1
00:00:24,000 --> 00:00:29,967
I will start today calculating
for you the normal mode
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00:00:29,967 --> 00:00:33,571
frequencies of a double
pendulum.
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00:00:33,571 --> 00:00:38,412
And then, I will drive that
double pendulum.
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00:00:38,412 --> 00:00:44,042
And then we will see very
dramatic things that have
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00:00:44,042 --> 00:00:48,096
changed.
So, let us start here with a
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00:00:48,096 --> 00:00:52,487
double pendulum.
This is the equilibrium
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00:00:52,487 --> 00:00:58,117
position if it hangs straight,
length L, mass is M.
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00:00:58,117 --> 00:01:06,000
This angle [be?] theta one.
And this angle is theta two.
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00:01:06,000 --> 00:01:13,913
I call this position X1,
and I call this position X2.
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00:01:13,913 --> 00:01:22,434
I'm going to introduce a
shorthand notation at omega zero
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00:01:22,434 --> 00:01:30,500
squared, G divided by L.
I want to remind you that the
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00:01:30,500 --> 00:01:40,086
sine of theta one is X1 divided
by L, and that the sine of theta
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00:01:40,086 --> 00:01:48,000
two is going to be X2 minus X1
divided by L.
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00:01:48,000 --> 00:01:54,279
For small angle approximation,
we do know the tangents.
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00:01:54,279 --> 00:01:59,279
Here we have a tangent which I
will call T2.
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00:01:59,279 --> 00:02:03,000
That is here,
of course, MG.
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00:02:03,000 --> 00:02:06,715
So, there is also a tangent,
T2 here.
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00:02:06,715 --> 00:02:11,565
Action equals minus reaction,
and then there is,
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here, the tangent T1.
And, here is MG.
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00:02:15,384 --> 00:02:20,234
But, there are the only forces
on these objects.
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00:02:20,234 --> 00:02:24,466
In small angles,
I do know that T1 must be
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00:02:24,466 --> 00:02:29,523
approximately 2MG because it's
carrying boy loads,
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00:02:29,523 --> 00:02:34,475
so to speak.
And we know that T2 is very
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00:02:34,475 --> 00:02:38,326
close to MG.
And we're going to use that in
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00:02:38,326 --> 00:02:42,819
our approximation.
So, let me first start with the
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00:02:42,819 --> 00:02:48,137
bottom one that has only two
forces on it so that maybe the
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easiest M2 X2 double dot.
And so, the only force that is
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00:02:53,180 --> 00:02:58,865
driving it back to equilibrium
in the small angle approximation
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00:02:58,865 --> 00:03:04,000
is, then, the horizontal
component of the T2.
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00:03:04,000 --> 00:03:12,547
So, that equals minus T2 times
the sine of theta two.
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00:03:12,547 --> 00:03:19,945
And, for that,
I can write minus MG divided by
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00:03:19,945 --> 00:03:27,671
L times theta two X2 minus X1.
We divided out M,
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and we use our shorthand
notation.
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And so, we get that X2 double
dot.
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00:03:40,000 --> 00:03:43,482
And now we have here minus MG
over LX2.
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00:03:43,482 --> 00:03:47,330
So, that comes in,
and that becomes a plus.
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00:03:47,330 --> 00:03:51,179
So, I got plus omega zero
squared times X2.
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00:03:51,179 --> 00:03:54,752
I have here plus X1.
When that comes in,
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00:03:54,752 --> 00:03:59,059
that becomes a minus.
So, I get minus omega zero
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squared X1, and that equals
zero.
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00:04:03,000 --> 00:04:07,249
So, this is my first
differential equation for the
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00:04:07,249 --> 00:04:10,978
second object.
So, now I'm going to my first
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00:04:10,978 --> 00:04:14,186
object.
So, now we get MX1 double dot,
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00:04:14,186 --> 00:04:19,216
and now there is one [SOUND
OFF/THEN ON] that is driving it
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00:04:19,216 --> 00:04:23,205
back to equilibrium,
and that is the horizontal
46
00:04:23,205 --> 00:04:27,108
component of T1.
But, the horizontal component
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00:04:27,108 --> 00:04:32,311
of T2 is driving it away from
equilibrium in the drawing that
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00:04:32,311 --> 00:04:38,399
I have made here.
So, you get minus T1 times the
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00:04:38,399 --> 00:04:43,950
sign of theta one plus T2 times
the sine of theta.
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00:04:43,950 --> 00:04:49,615
And so, I'm going to substitute
in there now sines,
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00:04:49,615 --> 00:04:55,280
and I'm going to substitute in
their T1 equals 2MG,
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00:04:55,280 --> 00:05:04,642
and T2 equals MG.
So, this becomes equal to minus
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00:05:04,642 --> 00:05:16,857
2MG times the sine of theta one,
which is X1 divided by L.
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00:05:16,857 --> 00:05:28,214
And then, I get plus T2,
which is MG times the sine X2
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00:05:28,214 --> 00:05:36,319
minus X1 divided by L.
And, I'm going to divide by M,
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00:05:36,319 --> 00:05:39,567
and I'm going to use my
shorthand notation.
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And, when I do that,
I will come out here,
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00:05:42,738 --> 00:05:44,904
X1 double dot.
It's this one,
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00:05:44,904 --> 00:05:48,306
here the N goes.
Go over L becomes omega zero
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squared.
But now, look closely.
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00:05:50,626 --> 00:05:55,189
There is here a two times X1.
And here, there is a one times
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00:05:55,189 --> 00:05:57,586
X1.
And, both have a minus sign.
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So, when they come out,
I get three omega zero squared.
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00:06:03,000 --> 00:06:08,647
So, I get plus three omega zero
squared times X1.
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00:06:08,647 --> 00:06:12,411
And then, I have to bring X2
out.
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00:06:12,411 --> 00:06:17,823
That becomes minus omega zero
squared times X2.
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00:06:17,823 --> 00:06:23,470
And, that is a zero.
And so, we have to solve now
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00:06:23,470 --> 00:06:30,176
this differential equation
coupled with X1 and X2 together
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00:06:30,176 --> 00:06:34,772
with this one.
My solution has to satisfy
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00:06:34,772 --> 00:06:37,090
both.
I want to go over that one to
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me sure that it is correct.
[SOUND OFF/THEN ON] pays off
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00:06:40,840 --> 00:06:43,431
because one sign wrong and
you're hung.
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00:06:43,431 --> 00:06:47,045
The whole thing falls apart.
You're dead in the water,
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00:06:47,045 --> 00:06:49,636
so it pays off to think about
it again.
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00:06:49,636 --> 00:06:53,045
So, we have an X2 double dot.
I can live with that,
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00:06:53,045 --> 00:06:56,795
plus omega zero squared X2.
That has the right smell for
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me, minus omega zero squared X1.
I can live with that
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00:07:01,503 --> 00:07:04,436
differential equation.
I go to this one.
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Of course, the three is
well-known in a system like this
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that you get to three.
I have an X1 double dot plus
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00:07:12,330 --> 00:07:16,616
three omega zero squared X1.
I think we are in good shape.
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00:07:16,616 --> 00:07:20,601
And so, now we're going to put
in our trial solutions.
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00:07:20,601 --> 00:07:24,812
X1 is C1 times cosine omega T.
And, X2 is C2 times cosine
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00:07:24,812 --> 00:07:27,669
omega T.
And, we're going to search for
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00:07:27,669 --> 00:07:31,053
the frequencies.
So, this omega,
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00:07:31,053 --> 00:07:33,231
we have to solve for this
omega.
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00:07:33,231 --> 00:07:36,814
This is not a given.
Since we are looking for normal
88
00:07:36,814 --> 00:07:39,836
mode solutions,
these two omegas must be the
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00:07:39,836 --> 00:07:42,224
same.
I cannot write down omega one
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00:07:42,224 --> 00:07:45,035
and omega two.
And, I don't have to worry
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00:07:45,035 --> 00:07:48,899
about any phase angles because
since we have no damping,
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00:07:48,899 --> 00:07:52,693
either they are in phase or
they are 180° out of phase.
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00:07:52,693 --> 00:07:56,135
And, 180° out of phase sibling
means a minus sign.
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00:07:56,135 --> 00:08:00,000
So, that's the great thing
about this.
95
00:08:00,000 --> 00:08:04,521
The signs will take care,
then, of the possible phase
96
00:08:04,521 --> 00:08:07,043
angles.
So, now we're going to
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00:08:07,043 --> 00:08:11,304
substitute this solution in
these two differential
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00:08:11,304 --> 00:08:14,869
equations.
And, I'm going to write it down
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in a form that I put the C1's to
the left, and the C2's to the
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right.
And so, I'm first going to this
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one.
That is my object number one.
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I take the second derivative.
So, I get C1 times minus omega
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00:08:32,790 --> 00:08:38,186
squared because the second
derivative here gives me a minus
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00:08:38,186 --> 00:08:41,906
omega squared.
I ditch the cosine omega T
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terms because I'm going to have
a cosine omega T everywhere.
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00:08:47,395 --> 00:08:52,046
So, I'm not going to write down
the cosine omega T.
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00:08:52,046 --> 00:08:56,697
I have here plus three omega
zero squared times C1.
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00:08:56,697 --> 00:09:02,000
I have here minus omega zero
squared times C2.
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00:09:02,000 --> 00:09:06,850
And, that is zero.
Notice I put the C1's here and
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the C2 there.
I'm going to the other
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00:09:10,386 --> 00:09:14,428
equation.
And, I'm going to put the C1 on
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00:09:14,428 --> 00:09:18,066
the left.
So, we get minus omega zero
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00:09:18,066 --> 00:09:22,815
squared times C1.
That's the term you have here.
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00:09:22,815 --> 00:09:28,979
And then, we have plus C2 times
omega zero squared minus omega
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00:09:28,979 --> 00:09:33,304
squared.
And, this minus omega squared
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00:09:33,304 --> 00:09:38,230
comes from the second derivative
of this one, second derivative
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00:09:38,230 --> 00:09:41,249
of this one minus omega squared,
right?
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00:09:41,249 --> 00:09:45,936
You get omega squared comes
out, and then the cosine goes to
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00:09:45,936 --> 00:09:48,797
a sine that gives you the minus
sign.
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00:09:48,797 --> 00:09:53,087
So, now we have here two
equations with three unknowns.
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00:09:53,087 --> 00:09:57,934
That is typical for normal mode
solutions of a system with two
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oscillations.
We don't know C1;
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we don't know C2,
and we do not know omega.
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00:10:04,119 --> 00:10:08,239
And so, we remember from last
time that you can always solve
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for the ratio C1 over C2,
and you can solve for omega.
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You can only find C1 if you
also know the initial
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00:10:15,291 --> 00:10:17,665
conditions, which I have not
given.
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Instead of solving it the high
school way that I did last week,
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the simple way,
the fast way,
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I'm going to do it now using
Kramer's rule.
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00:10:28,000 --> 00:10:33,073
And the reason why I want to do
this once, even though in this
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case it really is not necessary,
when you have three coupled
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oscillators or four,
there is just no way that the
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00:10:42,056 --> 00:10:45,134
high school method will do it
for you.
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You have to have a more general
approach.
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00:10:48,461 --> 00:10:52,037
I've sent this to you by
e-mail, all of you,
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00:10:52,037 --> 00:10:56,862
and I also assume that you have
worked on this a little bit
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00:10:56,862 --> 00:11:00,937
since I requested that you would
prepare for that.
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00:11:00,937 --> 00:11:04,514
So, I'm first going to write
down what D is,
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00:11:04,514 --> 00:11:08,589
which is the determinant of
[NOISE OBSCURES] C1's,
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00:11:08,589 --> 00:11:15,381
and we have the C2's.
So, let me write down here what
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00:11:15,381 --> 00:11:18,763
D is.
So, I'm going to get three
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omega zeroes squared.
[SOUND OFF/THEN ON] that is the
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C1.
Then I get here minus omega
145
00:11:27,709 --> 00:11:30,109
squared.
That's the C2.
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00:11:30,109 --> 00:11:36,000
And then I get here minus omega
zero squared.
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00:11:36,000 --> 00:11:41,345
And here I get omega zero
squared minus omega squared.
148
00:11:41,345 --> 00:11:45,178
And, the determinant of this,
that's D.
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00:11:45,178 --> 00:11:51,331
And, I presume you all know how
to get the determinant of this
150
00:11:51,331 --> 00:11:55,567
very simple matrix.
We'll do that together,
151
00:11:55,567 --> 00:11:59,400
of course.
So, following Kramer's rule,
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00:11:59,400 --> 00:12:02,325
then, my C1,
which is X there,
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00:12:02,325 --> 00:12:08,579
that's the one I want to solve
for, so my C1 I now have to take
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00:12:08,579 --> 00:12:12,640
this zero.
