phase angles other than the 180°
or zero because we have no

528
00:37:45,504 --> 00:37:48,245
damping.
We have taken the damping out.

529
00:37:48,245 --> 00:37:52,572
So, it is essential that when
you substitute that in Newton's

530
00:37:52,572 --> 00:37:57,115
Second Law that these omegas are
the same because they will then

531
00:37:57,115 --> 00:38:02,000
give you the normal modes.
OK, you ready?

532
00:38:02,000 --> 00:38:07,000



533
00:38:07,000 --> 00:38:12,861
I'm going to offset the first
one over a distance,

534
00:38:12,861 --> 00:38:15,732
X1.
So, they have mass M.

535
00:38:15,732 --> 00:38:19,559
They have length L to remind
you.

536
00:38:19,559 --> 00:38:26,736
And, we are going to have omega
zero squared equals G over L,

537
00:38:26,736 --> 00:38:32,000
and omega S squared equals K
over M.

538
00:38:32,000 --> 00:38:36,245
And, this one I offset over a
distance, X2.

539
00:38:36,245 --> 00:38:39,581
Notice, you don't have to do
that.

540
00:38:39,581 --> 00:38:43,826
I always offset them in the
same direction.

541
00:38:43,826 --> 00:38:48,779
And then, there is this spring
that connects them.

542
00:38:48,779 --> 00:38:53,025
I will just make it very thin,
this spring.

543
00:38:53,025 --> 00:39:00,000
Otherwise the picture becomes a
little bit too complicated.

544
00:39:00,000 --> 00:39:05,597
So let this angle be theta one.
And let this angle be theta

545
00:39:05,597 --> 00:39:08,300
two.
And so, we have here MG.

546
00:39:08,300 --> 00:39:11,388
We have here MG.
We have here the

547
00:39:11,388 --> 00:39:16,117
[tangent/tension?] is roughly MG
for small angles,

548
00:39:16,117 --> 00:39:20,557
and we have here a tangent
which is roughly MG.

549
00:39:20,557 --> 00:39:24,707
It's a little bit too big the
way I drew it,

550
00:39:24,707 --> 00:39:27,699
but that's not so important
now.

551
00:39:27,699 --> 00:39:32,139
And now, there is another force
that X on both,

552
00:39:32,139 --> 00:39:38,520
and which force is that?
That's the spring force.

553
00:39:38,520 --> 00:39:44,191
Now, do we agree what the
magnitude of the spring force

554
00:39:44,191 --> 00:39:47,552
is?
And then, we will argue about

555
00:39:47,552 --> 00:39:51,123
the direction.
[Nickel?] magnitude,

556
00:39:51,123 --> 00:39:55,009
look very closely.
I offset one by X2,

557
00:39:55,009 --> 00:39:59,000
and I offset the other by X1.


558
00:39:59,000 --> 00:40:04,000


559
00:40:04,000 --> 00:40:10,566
I was asking you a question.
You can't answer with a

560
00:40:10,566 --> 00:40:13,012
question.
Very good.

561
00:40:13,012 --> 00:40:19,450
The magnitude of that force is
K times X2 minus X1,

562
00:40:19,450 --> 00:40:22,025
nonnegotiable,
right?

563
00:40:22,025 --> 00:40:29,236
Now, if X2 is larger than X1,
do we agree that the spring

564
00:40:29,236 --> 00:40:35,290
force is in this direction?
And do we agree,

565
00:40:35,290 --> 00:40:40,048
then, that this spring force
must be in this direction if X2

566
00:40:40,048 --> 00:40:43,758
is larger than X1?
Well, that means I can leave

567
00:40:43,758 --> 00:40:48,516
everything the way it is now.
As long as I give this a minus

568
00:40:48,516 --> 00:40:53,032
sign that is a plus sign,
I'm OK because if X2 is smaller

569
00:40:53,032 --> 00:40:57,709
than X1, it will automatically
flip over, and my algebra is

570
00:40:57,709 --> 00:41:00,320
fine.
So, therefore,

571
00:41:00,320 --> 00:41:05,603
in my head I can just think of
X to being larger than X1,

572
00:41:05,603 --> 00:41:10,698
set up the differential
equations, and I no longer have

573
00:41:10,698 --> 00:41:16,169
to worry about the fact that
maybe X2 at certain moments in

574
00:41:16,169 --> 00:41:20,603
time is not larger than X1.
So, now I set up the

575
00:41:20,603 --> 00:41:24,094
differential equation,
MX1 double dot.

576
00:41:24,094 --> 00:41:28,622
So, that's this one.
Let's first do the pendulum.

577
00:41:28,622 --> 00:41:34,000
That is minus T times the sign
of theta one.

578
00:41:34,000 --> 00:41:38,765
T is MG, so it is minus MG
times X1 divided by L.

579
00:41:38,765 --> 00:41:43,829
Do we agree there is that
horizontal component here?

580
00:41:43,829 --> 00:41:48,794
The horizontal component of T
is driving it back to

581
00:41:48,794 --> 00:41:53,163
equilibrium minus sign.
FS, the spring force,

582
00:41:53,163 --> 00:41:56,340
is driving away from
equilibrium.

583
00:41:56,340 --> 00:42:02,000
So, it's going to get plus K
times X2 minus X1.

584
00:42:02,000 --> 00:42:07,182
Look at this.
Signs are very important now.

585
00:42:07,182 --> 00:42:11,131
Very important.
And the next one,

586
00:42:11,131 --> 00:42:14,587
X2 double dot,
it has, again,

587
00:42:14,587 --> 00:42:21,868
the restoring force due to the
pendulum which is minus MG X2

588
00:42:21,868 --> 00:42:29,395
divided by L because it's the
sign of theta two now that comes

589
00:42:29,395 --> 00:42:33,203
in.
And now it has the spring

590
00:42:33,203 --> 00:42:36,289
force, which now is a negative
sign.

591
00:42:36,289 --> 00:42:40,168
And so, now we get minus K
times X2 minus X1.

592
00:42:40,168 --> 00:42:44,223
And, when you have this down on
your next exam,

593
00:42:44,223 --> 00:42:48,983
you take a deep breath,
and you go over each individual

594
00:42:48,983 --> 00:42:53,303
step to make absolutely sure
that this is correct.

595
00:42:53,303 --> 00:42:58,151
If there's anything wrong with
this, you are dead in the

596
00:42:58,151 --> 00:43:01,148
waters.
The whole problem will fall

597
00:43:01,148 --> 00:43:05,021
apart.
It may not even be a simple

598
00:43:05,021 --> 00:43:08,690
harmonic oscillator.
You get something ridiculous.

