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PROFESSOR: Welcome back.
00:00:23.140 --> 00:00:29.546
Today, we will do two problems
involving many oscillators,
00:00:29.546 --> 00:00:34.420
or at least several oscillators,
coupled to each other.
00:00:34.420 --> 00:00:40.850
Now, you will not be surprised,
from past experience,
00:00:40.850 --> 00:00:45.340
that in the situation
with several oscillators,
00:00:45.340 --> 00:00:49.700
we are going to end up with
many equations and a lot
00:00:49.700 --> 00:00:55.270
of grindy, tough
solving of equations.
00:00:55.270 --> 00:00:58.510
Already, with a single
harmonic oscillator,
00:00:58.510 --> 00:01:01.820
for example, with
damping, the equations
00:01:01.820 --> 00:01:04.370
have been pretty horrendous.
00:01:04.370 --> 00:01:09.020
Now, we considered a
couple of oscillators
00:01:09.020 --> 00:01:13.440
with all the details, with
damping, driven, et cetera,
00:01:13.440 --> 00:01:16.900
and it really
becomes complicated.
00:01:16.900 --> 00:01:19.230
Now, what we are
trying to do by doing
00:01:19.230 --> 00:01:23.460
problems is to get a feel,
an understanding what goes on
00:01:23.460 --> 00:01:29.160
in a given situation, so one
wants to focus on new features
00:01:29.160 --> 00:01:32.180
and try to simplify
the mathematics as
00:01:32.180 --> 00:01:34.120
much as possible.
00:01:34.120 --> 00:01:37.330
And so in the problems
are we doing now,
00:01:37.330 --> 00:01:42.710
I am intentionally making them
even more ideal than before.
00:01:42.710 --> 00:01:45.640
Consider the first problem.
00:01:45.640 --> 00:01:52.660
Here is a highly
idealized system
00:01:52.660 --> 00:01:56.680
which has 3 degrees of freedom.
00:01:56.680 --> 00:02:03.750
What we have is we have a single
mass, 2m, two identical springs
00:02:03.750 --> 00:02:06.920
attached to two
masses, each of mass m.
00:02:09.690 --> 00:02:11.680
This is completely idealized.
00:02:11.680 --> 00:02:14.550
We are assuming
there's no friction,
00:02:14.550 --> 00:02:17.510
that these springs
are ideal springs.
00:02:17.510 --> 00:02:19.080
In other words,
they're massless,
00:02:19.080 --> 00:02:23.230
obey Hooke's laws
with constant k.
00:02:23.230 --> 00:02:25.910
We are assuming the
system is constrained,
00:02:25.910 --> 00:02:28.520
so it can only move
in one direction,
00:02:28.520 --> 00:02:33.670
it can't move up and down
or in or out from the board.
00:02:33.670 --> 00:02:40.480
So as I say, it's an
idealized situation.
00:02:40.480 --> 00:02:46.970
And the question we want to
answer is, for such a system,
00:02:46.970 --> 00:02:50.860
what are the normal
mode frequencies?
00:02:50.860 --> 00:02:55.880
And you know from lectures
given by Professor Walter Lewin
00:02:55.880 --> 00:03:00.710
that when you have coupled
oscillators, in this case, 3,
00:03:00.710 --> 00:03:09.060
all right, one finds that there
are very special oscillations,
00:03:09.060 --> 00:03:13.150
which we call normal mode
oscillations, in which
00:03:13.150 --> 00:03:20.980
every part of the
system is oscillating
00:03:20.980 --> 00:03:24.095
with the same
frequency and phase.
00:03:28.720 --> 00:03:32.680
So first, we want to find
out for this system, what
00:03:32.680 --> 00:03:36.505
are those three normal
mode frequencies?
00:03:39.240 --> 00:03:44.100
And secondly, suppose
I take this system,
00:03:44.100 --> 00:03:49.630
place these masses at some
arbitrary positions, some
00:03:49.630 --> 00:03:53.870
maybe even moving, and
let go, how would I
00:03:53.870 --> 00:04:00.380
predict where each mass
would be at some later time?
00:04:00.380 --> 00:04:02.070
So those are the
two questions I want
00:04:02.070 --> 00:04:05.730
to answer today
for this problem.
00:04:10.010 --> 00:04:18.750
Now, in general, coupled
oscillator problems
00:04:18.750 --> 00:04:21.850
are quite difficult.
00:04:21.850 --> 00:04:25.300
But there are
situations in which
00:04:25.300 --> 00:04:29.890
one can use logic
alone, well, logic
00:04:29.890 --> 00:04:33.920
and our understanding
of coupled oscillators,
00:04:33.920 --> 00:04:38.280
to do most of the
solution by hand-waving
00:04:38.280 --> 00:04:41.320
and not do any calculations.
00:04:41.320 --> 00:04:45.020
In this first example I'm
doing in this problem,
00:04:45.020 --> 00:04:52.930
since the system has
a lot of symmetry,
00:04:52.930 --> 00:04:56.670
I know from experience
that under those conditions
00:04:56.670 --> 00:05:02.910
it might be possible to solve
this, as I say, by logic alone
00:05:02.910 --> 00:05:06.230
and our understanding
of coupled oscillators.
00:05:06.230 --> 00:05:09.380
So the issue is,
can I guess what
00:05:09.380 --> 00:05:13.835
will be the motion
in the normal modes?
00:05:16.480 --> 00:05:19.200
And the answer is, actually,
if you stop and think
00:05:19.200 --> 00:05:23.370
for a second, at
least the first two
00:05:23.370 --> 00:05:26.770
modes you can figure
out rather easily.
00:05:26.770 --> 00:05:30.270
So here I'm showing
you schematically
00:05:30.270 --> 00:05:34.130
what would be one of the
modes of oscillation,
00:05:34.130 --> 00:05:35.780
one of the normal modes.
00:05:35.780 --> 00:05:41.120
Imagine that I first
hold this mass fixed
00:05:41.120 --> 00:05:45.390
and I displace these
symmetrically, one
00:05:45.390 --> 00:05:48.565
to the right, one to
the left, and I let go.
00:05:51.400 --> 00:05:54.720
What you will find
is, by symmetry,
00:05:54.720 --> 00:05:59.470
this spring will always be
stretched by the same amount
00:05:59.470 --> 00:06:06.500
as this is compressed, and
so the force on this mass
00:06:06.500 --> 00:06:07.550
will be 0.