This is also a zero,
155
00:12:12,640 --> 00:12:15,921
by the way.
I forgot to mention that this
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00:12:15,921 --> 00:12:18,710
is a zero.
This equation is a zero.
157
00:12:18,710 --> 00:12:22,156
So, this column,
now, zero-zero has to come
158
00:12:22,156 --> 00:12:24,699
first.
So, I get here zero-zero.
159
00:12:24,699 --> 00:12:29,539
And then, the second column is
the same as it was here minus
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00:12:29,539 --> 00:12:35,214
omega zero squared.
And then I get omega zero
161
00:12:35,214 --> 00:12:42,028
squared minus omega squared.
And, that's divided by D.
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00:12:42,028 --> 00:12:46,014
So, we now know that that is
C1.
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00:12:46,014 --> 00:12:52,442
And then, we go to C2.
So now, the zero-zero column
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00:12:52,442 --> 00:12:59,642
shifts towards the right,
goes here, and the first column
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00:12:59,642 --> 00:13:04,914
is unchanged.
So, we have here three omega
166
00:13:04,914 --> 00:13:12,885
zero squared minus omega squared
minus omega zero squared zero,
167
00:13:12,885 --> 00:13:18,781
zero, divided by D.
You've got to admit that the
168
00:13:18,781 --> 00:13:21,581
upstairs here,
the determinant of this matrix
169
00:13:21,581 --> 00:13:24,127
is zero because you have two
zeroes here.
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00:13:24,127 --> 00:13:26,290
And, it's also a zero for this
one.
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00:13:26,290 --> 00:13:29,918
But, clearly zero solutions for
C1 and C2 are meaningless.
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00:13:29,918 --> 00:13:33,163
They are not incorrect,
because you will get in this
173
00:13:33,163 --> 00:13:36,663
differential equation that zero
is zero, which is rather
174
00:13:36,663 --> 00:13:40,292
obvious.
So, we don't want the solutions
175
00:13:40,292 --> 00:13:44,213
that C1 and C2 are zero.
And, the only way that we can
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00:13:44,213 --> 00:13:48,724
avoid that is to demand that D
become zero because now you get
177
00:13:48,724 --> 00:13:52,349
zero divided by zero.
And that's a whole different
178
00:13:52,349 --> 00:13:54,863
story.
That's not necessarily zero.
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00:13:54,863 --> 00:13:58,192
And that, then,
is the idea behind getting the
180
00:13:58,192 --> 00:14:03,000
solutions to the searched-for
normal mode frequencies.
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00:14:03,000 --> 00:14:08,408
So, we must demand in this case
that this become zero.
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00:14:08,408 --> 00:14:13,612
Otherwise, you get trivial
solutions which are of no
183
00:14:13,612 --> 00:14:17,693
interest.
So, when we make D equals zero,
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00:14:17,693 --> 00:14:23,612
we do get that the determinant
of that matrix becomes three
185
00:14:23,612 --> 00:14:29,938
omega zero squared minus omega
squared times omega zero squared
186
00:14:29,938 --> 00:14:35,478
minus omega squared.
And then, minus the product,
187
00:14:35,478 --> 00:14:40,081
that becomes a minus,
not a plus, because minus-minus
188
00:14:40,081 --> 00:14:44,065
is already plus.
So, I get minus omega zero to
189
00:14:44,065 --> 00:14:48,137
the fourth equals zero.
And this equation is an
190
00:14:48,137 --> 00:14:52,652
equation in omega to the fourth.
You can solve that.
191
00:14:52,652 --> 00:14:55,839
You can solve that for omega
squared.
192
00:14:55,839 --> 00:15:00,000
I will leave you with that
solution.
193
00:15:00,000 --> 00:15:05,628
This is utterly trivial.
And, [NOISE OBSCURES] omega
194
00:15:05,628 --> 00:15:12,361
minus squared [NOISE OBSCURES]
is two minus the square root of
195
00:15:12,361 --> 00:15:18,762
two times omega zero squared,
and omega plus squared is two
196
00:15:18,762 --> 00:15:24,501
plus the square root of two
times omega zero squared.
197
00:15:24,501 --> 00:15:31,123
So, this step for you will take
you no more than maybe half a
198
00:15:31,123 --> 00:15:36,078
minute.
Be careful because you can
199
00:15:36,078 --> 00:15:39,039
easily slip up,
of course.
200
00:15:39,039 --> 00:15:46,144
So, we now have a solution that
omega minus is approximately,
201
00:15:46,144 --> 00:15:51,710
if I calculate the two minus
square root of two,
202
00:15:51,710 --> 00:15:56,447
it's approximately 0.77,
0.76 omega zero.
203
00:15:56,447 --> 00:16:02,123
Not intuitive at all.
So, it's lower than the
204
00:16:02,123 --> 00:16:05,815
resonance frequency of a single
pendulum.
205
00:16:05,815 --> 00:16:10,338
And then, I can substitute this
value for omega E,
206
00:16:10,338 --> 00:16:15,230
or back into my equations.
Or, I substitute it back in
207
00:16:15,230 --> 00:16:19,753
this if you want to.
You get zero divided by zero,
208
00:16:19,753 --> 00:16:25,384
which is not going to be zero.
And, that you're going to find,
209
00:16:25,384 --> 00:16:29,538
then that C2 divided by C1 in
that minus mode,
210
00:16:29,538 --> 00:16:34,984
in that lowest possible mode,
you will find that that is one
211
00:16:34,984 --> 00:16:40,061
plus the square root of two,
which is approximately plus
212
00:16:40,061 --> 00:16:44,168
2.4.
And, the omega plus solution
213
00:16:44,168 --> 00:16:48,118
gives you a frequency that is
about 0.85 omega zero.
214
00:16:48,118 --> 00:16:52,920
That solution you can put back
into your differential equations
215
00:16:52,920 --> 00:16:56,560
here, or not the differential
equations, I mean,
216
00:16:56,560 --> 00:17:00,045
into this one,
or you put it back into this if
217
00:17:00,045 --> 00:17:02,369
you want to.
And you will find,
218
00:17:02,369 --> 00:17:07,015
now, that C2 over C1 for that
plus mode; I call this the plus
219
00:17:07,015 --> 00:17:11,198
mode [with?] my shorthand
notation is going to be minus
220
00:17:11,198 --> 00:17:16,000
one divided by one plus the
square root of two.
221
00:17:16,000 --> 00:17:20,162
So, it's going to be minus one
divided by 2.4.
222
00:17:20,162 --> 00:17:23,308
So, it's going to be roughly
-0.41.
223
00:17:23,308 --> 00:17:27,378
So, those are,
then, the formal solutions for
224
00:17:27,378 --> 00:17:31,634
the normal modes.
And if I gave you the initial
225
00:17:31,634 --> 00:17:37,000
conditions, then of course you
could calculate C1.
226
00:17:37,000 --> 00:17:41,182
And then you know everything
because you know the ratio C2
227
00:17:41,182 --> 00:17:43,529
over C1.
But, without the initial
228
00:17:43,529 --> 00:17:47,638
conditions, you cannot do that.
Each of these normal mode
229
00:17:47,638 --> 00:17:51,086
solutions satisfies both
differential equations.
230
00:17:51,086 --> 00:17:54,241
So, therefore,
a linear superposition of the
231
00:17:54,241 --> 00:17:57,690
two normal mode solutions is a
general solution.
232
00:17:57,690 --> 00:18:01,872
So, when you start that system
off, at time T equals zero,
233
00:18:01,872 --> 00:18:05,983
you specify X1.
You specify the velocity of
234
00:18:05,983 --> 00:18:07,892
object one.
You specify X2.
235
00:18:07,892 --> 00:18:11,124
And, you specify the velocity
of that object.
236
00:18:11,124 --> 00:18:14,723
Then it's going to oscillate in
the superposition,
237
00:18:14,723 --> 00:18:18,615
the linear superposition of two
normal mode solutions.
238
00:18:18,615 --> 00:18:21,039
One, you see here has omega
minus.
239
00:18:21,039 --> 00:18:23,977
And this will be the ratio of
amplitudes.
240
00:18:23,977 --> 00:18:28,090
And this is the omega plus.
And that will be the ratio of
241
00:18:28,090 --> 00:18:32,810
amplitudes.
And that is nonnegotiable.
242
00:18:32,810 --> 00:18:38,972
That's what the system will do.
Now, we're going to make a
243
00:18:38,972 --> 00:18:43,837
dramatic change.
Now we are going to drive the
244
00:18:43,837 --> 00:18:47,945
system.
So, now this is the equilibrium
245
00:18:47,945 --> 00:18:54,324
position of the whole system.
I'm going to drive it back and
246
00:18:54,324 --> 00:18:58,000
forth holding in my hand like
this.
247
00:18:58,000 --> 00:19:04,630
And I'm going to shake it.
I call the position of my hand
248
00:19:04,630 --> 00:19:10,065
[eta?] equals eta zero times the
cosine [NOISE OBSCURES] is the
249
00:19:10,065 --> 00:19:12,958
amplitude of my motion of my
hand.
250
00:19:12,958 --> 00:19:17,254
That is what I'm going to do.
And so, when I look,
251
00:19:17,254 --> 00:19:20,410
now, at this equation,
at the figure,
252
00:19:20,410 --> 00:19:25,145
not the equation but at the
figure, I call this now X1.
253
00:19:25,145 --> 00:19:30,580
I call this now X2 because you
should always call X1 and X2 the
254
00:19:30,580 --> 00:19:36,258
distance from equilibrium.
And this is now equilibrium.
255
00:19:36,258 --> 00:19:40,209
If I don't shake at all,
I have the pendulum here,
256
00:19:40,209 --> 00:19:42,870
and so it's hanging straight
down.
257
00:19:42,870 --> 00:19:46,983
So, this is now equilibrium.
And this location here,
258
00:19:46,983 --> 00:19:50,048
at a random moment in time,
is now eta.
259
00:19:50,048 --> 00:19:53,112
What has changed?
Well, at first sight,
260
00:19:53,112 --> 00:19:57,548
very little has changed.
The only thing that has changed
261
00:19:57,548 --> 00:20:02,387
now is that the sine of theta
one is now X1 minus eta divided
262
00:20:02,387 --> 00:20:06,749
by L.
So, I have here a minus eta.
263
00:20:06,749 --> 00:20:11,952
It looks rather innocent with
the consequences will be
264
00:20:11,952 --> 00:20:16,076
unbelievable.
So, I can leave everything on
265
00:20:16,076 --> 00:20:19,218
the blackboard the way I have
it.
266
00:20:19,218 --> 00:20:24,519
All I have to do is to put
instead of the sine theta X1
267
00:20:24,519 --> 00:20:28,643
divided by L,
I have to put in X1 minus eta
268
00:20:28,643 --> 00:20:33,992
divided by L.
See, nature was very kind to
269
00:20:33,992 --> 00:20:37,092
me.
It already left some space
270
00:20:37,092 --> 00:20:39,123
there.
You see that?
271
00:20:39,123 --> 00:20:45,429
[SOUND OFF/THEN ON] anticipated
that I was going to do that.
272
00:20:45,429 --> 00:20:51,950
So, you have a minus eta there.
Well, if now you divide M out,
273
00:20:51,950 --> 00:20:56,866
and you're going to use the
shorthand notation,
274
00:20:56,866 --> 00:21:02,745
then leaving eta on the right
side, which is nice to do,
275
00:21:02,745 --> 00:21:07,769
you get minus eta.
So, this zero now becomes two
276
00:21:07,769 --> 00:21:13,969
omega zero squared times eta
zero times the cosine of omega
277
00:21:13,969 --> 00:21:18,096
T.
It becomes two,
278
00:21:18,096 --> 00:21:24,946
omega zero squared times eta.
[You see it's two?].
279
00:21:24,946 --> 00:21:31,489
So, it's no longer a zero.
And so, when we now substitute
280
00:21:31,489 --> 00:21:36,218
in there these trial functions,
they now have a completely
281
00:21:36,218 --> 00:21:40,200
different meaning.
Omega is no longer negotiable.
282
00:21:40,200 --> 00:21:42,938
Omega [as set?] by me is a
driver.
283
00:21:42,938 --> 00:21:46,008
So we're not going to solve for
omega.
284
00:21:46,008 --> 00:21:49,908
Omega is a given.
That means if omega is a given
285
00:21:49,908 --> 00:21:54,886
that we are going to end up not
with two equations with three
286
00:21:54,886 --> 00:21:57,126
unknowns, C1,
C2, and omega,
287
00:21:57,126 --> 00:22:02,270
but we're going to end up with
two equations with two unknowns,
288
00:22:02,270 --> 00:22:07,000
only C1 and C2 because omega is
a given.
289
00:22:07,000 --> 00:22:10,819
And so, in a steady state
solution, you get a number for
290
00:22:10,819 --> 00:22:14,361
C1 and you get a number for C2
in terms of eta zero,
291
00:22:14,361 --> 00:22:17,000
of course.