599
00:43:08,690 --> 00:43:11,461
Signs are crucial here.
So, therefore,

600
00:43:11,461 --> 00:43:13,857
look at this again.
Take a pause.

601
00:43:13,857 --> 00:43:18,425
This one, if there's a positive
offset is indeed in a negative

602
00:43:18,425 --> 00:43:21,270
direction, a component of T sine
theta.

603
00:43:21,270 --> 00:43:25,838
This one, if X2 is larger than
X1 is indeed in that direction.

604
00:43:25,838 --> 00:43:29,207
And you notice,
if X2 becomes smaller than X1,

605
00:43:29,207 --> 00:43:34,000
well then, this becomes
automatically a negative.

606
00:43:34,000 --> 00:43:36,947
So, are you OK?
So this is fine.

607
00:43:36,947 --> 00:43:41,702
This one has a restoring force
due to the pendulum,

608
00:43:41,702 --> 00:43:47,027
which is T sine theta two
negative sine and the restoring

609
00:43:47,027 --> 00:43:51,211
force of the spring always
opposite this one:

610
00:43:51,211 --> 00:43:53,208
I'm happy.
I am happy.

611
00:43:53,208 --> 00:43:58,438
So, now, we are going to
rearrange this a little bit and

612
00:43:58,438 --> 00:44:05,000
we're going to introduce the
omega zero squared notation.

613
00:44:05,000 --> 00:44:13,947
So, I'm going to divide M out,
and so I'm going to get X2,

614
00:44:13,947 --> 00:44:20,540
X1 double dot.
And then, I'm going to bring

615
00:44:20,540 --> 00:44:28,389
the X1's to the left.
And so, I get plus omega zero

616
00:44:28,389 --> 00:44:34,392
squared times X1.
But you have a minus KX1.

617
00:44:34,392 --> 00:44:38,221
You divide by M.
So, you get plus omega S

618
00:44:38,221 --> 00:44:41,762
squared.
And, that's [both?] times X1.

619
00:44:41,762 --> 00:44:45,686
You notice that?
You both have an X1 term.

620
00:44:45,686 --> 00:44:50,950
And now, the X2 comes out.
The X2 has a plus sign on the

621
00:44:50,950 --> 00:44:54,300
right side, so it gets a minus
sign.

622
00:44:54,300 --> 00:44:58,128
So, I get minus omega S squared
times X2.

623
00:44:58,128 --> 00:45:01,000
And that is zero.


624
00:45:01,000 --> 00:45:06,000


625
00:45:06,000 --> 00:45:13,461
And then we do the next one.
We get X2 double dot.

626
00:45:13,461 --> 00:45:21,532
Ah, I get the same terms.
I get an omega zero squared.

627
00:45:21,532 --> 00:45:27,167
And I get an omega S squared
times X2.

628
00:45:27,167 --> 00:45:36,000
And then I get this plus;
minus times minus is plus.

629
00:45:36,000 --> 00:45:40,055
So I get minus omega S squared
times X1 equals zero.

630
00:45:40,055 --> 00:45:44,428
When you have reached this
point, you take a deep breath

631
00:45:44,428 --> 00:45:49,278
and you go over each little term
to make sure that you haven't

632
00:45:49,278 --> 00:45:54,207
accidentally slipped on a minus
sign because you slip up on one

633
00:45:54,207 --> 00:45:56,990
minus sign you're dead in the
water.

634
00:45:56,990 --> 00:46:02,000
It may not even become a simple
harmonic oscillation.

635
00:46:02,000 --> 00:46:04,776
So let me do that.
I agree with that.

636
00:46:04,776 --> 00:46:07,553
I agree with this.
I agree with that.

637
00:46:07,553 --> 00:46:11,256
I agree with that.
Differential equation is fine.

638
00:46:11,256 --> 00:46:15,961
Notice that I already did point
number one, and that already I

639
00:46:15,961 --> 00:46:19,663
did even number two.
I've set up the differential

640
00:46:19,663 --> 00:46:23,983
equation, Newton's Second Law.
So now comes number three.

641
00:46:23,983 --> 00:46:28,225
And number three means and
going to substitute in there,

642
00:46:28,225 --> 00:46:34,395
X1 is C1.
Cosine omega T and X2 is C2

643
00:46:34,395 --> 00:46:37,000
cosine omega T.


644
00:46:37,000 --> 00:46:43,000


645
00:46:43,000 --> 00:46:46,822
We go slowly;
I want you to follow each step

646
00:46:46,822 --> 00:46:51,888
and I'm going to work on here so
that you can see it high.

647
00:46:51,888 --> 00:46:56,866
So, I'm going to substitute
this in here so the X1 double

648
00:46:56,866 --> 00:47:00,866
dot will give me a minus omega
squared, right,

649
00:47:00,866 --> 00:47:07,000
because you get twice omega out
minus omega squared times C1.

650
00:47:07,000 --> 00:47:09,520
And, to hell with cosine omega
T.

651
00:47:09,520 --> 00:47:11,962
Why to hell with cosine omega
T?

652
00:47:11,962 --> 00:47:15,349
Because every term will have
cosine omega T.

653
00:47:15,349 --> 00:47:19,208
So you might as well divide out
right away, right?

654
00:47:19,208 --> 00:47:24,092
Every term that you're going to
have in here will have a cosine

655
00:47:24,092 --> 00:47:27,636
omega T in it.
So, I'll leave the cosine omega

656
00:47:27,636 --> 00:47:34,610
T already out now.
So, I get minus omega squared

657
00:47:34,610 --> 00:47:39,220
C1.
Then I get plus omega zero

658
00:47:39,220 --> 00:47:45,102
squared plus omega S squared
times C1.

659
00:47:45,102 --> 00:47:52,733
That is this term,
minus omega S squared times C2

660
00:47:52,733 --> 00:48:01,000
equals zero because,
remember, X2 has the C2.

661
00:48:01,000 --> 00:48:04,099
Am I going to fast?
Beautiful.

662
00:48:04,099 --> 00:48:09,872
So, I go to the second one.
The second one is X2 double

663
00:48:09,872 --> 00:48:13,293
dot.
So, I also get a minus omega

664
00:48:13,293 --> 00:48:17,889
squared times C2.
And then, here I get a C2.

665
00:48:17,889 --> 00:48:23,555
So, I get plus omega zero
squared plus omega S squared

666
00:48:23,555 --> 00:48:27,617
times C2.
And then, I get minus omega S

667
00:48:27,617 --> 00:48:34,400
squared times C1 equals zero.
And, when you have reached this

668
00:48:34,400 --> 00:48:37,653
point in your exam,
you take a deep breath.