00:06:07.550 --> 00:06:13.210
This spring will be pushing this
mass exactly by the same amount
00:06:13.210 --> 00:06:16.290
as this one will be
pulling it and vice versa.
00:06:16.290 --> 00:06:19.145
So there will never be any
net force on this mass,
00:06:19.145 --> 00:06:22.420
and so they'll stay put.
00:06:22.420 --> 00:06:31.160
And these two will now behave
as a simple harmonic oscillator
00:06:31.160 --> 00:06:38.040
consisting of a spring attached
to a fixed wall and the mass m.
00:06:38.040 --> 00:06:40.420
So that's what this
one will be doing,
00:06:40.420 --> 00:06:42.070
and that's what
this will be doing.
00:06:42.070 --> 00:06:44.580
It will be out of
phase by 180 degrees.
00:06:44.580 --> 00:06:48.520
Or what one normally
says, it'll be in phase
00:06:48.520 --> 00:06:53.440
but the amplitude will
be minus the other one.
00:06:53.440 --> 00:06:56.490
This, by now, you've
had enough experience
00:06:56.490 --> 00:07:00.240
and I've done it here
before, I can just
00:07:00.240 --> 00:07:04.460
write what will be that
frequency of oscillation
00:07:04.460 --> 00:07:08.410
here, as we've seen many times.
00:07:08.410 --> 00:07:13.310
And, by the way, throughout
my solving the problems here,
00:07:13.310 --> 00:07:19.150
I do not differentiate frequency
and angular frequency, that you
00:07:19.150 --> 00:07:22.320
can see which I'm talking about
is whether I'm using omega
00:07:22.320 --> 00:07:25.500
or F. Omega is the
angular frequency,
00:07:25.500 --> 00:07:28.230
and there's just a fact
of 2 pi between the two.
00:07:28.230 --> 00:07:31.970
But I'll use, interchangeably,
I'll call this frequency.
00:07:31.970 --> 00:07:37.340
So this frequency of one of
the normal modes is just k/m.
00:07:37.340 --> 00:07:40.710
So we found one
normal mode frequency.
00:07:40.710 --> 00:07:42.850
Let's look for another one.
00:07:42.850 --> 00:07:48.376
We stare at this and we see,
sure, there is another one.
00:07:48.376 --> 00:07:55.760
Another one is where I take,
simply, these two masses
00:07:55.760 --> 00:07:59.540
and simultaneously pull
them out, in this mass,
00:07:59.540 --> 00:08:03.220
away and let go.
00:08:03.220 --> 00:08:07.730
Effectively, these two
will look like that.
00:08:07.730 --> 00:08:13.160
It's as if I had a mass of
2m attached to a mass of 2m
00:08:13.160 --> 00:08:18.790
by two springs, each
of spring constant k.
00:08:18.790 --> 00:08:23.730
And as I pull this back, this
will oscillate like this.
00:08:23.730 --> 00:08:26.470
Everything, both this
mass and this mass,
00:08:26.470 --> 00:08:31.570
will be oscillating with the
same frequency, same phase,
00:08:31.570 --> 00:08:33.789
although there's
this minus sign.
00:08:33.789 --> 00:08:39.600
I can always replace the
180-degree phase shift
00:08:39.600 --> 00:08:43.240
by saying the
amplitude is minus.
00:08:43.240 --> 00:08:47.170
So those are
identical statements.
00:08:47.170 --> 00:08:51.308
So you have these two masses
oscillating like that.
00:08:51.308 --> 00:08:52.670
OK?
00:08:52.670 --> 00:08:58.090
By the way, I've said
that in the normal mode
00:08:58.090 --> 00:09:02.040
I want everything to move
with the same frequency.
00:09:02.040 --> 00:09:04.170
Some of you may argue, hold on.
00:09:04.170 --> 00:09:05.680
Did I cheat on you?
00:09:05.680 --> 00:09:09.510
Here, this mass is
not oscillating.
00:09:09.510 --> 00:09:12.110
So am I contradicting myself?
00:09:12.110 --> 00:09:14.490
And the answer is no.
00:09:14.490 --> 00:09:16.230
It's a diabolical answer.
00:09:16.230 --> 00:09:18.970
But I'll say, no,
this is oscillating
00:09:18.970 --> 00:09:22.760
with this frequency,
but with 0 amplitude.
00:09:22.760 --> 00:09:24.970
And you can't tell
me that I'm wrong.
00:09:24.970 --> 00:09:27.810
It's not moving because
the amplitude is 0,
00:09:27.810 --> 00:09:32.360
but it is oscillating
with the same frequency.
00:09:32.360 --> 00:09:36.180
So now, what is the frequency
of this oscillation?
00:09:36.180 --> 00:09:38.550
Again, I can do it in my head.
00:09:38.550 --> 00:09:46.450
I can imagine that the middle
of this spring is not moving.
00:09:46.450 --> 00:09:52.650
And so this mass is a half
a length of the spring
00:09:52.650 --> 00:09:55.500
and a mass 2 attached
to it oscillating,
00:09:55.500 --> 00:09:57.440
and I can calculate
the frequency
00:09:57.440 --> 00:09:59.150
of oscillation of that.
00:09:59.150 --> 00:10:02.700
Or if you prefer, I
can do the following.
00:10:02.700 --> 00:10:09.180
I can move this mass out a
distance dx and this one dx
00:10:09.180 --> 00:10:12.030
to the other side
and calculate what
00:10:12.030 --> 00:10:14.450
will be the restoring force.
00:10:14.450 --> 00:10:19.890
Well, now each one has
a spring constant k.
00:10:19.890 --> 00:10:21.600
There is two of
them, so that's 2k.
00:10:25.140 --> 00:10:30.320
Although this mass has been
moved by dx, this by dx,
00:10:30.320 --> 00:10:35.030
the springs have been extended
by twice dx, so it's 2 dx.
00:10:35.030 --> 00:10:37.240
So that's the total
restoring force
00:10:37.240 --> 00:10:41.440
by Hooke's law, which
is equal minus 4k dx.
00:10:41.440 --> 00:10:42.610
All right?
00:10:42.610 --> 00:10:49.650
And so this will behave
like a single mass,
00:10:49.650 --> 00:10:52.880
like this system, a single
mass with a single spring
00:10:52.880 --> 00:10:56.770
where the spring has
a spring constant 4k.
00:10:56.770 --> 00:10:58.940
The mass is 2m.