You will see how that works.
292
00:22:17,000 --> 00:22:19,638
So, when we put in these
functions now,
293
00:22:19,638 --> 00:22:23,805
these omegas are fixed and no
longer something that we search
294
00:22:23,805 --> 00:22:25,402
for.
They are my omegas.
295
00:22:25,402 --> 00:22:28,597
I can make them zero.
I can make them infinity.
296
00:22:28,597 --> 00:22:32,000
I can make them anything I want
to.
297
00:22:32,000 --> 00:22:35,052
And that's what we want to
study.
298
00:22:35,052 --> 00:22:40,202
So, if we want to go back,
now, to these two equations,
299
00:22:40,202 --> 00:22:45,734
what changes here since we have
put in this trial function,
300
00:22:45,734 --> 00:22:50,789
the only thing that disappears
is this cosine omega T.
301
00:22:50,789 --> 00:22:55,176
So, we end up here with two
omega zero squared.
302
00:22:55,176 --> 00:22:59,086
Times eta zero.
And, nothing else changes.
303
00:22:59,086 --> 00:23:05,000
But now, we're looking now at
our solution for C1.
304
00:23:05,000 --> 00:23:12,240
And we're going to look at our
solution for C2 in steady state.
305
00:23:12,240 --> 00:23:14,576
Ah, it's easy,
right?
306
00:23:14,576 --> 00:23:20,883
We apply Kramer's rule,
and the only thing that changes
307
00:23:20,883 --> 00:23:26,138
is this one, which has to be
replaced by this,
308
00:23:26,138 --> 00:23:30,226
2 omega zero squared times eta
zero.
309
00:23:30,226 --> 00:23:37,000
Eta zero, remember,
is the amplitude of my hand.
310
00:23:37,000 --> 00:23:43,787
Eta is the displacement of my
hand at any moment in time.
311
00:23:43,787 --> 00:23:48,515
Eta zero is the amplitude.
And so, here,
312
00:23:48,515 --> 00:23:53,484
we get, then,
also, two omega zero squared
313
00:23:53,484 --> 00:23:56,757
times eta zero.
And so, now,
314
00:23:56,757 --> 00:24:02,383
we can solve for C1 and C2.
We get an answer,
315
00:24:02,383 --> 00:24:05,556
not just a ratio only.
We get an answer.
316
00:24:05,556 --> 00:24:10,275
We know exactly what C1 is
going to be, and we know exactly
317
00:24:10,275 --> 00:24:14,587
what C2 is going to be because
omega is nonnegotiable.
318
00:24:14,587 --> 00:24:17,842
Omega is now unknown.
When we solve this,
319
00:24:17,842 --> 00:24:22,480
we were searching for an omega,
and out came these omegas.
320
00:24:22,480 --> 00:24:25,897
That's not the case anymore.
We know omega.
321
00:24:25,897 --> 00:24:28,501
It's called omega,
and that's it.
322
00:24:28,501 --> 00:24:35,235
So, I can write down now C1.
So, I take the determinant
323
00:24:35,235 --> 00:24:41,457
there of the upstairs.
So, that gives me omega zero
324
00:24:41,457 --> 00:24:48,675
squared times eta zero times
omega zero squared minus omega
325
00:24:48,675 --> 00:24:52,408
squared.
That is this diagonal,
326
00:24:52,408 --> 00:24:58,631
and this one is zero.
And, I have to divide that by
327
00:24:58,631 --> 00:25:02,228
D.
Now, I could write D in that
328
00:25:02,228 --> 00:25:06,445
forum, and I could do that.
There's nothing wrong with
329
00:25:06,445 --> 00:25:09,151
that.
But, I can write it in a form
330
00:25:09,151 --> 00:25:12,175
which is a little bit more
transparent.
331
00:25:12,175 --> 00:25:16,870
We do know that omega minus and
omega plus will make D zero.
332
00:25:16,870 --> 00:25:20,848
So, you can write this,
then, in the following way.
333
00:25:20,848 --> 00:25:24,748
You can write here omega
squared minus omega minus
334
00:25:24,748 --> 00:25:30,000
squared times omega squared
minus omega plus squared.
335
00:25:30,000 --> 00:25:33,676
That must be the same as what I
have there because,
336
00:25:33,676 --> 00:25:36,397
you see, this one becomes omega
minus.
337
00:25:36,397 --> 00:25:39,926
And this goes to zero.
And then, this one becomes
338
00:25:39,926 --> 00:25:43,088
omega plus also.
So, it's a different way of
339
00:25:43,088 --> 00:25:45,882
writing.
It gives you a little bit more
340
00:25:45,882 --> 00:25:48,161
insight.
It reminds you that the
341
00:25:48,161 --> 00:25:52,132
downstairs will go to zero when
you hit those resonance
342
00:25:52,132 --> 00:25:55,220
frequencies.
And so, I will also write down
343
00:25:55,220 --> 00:25:57,132
C2, then.
This one is zero.
344
00:25:57,132 --> 00:26:02,379
I get minus this one.
So, I get plus two omega zero
345
00:26:02,379 --> 00:26:06,565
to the fourth times eta zero
divided by that same D.
346
00:26:06,565 --> 00:26:10,914
And, you can write this for it.
So, let me check that,
347
00:26:10,914 --> 00:26:15,428
see whether I'm happy with
that, two omega zero squared,
348
00:26:15,428 --> 00:26:20,188
eta zero, omega zero squared.
Well, it's omega squared that
349
00:26:20,188 --> 00:26:22,486
looked good.
This looks good.
350
00:26:22,486 --> 00:26:27,000
This looks good.
And, I have here I'm happy.
351
00:26:27,000 --> 00:26:33,000
352
00:26:33,000 --> 00:26:36,700
Our task, now,
is not to look at these
353
00:26:36,700 --> 00:26:40,099
equations, but to see through
them.
354
00:26:40,099 --> 00:26:45,700
And they are by no means
trivial what they're going to do
355
00:26:45,700 --> 00:26:50,299
as a function of omega.
It's an extraordinarily
356
00:26:50,299 --> 00:26:56,200
complicated dependence on omega.
And, I have plotted for you
357
00:26:56,200 --> 00:27:03,000
these values of C1 and C2 for
which you get an answer now.
358
00:27:03,000 --> 00:27:07,082
You also know C1 over C2 of
course because you know C1 and
359
00:27:07,082 --> 00:27:10,161
you know C2.
And, I'm going to show you this
360
00:27:10,161 --> 00:27:13,957
as a function of omega,
and then we will try to digest
361
00:27:13,957 --> 00:27:15,962
that together.
And this plot,
362
00:27:15,962 --> 00:27:20,259
like the other plot that I'll
show you later today will be on
363
00:27:20,259 --> 00:27:23,482
the 8.03 website.
They will be part of lecture
364
00:27:23,482 --> 00:27:25,631
notes.
So, you have to click on
365
00:27:25,631 --> 00:27:28,424
lecture notes.
And then you'll see these
366
00:27:28,424 --> 00:27:31,792
plots.
This is the first one,
367
00:27:31,792 --> 00:27:35,623
which is a double pendulum.
What is plotted here
368
00:27:35,623 --> 00:27:39,127
horizontally is omega divided by
omega zero.
369
00:27:39,127 --> 00:27:43,121
So, that's the omega zero that
we mentioned there.
370
00:27:43,121 --> 00:27:47,603
And so you see the first
resonance here is about at 0.76
371
00:27:47,603 --> 00:27:50,944
omega zero.
And, the second resonance here
372
00:27:50,944 --> 00:27:55,101
is at 1.85 omega zero.
And, what we plot here is the
373
00:27:55,101 --> 00:27:58,361
amplitude, C,
divided by eta zero because
374
00:27:58,361 --> 00:28:03,251
obviously if you make eta zero
larger, you expect that has an
375
00:28:03,251 --> 00:28:07,000
effect on the C's,
of course.
376
00:28:07,000 --> 00:28:10,702
You know, if you drive it to
the larger amplitude,
377
00:28:10,702 --> 00:28:14,178
of course object will also
respond accordingly.
378
00:28:14,178 --> 00:28:17,881
And so, therefore,
we have it as a function of eta
379
00:28:17,881 --> 00:28:20,224
zero.
You see the eta zero here?
380
00:28:20,224 --> 00:28:23,322
C1 is linearly proportional
with eta zero.
381
00:28:23,322 --> 00:28:26,269
C2 is linearly proportional to
eta zero.
382
00:28:26,269 --> 00:28:29,745
That's no surprise.
So, we divided by eta zero.
383
00:28:29,745 --> 00:28:33,901
When we plot it up there,
it means that the amplitude is
384
00:28:33,901 --> 00:28:39,652
in phase with the driver.
That's the way we have written
385
00:28:39,652 --> 00:28:42,304
it.
And if it is below the zero
386
00:28:42,304 --> 00:28:46,990
line, it means that the
amplitude to the object is out
387
00:28:46,990 --> 00:28:50,969
of phase with the driver.
That's all it means.
388
00:28:50,969 --> 00:28:56,097
That's the sine convention.
So, let us now look at this and
389
00:28:56,097 --> 00:28:59,899
try to digest it,
and use to some degree our
390
00:28:59,899 --> 00:29:06,000
intuition, and see whether that
agrees with our intuition.
391
00:29:06,000 --> 00:29:09,047
And, let us start when omega
goes to zero.
392
00:29:09,047 --> 00:29:11,574
And don't look now at the
solution.
393
00:29:11,574 --> 00:29:15,067
If omega is zero,
I have a double pendulum in my
394
00:29:15,067 --> 00:29:17,520
hand.
And, I'm going to move it to
395
00:29:17,520 --> 00:29:19,898
the left.
And, 25 years from now,
396
00:29:19,898 --> 00:29:23,466
I'm going to go back.
And, I move it to the right
397
00:29:23,466 --> 00:29:25,844
again.
So, the pendulum is always
398
00:29:25,844 --> 00:29:28,743
straight.
And it is clear that C1 and C2
399
00:29:28,743 --> 00:29:32,459
both must be eta zero.
And, you'd better believe it
400
00:29:32,459 --> 00:29:37,141
that if you substitute in there
omega zero, that's what you will
401
00:29:37,141 --> 00:29:40,544
find.
So, C1 must be C2,
402
00:29:40,544 --> 00:29:44,835
and must be eta zero.
And, it must even be plus eta
403
00:29:44,835 --> 00:29:47,667
zero.
It must be in phase with the
404
00:29:47,667 --> 00:29:49,727
driver.
And you see that,
405
00:29:49,727 --> 00:29:53,589
not the ratio,
but just C divided by eta zero.
406
00:29:53,589 --> 00:29:57,451
That is a ratio that is plotted
that plus one.
407
00:29:57,451 --> 00:30:02,000
And so, both are in phase with
the driver.
408
00:30:02,000 --> 00:30:07,225
So that's a trivial result.
It means that the pendulum,
409
00:30:07,225 --> 00:30:11,096
which is here,
25 years from now is here.
410
00:30:11,096 --> 00:30:14,000
That's omega zero,
almost zero.
411
00:30:14,000 --> 00:30:18,741
So, notice that we now know the
ratio, C2 over C1,
412
00:30:18,741 --> 00:30:23,483
which is plus one.
The ratio C1 over C2 or C2 over
413
00:30:23,483 --> 00:30:26,967
C1 are now entirely dictated by
this.
414
00:30:26,967 --> 00:30:32,000
Nothing to do with normal modes
anymore.
415
00:30:32,000 --> 00:30:35,886
Don't be surprised that C1 over
C2 is now plus one.
416
00:30:35,886 --> 00:30:40,082
Now, look what happens.
I'm going to increase my omega.
417
00:30:40,082 --> 00:30:44,590
And what you see is that the
red curve, which is the second
418
00:30:44,590 --> 00:30:49,020
object, it's the lowest one of
the two, is going to have a
419
00:30:49,020 --> 00:30:51,507
large amplitude than the top
one.
420
00:30:51,507 --> 00:30:55,160
You see, it already begins to
grow very rapidly.
421
00:30:55,160 --> 00:30:59,746
And, when you reach resonance,
the ratio is going to be plus
422
00:30:59,746 --> 00:31:02,989
2.4, of course.
That's obvious.
423
00:31:02,989 --> 00:31:06,087
Now, at resonance,
you get an infinite amplitude
424
00:31:06,087 --> 00:31:07,736
for each.
That, of course,
425
00:31:07,736 --> 00:31:10,571
is nonsense.
There is physically no meaning.
426
00:31:10,571 --> 00:31:14,329
So, you shouldn't really think
of it as going to infinity.