669
00:48:37,653 --> 00:48:41,526
And, you make sure that all
your signs are correct.

670
00:48:41,526 --> 00:48:45,940
If not, you're dead in the
water and the problem will fall

671
00:48:45,940 --> 00:48:47,799
apart.
So, let's do that.

672
00:48:47,799 --> 00:48:51,207
Minus omega squared C1,
I can live with that,

673
00:48:51,207 --> 00:48:53,298
plus that term,
which is C1,

674
00:48:53,298 --> 00:48:57,945
yes, minus omega S squared C2.
Even if you make a little slip

675
00:48:57,945 --> 00:49:02,747
of the pen, and you change this
into a one and this into a two,

676
00:49:02,747 --> 00:49:06,000
it's all over,
of course.

677
00:49:06,000 --> 00:49:09,745
And then, the next equation,
minus omega squared C2.

678
00:49:09,745 --> 00:49:13,711
Then I have this times C2,
and then at minus that times

679
00:49:13,711 --> 00:49:15,327
C1.
We're almost there,

680
00:49:15,327 --> 00:49:18,632
even though it doesn't look
that way, does it?

681
00:49:18,632 --> 00:49:22,378
Remember what I said.
You cannot solve two equations

682
00:49:22,378 --> 00:49:26,858
with three unknowns there is no
way that you can solve for C1,

683
00:49:26,858 --> 00:49:30,531
C2, and for this omega,
which is what you're really

684
00:49:30,531 --> 00:49:34,252
after.
This is the omega you are

685
00:49:34,252 --> 00:49:37,405
after.
But you cancel for C1 divided

686
00:49:37,405 --> 00:49:42,810
by C2 and omega knowing that,
already, I'm going to eliminate

687
00:49:42,810 --> 00:49:46,504
C1 over C2.
And, you'll see how I do that.

688
00:49:46,504 --> 00:49:50,468
C1 divided by C2:
I go to the first equation.

689
00:49:50,468 --> 00:49:53,531
This is not so hard what I'm
doing.

690
00:49:53,531 --> 00:49:57,045
I put in my mind this on the
right side.

691
00:49:57,045 --> 00:50:02,000
So, I get a plus omega S
squared times C2.

692
00:50:02,000 --> 00:50:06,906
And then, I divide C1 by C2.
And so, when I get then is the

693
00:50:06,906 --> 00:50:10,374
following.
I get omega S squared up stairs

694
00:50:10,374 --> 00:50:14,435
which is this one.
And all this comes downstairs,

695
00:50:14,435 --> 00:50:19,087
minus omega squared plus omega
zero squared plus omega S

696
00:50:19,087 --> 00:50:20,779
squared.
Do we agree?

697
00:50:20,779 --> 00:50:25,347
That's the first equation.
I've simply written it as C1

698
00:50:25,347 --> 00:50:28,308
divided by C2.
I can always do that,

699
00:50:28,308 --> 00:50:32,531
right?
Bring the C1's to one side,

700
00:50:32,531 --> 00:50:36,000
the C2's to one side,
and divide them.

701
00:50:36,000 --> 00:50:39,843
That's this.
And, I'm going to do the same

702
00:50:39,843 --> 00:50:44,625
with the next equation.
I bring this up to the right

703
00:50:44,625 --> 00:50:48,562
side, that I have C1's there and
C2's here.

704
00:50:48,562 --> 00:50:54,187
And now, this becomes upstairs.
So, I get minus omega squared

705
00:50:54,187 --> 00:50:58,312
plus omega zero squared plus
omega S squared,

706
00:50:58,312 --> 00:51:03,000
and I get downstairs omega S
squared.

707
00:51:03,000 --> 00:51:07,614
And now, I have eliminated C1
and C2 because if I solve this

708
00:51:07,614 --> 00:51:10,586
equation, I get my solutions for
omega.

709
00:51:10,586 --> 00:51:13,871
This is one equation.
There is one unknown.

710
00:51:13,871 --> 00:51:16,452
That's omega.
You've got to admit,

711
00:51:16,452 --> 00:51:19,033
and there better be two
solutions.

712
00:51:19,033 --> 00:51:23,648
There better be an omega minus,
and there better be an omega

713
00:51:23,648 --> 00:51:27,871
plus coming out of this
untouched by young human hands.

714
00:51:27,871 --> 00:51:34,369
I must find two solutions.
Let's first make sure that this

715
00:51:34,369 --> 00:51:37,956
is correct, and the answer is
yes.

716
00:51:37,956 --> 00:51:43,173
So, what do I do now?
Well, I'm going to simplify

717
00:51:43,173 --> 00:51:48,065
this one step further,
which is now very easy.

718
00:51:48,065 --> 00:51:52,739
I multiply this by this,
and this with this.

719
00:51:52,739 --> 00:51:59,043
So, we get omega S through the
power of four is minus omega

720
00:51:59,043 --> 00:52:06,000
squared plus omega zero squared
plus omega S squared.

721
00:52:06,000 --> 00:52:13,033
One equation with one unknown;
omega is the only unknown.

722
00:52:13,033 --> 00:52:17,429
Take the square root left and
right.

723
00:52:17,429 --> 00:52:24,589
So, I get minus omega squared
plus omega zero squared plus

724
00:52:24,589 --> 00:52:32,000
omega S squared equals plus or
minus omega S squared.

725
00:52:32,000 --> 00:52:38,279
Do not forget the plus or minus
because the square root of omega

726
00:52:38,279 --> 00:52:43,064
S to the fourth is plus or minus
omega S squared.

727
00:52:43,064 --> 00:52:47,948
And now, you're going to see
the light of the day.

728
00:52:47,948 --> 00:52:53,032
You won't believe this.
I bring omega squared to the

729
00:52:53,032 --> 00:52:59,012
other side, and I get omega zero
squared plus omega S squared

730
00:52:59,012 --> 00:53:05,106
minus or plus omega S squared.
And, the simplicity is

731
00:53:05,106 --> 00:53:09,127
overpowering.
I feel it over my whole body.

732
00:53:09,127 --> 00:53:13,914
It is unbelievable.
When I have a minus sign here I

733
00:53:13,914 --> 00:53:18,893
find that omega is omega zero.
That's my omega minus.

734
00:53:18,893 --> 00:53:22,436
That's the one.
That's my omega minus.