00:10:58.940 --> 00:11:05.450
And so the angular frequency, or
frequency of this mode is 2k/m.
00:11:05.450 --> 00:11:07.660
So we've found two of them.
00:11:07.660 --> 00:11:12.330
Now, we know from our studies
of coupled oscillators
00:11:12.330 --> 00:11:15.685
that for a system of
3 degrees of freedom,
00:11:15.685 --> 00:11:20.940
so it would be three masses,
there are three normal modes.
00:11:20.940 --> 00:11:24.370
What is the third normal mode?
00:11:24.370 --> 00:11:27.210
And I intentionally
did not write it
00:11:27.210 --> 00:11:30.955
down here because I wanted
you for a second to think.
00:11:30.955 --> 00:11:35.920
Well, how else can
this system move such
00:11:35.920 --> 00:11:45.591
that every mass is moving in
phase with the same frequency?
00:11:45.591 --> 00:11:47.110
All right?
00:11:47.110 --> 00:11:50.810
And I tell you,
when I first saw it,
00:11:50.810 --> 00:11:54.290
it took me a long
time to figure out,
00:11:54.290 --> 00:12:00.030
and most people don't find it.
00:12:00.030 --> 00:12:03.400
And yet the answer
is extremely simple.
00:12:03.400 --> 00:12:08.430
This answer is there is
one more normal mode.
00:12:08.430 --> 00:12:17.600
It is one of almost infinite
amplitude, but 0 frequency.
00:12:17.600 --> 00:12:24.290
Just the opposite to here,
where this was 0 amplitude, OK,
00:12:24.290 --> 00:12:26.700
and the angular frequency
doesn't even matter.
00:12:26.700 --> 00:12:32.480
Here it has very large
amplitude, infinite in fact,
00:12:32.480 --> 00:12:34.100
but 0 frequency.
00:12:34.100 --> 00:12:38.770
What does such motion look
like at any instant of time?
00:12:38.770 --> 00:12:41.530
It's moving with
the same velocity.
00:12:41.530 --> 00:12:54.770
So if I take and if I
consider the motion where
00:12:54.770 --> 00:13:01.100
each one of these is moving
uniformly to the right
00:13:01.100 --> 00:13:04.240
with a constant
velocity, that is
00:13:04.240 --> 00:13:10.200
a normal mode with 0 frequency.
00:13:10.200 --> 00:13:18.240
So the last one is
omega C is equal to 0.
00:13:21.010 --> 00:13:21.920
OK?
00:13:21.920 --> 00:13:28.800
So we have now found
the three normal modes.
00:13:28.800 --> 00:13:31.000
OK.
00:13:31.000 --> 00:13:31.650
All right.
00:13:31.650 --> 00:13:38.960
Now that I've got a piece
of chalk, let's continue.
00:13:38.960 --> 00:13:41.980
So we've answered the first
part of the question, what
00:13:41.980 --> 00:13:44.490
are the normal mode
frequencies of the system?
00:13:44.490 --> 00:13:47.340
And I found you those
three normal modes.
00:13:47.340 --> 00:13:49.970
Now, the next question
we want to answer
00:13:49.970 --> 00:13:54.340
is, if, at any given
instant of time,
00:13:54.340 --> 00:13:57.930
I know the positions and the
velocities of the three masses,
00:13:57.930 --> 00:14:01.380
can I predict what
will happen at the end?
00:14:01.380 --> 00:14:04.690
That's equivalent
of saying, can I
00:14:04.690 --> 00:14:09.570
write equations for the
positions of the masses
00:14:09.570 --> 00:14:11.400
as a function of time?
00:14:11.400 --> 00:14:15.700
For that I need to define for
myself a coordinate system.
00:14:15.700 --> 00:14:18.585
So I'll say those are
those three masses.
00:14:23.120 --> 00:14:26.510
All the motion is
along the x direction,
00:14:26.510 --> 00:14:35.900
so there's just one variable,
one variable for each mass, x.
00:14:35.900 --> 00:14:38.080
So the position
of the first mass
00:14:38.080 --> 00:14:42.090
I'll call x1, position
of the second mass x2,
00:14:42.090 --> 00:14:43.760
and of the third is x3.
00:14:43.760 --> 00:14:46.960
And the origin of
coordinates I will
00:14:46.960 --> 00:14:51.070
take to be through the
center of each mass
00:14:51.070 --> 00:14:54.580
at a time when these
are in equilibrium.
00:14:54.580 --> 00:14:57.990
In other words, the spring
is unstretched, et cetera.
00:14:57.990 --> 00:14:58.880
OK.
00:14:58.880 --> 00:15:07.630
So I can now, using the
information we've obtained,
00:15:07.630 --> 00:15:17.060
write down the description
of each mass in each mode.
00:15:17.060 --> 00:15:23.360
So in mode A, I know that
the x1 is 0 all the time.
00:15:23.360 --> 00:15:26.690
It's not moving, OK?
00:15:26.690 --> 00:15:34.830
Now, in a normal mode, that's
what we mean by a normal mode,
00:15:34.830 --> 00:15:38.020
each mass is moving
with the same frequency
00:15:38.020 --> 00:15:42.240
and the same phase,
and it's oscillating.
00:15:42.240 --> 00:15:49.900
So there will be some sinusoidal
function of the mode frequency
00:15:49.900 --> 00:15:54.730
t plus this phase times
some arbitrary amplitude.
00:15:57.480 --> 00:16:00.960
The other one, the
last one, will also
00:16:00.960 --> 00:16:03.330
be oscillating in
the normal mode,
00:16:03.330 --> 00:16:05.025
so everything will be the same.
00:16:05.025 --> 00:16:07.930
It will be cosine of
omega A t plus phi
00:16:07.930 --> 00:16:11.320
A with some other amplitude.
00:16:11.320 --> 00:16:18.350
But from our analysis
here, we know
00:16:18.350 --> 00:16:23.810
that these two are
not both arbitrary.
00:16:23.810 --> 00:16:27.510
Because we said that
in the normal mode,
00:16:27.510 --> 00:16:35.360
it was the case where the
2m mass was stationary,
00:16:35.360 --> 00:16:37.970
but these were going like this.
00:16:37.970 --> 00:16:40.340
That was the first mode.
00:16:40.340 --> 00:16:46.080
So whenever this one was going
to the right a distance A,
00:16:46.080 --> 00:16:49.440
this one was going to
the left a distance A.