427
00:31:14,329 --> 00:31:16,703
For one thing,
if C1 became infinity,
428
00:31:16,703 --> 00:31:20,329
that means if this point here
ends up there in the hole,
429
00:31:20,329 --> 00:31:23,560
it's hard to argue that it's a
small angle, right?
430
00:31:23,560 --> 00:31:26,593
So in any case,
the solution wouldn't even hold
431
00:31:26,593 --> 00:31:29,890
apart from the fact,
of course, that there's always
432
00:31:29,890 --> 00:31:34,282
some damping.
And so, we never go completely
433
00:31:34,282 --> 00:31:38,163
off scale so to speak.
However, you'll see that when
434
00:31:38,163 --> 00:31:41,282
you drive the system,
the double pendulum,
435
00:31:41,282 --> 00:31:45,543
and you approach resonance,
but you will very quickly see
436
00:31:45,543 --> 00:31:50,260
that the racial of the amplitude
of the second one over the top
437
00:31:50,260 --> 00:31:53,532
one will grow,
will become one and one half,
438
00:31:53,532 --> 00:31:57,032
will become two,
and then in the limiting case,
439
00:31:57,032 --> 00:32:01,903
you will get the plus 2.4.
So, as omega goes up,
440
00:32:01,903 --> 00:32:06,980
you're going to see that C2
over C1 is going to be larger
441
00:32:06,980 --> 00:32:09,699
than one right there,
this one.
442
00:32:09,699 --> 00:32:13,144
And then, ultimately it will
reach 2.4.
443
00:32:13,144 --> 00:32:16,951
But that is that extreme case
of resonance.
444
00:32:16,951 --> 00:32:21,212
And when you look at the motion
of the pendulum,
445
00:32:21,212 --> 00:32:26,470
if you drive it somewhere here.
Notice that they're both in
446
00:32:26,470 --> 00:32:32,000
phase with the driver,
and C2 is larger than C1.
447
00:32:32,000 --> 00:32:35,553
So, what you'll see is this is
C1.
448
00:32:35,553 --> 00:32:40,507
And then, this is C2.
So, C2 is larger than C1.
449
00:32:40,507 --> 00:32:44,599
And, they are in phase with the
driver.
450
00:32:44,599 --> 00:32:50,307
And so, when they return,
you will see a pendulum like
451
00:32:50,307 --> 00:32:54,076
this.
That's the [sweet?] thing that
452
00:32:54,076 --> 00:33:00,000
you will see.
Now is something truly bizarre.
453
00:33:00,000 --> 00:33:02,727
I go a little higher in
frequency.
454
00:33:02,727 --> 00:33:07,190
I go over the resonance,
and I see here a point whereby
455
00:33:07,190 --> 00:33:11,570
the frequency happens to be
exactly the frequency of a
456
00:33:11,570 --> 00:33:15,371
single pendulum.
Omega divided by omega zero is
457
00:33:15,371 --> 00:33:17,851
one.
And the top one refuses to
458
00:33:17,851 --> 00:33:20,247
move.
But the bottom one does.
459
00:33:20,247 --> 00:33:24,710
The bottom one here has an
amplitude which is twice the
460
00:33:24,710 --> 00:33:30,000
amplitude of the driver.
Look, you see a two here.
461
00:33:30,000 --> 00:33:33,913
And, it is out of phase with
the driver.
462
00:33:33,913 --> 00:33:39,030
That is unimaginable.
It is unimaginable what you're
463
00:33:39,030 --> 00:33:43,043
going to see.
So, in omega is omega zero.
464
00:33:43,043 --> 00:33:47,658
C1 becomes zero.
C2 becomes minus two eta zero.
465
00:33:47,658 --> 00:33:52,073
And so, this pendulum is going
to look, then,
466
00:33:52,073 --> 00:33:55,685
as follows.
I'll make a drawing here.
467
00:33:55,685 --> 00:34:03,492
I have a little bit more space.
So, if my hand is here at eta
468
00:34:03,492 --> 00:34:08,069
zero, then number one will stand
still.
469
00:34:08,069 --> 00:34:13,850
It won't do anything.
But number two will have an
470
00:34:13,850 --> 00:34:20,474
amplitude which is twice eta
zero, but it's on the other
471
00:34:20,474 --> 00:34:23,967
side.
So this is two eta zero.
472
00:34:23,967 --> 00:34:28,905
And then, if you look half a
period later,
473
00:34:28,905 --> 00:34:35,513
it will look like this.
So, this is eta zero,
474
00:34:35,513 --> 00:34:42,335
and then this is two eta zero.
And, this one doesn't move.
475
00:34:42,335 --> 00:34:48,440
That's what it tells you.
How on Earth can the lower
476
00:34:48,440 --> 00:34:54,185
pendulum oscillate if the upper
one stands still?
477
00:34:54,185 --> 00:35:01,007
I want you to think about that.
I'm still having sleepless
478
00:35:01,007 --> 00:35:06,127
nights about it.
And so, maybe you'll have some
479
00:35:06,127 --> 00:35:08,396
too.
And, if you have some clever
480
00:35:08,396 --> 00:35:12,083
ideas, come to see me.
But the logical consequence of
481
00:35:12,083 --> 00:35:15,273
what we did is that this one
will stand still,
482
00:35:15,273 --> 00:35:19,315
and the other one will still
oscillate and be driven by my
483
00:35:19,315 --> 00:35:21,584
hand.
I have to keep moving this,
484
00:35:21,584 --> 00:35:23,853
otherwise this will go to
pieces.
485
00:35:23,853 --> 00:35:28,178
I must keep doing this all the
time at that frequency which is
486
00:35:28,178 --> 00:35:33,000
exactly the resonance frequency
of a single pendulum.
487
00:35:33,000 --> 00:35:36,783
It is that omega zero.
I must keep doing this,
488
00:35:36,783 --> 00:35:41,323
and this one does nothing.
And this one has doubled the
489
00:35:41,323 --> 00:35:44,097
swing of this,
it is out of phase.
490
00:35:44,097 --> 00:35:48,889
If, then, you go even higher,
then you'll see that the two
491
00:35:48,889 --> 00:35:53,849
objects will go out of phase.
The upper one will be in phase
492
00:35:53,849 --> 00:35:57,632
with the driver,
and the lower one will be out
493
00:35:57,632 --> 00:36:02,088
of phase with the driver.
And then you get that second
494
00:36:02,088 --> 00:36:05,619
resonance when things will get
out of hand.
495
00:36:05,619 --> 00:36:11,000
And you get that ratio,
-0.42 [back?] of course.
496
00:36:11,000 --> 00:36:16,488
I will want to demonstrate to
you the situation here and the
497
00:36:16,488 --> 00:36:20,209
situation to see whether they
make sense.
498
00:36:20,209 --> 00:36:23,651
And so, now,
I'm going to use a double
499
00:36:23,651 --> 00:36:27,744
pendulum, and drive it with my
own frequency,
500
00:36:27,744 --> 00:36:30,627
which I determine.
I'm the boss.
501
00:36:30,627 --> 00:36:35,186
I determine omega.
I'm not looking for normal mode
502
00:36:35,186 --> 00:36:39,000
solutions.
I determine omega.
503
00:36:39,000 --> 00:36:45,064
And I'm going to first drive it
with omega zero.
504
00:36:45,064 --> 00:36:49,580
Omega is about zero.
In other words,
505
00:36:49,580 --> 00:36:56,677
I'm going to drive it here.
What you're going to see is,
506
00:36:56,677 --> 00:37:03,000
of course, fantastic,
absolutely fantastic.
507
00:37:03,000 --> 00:37:07,892
It's hanging straight down now.
And I'm moving it over a
508
00:37:07,892 --> 00:37:11,806
distance of 1 foot.
So, eta zero is one foot.
509
00:37:11,806 --> 00:37:15,276
That C1 is one foot,
and C2 is one foot.
510
00:37:15,276 --> 00:37:18,656
And, they are in phase with the
driver.
511
00:37:18,656 --> 00:37:22,214
Physics works.
If I go a little higher up
512
00:37:22,214 --> 00:37:26,751
here, so I go somewhere here,
approaching resonance,
513
00:37:26,751 --> 00:37:32,000
then you'll see that C2 becomes
larger than C1.
514
00:37:32,000 --> 00:37:37,866
This is C1 and this is C1.
And, you get to see a picture
515
00:37:37,866 --> 00:37:42,880
which is very much like this.
And I'll try that.
516
00:37:42,880 --> 00:37:48,106
So, I'll drive it below
resonance, but not too far
517
00:37:48,106 --> 00:37:50,986
below.
And there you see it.
518
00:37:50,986 --> 00:37:56,746
You see, C1 is smaller than C2.
No longer one plus one.
519
00:37:56,746 --> 00:38:04,000
This was C1 over C2 is one.
That's no longer the case now.
520
00:38:04,000 --> 00:38:07,276
You really see that C2 is
getting ahead,
521
00:38:07,276 --> 00:38:11,393
ahead not in terms of phase
ahead, but in terms of
522
00:38:11,393 --> 00:38:15,762
amplitude, very clear.
Now I'm going to attempt to do
523
00:38:15,762 --> 00:38:19,627
the impossible.
And the impossible is to try to
524
00:38:19,627 --> 00:38:23,491
hit this point.
The point whereby the upper one
525
00:38:23,491 --> 00:38:27,104
stands still,
and whereby the lower one will
526
00:38:27,104 --> 00:38:32,061
have an amplitude which is twice
that of my hands but out of
527
00:38:32,061 --> 00:38:36,178
phase with my hands,
how on Earth can I ever drive
528
00:38:36,178 --> 00:38:40,883
this system with that frequency,
omega zero, which is the
529
00:38:40,883 --> 00:38:46,752
frequency of a single pendulum?
Well, maybe I can't.
530
00:38:46,752 --> 00:38:50,608
But I will try.
And the way that I'm going to
531
00:38:50,608 --> 00:38:55,077
try this is the following.
I know what the resonance
532
00:38:55,077 --> 00:38:57,969
frequency of a single pendulum
is.
533
00:38:57,969 --> 00:39:02,000
That's this.
I can feel it in my hands.
534
00:39:02,000 --> 00:39:06,204
I can feel it in my stomach.
I can feel it in my brain.
535
00:39:06,204 --> 00:39:10,643
I feel it all over my body.
I can burn this frequency into
536
00:39:10,643 --> 00:39:14,069
my chips here.
Then I can close my eyes while
537
00:39:14,069 --> 00:39:17,418
you are looking,
and generate that frequency
538
00:39:17,418 --> 00:39:22,323
which was burned here and drive
the system as a double pendulum.
539
00:39:22,323 --> 00:39:25,750
If I succeed,
you'll see that upper one stand
540
00:39:25,750 --> 00:39:28,397
still.
And the bottom one will have
541
00:39:28,397 --> 00:39:33,303
twice the amplitude of my hands.
So, the success of this depends
542
00:39:33,303 --> 00:39:38,131
exclusively on how accurately I
can burn this frequency into my
543
00:39:38,131 --> 00:39:43,020
chips.
So, you have to be quiet.
544
00:39:43,020 --> 00:39:47,610
So I'm going to count:
one, two, three,
545
00:39:47,610 --> 00:39:51,476
I'm burning now.
One, two, three,
546
00:39:51,476 --> 00:39:54,738
four, one, two,
three, four.
547
00:39:54,738 --> 00:40:01,022
I'm closing my eyes.
One, two, three,
548
00:40:01,022 --> 00:40:06,122
four, one, two,
three, four,
549
00:40:06,122 --> 00:40:11,977
one, two, three,
four, one, two,
550
00:40:11,977 --> 00:40:15,000
three, four,
one.
551
00:40:15,000 --> 00:40:24,444
You're not saying anything.
Didn't you see this one
552
00:40:24,444 --> 00:40:32,000
standing still?
Did you see it?
553
00:40:32,000 --> 00:40:36,121
Did you also see that the other
one had twice the amplitude of
554
00:40:36,121 --> 00:40:38,756
my hands and out of phase with
my hands?
555
00:40:38,756 --> 00:40:40,513
You didn't see that,
right?
556
00:40:40,513 --> 00:40:43,081
Admit it.
You didn't see it because you
557
00:40:43,081 --> 00:40:45,986
were not looking for it.
Especially for you,
558
00:40:45,986 --> 00:40:48,959
I'll do it again.
So, you really have to see,
559
00:40:48,959 --> 00:40:52,608
number one, that this one will
practically stand still,
560
00:40:52,608 --> 00:40:56,662
number two, but this one has
doubled the amplitude of my hand
561
00:40:56,662 --> 00:40:59,702
but out of phase.
The decay time of burning is
562
00:40:59,702 --> 00:41:06,153
only one minute.
So I have to burn it again.
563
00:41:06,153 --> 00:41:11,076
One, two, three,
four, five, one,
564
00:41:11,076 --> 00:41:16,307
two, three, four,
five, one, oh-oh.