735
00:53:22,436 --> 00:53:28,276
When there is a plus sign here,
I find exactly what we predict

736
00:53:28,276 --> 00:53:32,916
before.
I find that omega plus is the

737
00:53:32,916 --> 00:53:38,648
square root of omega zero
squared plus two omega S squared

738
00:53:38,648 --> 00:53:42,167
because the plus sign make this
two.

739
00:53:42,167 --> 00:53:45,988
Isn't that elegant?
Isn't it beautiful?

740
00:53:45,988 --> 00:53:51,117
So, they just pop out.
And I predict that there will

741
00:53:51,117 --> 00:53:54,737
be two solutions.
You have them both.

742
00:53:54,737 --> 00:54:00,368
So, by substituting this in
there, automatically come out

743
00:54:00,368 --> 00:54:06,000
two solutions.
But, what now is C1 over C2?

744
00:54:06,000 --> 00:54:12,620
We haven't solved for that yet.
But I promise you can always

745
00:54:12,620 --> 00:54:18,567
solve it in terms of omega and
in terms of C1 over C2.

746
00:54:18,567 --> 00:54:21,933
Any volunteers?
Any volunteers?

747
00:54:21,933 --> 00:54:27,881
Look at the blackboard.
It's somewhere hidden in those

748
00:54:27,881 --> 00:54:31,471
equations.
I said we are going to

749
00:54:31,471 --> 00:54:37,082
eliminate C1 over C2.
You think C1 over C2 likes to

750
00:54:37,082 --> 00:54:41,884
be eliminated?
Let's call it back.

751
00:54:41,884 --> 00:54:47,242
And look at this equation.
Substitute in here for omega

752
00:54:47,242 --> 00:54:50,615
squared.
Substitute in omega minus.

753
00:54:50,615 --> 00:54:54,285
Then you get an answer for C1
over C2.

754
00:54:54,285 --> 00:54:57,460
What do you think that answer
is?

755
00:54:57,460 --> 00:55:00,634
What do you think that answer
is?

756
00:55:00,634 --> 00:55:02,817
Plus one.
But even one.

757
00:55:02,817 --> 00:55:07,300
Plus one.
So, you get C1 over C2 is plus

758
00:55:07,300 --> 00:55:09,984
one.
If you take this omega zero and

759
00:55:09,984 --> 00:55:13,819
you put it in here,
you get omega zero squared with

760
00:55:13,819 --> 00:55:16,119
a minus sign,
with a plus sign.

761
00:55:16,119 --> 00:55:20,030
You have omega S squared
divided by omega S squared,

762
00:55:20,030 --> 00:55:23,558
and it's plus one.
And now, you take the second

763
00:55:23,558 --> 00:55:26,012
solution, and you put it in
here.

764
00:55:26,012 --> 00:55:30,000
What do you think you're going
to find?

765
00:55:30,000 --> 00:55:33,518
Minus one.
C1 over C2 is now minus one.

766
00:55:33,518 --> 00:55:39,259
And so, what you have seen now
is that the general recipe comes

767
00:55:39,259 --> 00:55:43,981
up with frequencies,
comes up with the ratios of the

768
00:55:43,981 --> 00:55:48,888
amplitudes, not with the
individual amplitudes because

769
00:55:48,888 --> 00:55:52,222
you don't know the initial
condition.

770
00:55:52,222 --> 00:55:56,296
It comes up with the ratio of
the amplitudes,

771
00:55:56,296 --> 00:56:01,574
and now you can write any
solution, provided that you know

772
00:56:01,574 --> 00:56:06,842
the initial conditions.
If you know the initial

773
00:56:06,842 --> 00:56:09,807
conditions, then you can also
find C1.

774
00:56:09,807 --> 00:56:12,932
And therefore,
since you know the ratio,

775
00:56:12,932 --> 00:56:17,099
you automatically have C2.
And, you can find this C1.

776
00:56:17,099 --> 00:56:21,666
So, I'll give this one a minus
sign, and this a plus sign.

777
00:56:21,666 --> 00:56:24,951
And then, you know the ratio in
that mode.

778
00:56:24,951 --> 00:56:30,000
So, now you may think
erroneously that life is easy.

779
00:56:30,000 --> 00:56:35,935
That's far from the truth.
Suppose you have something as

780
00:56:35,935 --> 00:56:38,741
simple as this:
L, M, L, M.

781
00:56:38,741 --> 00:56:44,244
It's called a double pendulum.
There is no symmetry.

782
00:56:44,244 --> 00:56:50,179
I want to test your intuition.
There are going to be two

783
00:56:50,179 --> 00:56:55,791
modes, two normal modes:
a lower one and a higher one

784
00:56:55,791 --> 00:57:02,440
because they are two objects.
In the lowest frequency,

785
00:57:02,440 --> 00:57:08,135
use your hands in your legs.
What will this pendulum look

786
00:57:08,135 --> 00:57:11,593
like?
You are all doing sort of the

787
00:57:11,593 --> 00:57:15,254
right thing.
Would it look like this,

788
00:57:15,254 --> 00:57:19,525
or would it look like this?
In other words,

789
00:57:19,525 --> 00:57:25,627
let me make this one a little
bit, let me make the difference

790
00:57:25,627 --> 00:57:29,694
a little larger.
Would it look like this,

791
00:57:29,694 --> 00:57:36,000
or would it look like this?
Slightly exaggerated.

792
00:57:36,000 --> 00:57:41,636
Who is for this one?
OK, that means that the ratio,

793
00:57:41,636 --> 00:57:46,370
C2 over C1, is going to be plus
two, right?

794
00:57:46,370 --> 00:57:50,992
Who is for this one?
That's the way it is,

795
00:57:50,992 --> 00:57:56,290
believe it or not.
C2 over C1 is going to be one

796
00:57:56,290 --> 00:58:02,152
plus the square root of two.
And, it will take you 30

797
00:58:02,152 --> 00:58:07,000
minutes of grinding to find
that.

798
00:58:07,000 --> 00:58:12,235
And I'm not exaggerating when I
say 30 minutes.

799
00:58:12,235 --> 00:58:18,951
You have to go to the whole
procedure and then you will find

800
00:58:18,951 --> 00:58:24,528
that this ratio is 2.4.
And I will demonstrate it.

801
00:58:24,528 --> 00:58:29,536
The highest mode must be
something like this,

802
00:58:29,536 --> 00:58:33,292
right?
Because they must be out of

803
00:58:33,292 --> 00:58:37,947
pace.
Any idea of the ratio,

804
00:58:37,947 --> 00:58:41,832
C1 over C2?
I'll call this C1,

805
00:58:41,832 --> 00:58:45,583
call this C2?
Any one of you,

806
00:58:45,583 --> 00:58:49,066
any intuition?
Minus, yeah?