00:16:49.440 --> 00:16:52.830
So the overall
normalization is arbitrary.
00:16:52.830 --> 00:16:54.620
It can be anything.
00:16:54.620 --> 00:16:58.210
But these are not
independent of the other,
00:16:58.210 --> 00:17:01.240
because we discovered
that in that mode,
00:17:01.240 --> 00:17:06.589
this was the frequency,
and the amplitudes were
00:17:06.589 --> 00:17:09.849
opposite to each other.
00:17:09.849 --> 00:17:17.630
So this describes the
situation for this system
00:17:17.630 --> 00:17:22.650
if it's oscillating in
the first normal mode.
00:17:22.650 --> 00:17:26.540
It's the most general
description of that.
00:17:26.540 --> 00:17:29.490
Next, I will
describe what it will
00:17:29.490 --> 00:17:32.350
do in the second normal mode.
00:17:32.350 --> 00:17:34.880
And now I can go much quicker.
00:17:34.880 --> 00:17:37.270
This time, it'll have
the second normal mode
00:17:37.270 --> 00:17:40.750
frequency, different phase.
00:17:40.750 --> 00:17:44.980
But these have to be the same
because all of these masses
00:17:44.980 --> 00:17:49.910
are moving in the same normal
mode, same normal frequency.
00:17:49.910 --> 00:17:54.560
These amplitudes,
overall, it's arbitrary.
00:17:54.560 --> 00:17:55.960
I can make it anything.
00:17:55.960 --> 00:17:58.600
It'll satisfy the equations.
00:17:58.600 --> 00:18:03.520
But I know that in the
second normal mode,
00:18:03.520 --> 00:18:06.820
this one is moving in that
direction and those two
00:18:06.820 --> 00:18:10.220
in the opposite direction
with the same magnitude.
00:18:10.220 --> 00:18:14.220
So these two have to have
the same size and magnitude,
00:18:14.220 --> 00:18:16.620
but this has to be opposite.
00:18:16.620 --> 00:18:21.460
So this describes the
motion in the second mode.
00:18:21.460 --> 00:18:25.090
And finally, in the
third normal mode,
00:18:25.090 --> 00:18:29.160
we said that it was 0 frequency.
00:18:29.160 --> 00:18:30.280
All right?
00:18:30.280 --> 00:18:32.990
And all that was
happening, the three masses
00:18:32.990 --> 00:18:37.780
were moving with
uniform velocity.
00:18:37.780 --> 00:18:41.850
So the positions
will be some constant
00:18:41.850 --> 00:18:44.265
plus the velocity
of that system.
00:18:51.310 --> 00:18:55.030
So this describes the
system in the third mode,
00:18:55.030 --> 00:18:58.650
and these are the other two.
00:18:58.650 --> 00:19:17.870
Now, I know that, in general,
each one of these masses
00:19:17.870 --> 00:19:25.990
can move in a superposition
of the normal modes.
00:19:25.990 --> 00:19:37.390
So the most general
expression for x1 of t
00:19:37.390 --> 00:19:45.430
has to be of the form which is
the sum of its possible motion
00:19:45.430 --> 00:19:47.360
in each of the modes.
00:19:47.360 --> 00:19:51.660
So x1, so this is
the mass 2m, will
00:19:51.660 --> 00:19:56.320
have possible motion in
the first mode, which is 0.
00:19:56.320 --> 00:19:57.520
All right?
00:19:57.520 --> 00:20:13.570
So I'll write 0 plus B times
cosine omega B t plus phi
00:20:13.570 --> 00:20:21.920
B, all right, plus
C plus v. So that
00:20:21.920 --> 00:20:26.940
will be the most general
description of the motion
00:20:26.940 --> 00:20:30.880
of the big mass, the 2m one.
00:20:30.880 --> 00:20:34.310
The B is arbitrary,
this phase is arbitrary,
00:20:34.310 --> 00:20:37.990
this is arbitrary, and
v. Those quantities
00:20:37.990 --> 00:20:42.986
will be determined by
the initial conditions.
00:20:42.986 --> 00:20:44.280
Well, how about x2?
00:20:48.340 --> 00:21:08.190
Well, I know that if x1 is
0, x2 is A cosine omega A t
00:21:08.190 --> 00:21:26.250
plus phi A. When the x1 is this,
then x2 is minus B cosine omega
00:21:26.250 --> 00:21:37.270
B t plus phi B, OK,
and plus C plus v. OK?
00:21:37.270 --> 00:21:45.210
And finally, x3 of t,
again now just following
00:21:45.210 --> 00:21:50.360
the same pattern, this is
minus A cosine omega A t
00:21:50.360 --> 00:21:59.620
plus phi A. This one is
minus B cosine omega B
00:21:59.620 --> 00:22:15.940
t plus phi B plus C plus v.
00:22:15.940 --> 00:22:19.780
And I'm almost home.
00:22:19.780 --> 00:22:22.650
I have described the
motion of each one.
00:22:27.015 --> 00:22:31.620
For each particle,
it is oscillating
00:22:31.620 --> 00:22:34.410
in a superposition
of its normal modes.
00:22:38.230 --> 00:22:45.460
And the amplitudes are
arbitrary within the constraint
00:22:45.460 --> 00:22:50.650
that the ratios between
them is determined
00:22:50.650 --> 00:22:56.070
by the constraints
of the system.
00:22:56.070 --> 00:23:00.450
So now the question is,
can I predict the future?
00:23:00.450 --> 00:23:02.720
Well, in order to
predict the future,
00:23:02.720 --> 00:23:06.645
say, where will
particle 3 be at time t?
00:23:06.645 --> 00:23:11.876
I need to know the value of
these arbitrary constants.
00:23:14.420 --> 00:23:16.640
Omega is not an
arbitrary constant.
00:23:16.640 --> 00:23:17.740
We found it.
00:23:17.740 --> 00:23:19.630
Omega A is here.
00:23:19.630 --> 00:23:20.580
OK?
00:23:20.580 --> 00:23:26.370
This is arbitrary, this is
arbitrary, this, and that.
00:23:26.370 --> 00:23:30.670
So there are six
arbitrary constants.
00:23:30.670 --> 00:23:32.280
How do I find them?
00:23:32.280 --> 00:23:36.730
They are determined by
the initial conditions.
00:23:36.730 --> 00:23:40.420
And you know,
fortunately, or else it
00:23:40.420 --> 00:23:44.660
would mean we've made a mistake,
we have three conditions.