565
00:41:16,307 --> 00:41:25,076
Yeah, these things happen.
You have to start all over with
566
00:41:25,076 --> 00:41:30,000
the burning.
There we go.
567
00:41:30,000 --> 00:41:37,946
One, two, three,
four, five, one,
568
00:41:37,946 --> 00:41:45,892
two, three, four,
five, one, two,
569
00:41:45,892 --> 00:41:52,597
three, four,
five, one, two,
570
00:41:52,597 --> 00:41:59,798
three, are you seeing it?
Yes.
571
00:41:59,798 --> 00:42:07,000
All right.
[APPLAUSE]
572
00:42:07,000 --> 00:42:09,917
This is an ideal moment for the
break.
573
00:42:09,917 --> 00:42:14,413
I'm going to hand out the mini
quiz and we will reconvene.
574
00:42:14,413 --> 00:42:18,987
I'll give you six minutes this
time so you can even stretch
575
00:42:18,987 --> 00:42:23,324
your legs, and I would like some
help handing these out.
576
00:42:23,324 --> 00:42:27,977
And then you bring it back and
I will put boxes out there so
577
00:42:27,977 --> 00:42:32,000
that you can help me handing it
out here.
578
00:42:32,000 --> 00:42:36,291
You can start right away,
hand these out here.
579
00:42:36,291 --> 00:42:42,013
For those of you who have no
seats, come forward and get some
580
00:42:42,013 --> 00:42:45,256
seats.
Nicole, we're still friends,
581
00:42:45,256 --> 00:42:48,498
right?
So, why don't you hand those
582
00:42:48,498 --> 00:42:51,645
out?
And, why don't you hand these
583
00:42:51,645 --> 00:42:55,365
out here?
You can also give it to people
584
00:42:55,365 --> 00:42:58,226
here.
You can start right away.
585
00:42:58,226 --> 00:43:03,566
[SOUND OFF/THEN ON] I'm now
going to do something perhaps
586
00:43:03,566 --> 00:43:09,692
even more ambitious.
And I'm going to now couple
587
00:43:09,692 --> 00:43:13,282
three oscillators,
not pendulums yet.
588
00:43:13,282 --> 00:43:19,165
But I'm going to couple three
oscillators, which I connected
589
00:43:19,165 --> 00:43:23,552
with four springs.
I'm going to work on this.
590
00:43:23,552 --> 00:43:27,441
Three masses equal masses.
Four springs,
591
00:43:27,441 --> 00:43:32,028
spring constant K,
and the spring constants are
592
00:43:32,028 --> 00:43:35,618
the same.
And I'm going to drive that
593
00:43:35,618 --> 00:43:37,911
system.
One, two, three,
594
00:43:37,911 --> 00:43:43,785
four, and this is the end.
In other words,
595
00:43:43,785 --> 00:43:49,619
I have here a spring.
And, here is the first mass,
596
00:43:49,619 --> 00:43:54,142
second mass,
third mass, and here it is
597
00:43:54,142 --> 00:43:58,428
fixed.
And I'm going to drive it here
598
00:43:58,428 --> 00:44:04,142
with a displacement,
eta, which is eta zero times
599
00:44:04,142 --> 00:44:10,030
cosine omega T.
So, at a random moment in time,
600
00:44:10,030 --> 00:44:14,428
this is where my hand will be.
So this is eta.
601
00:44:14,428 --> 00:44:17,947
This is where the first mass
will be.
602
00:44:17,947 --> 00:44:23,127
Remember, you always call the
displacement X1 from its
603
00:44:23,127 --> 00:44:26,646
equilibrium.
That is its dotted line.
604
00:44:26,646 --> 00:44:30,165
Here is the spring.
This one is here.
605
00:44:30,165 --> 00:44:34,529
So I call this X2.
Here is the spring.
606
00:44:34,529 --> 00:44:37,428
And this one is here.
So, this is X3.
607
00:44:37,428 --> 00:44:40,971
So, here's the spring and here
is the spring.
608
00:44:40,971 --> 00:44:45,802
You may have noticed more than
once now that I have a certain
609
00:44:45,802 --> 00:44:50,472
discipline that I always offset
them in the same direction.
610
00:44:50,472 --> 00:44:52,566
Do you have to do that?
No.
611
00:44:52,566 --> 00:44:56,592
If you don't do it,
your chance of a mistake on the
612
00:44:56,592 --> 00:45:01,584
sign slip is much larger than if
you always set them off in the
613
00:45:01,584 --> 00:45:06,267
same direction.
You'll see shortly why.
614
00:45:06,267 --> 00:45:11,000
So that's certainly something
that is not a must.
615
00:45:11,000 --> 00:45:16,816
But, it's a smart thing to do.
I do find this is my positive
616
00:45:16,816 --> 00:45:19,774
direction.
But that, of course,
617
00:45:19,774 --> 00:45:24,408
is [free?] choice.
Now, at the situation at this
618
00:45:24,408 --> 00:45:28,450
moment in time,
let X1 be larger than eta.
619
00:45:28,450 --> 00:45:34,565
Let X2 be larger than X1.
And let X3 be larger than X2.
620
00:45:34,565 --> 00:45:39,343
And, this assumption will have
no consequences for what
621
00:45:39,343 --> 00:45:43,943
follows, at least not for the
differential equations.
622
00:45:43,943 --> 00:45:48,544
If X1 is larger than eta,
that first spring is longer
623
00:45:48,544 --> 00:45:53,764
than it wants to be because I
have assumed that X1 is larger
624
00:45:53,764 --> 00:45:57,126
than eta.
And so, that means there will
625
00:45:57,126 --> 00:46:02,080
be a force in this direction
because the spring is longer
626
00:46:02,080 --> 00:46:07,434
than it wants to be.
If X2 is larger than X1,
627
00:46:07,434 --> 00:46:12,000
the spring is also longer than
it wants to be.
628
00:46:12,000 --> 00:46:16,463
[NOISE OBSCURES] So,
there is a force in this
629
00:46:16,463 --> 00:46:20,521
direction.
And so, I can write down now a
630
00:46:20,521 --> 00:46:24,681
differential equation for my
first object.
631
00:46:24,681 --> 00:46:28,536
So that's going to be N,
X1 double dot.
632
00:46:28,536 --> 00:46:34,217
That equals minus K times X1
minus eta because that's the
633
00:46:34,217 --> 00:46:40,000
amount by which it is longer it
wants to be.
634
00:46:40,000 --> 00:46:45,631
So, times X1 minus eta,
that is this force.
635
00:46:45,631 --> 00:46:51,530
And, this force is now in the
plus direction.
636
00:46:51,530 --> 00:46:57,966
It's plus K times,
the spring here is longer than
637
00:46:57,966 --> 00:47:04,000
a wants to be by an amount X2
minus X1.
638
00:47:04,000 --> 00:47:05,844
Not omega one,
but X1.
639
00:47:05,844 --> 00:47:10,413
That's my differential equation
for the first object.
640
00:47:10,413 --> 00:47:15,596
And this one is always correct
even if X1 is not larger than
641
00:47:15,596 --> 00:47:21,043
eta because if X1 is not larger
than eta, then this force flips
642
00:47:21,043 --> 00:47:24,206
over.
Well, this will also flip over.
643
00:47:24,206 --> 00:47:28,950
So, that's why it's always
kosher and advisable to make
644
00:47:28,950 --> 00:47:32,728
that ascension to start with
because, again,
645
00:47:32,728 --> 00:47:38,000
it reduces the probability of
making mistakes.
646
00:47:38,000 --> 00:47:40,826
That's all.
There's nothing else to add.
647
00:47:40,826 --> 00:47:43,507
Just reduce the chance of
slipping up.
648
00:47:43,507 --> 00:47:45,753
So, let's now go to this
object.
649
00:47:45,753 --> 00:47:49,739
If this spring is longer than a
wants to be, it wants to
650
00:47:49,739 --> 00:47:52,202
contract.
So, this object we'll see
651
00:47:52,202 --> 00:47:55,608
forced to the left.
But, if the spring is longer
652
00:47:55,608 --> 00:47:59,086
than a wants to be,
because X3 is larger than X2,
653
00:47:59,086 --> 00:48:03,000
it will experience a force to
the right.
654
00:48:03,000 --> 00:48:06,840
So, I can write down,
now, a differential equation
655
00:48:06,840 --> 00:48:09,897
for object number two and X2
double dot.
656
00:48:09,897 --> 00:48:14,522
Notice that the one that is
here to the left is the same one
657
00:48:14,522 --> 00:48:18,833
that is here to the right.
Action equals minus reaction.
658
00:48:18,833 --> 00:48:21,498
This pool is the same as this
pool.
659
00:48:21,498 --> 00:48:26,123
So, it is going to be this term
which is now the minus sign.
660
00:48:26,123 --> 00:48:30,669
And, you always see that in
coupled oscillators that was as
661
00:48:30,669 --> 00:48:36,000
a plus here is going to come out
here as a minus sign.
662
00:48:36,000 --> 00:48:41,385
You see that comes out nicely
because the spring is longer
663
00:48:41,385 --> 00:48:47,338
than a wants to be by an amount
X2 minus X1, and the force is in
664
00:48:47,338 --> 00:48:52,157
the minus direction.
And this one is now going to be
665
00:48:52,157 --> 00:48:57,448
plus K times X3 minus X2.
So, now I go to the next spring
666
00:48:57,448 --> 00:49:02,291
to the next object.
So, this object here will
667
00:49:02,291 --> 00:49:07,791
experience a force to the left.
This spring is longer than it
668
00:49:07,791 --> 00:49:11,183
wants to be.
So, it wants to contract.
669
00:49:11,183 --> 00:49:14,758
But, this one is pushing.
So, therefore,
670
00:49:14,758 --> 00:49:19,891
the force due to the spring is
now also in this direction
671
00:49:19,891 --> 00:49:25,025
because the [end?] is fixed.
And so, we get for the third
672
00:49:25,025 --> 00:49:30,066
object M X3 double dot equals
minus K times X3 minus X2,
673
00:49:30,066 --> 00:49:35,005
which is this term.
But it switches signs.
674
00:49:35,005 --> 00:49:39,108
And then, in addition,
I get minus K times X3.
675
00:49:39,108 --> 00:49:43,393
When you reach this point on an
exam, you pause.
676
00:49:43,393 --> 00:49:48,225
You take a deep breath,
and you go over every term and
677
00:49:48,225 --> 00:49:51,689
every sign.
If you slip up on one sign,
678
00:49:51,689 --> 00:49:56,794
one casual mistake that you
just, even though you know it
679
00:49:56,794 --> 00:50:00,168
you casually write here,
for instance,
680
00:50:00,168 --> 00:50:05,000
a one instead of a three,
it's all over.
681
00:50:05,000 --> 00:50:08,681
You are dead in the water.
The problem will fall apart,
682
00:50:08,681 --> 00:50:12,363
and it may not even oscillate
in a simple harmonic way.
683
00:50:12,363 --> 00:50:14,477
So, therefore,
let's look at it.
684
00:50:14,477 --> 00:50:17,000
MX1 double dot.
X1 is larger than eta.
685
00:50:17,000 --> 00:50:19,795
Therefore, the force is in this
direction.
686
00:50:19,795 --> 00:50:21,431
I love it.
The other one:
687
00:50:21,431 --> 00:50:24,090
in this direction.
Perfect, X2 minus X1.
688
00:50:24,090 --> 00:50:27,840
That same force here is going
to pull on the second one.
689
00:50:27,840 --> 00:50:32,000
So, if this is correct,
this is also correct.
690
00:50:32,000 --> 00:50:35,913
This one is driving it away
from equilibrium,
691
00:50:35,913 --> 00:50:38,849
X3 minus X2.
It's got to be right.
692
00:50:38,849 --> 00:50:43,830
This term shows up here with a
minus sign, can't go wrong
693
00:50:43,830 --> 00:50:47,121
there.
And since this spring is always
694
00:50:47,121 --> 00:50:50,768
shorter here,
if it's pushed to the right,
695
00:50:50,768 --> 00:50:54,415
I am happy with my differential
equations.
696
00:50:54,415 --> 00:50:58,862
So now, we are going to
substitute in here X1 is C1
697
00:50:58,862 --> 00:51:05,000
cosine omega T trial functions
X2 is C2 cosine omega T.
698
00:51:05,000 --> 00:51:09,637
And, X3 is C3 cosine omega T.
I was looking for omegas,
699
00:51:09,637 --> 00:51:12,556
oh no.
Oh no, omega is given by me.
700
00:51:12,556 --> 00:51:17,623
I'm telling you what omega is.
You're not going to negotiate
701
00:51:17,623 --> 00:51:21,058
that with me.
We are only solving for C1,
702
00:51:21,058 --> 00:51:24,063
C2, and C3.
And in the steady state,
703
00:51:24,063 --> 00:51:29,130
you won't be able to do that
because omega is nonnegotiable.