807
00:58:49,066 --> 00:58:52,952
Boy, you're good.
You're good.

808
00:58:52,952 --> 00:58:58,980
It is minus 2.4.
This one will be much further

809
00:58:58,980 --> 00:59:05,148
away than this one.
Do any one of you want to make

810
00:59:05,148 --> 00:59:09,086
a guess what omega minus is?
Anyone want to make a guess

811
00:59:09,086 --> 00:59:12,237
what omega plus is?
No way on Earth have you,

812
00:59:12,237 --> 00:59:15,459
or I, or anyone else can look
at this and say,

813
00:59:15,459 --> 00:59:18,180
oh yes, of course,
omega minus is this.

814
00:59:18,180 --> 00:59:21,259
It's a long road.
30 minutes of calculating,

815
00:59:21,259 --> 00:59:24,481
and then out of that popped
these frequencies.

816
00:59:24,481 --> 00:59:27,202
And out of that,
then, pops the ratios.

817
00:59:27,202 --> 00:59:32,000
And now I'm going to
demonstrate you just this case.

818
00:59:32,000 --> 00:59:35,780
And the way I'm going to do
that, even though we have not

819
00:59:35,780 --> 00:59:39,493
discussed driven coupled
oscillators, in order to set it

820
00:59:39,493 --> 00:59:42,869
off in these normal modes,
a normal modes is also a

821
00:59:42,869 --> 00:59:45,097
resonance frequency.
We call that,

822
00:59:45,097 --> 00:59:48,742
often, natural frequency.
It's what the system likes to

823
00:59:48,742 --> 00:59:51,983
do if you leave it alone.
And, it is easy for me,

824
00:59:51,983 --> 00:59:55,628
when I just tease it a little
bit, to excite it in that

825
00:59:55,628 --> 00:59:57,586
resonance mode.
So, therefore,

826
00:59:57,586 --> 1:00:00,624
I am going to drive it,
but only very briefly.

827
1:00:00,624 --> 1:00:04,000
And then, I will leave it
alone.

828
1:00:04,000 --> 1:00:08,302
And I will get it into the
state that you will see this

829
1:00:08,302 --> 1:00:11,170
that is [really?] in the normal
mode.

830
1:00:11,170 --> 1:00:13,799
There's one at only one
frequency.

831
1:00:13,799 --> 1:00:17,464
They come to a halt at the same
moment in time.

832
1:00:17,464 --> 1:00:21,208
They are in phase,
but you will clearly see that

833
1:00:21,208 --> 1:00:24,554
this distance is more than
double this one.

834
1:00:24,554 --> 1:00:27,024
And then, I will go to this
one.

835
1:00:27,024 --> 1:00:30,450
I will try that.
All right, double pendulum.

836
1:00:30,450 --> 1:00:41,000
Oh, my goodness.
This is a double pendulum. Yeah.

837
1:00:41,000 --> 1:00:48,000



838
1:00:48,000 --> 1:00:51,076
So, I have to drive it a
little.

839
1:00:51,076 --> 1:00:55,541
But it's really only very
little at resonance.

840
1:00:55,541 --> 1:00:58,717
And then, I will stop writing
it.

841
1:00:58,717 --> 1:01:03,412
This is the mode.
Do you see that the bottom one

842
1:01:03,412 --> 1:01:06,469
is further away than twice the
top one?

843
1:01:06,469 --> 1:01:10,410
This is a normal mode.
OK, you see that it's not a

844
1:01:10,410 --> 1:01:13,868
straight line?
Now, it's hard to say that it

845
1:01:13,868 --> 1:01:17,487
is one plus the square root of
two, of course,

846
1:01:17,487 --> 1:01:20,544
from your seats.
But that's what it is.

847
1:01:20,544 --> 1:01:24,163
And now, I will excite the
second normal mode,

848
1:01:24,163 --> 1:01:26,656
which is a resonance.
That's it.

849
1:01:26,656 --> 1:01:31,000
And now, I can almost stop
swinging it.

850
1:01:31,000 --> 1:01:35,570
You notice that your upper one
has a much larger amplitude than

851
1:01:35,570 --> 1:01:38,739
the lower one,
and the ratio is one plus the

852
1:01:38,739 --> 1:01:42,351
square root of two.
Did you notice they are out of

853
1:01:42,351 --> 1:01:44,267
phase?
It's the minus sign.

854
1:01:44,267 --> 1:01:47,511
They are out of phase.
And in the other mode,

855
1:01:47,511 --> 1:01:51,196
they were in phase.
And so, all of that will follow

856
1:01:51,196 --> 1:01:54,439
from the recipe:
if I gave this problem on an

857
1:01:54,439 --> 1:01:58,862
exam, you would have reasons to
kill me because it would take

858
1:01:58,862 --> 1:02:02,536
you too long.
I would certainly wear a

859
1:02:02,536 --> 1:02:06,119
bulletproof vest on campus.
That is easy because you have

860
1:02:06,119 --> 1:02:08,103
symmetry.
This doesn't have that

861
1:02:08,103 --> 1:02:09,000
symmetry.


862
1:02:09,000 --> 1:02:14,000


863
1:02:14,000 --> 1:02:16,345
Yeah, C2 over C1.
This is C2.

864
1:02:16,345 --> 1:02:19,947
I haven't specified what I call
one and two.

865
1:02:19,947 --> 1:02:24,219
Is that the problem?
I call this one and I call this

866
1:02:24,219 --> 1:02:26,816
two.
So, C2 over C1 is plus 2.4.

867
1:02:26,816 --> 1:02:32,011
It's good that you asked that.
And here, I changed the

868
1:02:32,011 --> 1:02:36,997
sequence because I knew that
this one was larger than this

869
1:02:36,997 --> 1:02:40,058
one.
So here, you have C1 over C2 is

870
1:02:40,058 --> 1:02:43,294
minus 2.4.
The minus sign means out of

871
1:02:43,294 --> 1:02:46,530
phase.
Yeah: good that you asked that.

872
1:02:46,530 --> 1:02:48,892
Very good.
Well, I'm slowly,

873
1:02:48,892 --> 1:02:54,227
slowly going to turn you into
experts, and now we are going to

874
1:02:54,227 --> 1:03:00,000
try something else to see how
good your intuition is.

875
1:03:00,000 --> 1:03:03,458
I want you to understand,
though, that your intuition is

876
1:03:03,458 --> 1:03:06,413
no better, no worse than my own.
In other words,

877
1:03:06,413 --> 1:03:09,997
there is no way that I could
have done any better than you

878
1:03:09,997 --> 1:03:11,884
did.
When I was in high school,

879
1:03:11,884 --> 1:03:14,902
or whatever it was,
maybe in college when I first

880
1:03:14,902 --> 1:03:17,920
saw this, my first impression
was, what the hell?