00:23:47.280 --> 00:23:51.560
At the beginning,
someone has to tell me
00:23:51.560 --> 00:23:58.000
where each mass was, the
location, and with what
00:23:58.000 --> 00:24:01.810
velocity is it
moving at that time.
00:24:01.810 --> 00:24:09.290
So I can write these
three equations at, say,
00:24:09.290 --> 00:24:14.920
with t equals 0, at t
equals 0, and equate this
00:24:14.920 --> 00:24:21.220
to the position of x1 of t
equals 0, wherever it is.
00:24:21.220 --> 00:24:26.220
When the t is 0, I make it
equal to the position of x2 at t
00:24:26.220 --> 00:24:28.170
equals 0.
00:24:28.170 --> 00:24:32.890
When t is 0, this tells me has
to be equal to where x3 is.
00:24:32.890 --> 00:24:34.650
That's three equations.
00:24:34.650 --> 00:24:36.680
I can differentiate
each with respect
00:24:36.680 --> 00:24:40.650
to time to get the
velocities, and likewise,
00:24:40.650 --> 00:24:42.440
get three more equations.
00:24:42.440 --> 00:24:46.500
So I'm going to end up with
six algebraic equations.
00:24:46.500 --> 00:24:49.210
And you, as well as
I, know that if we
00:24:49.210 --> 00:24:53.540
have six algebraic equations,
I can solve for six unknowns.
00:24:53.540 --> 00:24:56.550
And that will give me
those six unknowns.
00:24:56.550 --> 00:25:00.010
So once I've used the
initial conditions,
00:25:00.010 --> 00:25:03.880
I can then find these
arbitrary constants.
00:25:03.880 --> 00:25:07.420
They're no longer arbitrary
for a specific situation.
00:25:07.420 --> 00:25:11.960
And I can predict the future,
what will happen in this case.
00:25:11.960 --> 00:25:12.610
OK?
00:25:12.610 --> 00:25:19.430
So that's the way one
would solve a problem when
00:25:19.430 --> 00:25:21.570
you are lucky
enough that there is
00:25:21.570 --> 00:25:25.340
sufficient symmetry in
the original situation
00:25:25.340 --> 00:25:29.510
so you can guess
the normal modes.
00:25:29.510 --> 00:25:34.730
The next problem I'll do, I will
do one where you cannot guess
00:25:34.730 --> 00:25:36.110
the normal modes.
00:25:36.110 --> 00:25:38.160
What do you do under
those circumstances?
00:25:38.160 --> 00:25:40.180
And you won't be
surprised that it's
00:25:40.180 --> 00:25:46.350
going to be mathematically much,
much harder, but conceptually
00:25:46.350 --> 00:25:47.500
no harder.
00:25:47.500 --> 00:25:50.950
So now we'll move
to the next problem.
00:25:50.950 --> 00:25:51.720
Here it is.
00:25:51.720 --> 00:25:54.050
But in order to be able
to do this problem,
00:25:54.050 --> 00:25:55.600
I need more board space.
00:25:55.600 --> 00:25:59.281
So we're going to erase the
boards and continue from there.
00:25:59.281 --> 00:25:59.780
OK.
00:25:59.780 --> 00:26:03.130
So now let's go and
do the second problem.
00:26:03.130 --> 00:26:07.190
And it's going to be a
problem with no symmetry,
00:26:07.190 --> 00:26:10.760
so we can't guess the
answer like we did before.
00:26:10.760 --> 00:26:12.870
And I'll tell you
what I decided to do.
00:26:12.870 --> 00:26:15.770
You remember
Professor Walter Lewin
00:26:15.770 --> 00:26:21.660
in his lectures discussed
a double pendulum.
00:26:21.660 --> 00:26:23.400
In other words,
a situation where
00:26:23.400 --> 00:26:27.990
you had a string, a mass,
another string, and mass.
00:26:27.990 --> 00:26:30.200
He gave you the answer,
but he didn't prove it.
00:26:30.200 --> 00:26:32.590
He said it's hard and
lengthy, et cetera.
00:26:32.590 --> 00:26:35.140
And I thought that would
be a perfect example
00:26:35.140 --> 00:26:39.660
to do here from
first principles.
00:26:39.660 --> 00:26:44.930
So let me just
describe in detail
00:26:44.930 --> 00:26:47.230
the problem we're
trying to solve.
00:26:47.230 --> 00:26:51.230
What we have is, again,
an idealized situation.
00:26:51.230 --> 00:26:57.250
We have a simple pendulum
with a string of length L,
00:26:57.250 --> 00:26:59.960
attached to it a mass m.
00:26:59.960 --> 00:27:03.850
Attached to this mass at one
end is another string of length
00:27:03.850 --> 00:27:06.930
L to another mass m.
00:27:06.930 --> 00:27:09.930
We'll again idealize
the situation.
00:27:09.930 --> 00:27:12.200
So the assumptions
we are making,
00:27:12.200 --> 00:27:18.460
that this string and this string
is massless, which, by the way,
00:27:18.460 --> 00:27:22.530
is equivalent to saying that
it's taut and straight all
00:27:22.530 --> 00:27:23.310
the time.
00:27:23.310 --> 00:27:25.970
You can think about
that for yourself.
00:27:25.970 --> 00:27:31.350
Next, these masses
are point masses.
00:27:31.350 --> 00:27:34.330
And we're assuming no friction.
00:27:34.330 --> 00:27:37.980
Furthermore, as we've
done in the past,
00:27:37.980 --> 00:27:41.500
we'll assume that
all displacements
00:27:41.500 --> 00:27:47.260
when this is oscillating, at all
times the angle that the string
00:27:47.260 --> 00:27:52.220
makes with the vertical is
always such that we can ignore
00:27:52.220 --> 00:27:56.520
the difference between
a sine theta, theta,
00:27:56.520 --> 00:28:01.695
or [INAUDIBLE] can make the
assumption that cosine theta is
00:28:01.695 --> 00:28:03.090
0.
00:28:03.090 --> 00:28:06.410
By the way, the fact that
we'll make this assumption
00:28:06.410 --> 00:28:09.290
is equivalent to
saying that we'll
00:28:09.290 --> 00:28:13.080
ignore vertical motion
of these masses.
00:28:13.080 --> 00:28:16.780
The motion is sufficiently
small that vertically the masses
00:28:16.780 --> 00:28:18.210
are not moving.