704
00:51:29,130 --> 00:51:33,853
You're going to get three
equations with three unknowns:
705
00:51:33,853 --> 00:51:37,820
C1, C2, C3.
You don't have to settle to
706
00:51:37,820 --> 00:51:40,788
only calculate the ratios with
the amplitude.
707
00:51:40,788 --> 00:51:43,755
No, you're going to get a real
answer for C1,
708
00:51:43,755 --> 00:51:47,397
for C2, and for C3 which,
of course, will depend on eta
709
00:51:47,397 --> 00:51:49,353
zero.
Sure, you know eta zero.
710
00:51:49,353 --> 00:51:51,713
Do we worry about phase angles
here?
711
00:51:51,713 --> 00:51:53,804
No, because there is no
damping.
712
00:51:53,804 --> 00:51:57,783
And, if there is no damping,
either the objects are in phase
713
00:51:57,783 --> 00:52:01,830
or they are out of phase because
it is the damping that gives
714
00:52:01,830 --> 00:52:07,072
these phase angles in between.
And 180° out of phase is a
715
00:52:07,072 --> 00:52:09,863
minus sign.
So, we have the power to
716
00:52:09,863 --> 00:52:13,451
introduce some [NOISE OBSCURES]
phase changes.
717
00:52:13,451 --> 00:52:17,198
And, zero phase,
for that we have plus and minus
718
00:52:17,198 --> 00:52:19,988
signs.
Now, you are going to do some
719
00:52:19,988 --> 00:52:22,539
grinding.
I did all the grinding,
720
00:52:22,539 --> 00:52:25,330
every detail on the double
pendulum.
721
00:52:25,330 --> 00:52:28,200
Now, you're going to do the
grinding.
722
00:52:28,200 --> 00:52:32,585
However, I want to make sure
that if you go through that
723
00:52:32,585 --> 00:52:37,289
effort to make the grinding,
that you indeed end up with the
724
00:52:37,289 --> 00:52:41,562
right solution.
So, in that sense,
725
00:52:41,562 --> 00:52:46,250
I'm going to help you a little
bit by giving you the D,
726
00:52:46,250 --> 00:52:49,115
which is the D that we have
there.
727
00:52:49,115 --> 00:52:52,067
But, you have to bring me to
the D.
728
00:52:52,067 --> 00:52:56,929
And so, the D is going to be,
so you have to divide by M.
729
00:52:56,929 --> 00:53:00,575
You have to also,
omega S squared K over M,
730
00:53:00,575 --> 00:53:04,916
shorthand notation,
some of you may want to call it
731
00:53:04,916 --> 00:53:09,006
omega zero.
That's fine because there's
732
00:53:09,006 --> 00:53:12,126
only one spring.
There are no pendulums and
733
00:53:12,126 --> 00:53:16,659
springs, but I still call it an
S to remind you that it is the
734
00:53:16,659 --> 00:53:19,482
resonance frequency of a single
spring.
735
00:53:19,482 --> 00:53:22,306
Then, my D becomes minus omega
squared.
736
00:53:22,306 --> 00:53:26,021
That's always the result of
that second derivative,
737
00:53:26,021 --> 00:53:28,993
remember?
You always get that minus omega
738
00:53:28,993 --> 00:53:33,716
squared route.
And, you get plus two omega S
739
00:53:33,716 --> 00:53:36,900
squared.
Then you get in the second
740
00:53:36,900 --> 00:53:42,519
column minus omega S squared,
and in the third column you get
741
00:53:42,519 --> 00:53:46,078
a zero.
No surprise that you get a zero
742
00:53:46,078 --> 00:53:51,885
in the third column because the
first differential equation has
743
00:53:51,885 --> 00:53:57,410
no connection with X3 at all.
And so, you never see anything
744
00:53:57,410 --> 00:54:02,000
in the third column that will be
zero.
745
00:54:02,000 --> 00:54:07,239
But, if you look at the second
differential equation that has
746
00:54:07,239 --> 00:54:12,129
an X1, X2, and an X3 in it.
So, now you don't see zeroes.
747
00:54:12,129 --> 00:54:17,107
And so, what you're going to
see is minus omega S squared.
748
00:54:17,107 --> 00:54:22,259
That's going to be the C1 term.
And then, you get here minus
749
00:54:22,259 --> 00:54:26,101
omega squared.
I think it is plus two omega S
750
00:54:26,101 --> 00:54:29,507
squared.
And, your last problem is going
751
00:54:29,507 --> 00:54:36,115
to be minus omega S squared.
Now, the third differential
752
00:54:36,115 --> 00:54:42,576
equation, there is no X1.
Therefore, that is a zero here.
753
00:54:42,576 --> 00:54:47,076
And then, you get minus omega S
squared.
754
00:54:47,076 --> 00:54:53,769
And then, here you get minus
omega squared plus two omega S
755
00:54:53,769 --> 00:54:57,576
squared.
And, you have to take the
756
00:54:57,576 --> 00:55:03,000
determinant of this matrix.
That is D.
757
00:55:03,000 --> 00:55:06,921
Let me check it to make sure
that I didn't slip up with a
758
00:55:06,921 --> 00:55:10,702
minus sign, so when you get home
that you don't wonder,
759
00:55:10,702 --> 00:55:15,043
why didn't you get that result?
And, I think that looks good to
760
00:55:15,043 --> 00:55:17,004
me.
Remember, all those omega
761
00:55:17,004 --> 00:55:21,205
squares always come from those
second derivatives because you
762
00:55:21,205 --> 00:55:24,846
have to take the second
derivative of cosine omega T.
763
00:55:24,846 --> 00:55:28,067
That always brings out the
minus omega squared.
764
00:55:28,067 --> 00:55:32,129
So, no surprise that this is a
minus, that this is a minus,
765
00:55:32,129 --> 00:55:38,774
and that that is a minus.
So, now we want to know what C1
766
00:55:38,774 --> 00:55:42,306
is.
And, the first column will
767
00:55:42,306 --> 00:55:47,664
reflect this eta because the
right side, now,
768
00:55:47,664 --> 00:55:53,630
is not going to be zero,
remember, like the double
769
00:55:53,630 --> 00:55:58,501
pendulum.
So, you're going to get here in
770
00:55:58,501 --> 00:56:03,859
the first column,
you're going to get omega S
771
00:56:03,859 --> 00:56:12,982
squared times eta zero.
And then, you're going to get a
772
00:56:12,982 --> 00:56:19,892
zero and a zero.
And then, this column comes
773
00:56:19,892 --> 00:56:25,035
here.
And this column comes here.
774
00:56:25,035 --> 00:56:30,178
That is the determinant
upstairs.
775
00:56:30,178 --> 00:56:37,307
And you divide it by D.
That, then, is C1.
776
00:56:37,307 --> 00:56:44,615
And of course I will only go
one step further to go to C2,
777
00:56:44,615 --> 00:56:49,230
but that becomes a little
boring now.
778
00:56:49,230 --> 00:56:53,589
C2, then, you get,
that's this one.
779
00:56:53,589 --> 00:57:01,153
And then, the second column
[NOISE OBSCURES] omega S squared
780
00:57:01,153 --> 00:57:05,000
eta zero, zero,
zero.
781
00:57:05,000 --> 00:57:10,464
And then the third one is going
to be this.
782
00:57:10,464 --> 00:57:15,408
And then you go,
and we divide it by D,
783
00:57:15,408 --> 00:57:20,353
of course.
And then, you can write down
784
00:57:20,353 --> 00:57:23,215
C3.
You're on your own.
785
00:57:23,215 --> 00:57:27,769
I'll help you.
So, when you do this,
786
00:57:27,769 --> 00:57:35,055
you could, if you wanted to,
first solve for what we call
787
00:57:35,055 --> 00:57:41,841
the resonance frequencies.
The resonance frequencies are
788
00:57:41,841 --> 00:57:44,761
the ones which are the normal
mode frequencies.
789
00:57:44,761 --> 00:57:47,873
That's a resonance.
And so, you may want to put in
790
00:57:47,873 --> 00:57:50,603
D equals zero.
So, you made that determinant
791
00:57:50,603 --> 00:57:54,666
equal zero, which gives you then
the three resonance frequencies,
792
00:57:54,666 --> 00:57:58,285
which earlier we would have
called normal modes in case we
793
00:57:58,285 --> 00:58:02,877
are not driving.
And so, for those of you who
794
00:58:02,877 --> 00:58:06,680
worked this out,
the lowest frequency,
795
00:58:06,680 --> 00:58:11,305
resonance frequency,
which is the normal mode,
796
00:58:11,305 --> 00:58:17,163
is two minus the square root of
two times omega S squared.
797
00:58:17,163 --> 00:58:21,891
The one that follows is two
omega zero squared.
798
00:58:21,891 --> 00:58:27,750
And, the one that follows then,
which I will call it omega
799
00:58:27,750 --> 00:58:33,402
plus-plus is going to be two
plus the square root of two
800
00:58:33,402 --> 00:58:38,412
times omega S squared.
None of these are,
801
00:58:38,412 --> 00:58:42,259
of course, [NOISE OBSCURES].
None of the resonance
802
00:58:42,259 --> 00:58:46,969
frequencies are coupled double
pendulums or intuitive because
803
00:58:46,969 --> 00:58:51,129
there's no way that you can even
look at this and say,
804
00:58:51,129 --> 00:58:53,406
oh yeah, of course.
Excuse me?
805
00:58:53,406 --> 00:58:56,938
What did I do wrong?
Yes, thank you very much.
806
00:58:56,938 --> 00:59:02,148
I have an omega zero square.
You deserve extra credit.
807
00:59:02,148 --> 00:59:06,790
I've called that omega S.
If you want to change all the
808
00:59:06,790 --> 00:59:09,885
omega S's in omega zero,
that's fine.
809
00:59:09,885 --> 00:59:15,214
But, you cannot have one omega
zero, and the other one omega S.
810
00:59:15,214 --> 00:59:20,458
Yes, so this is the square of
the frequency of a single spring
811
00:59:20,458 --> 00:59:24,068
oscillating mass M.
Thank you very much for
812
00:59:24,068 --> 00:59:28,280
pointing that out.
Ah, boy, you also deserve extra
813
00:59:28,280 --> 00:59:32,325
credit.
I really set you up with this,
814
00:59:32,325 --> 00:59:35,348
didn't I?
I wanted to people to get some
815
00:59:35,348 --> 00:59:37,596
extra credit.
You deserve one;
816
00:59:37,596 --> 00:59:41,317
come and see me later.
And who was the other one?
817
00:59:41,317 --> 00:59:43,023
You did one.
Thank you.
818
00:59:43,023 --> 00:59:47,596
Yeah, there is a square here.
There are more people than one
819
00:59:47,596 --> 00:59:51,627
who claimed this one.
All right, thank you very much.
820
00:59:51,627 --> 00:59:56,511
OK, so now, for any given value
of omega, for any given value of
821
00:59:56,511 --> 1:00:01,882
omega, you find three answers.
You get C1, you get C2,
822
1:00:01,882 --> 1:00:05,402
and you get C3 in the steady
state solution.
823
1:00:05,402 --> 1:00:10,068
Of course, you get infinite
amplitude, which is physically
824
1:00:10,068 --> 1:00:13,506
meaningless.
So, you've got to stay away in
825
1:00:13,506 --> 1:00:16,289
your solutions from the
infinities.
826
1:00:16,289 --> 1:00:21,037
But, you will see if you don't
come too close to infinities
827
1:00:21,037 --> 1:00:25,785
that the results that you get
are quite accurate for high Q
828
1:00:25,785 --> 1:00:27,422
systems.
And so, now,
829
1:00:27,422 --> 1:00:32,088
I will show you the three
amplitudes for this system which
830
1:00:32,088 --> 1:00:37,000
will also be put on the Web this
afternoon.
831
1:00:37,000 --> 1:00:41,568
And so that is the three-car
system which we have set up
832
1:00:41,568 --> 1:00:44,392
there.
The only difference with the
833
1:00:44,392 --> 1:00:49,043
previous one is we don't plot
here omega divided by omega
834
1:00:49,043 --> 1:00:53,445
zero, but the squares.
This plot was provided to me by
835
1:00:53,445 --> 1:00:57,681
Professor [Witslaws?],
who lectured 8.03 a year ago.
836
1:00:57,681 --> 1:01:02,000
I was a recitation instructor
at the time.
837
1:01:02,000 --> 1:01:05,007
And very kind of me that he
gave me this plot.
838
1:01:05,007 --> 1:01:08,550
He even added this at my
request, which was very nice.
839
1:01:08,550 --> 1:01:12,426
[Car?] one is the first car,
green, and then the car two is
840
1:01:12,426 --> 1:01:16,236
the red one, is the second car.