881
1:03:17,920 --> 1:03:20,624
A straight line?
But if you give it a little

882
1:03:20,624 --> 1:03:23,957
thought, you will come to the
conclusion it cannot be.

883
1:03:23,957 --> 1:03:26,661
Then, it ultimately comes out
this solution.

884
1:03:26,661 --> 1:03:31,000
So, don't feel bad if your
intuition lets you down.

885
1:03:31,000 --> 1:03:35,562
I'm as bad as you are when it
comes to that.

886
1:03:35,562 --> 1:03:42,247
So, now we're going to evaluate
the system, which would take you

887
1:03:42,247 --> 1:03:47,765
an hour and a half to work out
with a general recipe.

888
1:03:47,765 --> 1:03:51,903
And that is four springs and
three cars.

889
1:03:51,903 --> 1:03:56,678
One car, two cars,
three cars, and all springs

890
1:03:56,678 --> 1:04:01,028
have spring constant,
K, nicely symmetric,

891
1:04:01,028 --> 1:04:07,436
you would think.
And all objects have mass M.

892
1:04:07,436 --> 1:04:13,449
And, I want to know,
I want to see whether we have

893
1:04:13,449 --> 1:04:20,445
any intuition without being too
quantitative for the three

894
1:04:20,445 --> 1:04:26,581
normal mode solutions.
There must be an omega minus

895
1:04:26,581 --> 1:04:33,074
where all three are in phase.
Then, there must be an omega

896
1:04:33,074 --> 1:04:37,370
plus, which is more complicated.
And then, there must be an

897
1:04:37,370 --> 1:04:40,629
omega plus, plus,
which is the highest of all

898
1:04:40,629 --> 1:04:42,777
three.
In the highest possible

899
1:04:42,777 --> 1:04:46,851
frequency, every object,
two adjacent objects are always

900
1:04:46,851 --> 1:04:49,962
out of phase.
So, if you have 20 objects in

901
1:04:49,962 --> 1:04:53,518
the highest mode,
number one is out of phase with

902
1:04:53,518 --> 1:04:56,629
number two.
Number two is out of phase with

903
1:04:56,629 --> 1:04:59,666
number three.
Number three is out of phase

904
1:04:59,666 --> 1:05:03,000
with number four,
and so on.

905
1:05:03,000 --> 1:05:05,569
Let's first look at omega
minus.

906
1:05:05,569 --> 1:05:10,292
I will even dare asking you
whether you have any idea what

907
1:05:10,292 --> 1:05:13,193
omega minus is.
You don't have that;

908
1:05:13,193 --> 1:05:17,171
I don't have that.
Where would your object be if,

909
1:05:17,171 --> 1:05:20,734
for instance,
I displaced the first one over

910
1:05:20,734 --> 1:05:24,795
a distance which I just
normalized to be plus one?

911
1:05:24,795 --> 1:05:28,856
I call that plus one.
Any idea where this one will

912
1:05:28,856 --> 1:05:31,425
be?
Any idea where that one will

913
1:05:31,425 --> 1:05:35,453
be?
What to think the second one

914
1:05:35,453 --> 1:05:36,942
will be?
Plus one.

915
1:05:36,942 --> 1:05:39,658
Will be the third one?
Plus one.

916
1:05:39,658 --> 1:05:42,725
Wrong.
Why does it have to be wrong?

917
1:05:42,725 --> 1:05:48,157
Suppose this one is a plus one.
And, suppose this one is also a

918
1:05:48,157 --> 1:05:51,574
plus one.
Then, this spring is no longer

919
1:05:51,574 --> 1:05:55,779
than it wants to be,
and this spring is no longer

920
1:05:55,779 --> 1:06:00,072
than it wants to be.
So, there is no force on this

921
1:06:00,072 --> 1:06:03,866
object.
So, that's not possible.

922
1:06:03,866 --> 1:06:08,533
The object cannot go anywhere.
Any suggestions which of those

923
1:06:08,533 --> 1:06:10,477
is wrong?
Is this correct,

924
1:06:10,477 --> 1:06:13,822
and it's plus one?
How about the middle one?

925
1:06:13,822 --> 1:06:17,477
Where do you want it,
a little further away or a

926
1:06:17,477 --> 1:06:20,277
little less?
All have to be in phase.

927
1:06:20,277 --> 1:06:24,944
Don't use the word minus sign.
If you use the word minus sign

928
1:06:24,944 --> 1:06:30,000
then, how can you do that?
They have to be in phase.

929
1:06:30,000 --> 1:06:37,690
They must be on this side.
Do we make the distance larger

930
1:06:37,690 --> 1:06:40,849
or smaller?
It's larger.

931
1:06:40,849 --> 1:06:48,815
Should I tell you what it is?
Should I tell you what it is?

932
1:06:48,815 --> 1:06:52,248
You want to know?
Tell me.

933
1:06:52,248 --> 1:06:56,369
You really want to know,
right?

934
1:06:56,369 --> 1:06:59,939
I knew it.
Is that obvious?

935
1:06:59,939 --> 1:07:04,276
No.
Now, the next one.

936
1:07:04,276 --> 1:07:09,588
The next one is interesting.
I will, again,

937
1:07:09,588 --> 1:07:16,039
put this one in plus one
because that is my starting

938
1:07:16,039 --> 1:07:20,592
point.
I always put that at plus one.

939
1:07:20,592 --> 1:07:24,640
Now what?
This one you can guess.

940
1:07:24,640 --> 1:07:30,458
You really can.
Think now about the symmetry of

941
1:07:30,458 --> 1:07:34,000
the system.
Yeah?

942
1:07:34,000 --> 1:07:36,006
Very good.
Very good.

943
1:07:36,006 --> 1:07:41,422
The middle one stays still,
and this one comes to minus

944
1:07:41,422 --> 1:07:43,629
one.
Now the third one.

945
1:07:43,629 --> 1:07:49,045
Remember, the highest mode,
neighbors are always out of

946
1:07:49,045 --> 1:07:53,960
phase with each other no matter
how many you have.

947
1:07:53,960 --> 1:07:59,778
So, it's already certain that
this one must be on the side,

948
1:07:59,778 --> 1:08:07,000
and it's already certain that
this one must be on that side.