00:28:18.210 --> 00:28:22.780
We can assume the acceleration
vertically is 0, et cetera.
00:28:22.780 --> 00:28:23.340
OK?
00:28:23.340 --> 00:28:24.850
This is in the vertical plane.
00:28:24.850 --> 00:28:26.420
Gravity is down.
00:28:26.420 --> 00:28:27.180
OK?
00:28:27.180 --> 00:28:32.440
And what we are trying
to derive for this,
00:28:32.440 --> 00:28:37.980
predict what are the normal
mode frequencies of this.
00:28:37.980 --> 00:28:40.170
And once we do
that, of course, we
00:28:40.170 --> 00:28:43.610
can use the same kind of
technique as we did before.
00:28:43.610 --> 00:28:45.800
Once we've managed to
find the normal modes
00:28:45.800 --> 00:28:47.810
and the frequencies,
we can always
00:28:47.810 --> 00:28:50.450
write the most
general expression.
00:28:50.450 --> 00:28:53.920
And then using the boundary
conditions, initial conditions,
00:28:53.920 --> 00:28:57.440
predict what this will
do as a function of time.
00:28:57.440 --> 00:28:58.774
OK, so let's get going.
00:29:05.280 --> 00:29:05.780
OK.
00:29:05.780 --> 00:29:07.660
Now, we are not guessing.
00:29:07.660 --> 00:29:09.120
We are not using logic.
00:29:09.120 --> 00:29:11.950
We are following the
kind of prescription
00:29:11.950 --> 00:29:14.450
I've told you in the past.
00:29:14.450 --> 00:29:17.020
This is our description
of the situation.
00:29:17.020 --> 00:29:19.700
And all the new words
and ordinary language, we
00:29:19.700 --> 00:29:23.820
must now translate
it into mathematics.
00:29:23.820 --> 00:29:26.230
We've got to
describe this problem
00:29:26.230 --> 00:29:29.580
in terms of
mathematical equations.
00:29:29.580 --> 00:29:33.600
So step one is we
redraw this and define
00:29:33.600 --> 00:29:35.380
some coordinate system.
00:29:35.380 --> 00:29:39.690
So we will say that the
angle this first string makes
00:29:39.690 --> 00:29:43.200
with respect to the
vertical is theta 1.
00:29:43.200 --> 00:29:45.980
This angle the second
string makes with respect
00:29:45.980 --> 00:29:48.420
to the vertical is theta 2.
00:29:48.420 --> 00:29:51.420
We will take the vertical
through this pivot
00:29:51.420 --> 00:29:56.140
here as our origin
of coordinate x.
00:29:56.140 --> 00:30:02.870
x equals 0 here,
is defined here.
00:30:02.870 --> 00:30:05.190
From the point of your
motion of the masses,
00:30:05.190 --> 00:30:06.960
it's a one-dimensional problem.
00:30:06.960 --> 00:30:10.750
The masses only move
along the x-axis.
00:30:10.750 --> 00:30:13.430
So we'll define
this distance as x1,
00:30:13.430 --> 00:30:17.440
and we'll define
this distance as x2.
00:30:17.440 --> 00:30:22.720
Now we have to use the laws of
physics, Newtonian mechanics,
00:30:22.720 --> 00:30:29.390
to derive the equations of
motion for the two masses.
00:30:29.390 --> 00:30:36.010
So I draw a force diagram
separately for the two masses.
00:30:36.010 --> 00:30:39.930
So the mass is there, what
forces act on that mass?
00:30:39.930 --> 00:30:43.790
Well, I have to look at the mass
and see what's attached to it.
00:30:43.790 --> 00:30:46.670
There is a string
here attached to it.
00:30:46.670 --> 00:30:49.280
It's taut, so there
will be a tension in it.
00:30:49.280 --> 00:30:53.890
So along this string,
there will be tension T1,
00:30:53.890 --> 00:30:58.690
which will exert a force
T1 along this string.
00:30:58.690 --> 00:31:02.100
This string is
attached to that mass.
00:31:02.100 --> 00:31:06.030
It has a tension, so it's
pulling on this mass.
00:31:06.030 --> 00:31:08.630
That tension I call
T2, so there will
00:31:08.630 --> 00:31:13.990
be a force along
that direction T2.
00:31:13.990 --> 00:31:16.820
This sits in a
gravitational field.
00:31:16.820 --> 00:31:19.360
Gravity acts on this mass.
00:31:19.360 --> 00:31:23.330
There is a force downwards
due to gravity, Fg.
00:31:26.330 --> 00:31:29.660
The sum of these forces,
by Newton's laws,
00:31:29.660 --> 00:31:33.460
must equal to the mass
times the acceleration.
00:31:33.460 --> 00:31:37.810
That acceleration, in this
approximation that is not
00:31:37.810 --> 00:31:40.550
moving up, is
horizontally, and it's
00:31:40.550 --> 00:31:43.920
the second derivative of x1.
00:31:43.920 --> 00:31:50.160
Similarly, for the second mass,
again, I'll go now faster.
00:31:50.160 --> 00:31:52.390
There is this mass
m, and this tension
00:31:52.390 --> 00:31:57.636
exerts a force here of T2
and the gravity on it Fg.
00:31:57.636 --> 00:32:00.310
The only subtlety
is, why did I say
00:32:00.310 --> 00:32:03.350
this force is equal
to that force?
00:32:03.350 --> 00:32:04.650
Think about it for a second.
00:32:04.650 --> 00:32:08.490
Why should the two
forces be equal?
00:32:08.490 --> 00:32:12.870
Why is the tension at both
ends of the string equal?
00:32:12.870 --> 00:32:16.756
And the answer is,
actually, a subtle one.
00:32:16.756 --> 00:32:20.740
It's equal because we made the
assumption that the string has
00:32:20.740 --> 00:32:28.240
no mass, and there can be no net
force on an object of 0 mass.
00:32:28.240 --> 00:32:31.190
Because if there was, that
object would disappear,
00:32:31.190 --> 00:32:33.650
would have an
infinite acceleration.
00:32:33.650 --> 00:32:37.620
So if you treat the
string as a mass,
00:32:37.620 --> 00:32:39.930
there cannot be net force on it.
00:32:39.930 --> 00:32:44.780
And so the force of
each of these masses
00:32:44.780 --> 00:32:49.460
must be pulling on this string
with exactly the same force
00:32:49.460 --> 00:32:51.031
equal and opposite.