And this is supposed to be
841
1:01:16,236 --> 1:01:17,706
blue.
You don't see it,
842
1:01:17,706 --> 1:01:20,781
but in any case,
if you think it's black that's
843
1:01:20,781 --> 1:01:22,051
fine.
That is, then,
844
1:01:22,051 --> 1:01:24,190
the black line.
So, horizontally,
845
1:01:24,190 --> 1:01:28,000
it is the ratio of the
frequency squared.
846
1:01:28,000 --> 1:01:31,605
So, you see that the second
resonance is,
847
1:01:31,605 --> 1:01:36,293
indeed, at a two here.
[SOUND OFF/THEN ON] have there
848
1:01:36,293 --> 1:01:40,530
on the blackboard.
If we plot it above the zero,
849
1:01:40,530 --> 1:01:44,496
it means that it is in phase
with the driver.
850
1:01:44,496 --> 1:01:49,905
If we plotted below the zero,
it means that it's out of phase
851
1:01:49,905 --> 1:01:53,511
with the driver.
If now you look at this,
852
1:01:53,511 --> 1:01:58,649
then already in omega zero you
see something that is by no
853
1:01:58,649 --> 1:02:01,444
means intuitive.
Notice that C1,
854
1:02:01,444 --> 1:02:05,591
C2, and C3 are all
substantially lower than eta
855
1:02:05,591 --> 1:02:11,000
zero because this is in units of
eta zero.
856
1:02:11,000 --> 1:02:15,172
And they are not even the same.
They are all three different.
857
1:02:15,172 --> 1:02:17,466
Would I have anticipated that?
No.
858
1:02:17,466 --> 1:02:20,387
I wouldn't.
Maybe two would be the same for
859
1:02:20,387 --> 1:02:24,072
me, but not all three different.
And, that's the case.
860
1:02:24,072 --> 1:02:28,036
They're all three different.
When we approached resonance,
861
1:02:28,036 --> 1:02:32,000
things got out of hand.
All three are in phase.
862
1:02:32,000 --> 1:02:35,885
And when you just cross over
the first resonance,
863
1:02:35,885 --> 1:02:40,579
they are all three again in
phase but out of phase with the
864
1:02:40,579 --> 1:02:43,655
driver.
That's not so surprising all by
865
1:02:43,655 --> 1:02:45,921
itself.
But now, look at this
866
1:02:45,921 --> 1:02:49,806
ridiculous point.
If I drive that system with the
867
1:02:49,806 --> 1:02:54,420
resonance frequency of the
individual spring with one mass
868
1:02:54,420 --> 1:02:57,819
on it because,
remember, if this squared is
869
1:02:57,819 --> 1:03:03,000
one, then omega divided by omega
zero is also one.
870
1:03:03,000 --> 1:03:06,944
So that's exactly the square
root of K over M.
871
1:03:06,944 --> 1:03:09,662
Then, this one will stand
still.
872
1:03:09,662 --> 1:03:13,694
And, these two have roughly the
same amplitude.
873
1:03:13,694 --> 1:03:16,587
You could sort of eyeball it
here.
874
1:03:16,587 --> 1:03:20,269
It looks like it almost crosses
over there.
875
1:03:20,269 --> 1:03:25,353
And it's about minus eta zero
because it's about minus one.
876
1:03:25,353 --> 1:03:30,000
So, let me write that on the
blackboards.
877
1:03:30,000 --> 1:03:33,250
That's a quite bizarre
situation.
878
1:03:33,250 --> 1:03:39,041
So, at that one frequency,
so omega equals the square root
879
1:03:39,041 --> 1:03:42,393
of K over M.
I get C1 equals zero.
880
1:03:42,393 --> 1:03:46,965
And, my C2 is about C3,
maybe even exactly C3.
881
1:03:46,965 --> 1:03:52,044
I never checked that,
and that is roughly minus eta
882
1:03:52,044 --> 1:03:56,311
one, eta zero.
You can just see that there.
883
1:03:56,311 --> 1:04:02,000
So, it means that this car will
stand still.
884
1:04:02,000 --> 1:04:07,294
It's closest to the driver.
These two are in phase with
885
1:04:07,294 --> 1:04:10,725
each other, have the same
amplitude.
886
1:04:10,725 --> 1:04:14,745
But they are out of phase with
the driver.
887
1:04:14,745 --> 1:04:17,588
Crazy, got to be wrong,
right?
888
1:04:17,588 --> 1:04:22,000
I mean, how on Earth can this
one stand still?
889
1:04:22,000 --> 1:04:25,921
And these happily go
hand-in-hand and go,
890
1:04:25,921 --> 1:04:30,725
it can't be right.
But I will demonstrate you that
891
1:04:30,725 --> 1:04:37,000
it is right at lease to a high
degree of accuracy.
892
1:04:37,000 --> 1:04:40,759
I'm going to concentrate on two
points in that graph.
893
1:04:40,759 --> 1:04:44,879
The first point that I want to
concentrate on is this one.
894
1:04:44,879 --> 1:04:47,120
It's the hardest one of the
two.
895
1:04:47,120 --> 1:04:51,530
I'm going to drive this system
at a frequency that we think is
896
1:04:51,530 --> 1:04:54,566
almost there.
I'm going to turn it on right
897
1:04:54,566 --> 1:04:58,397
now because it will take three
to four minutes for the
898
1:04:58,397 --> 1:05:03,301
[trangents?] to die out.
And then in the mean time I'll
899
1:05:03,301 --> 1:05:05,755
explain what you're going to
see.
900
1:05:05,755 --> 1:05:08,977
So, I'm driving it now with
that frequency.
901
1:05:08,977 --> 1:05:12,122
So don't look at it now.
It looks chaotic.
902
1:05:12,122 --> 1:05:15,113
It's causing a ridiculous
trangent mode.
903
1:05:15,113 --> 1:05:18,795
This one is oscillating.
And it's not supposed to
904
1:05:18,795 --> 1:05:20,406
oscillate.
Look there.
905
1:05:20,406 --> 1:05:23,321
It's oscillating,
and it looks chaotic.
906
1:05:23,321 --> 1:05:28,000
The amplitudes are changing.
Just let it cook.
907
1:05:28,000 --> 1:05:33,072
Now, what are we going to see?
If the trangents die out,
908
1:05:33,072 --> 1:05:38,420
which can really take three
minutes, you will see that this
909
1:05:38,420 --> 1:05:41,740
is my eta zero.
This is two eta zero.
910
1:05:41,740 --> 1:05:45,429
You see this arrow?
That is two eta zero.
911
1:05:45,429 --> 1:05:49,394
That's the driver.
So, you can calibrate the
912
1:05:49,394 --> 1:05:52,345
amplitude eta zero.
You will see,
913
1:05:52,345 --> 1:05:57,141
then, that these two cars have
the same displacement,
914
1:05:57,141 --> 1:06:02,441
but out of phase.
So when this one goes to the
915
1:06:02,441 --> 1:06:05,929
right for me,
they go to the left for me,
916
1:06:05,929 --> 1:06:09,765
and vice versa.
They have an amplitude that's
917
1:06:09,765 --> 1:06:14,299
very closely the same.
And this one is going to stand
918
1:06:14,299 --> 1:06:17,525
still.
What you're going to see is not
919
1:06:17,525 --> 1:06:22,234
something truly spectacular,
but it's extremely subtle.
920
1:06:22,234 --> 1:06:27,553
You have never seen it before,
and I don't think this has ever
921
1:06:27,553 --> 1:06:32,000
been demonstrated in any lecture
hall.
922
1:06:32,000 --> 1:06:35,125
It's extremely subtle.
First of all,
923
1:06:35,125 --> 1:06:40,482
to have the patience to wait
four minutes for this to happen,
924
1:06:40,482 --> 1:06:46,017
number one, and the number two
to make such a daring prediction
925
1:06:46,017 --> 1:06:51,642
that this one can almost come to
a halt, and that the other will
926
1:06:51,642 --> 1:06:57,089
have an amplitude which is the
same as the driver but 180° out
927
1:06:57,089 --> 1:07:00,571
of phase.
So, be patient and it will pay
928
1:07:00,571 --> 1:07:03,993
off.
Don't fall asleep now.
929
1:07:03,993 --> 1:07:09,430
So let's look at this middle
one to see whether its amplitude
930
1:07:09,430 --> 1:07:14,051
is becoming constant.
As long as they amplitudes are
931
1:07:14,051 --> 1:07:17,495
not constant,
they are still trangents.
932
1:07:17,495 --> 1:07:21,663
Well, these two are already
going hand-in-hand.
933
1:07:21,663 --> 1:07:25,559
You see that?
And I would say very much with
934
1:07:25,559 --> 1:07:30,000
the same amplitudes.
And look at this.
935
1:07:30,000 --> 1:07:34,401
Hey, they are really out of
phase with each other.
936
1:07:34,401 --> 1:07:36,736
You see that?
Now this one.
937
1:07:36,736 --> 1:07:41,856
Oh, boy, that amplitude is
nowhere nearly as large as this
938
1:07:41,856 --> 1:07:44,730
one.
And yet, when we started it,
939
1:07:44,730 --> 1:07:49,670
it was even large isn't this
one because of the trangent
940
1:07:49,670 --> 1:07:52,185
phenomenon.
Look at this one.
941
1:07:52,185 --> 1:07:55,239
It's almost standing still
already.
942
1:07:55,239 --> 1:07:58,293
Can we find that precise
frequency?
943
1:07:58,293 --> 1:08:01,829
No.
We have to set the dial
944
1:08:01,829 --> 1:08:04,215
somehow.
We do the best we can.
945
1:08:04,215 --> 1:08:07,715
I think it's fantastic.
It's blowing my mind.
946
1:08:07,715 --> 1:08:10,738
I can't believe it.
Physics is working.
947
1:08:10,738 --> 1:08:14,159
This one is [UNINTELLIGIBLE]
standing still.
948
1:08:14,159 --> 1:08:18,295
If I had to estimate the
amplitude, I would say maybe
949
1:08:18,295 --> 1:08:21,875
1/10 of eta zero.
And these, you see the marks
950
1:08:21,875 --> 1:08:24,500
here.
This one is really eta zero.
951
1:08:24,500 --> 1:08:29,431
So, those of you who see these
marks, that is the same as this.
952
1:08:29,431 --> 1:08:35,000
It's really eta zero.
And, boy, they go hand in hand.
953
1:08:35,000 --> 1:08:40,296
Aren't you thrilled?
[APPLAUSE] Now I'm going to try
954
1:08:40,296 --> 1:08:45,800
to make you see this point.
So, I'm going to start it,
955
1:08:45,800 --> 1:08:51,719
and that I'll explain to you
what is so special about that
956
1:08:51,719 --> 1:08:56,080
point because,
again, we have to be patient
957
1:08:56,080 --> 1:09:02,000
and we have to wait for the
trangents to die out.
958
1:09:02,000 --> 1:09:07,388
I'm going to drive it very
close to resonance right below
959
1:09:07,388 --> 1:09:11,621
the resonance.
You get a huge amplitude of C1
960
1:09:11,621 --> 1:09:15,085
and C3.
But, look what C2 is going to
961
1:09:15,085 --> 1:09:17,491
do.
C2 is going to have an
962
1:09:17,491 --> 1:09:22,302
amplitude which you and I may
have thought is zero.
963
1:09:22,302 --> 1:09:28,171
Namely, the outer ones do this
in the middle ones stand still.
964
1:09:28,171 --> 1:09:30,564
No, no.
No, no.
965
1:09:30,564 --> 1:09:33,224
Look at one and three,
by the way.
966
1:09:33,224 --> 1:09:37,254
They go nuts already.
You see, this is one and this
967
1:09:37,254 --> 1:09:39,914
is three.
Now you see the trangent
968
1:09:39,914 --> 1:09:43,138
phenomenon.
And now it's picking up again
969
1:09:43,138 --> 1:09:47,329
so we have to be patient.
But there is something very
970
1:09:47,329 --> 1:09:51,037
special about this C2.
Why is this C2 not zero,
971
1:09:51,037 --> 1:09:56,035
which is what you would expect?
You would expect that the outer
972
1:09:56,035 --> 1:10:02,000
two go like this and the middle
one would just stand still.
973
1:10:02,000 --> 1:10:05,413
That's one answer.
Yes, of course.
974
1:10:05,413 --> 1:10:09,344
But isn't it so that this must
be zero?
975
1:10:09,344 --> 1:10:15,137
You don't believe that this
upstairs is going to be zero?
976
1:10:15,137 --> 1:10:19,275
Just say you don't believe it.
This zero.
977
1:10:19,275 --> 1:10:23,517
Why now, even though the
upstairs is zero,
978
1:10:23,517 --> 1:10:29,413
why do we get an amplitude
which is half eta zero but with
979
1:10:29,413 --> 1:10:34,346
a minus sign?