949
1:08:07,000 --> 1:08:11,847
Any volunteers?
Would you like to try that

950
1:08:11,847 --> 1:08:16,221
again?
Would you do plus one and minus

951
1:08:16,221 --> 1:08:20,004
one?
Help me put this one at plus

952
1:08:20,004 --> 1:08:25,561
one, sorry, minus one,
and this one at plus one,

953
1:08:25,561 --> 1:08:31,000
or shall we change the minus
one here?

954
1:08:31,000 --> 1:08:35,380
Don't feel bad.
I wouldn't have been able to do

955
1:08:35,380 --> 1:08:39,000
either.
Shall I tell you what it is?

956
1:08:39,000 --> 1:08:45,000



957
1:08:45,000 --> 1:08:49,299
Now, I can demonstrate this.
And the way I'm going to

958
1:08:49,299 --> 1:08:53,846
demonstrate this is with the air
track that I have here.

959
1:08:53,846 --> 1:08:58,807
See, the beauty with this air
track is that I can place these

960
1:08:58,807 --> 1:09:02,940
cars wherever I want,
and they're not going to move

961
1:09:02,940 --> 1:09:08,071
because I don't have any air.
Then, I turn on the air and

962
1:09:08,071 --> 1:09:10,757
they are and immediately free to
go.

963
1:09:10,757 --> 1:09:13,826
In other words,
if I offset this one in a

964
1:09:13,826 --> 1:09:16,818
positive direction over a
distance, one,

965
1:09:16,818 --> 1:09:19,196
we've just called this
positive.

966
1:09:19,196 --> 1:09:23,109
I apologize for that because I
called this positive.

967
1:09:23,109 --> 1:09:27,099
And the other one has to be
offset by one in the same

968
1:09:27,099 --> 1:09:31,473
direction, which I just did.
But this one has to be offset

969
1:09:31,473 --> 1:09:36,000
by the square root of two,
which I just did.

970
1:09:36,000 --> 1:09:40,095
And now, turn on the air,
and I just have exactly the

971
1:09:40,095 --> 1:09:43,089
right initial conditions for
this mode.

972
1:09:43,089 --> 1:09:47,578
So, it is going to oscillate in
the superposition of three

973
1:09:47,578 --> 1:09:50,335
normal modes.
But there is only one,

974
1:09:50,335 --> 1:09:54,825
and that's the one you're going
to see because I chose the

975
1:09:54,825 --> 1:09:59,000
initial conditions.
Are you ready for this?

976
1:09:59,000 --> 1:10:01,974
Now, admit it.
This is fantastic.

977
1:10:01,974 --> 1:10:06,901
They are all in phase.
That was a criterion for normal

978
1:10:06,901 --> 1:10:09,039
mode.
They are in phase.

979
1:10:09,039 --> 1:10:13,315
They come to a halt at the same
moment in time.

980
1:10:13,315 --> 1:10:18,335
They all have the same
frequency, and amplitude here is

981
1:10:18,335 --> 1:10:23,354
the square of two times larger.
That is not so obvious,

982
1:10:23,354 --> 1:10:28,002
but you see this is the
criterion for normal modes.

983
1:10:28,002 --> 1:10:32,000
Same frequency,
and in phase.

984
1:10:32,000 --> 1:10:35,297
This one, OK,
we are going to put this one at

985
1:10:35,297 --> 1:10:37,770
zero.
I'm going to put this one at

986
1:10:37,770 --> 1:10:40,617
plus one.
And I'm going to put this one

987
1:10:40,617 --> 1:10:43,839
at minus one.
Now, this one is offset by one

988
1:10:43,839 --> 1:10:47,286
in this direction.
This one is offset by one in

989
1:10:47,286 --> 1:10:50,434
this direction,
and this one is not offset.

990
1:10:50,434 --> 1:10:54,031
I turn on the air;
what do you think will happen?

991
1:10:54,031 --> 1:10:57,103
That's right,
it's going to oscillate in a

992
1:10:57,103 --> 1:11:01,000
superposition of three normal
modes.

993
1:11:01,000 --> 1:11:06,000
But there is only one.
That's the omega plus.

994
1:11:06,000 --> 1:11:11,000



995
1:11:11,000 --> 1:11:16,588
Impressive, isn't it?
Next one, this one.

996
1:11:16,588 --> 1:11:22,455
So, we are first going to put
this at zero,

997
1:11:22,455 --> 1:11:28,463
this at zero.
And we're going to put this at

998
1:11:28,463 --> 1:11:32,182
zero.
And now, what we have to do,

999
1:11:32,182 --> 1:11:35,379
this has to be put over plus
one plus one.

1000
1:11:35,379 --> 1:11:39,744
The other one has to be put
minus the square root of two.

1001
1:11:39,744 --> 1:11:44,422
And we have measured that here.
We've come so close that this

1002
1:11:44,422 --> 1:11:46,916
spring is not going to like
that.

1003
1:11:46,916 --> 1:11:51,594
And now, this one is going to
be, again, plus one in the same

1004
1:11:51,594 --> 1:11:55,569
direction as this one.
That's the highest frequency.

1005
1:11:55,569 --> 1:11:59,000
And there we go.
Are you ready?

1006
1:11:59,000 --> 1:12:06,000



1007
1:12:06,000 --> 1:12:07,919
Three, two, one,
zero.

1008
1:12:07,919 --> 1:12:13,220
One, and only one frequency.
And, every adjacent one is out

1009
1:12:13,220 --> 1:12:17,516
of phase with the other.
These are out of phase.

1010
1:12:17,516 --> 1:12:23,000
But these are in phase because
these are out of phase.

1011
1:12:23,000 --> 1:12:34,000



1012
1:12:34,000 --> 1:12:37,623
Now, I had one student
calculate this for me.

1013
1:12:37,623 --> 1:12:41,823
And I'll be frank with you,
I even paid him for that

1014
1:12:41,823 --> 1:12:44,952
because I was too lazy to do it
myself.

1015
1:12:44,952 --> 1:12:49,729
I put the job on the market,
and he came within a few days.

1016
1:12:49,729 --> 1:12:54,670
There were a few people that
said, oh, and an A plus for 8.03

1017
1:12:54,670 --> 1:12:58,376
and he was going to do that in
no time at all.

1018
1:12:58,376 --> 1:13:02,000
20 hours later,
he had it right.

1019
1:13:02,000 --> 1:13:07,665
It cost me a fortune.
Now, I'm going to test you once

1020
1:13:07,665 --> 1:13:12,459
more to show you that these
things are not so

1021
1:13:12,459 --> 1:13:17,143
straightforward,
not to make you feel bad at

1022
1:13:17,143 --> 1:13:19,322
all.
On the contrary,

1023
1:13:19,322 --> 1:13:24,988
my intuition is no better.
I'm going to make a triple

1024
1:13:24,988 --> 1:13:30,000
pendulum.
This is a triple pendulum.