00:32:51.031 --> 00:32:51.530
OK?
00:32:51.530 --> 00:32:55.020
And we're using the third
law to equate those forces.
00:32:55.020 --> 00:32:58.770
So the net result is this
T2 is the same as that,
00:32:58.770 --> 00:33:02.850
but in opposite direction,
and this is the mass of that.
00:33:02.850 --> 00:33:03.710
OK.
00:33:03.710 --> 00:33:07.410
So these are the force diagrams.
00:33:07.410 --> 00:33:11.020
Using now Newton's
laws of motion,
00:33:11.020 --> 00:33:15.220
I can translate this
into equations of motion.
00:33:15.220 --> 00:33:18.810
So let me consider
the horizontal motion
00:33:18.810 --> 00:33:20.950
of each mass separately.
00:33:20.950 --> 00:33:26.050
First this mass, and so
its mass times acceleration
00:33:26.050 --> 00:33:31.320
is equal to the
horizontal force of T2.
00:33:31.320 --> 00:33:35.690
And remember that this
angle here is theta 2.
00:33:35.690 --> 00:33:38.600
And we see that the
force due to T2,
00:33:38.600 --> 00:33:40.930
and there's a T2 sine theta 2.
00:33:40.930 --> 00:33:45.150
And the force due to T1
is minus T1 sine theta 1
00:33:45.150 --> 00:33:47.530
because it's in the
opposite direction.
00:33:47.530 --> 00:33:50.190
For this, the only
horizontal component
00:33:50.190 --> 00:33:53.180
is the force horizontal
component of T2.
00:33:53.180 --> 00:33:58.800
And x2 double dot is equal
to minus T2 sine theta 2.
00:33:58.800 --> 00:34:01.350
OK, that's horizontal motion.
00:34:01.350 --> 00:34:06.190
Applying Newton's laws of
motion to the vertical motion,
00:34:06.190 --> 00:34:12.170
we said that, because cosine
theta is 0 or approximately 0,
00:34:12.170 --> 00:34:14.250
the masses are not
moving up and down.
00:34:14.250 --> 00:34:16.739
There's no acceleration
of the masses.
00:34:16.739 --> 00:34:20.800
So the vertical acceleration
of the first mass
00:34:20.800 --> 00:34:28.570
is 0 must be equal to the
vertical component of T1, which
00:34:28.570 --> 00:34:33.219
is T1 cosine theta 1, minus
the vertical component of this,
00:34:33.219 --> 00:34:40.940
which is minus T2
cosine theta 2 minus mg.
00:34:40.940 --> 00:34:41.440
All right.
00:34:41.440 --> 00:34:47.219
Similarly for the second
mass, it's 0 is equal to this.
00:34:47.219 --> 00:34:51.850
0 must equal to the
vertical component
00:34:51.850 --> 00:34:58.390
of T1 minus the
vertical component of T2
00:34:58.390 --> 00:35:02.170
minus the force of gravity down.
00:35:02.170 --> 00:35:06.040
And similarly for
the second mass,
00:35:06.040 --> 00:35:09.170
we know that vertically
the acceleration is 0.
00:35:09.170 --> 00:35:12.560
That must be equal to the
vertical component of T2
00:35:12.560 --> 00:35:18.650
minus the gravitational
force pulling down.
00:35:18.650 --> 00:35:26.790
Now I can make use of what we
made the assumption-- oops,
00:35:26.790 --> 00:35:29.130
I see there is an error.
00:35:29.130 --> 00:35:31.540
If you were in this
room, I'm sure you
00:35:31.540 --> 00:35:35.020
would have corrected me.
00:35:35.020 --> 00:35:38.540
For very small angles,
the sine of an angle
00:35:38.540 --> 00:35:40.790
is approximately
equal to the angle.
00:35:40.790 --> 00:35:44.740
But for small angles,
the cosine is 1.
00:35:44.740 --> 00:35:48.250
This is an error, and
I apologize for that.
00:35:48.250 --> 00:35:50.450
But that's the
approximation we are making.
00:35:50.450 --> 00:35:54.580
And it is this
approximation which
00:35:54.580 --> 00:36:00.810
is equivalent to saying that we
can ignore the vertical motion.
00:36:00.810 --> 00:36:02.200
All right.
00:36:02.200 --> 00:36:07.230
So with the assumption
that the cosines are all 1,
00:36:07.230 --> 00:36:12.470
I can, from these two
equations, I clearly
00:36:12.470 --> 00:36:18.430
derive that T2 must equal to
mg, and T1 equals twice mg.
00:36:18.430 --> 00:36:20.580
Actually, it makes sense.
00:36:20.580 --> 00:36:23.770
Imagine this is hanging
completely vertically.
00:36:23.770 --> 00:36:28.210
Obviously, this string, the
upper part of the string,
00:36:28.210 --> 00:36:30.875
is supporting not only
this mass but also that.
00:36:30.875 --> 00:36:35.530
It's supporting 2m masses,
while this string is only
00:36:35.530 --> 00:36:36.640
supporting 1.
00:36:36.640 --> 00:36:40.440
So that's consistent
with what we see,
00:36:40.440 --> 00:36:45.230
that the top string has twice
the tension and the lower
00:36:45.230 --> 00:36:47.780
one half the tension.
00:36:47.780 --> 00:36:50.640
Having determined T1
and T2, we can now
00:36:50.640 --> 00:36:56.170
go back into our equation,
replace the T1 and T2.
00:36:56.170 --> 00:36:58.830
We also can replace--
we know what
00:36:58.830 --> 00:37:02.440
sine theta 1 and
sine theta 2 are.
00:37:02.440 --> 00:37:04.440
We can replace those.
00:37:04.440 --> 00:37:06.990
And we end up--
the two equations
00:37:06.990 --> 00:37:10.750
of motions are written here.
00:37:10.750 --> 00:37:11.770
All right?
00:37:11.770 --> 00:37:15.750
So here is the equation
of motion for x1.
00:37:15.750 --> 00:37:19.380
And the second one, the
equation for motion of x2,
00:37:19.380 --> 00:37:21.380
is written here.
00:37:21.380 --> 00:37:22.140
OK?
00:37:22.140 --> 00:37:27.200
So these are the
equations of motion,
00:37:27.200 --> 00:37:37.930
all right, which
we have to solve.
00:37:37.930 --> 00:37:46.300
Now, to simplify the algebra,
let me define the quantity g/l
00:37:46.300 --> 00:37:49.525
by omega squared over 2.