Because we get zero divided by
980
1:10:34,346 --> 1:10:37,553
zero because right at that
frequency here,
981
1:10:37,553 --> 1:10:41,072
there is a resonance frequency.
So, D is zero.
982
1:10:41,072 --> 1:10:45,296
So, you get an amazing
coincidence that the upstairs is
983
1:10:45,296 --> 1:10:49,832
zero, the downstairs is zero,
but that's what we have 18.01
984
1:10:49,832 --> 1:10:51,865
for.
The ratio is not zero.
985
1:10:51,865 --> 1:10:54,681
The ratio is minus one half eta
zero.
986
1:10:54,681 --> 1:11:00,000
And, that's what you see.
It's minus one half eta zero.
987
1:11:00,000 --> 1:11:02,432
Now, look, one in three are
happy.
988
1:11:02,432 --> 1:11:06,191
They oscillate happily out of
phase with each other.
989
1:11:06,191 --> 1:11:09,508
Number one has to be in phase
with the driver.
990
1:11:09,508 --> 1:11:11,646
It's doing that.
Number three,
991
1:11:11,646 --> 1:11:14,963
out of phase with the driver.
It's doing that.
992
1:11:14,963 --> 1:11:19,238
And look at the middle one.
The middle one for those of you
993
1:11:19,238 --> 1:11:23,218
who can see, the arrows are
given eta zero to eta zero.
994
1:11:23,218 --> 1:11:26,167
Its right half of it.
Just walk up to it,
995
1:11:26,167 --> 1:11:28,304
and you can see.
It's amazing,
996
1:11:28,304 --> 1:11:32,222
this.
We hit that right on the nose.
997
1:11:32,222 --> 1:11:34,888
The reason why this is easy is,
look.
998
1:11:34,888 --> 1:11:38,000
My omega doesn't have to be
precisely here.
999
1:11:38,000 --> 1:11:41,777
Even if it's a little bit to
the left, I'm still OK.
1000
1:11:41,777 --> 1:11:45,777
That amplitude of number two is
not changing very much.
1001
1:11:45,777 --> 1:11:48,888
You see that?
This is so nicely horizontal.
1002
1:11:48,888 --> 1:11:51,481
So, this for me was a piece of
cake.
1003
1:11:51,481 --> 1:11:53,555
This was hard.
Piece of cake.
1004
1:11:53,555 --> 1:11:56,000
So, you see there?
1005
1:11:56,000 --> 1:12:01,000
1006
1:12:01,000 --> 1:12:05,292
Now, I'm going to move to the
triple pendulum.
1007
1:12:05,292 --> 1:12:09,680
In other words,
we have now seen this structure
1008
1:12:09,680 --> 1:12:12,923
of my talk.
We first did the double
1009
1:12:12,923 --> 1:12:16,070
pendulum.
The double pendulum was,
1010
1:12:16,070 --> 1:12:19,123
I worked out all the way for
you.
1011
1:12:19,123 --> 1:12:23,701
Then we did the three cars with
the four springs.
1012
1:12:23,701 --> 1:12:28,089
I set it up for you so that you
can't go wrong,
1013
1:12:28,089 --> 1:12:34,923
go wrong, go wrong anymore.
I gave you the final solution,
1014
1:12:34,923 --> 1:12:39,210
and I demonstrated that indeed
it is working,
1015
1:12:39,210 --> 1:12:44,374
the way we calculate it.
Now, I'm going to simply show
1016
1:12:44,374 --> 1:12:47,784
you the result of a triple
pendulum.
1017
1:12:47,784 --> 1:12:53,630
And, the plots that I'm going
to show you will be on the Web,
1018
1:12:53,630 --> 1:12:58,892
but no calculations at all.
So, that triple pendulum or
1019
1:12:58,892 --> 1:13:04,738
that, yeah, let me put these
down so that we have not so much
1020
1:13:04,738 --> 1:13:10,000
[shadow?] on this board,
triple pendulum.
1021
1:13:10,000 --> 1:13:19,000
This is a triple pendulum.
And we are going to drive it
1022
1:13:19,000 --> 1:13:27,833
here: eta equals eta zero times
the cosine of omega T.
1023
1:13:27,833 --> 1:13:34,000
The top one is going to be
green.
1024
1:13:34,000 --> 1:13:37,530
The middle one is going to be
red.
1025
1:13:37,530 --> 1:13:44,164
And, the bottom one is going to
be blue even though it may look
1026
1:13:44,164 --> 1:13:47,374
black there.
I'll make it blue.
1027
1:13:47,374 --> 1:13:52,403
That is the color code that we
have on our plot,
1028
1:13:52,403 --> 1:13:57,004
which will be put on the Web
this afternoon.
1029
1:13:57,004 --> 1:14:00,957
So, here it comes.
There it is.
1030
1:14:00,957 --> 1:14:04,386
Horizontally,
we plot again omega divided by
1031
1:14:04,386 --> 1:14:09,012
omega zero where omega zero is
the square root of G over L,
1032
1:14:09,012 --> 1:14:13,000
the resonance frequency of a
single pendulum.
1033
1:14:13,000 --> 1:14:19,000
1034
1:14:19,000 --> 1:14:22,252
And vertically we do the same
thing.
1035
1:14:22,252 --> 1:14:27,176
We want C divided by eta zero.
Everything has the same
1036
1:14:27,176 --> 1:14:32,938
meaning, namely plus one means
in phase with the driver and the
1037
1:14:32,938 --> 1:14:38,234
same amplitude as eta zero,
and minus is out of phase with
1038
1:14:38,234 --> 1:14:42,844
the driver.
Now, if you take this pendulum,
1039
1:14:42,844 --> 1:14:47,706
and you move it with zero
frequency, you don't have to
1040
1:14:47,706 --> 1:14:52,293
know any differential equations.
That will be here.
1041
1:14:52,293 --> 1:14:55,779
And 20 years from now it will
be there.
1042
1:14:55,779 --> 1:15:00,000
And this separation is,
then, eta zero.
1043
1:15:00,000 --> 1:15:04,368
And so you expect that C1,
C2, and C3 will all three be
1044
1:15:04,368 --> 1:15:07,199
eta zero.
And, they will be plus eta
1045
1:15:07,199 --> 1:15:10,678
zero, not minus,
but plus, in phase with the
1046
1:15:10,678 --> 1:15:11,973
driver.
And look.
1047
1:15:11,973 --> 1:15:16,017
That's what you see there.
So that is by no means a
1048
1:15:16,017 --> 1:15:18,930
surprise.
You go closer to resonance,
1049
1:15:18,930 --> 1:15:23,298
and you see that the bottom
one, which is the blue one,
1050
1:15:23,298 --> 1:15:27,020
it looks black here,
is picking up an amplitude
1051
1:15:27,020 --> 1:15:31,307
which is larger than the middle
one, than the red one,
1052
1:15:31,307 --> 1:15:36,000
and the red one larger than the
top one.
1053
1:15:36,000 --> 1:15:40,577
So, you are going,
then, into the domain where you
1054
1:15:40,577 --> 1:15:43,941
are going to see something like
this.
1055
1:15:43,941 --> 1:15:47,771
So, this one:
C1, C2, C3, and then finally
1056
1:15:47,771 --> 1:15:52,629
when you hit resonance,
I do not know what the ratios
1057
1:15:52,629 --> 1:15:55,899
are.
And everything gets out of hand
1058
1:15:55,899 --> 1:15:59,262
anyhow.
Now there comes a point which
1059
1:15:59,262 --> 1:16:06,151
boggles the mind right here.
The top one is not moving.
1060
1:16:06,151 --> 1:16:11,727
It stands still.
And there is nothing really so
1061
1:16:11,727 --> 1:16:17,060
special about that frequency.
That frequency,
1062
1:16:17,060 --> 1:16:23,727
for which the top one stands
still, so one goes to zero,
1063
1:16:23,727 --> 1:16:30,636
that omega, if I try to eyeball
it, I would say it's about
1064
1:16:30,636 --> 1:16:37,327
0.75-0.77 times omega zero.
And somehow at that frequency
1065
1:16:37,327 --> 1:16:41,339
the top one will not move.
The other two will move.
1066
1:16:41,339 --> 1:16:44,870
They have an amplitude which is
not stunning.
1067
1:16:44,870 --> 1:16:48,000
It's nothing to write your
mother about.
1068
1:16:48,000 --> 1:16:51,530
But it is about eta zero.
It is a minus sign.
1069
1:16:51,530 --> 1:16:54,500
So, it's out of phase with the
driver.
1070
1:16:54,500 --> 1:16:57,950
And this one is a little more.
It's bizarre.
1071
1:16:57,950 --> 1:17:03,875
Could I demonstrate this?
No way on Earth can I generate
1072
1:17:03,875 --> 1:17:08,763
0.77 times omega zero.
I can burn into my chip omega
1073
1:17:08,763 --> 1:17:12,118
zero as I did.
But I cannot do 0.77.
1074
1:17:12,118 --> 1:17:14,897
There's no way.
So, forget it.
1075
1:17:14,897 --> 1:17:19,977
I have to disappoint you.
But now comes the good news.
1076
1:17:19,977 --> 1:17:24,674
Look at this point here.
That is a point where the
1077
1:17:24,674 --> 1:17:31,000
middle one will stand still.
And that is truly amazing.
1078
1:17:31,000 --> 1:17:37,372
The middle one stands still.
The top one has an amplitude of
1079
1:17:37,372 --> 1:17:41,585
about 0.7 eta zero.
I just eyeball that.
1080
1:17:41,585 --> 1:17:47,742
And the bottom one has an
amplitude of about minus 1.5 eta
1081
1:17:47,742 --> 1:17:51,630
zero.
Let me try to make a drawing of
1082
1:17:51,630 --> 1:17:55,519
that.
So, this now is an omega equals
1083
1:17:55,519 --> 1:17:58,759
omega zero.
So my hand is here,
1084
1:17:58,759 --> 1:18:05,843
which is eta zero.
The top one has an amplitude
1085
1:18:05,843 --> 1:18:10,921
which is about 0.7 eta zero,
this one.
1086
1:18:10,921 --> 1:18:15,176
So, it is roughly here in
phase.
1087
1:18:15,176 --> 1:18:21,352
This is connected.
So, this is roughly 0.7 eta
1088
1:18:21,352 --> 1:18:25,607
zero.
The next one stands still,
1089
1:18:25,607 --> 1:18:33,109
believe it or not here.
And then the next one is out of
1090
1:18:33,109 --> 1:18:39,027
phase with me and has roughly an
amplitude of one and a half
1091
1:18:39,027 --> 1:18:42,738
times this.
So, that's one and a half.
1092
1:18:42,738 --> 1:18:46,249
So, I call that -1.5 times eta
zero.
1093
1:18:46,249 --> 1:18:50,762
And then half a period later,
my hand is here.
1094
1:18:50,762 --> 1:18:55,778
And this object is here.
And this one stands still.
1095
1:18:55,778 --> 1:19:00,592
And this one is here.
Truly, these are almost not
1096
1:19:00,592 --> 1:19:05,363
believable.
And then, there is another
1097
1:19:05,363 --> 1:19:09,181
frequency whereby the top one
stands still.
1098
1:19:09,181 --> 1:19:13,363
I will attempt,
I will make a daring attempt to
1099
1:19:13,363 --> 1:19:17,909
aim for this solution.
And the reason why I can try
1100
1:19:17,909 --> 1:19:23,363
that is because the frequency is
omega zero, which I can burn
1101
1:19:23,363 --> 1:19:27,363
into my chips.
That's the last attempt I will
1102
1:19:27,363 --> 1:19:32,000
make today.
This is the triple pendulum.
1103
1:19:32,000 --> 1:19:37,887
So, I have to go through the
exercise of learning again,
1104
1:19:37,887 --> 1:19:44,416
the period, and then as I close
my eyes and drive it with this
1105
1:19:44,416 --> 1:19:51,267
frequency, the idea then is that
you will see the top one move in
1106
1:19:51,267 --> 1:19:56,191
phase with my hand.
The one below that wouldn't
1107
1:19:56,191 --> 1:20:00,366
stand still.
And then the one below that
1108
1:20:00,366 --> 1:20:04,005
will have an even larger
amplitude.
1109
1:20:04,005 --> 1:20:06,681
So, one, two,
three, four,
1110
1:20:06,681 --> 1:20:11,756
five, six, seven.
One, two, three,
1111
1:20:11,756 --> 1:20:14,721
four, five.
One, two, three,
1112
1:20:14,721 --> 1:20:17,686
four, five.
One, you see it?
1113
1:20:17,686 --> 1:20:22,518
Three, you see it?
Did you see that one stand
1114
1:20:22,518 --> 1:20:26,361
still?
Isn't it amazing that physics
1115
1:20:26,361 --> 1:20:27,020
works?
OK, see you Thursday.