1025
1:13:30,000 --> 1:13:37,000



1026
1:13:37,000 --> 1:13:41,203
Omega minus,
what do you think it would look

1027
1:13:41,203 --> 1:13:44,428
like?
My favorite student is here.

1028
1:13:44,428 --> 1:13:48,240
Do you think it will be a
straight line?

1029
1:13:48,240 --> 1:13:53,323
I don't think so either.
I think what you're going to

1030
1:13:53,323 --> 1:13:58,308
see without making any
predictions about amplitudes,

1031
1:13:58,308 --> 1:14:03,000
exaggerated,
you're going to see this.

1032
1:14:03,000 --> 1:14:13,064
That's what I think.
No idea what it is.

1033
1:14:13,064 --> 1:14:19,000
How about omega plus?


1034
1:14:19,000 --> 1:14:26,000


1035
1:14:26,000 --> 1:14:31,816
This is not going to work.
It's a whole different system.

1036
1:14:31,816 --> 1:14:36,489
Two of them are probably going
to be in phase,

1037
1:14:36,489 --> 1:14:41,474
and one out of phase.
Which two will be in phase?

1038
1:14:41,474 --> 1:14:46,252
Will you see this?
Now, these two are in phase,

1039
1:14:46,252 --> 1:14:51,341
and this is out of phase.
Or really, you see this.

1040
1:14:51,341 --> 1:14:57,261
Now, these two are in phase,
and this one is out of phase.

1041
1:14:57,261 --> 1:15:03,588
Who is in favor of this?
Who is in favor of that?

1042
1:15:03,588 --> 1:15:09,303
I want to see the hands long.
Who is in favor of this?

1043
1:15:09,303 --> 1:15:14,264
Who is in favor of that?
Yeah, well, we'll see.

1044
1:15:14,264 --> 1:15:18,254
[LAUGHTER] How about the
highest mode?

1045
1:15:18,254 --> 1:15:24,401
That's the easiest one because
now they must go like this.

1046
1:15:24,401 --> 1:15:28,392
Now, of course,
the ratio, C1 over C2,

1047
1:15:28,392 --> 1:15:34,000
and C1 over C3,
that's a different story.

1048
1:15:34,000 --> 1:15:38,812
That is your 20 hours of
grinding if you get paid.

1049
1:15:38,812 --> 1:15:43,821
If you don't get paid,
you can probably do it in two

1050
1:15:43,821 --> 1:15:47,160
hours.
So, I'm going to demonstrate

1051
1:15:47,160 --> 1:15:51,187
this to you.
I'm going to find these three

1052
1:15:51,187 --> 1:15:54,526
modes.
Again, I do that by a little

1053
1:15:54,526 --> 1:15:58,651
bit of driving because of their
resonances.

1054
1:15:58,651 --> 1:16:02,973
And then, I want you to eyeball
these ratios,

1055
1:16:02,973 --> 1:16:08,431
C1 over C2, and C1 over C3.
Oh, no problem.

1056
1:16:08,431 --> 1:16:10,625
No problem.
There we go.

1057
1:16:10,625 --> 1:16:15,301
All right, so now,
this is not where the length is

1058
1:16:15,301 --> 1:16:18,832
the same.
So, to get this one into the

1059
1:16:18,832 --> 1:16:23,412
lowest mode is easy for me.
Just give it a swing,

1060
1:16:23,412 --> 1:16:26,656
and let it go.
And just do nothing.

1061
1:16:26,656 --> 1:16:29,805
Then, I will stop even driving
it.

1062
1:16:29,805 --> 1:16:32,000
That's it.


1063
1:16:32,000 --> 1:16:38,000


1064
1:16:38,000 --> 1:16:42,461
Now, it's nicely in that mode.
You see, the bottom one [has

1065
1:16:42,461 --> 1:16:45,230
it?].
Notice that the lines are never

1066
1:16:45,230 --> 1:16:48,846
in the same direction.
You see the offset of the

1067
1:16:48,846 --> 1:16:51,000
angles?
Can you all see that?

1068
1:16:51,000 --> 1:16:55,384
So, our first picture on the
blackboard is quite accurate.

1069
1:16:55,384 --> 1:17:00,000
Now, the second one,
that's harder for me to find.

1070
1:17:00,000 --> 1:17:04,070
I mean, we first make sure that
they are not moving.

1071
1:17:04,070 --> 1:17:07,262
OK, so I'm going to try to
search for it.

1072
1:17:07,262 --> 1:17:08,859
I got it.
This is it.

1073
1:17:08,859 --> 1:17:12,131
So, the first solution is the
correct one.

1074
1:17:12,131 --> 1:17:15,802
The upper two are in phase,
and the bottom one,

1075
1:17:15,802 --> 1:17:19,394
look at the large amplitude of
the bottom one.

1076
1:17:19,394 --> 1:17:23,704
That's not so intuitive.
The bottom one is out of phase

1077
1:17:23,704 --> 1:17:26,417
with the upper two.
Can you see it?

1078
1:17:26,417 --> 1:17:29,291
Normal mode,
why is it a normal mode?

1079
1:17:29,291 --> 1:17:34,000
Because they all have the same
frequency.

1080
1:17:34,000 --> 1:17:38,904
And, they are in phase or they
are out of phase.

1081
1:17:38,904 --> 1:17:44,643
Now, the highest frequency,
let me see whether I can get

1082
1:17:44,643 --> 1:17:46,000
that one.


1083
1:17:46,000 --> 1:17:52,000


1084
1:17:52,000 --> 1:18:05,063
Got it.
That's it.

1085
1:18:05,063 --> 1:18:46,557
Now, look how small the
amplitude of the bottom one is.

1086
1:18:46,557 --> 1:19:10,378
And, that can all be
calculated.

1087
1:19:10,378 --> 1:19:41,883
They are beginning to rotate a
little bit.

1088
1:19:41,883 --> 1:20:08,009
There we go.
That's the third mode.

1089
1:20:08,009 --> 1:20:48,735
So, now you know everything
there is to be known about

1090
1:20:48,735 --> 1:21:26,387
coupled oscillators.
The most important thing that

1091
1:21:26,387 --> 1:22:10,955
you probably have learned is
that if you really want to get

1092
1:22:10,955 --> 1:22:49,376
the general solution for a
system, it is hard work.

1093
1:22:49,376 --> 1:23:05,513
See you next Tuesday.
Have a good weekend.