00:37:49.525 --> 00:37:51.140
And you'll recognize this.
00:37:51.140 --> 00:37:52.240
This is not [INAUDIBLE].
00:37:52.240 --> 00:37:54.010
I'm using that terminology.
00:37:54.010 --> 00:37:56.620
This is the frequency
of oscillation
00:37:56.620 --> 00:38:01.140
of a mass on a
string of length l.
00:38:01.140 --> 00:38:02.260
All right?
00:38:02.260 --> 00:38:08.930
So here are our two differential
equations, the two equations
00:38:08.930 --> 00:38:14.620
of motion, one for
x1 and one for x2.
00:38:14.620 --> 00:38:17.980
When we had one mass,
one harmonic oscillator,
00:38:17.980 --> 00:38:21.420
we had a single second-order
differential equation.
00:38:21.420 --> 00:38:25.070
We now have two
masses, so we have
00:38:25.070 --> 00:38:27.740
two second-order
differential equations.
00:38:27.740 --> 00:38:28.490
These are it.
00:38:28.490 --> 00:38:30.970
They are coupled
differential equations.
00:38:30.970 --> 00:38:38.860
You see, the derivatives of x1
is related both to x1 and x2.
00:38:38.860 --> 00:38:43.870
The second derivative of x2 is
related both to x1 and to x2.
00:38:43.870 --> 00:38:47.210
So these are two coupled
differential equations.
00:38:47.210 --> 00:38:48.220
OK?
00:38:48.220 --> 00:38:51.655
This is the end of step 1.
00:38:51.655 --> 00:38:55.370
What we succeeded
in, we've translated
00:38:55.370 --> 00:39:00.610
the physical situation
into mathematics.
00:39:00.610 --> 00:39:06.580
These two second-order
differential equations
00:39:06.580 --> 00:39:11.285
describe exactly that
idealized situation we had.
00:39:14.050 --> 00:39:17.280
So if I now want to
answer the question, what
00:39:17.280 --> 00:39:19.770
will be the motion
of those masses,
00:39:19.770 --> 00:39:22.240
I have to go into the
world of mathematics.
00:39:22.240 --> 00:39:25.630
I have to solve these equations.
00:39:25.630 --> 00:39:28.290
And life is not
as easy as it was
00:39:28.290 --> 00:39:31.600
for a single second-order
differential equation.
00:39:31.600 --> 00:39:33.720
It's more complicated.
00:39:33.720 --> 00:39:39.060
But here I will use
my general knowledge
00:39:39.060 --> 00:39:44.040
of what happens when you
have coupled oscillators.
00:39:44.040 --> 00:39:51.380
I know that the general
solution of coupled oscillations
00:39:51.380 --> 00:39:57.410
is a superposition
of normal modes.
00:39:57.410 --> 00:40:00.620
Here I have 2
degrees of freedom,
00:40:00.620 --> 00:40:03.020
two second-order
differential equations.
00:40:03.020 --> 00:40:05.940
There will be two normal modes.
00:40:05.940 --> 00:40:09.490
If I succeed in
finding them, I will
00:40:09.490 --> 00:40:13.890
have found the most general
solution to this problem,
00:40:13.890 --> 00:40:17.220
because it will be the sum
of the two normal modes.
00:40:17.220 --> 00:40:24.900
So I will now do it by trial.
00:40:24.900 --> 00:40:30.630
In a normal mode,
we know that x1
00:40:30.630 --> 00:40:34.230
will be oscillating
with a single frequency
00:40:34.230 --> 00:40:39.500
omega, some phase phi,
some amplitude A1.
00:40:39.500 --> 00:40:41.820
This will be a solution
to these equations.
00:40:44.370 --> 00:40:47.250
If this is a solution
to those equations,
00:40:47.250 --> 00:40:51.280
the other mass must
also be oscillating
00:40:51.280 --> 00:40:55.165
in the same normal mode,
meaning with the same frequency
00:40:55.165 --> 00:40:58.080
and same phase.
00:40:58.080 --> 00:41:00.260
It'll have some
arbitrary amplitude.
00:41:02.860 --> 00:41:05.820
So I don't know what A1 is,
and I don't know what omega is.
00:41:05.820 --> 00:41:07.030
I don't know what phi is.
00:41:07.030 --> 00:41:12.600
But I know that the
solutions to these equations
00:41:12.600 --> 00:41:16.980
must be of this form.
00:41:16.980 --> 00:41:18.140
OK?
00:41:18.140 --> 00:41:29.780
So let me now try to find
these various constants.
00:41:29.780 --> 00:41:37.274
Well, if this and that
satisfies these equations--
00:41:37.274 --> 00:41:39.190
and it has to because
that's what it's saying,
00:41:39.190 --> 00:41:41.470
these are the solutions
of those equations--
00:41:41.470 --> 00:41:49.440
then I can take these, x1 and
x2, plug it into this equation.
00:41:49.440 --> 00:41:51.800
In other words, calculate
the second derivative
00:41:51.800 --> 00:41:54.780
of x1, et cetera,
calculate second,
00:41:54.780 --> 00:41:58.750
and I'll end up
with two equations.
00:42:02.415 --> 00:42:02.915
OK?
00:42:06.640 --> 00:42:11.270
So I won't bother to go
in great detail here.
00:42:11.270 --> 00:42:17.560
But just to start off
with, x1 double dot,
00:42:17.560 --> 00:42:23.010
I have to differentiate
A1 cosine omega t twice.
00:42:23.010 --> 00:42:34.830
So I get minus omega squared A1
times cosine omega t plus phi.
00:42:34.830 --> 00:42:40.220
If I take the next term, I get
this, and the third term that.
00:42:40.220 --> 00:42:44.470
And in each case, if
you notice, since it'll
00:42:44.470 --> 00:42:47.730
be multiplied by the
same cosine function,
00:42:47.730 --> 00:42:51.210
I've just canceled in my head
the cosine function there.
00:42:51.210 --> 00:42:52.270
All right?
00:42:52.270 --> 00:42:53.540
Just saving time.
00:42:53.540 --> 00:42:58.360
Similarly, next equation,
I end up with this.
00:42:58.360 --> 00:43:04.190
So if my guessed functions,
and I'm returning to this,
00:43:04.190 --> 00:43:08.810
if these two satisfy
those equations,
00:43:08.810 --> 00:43:16.390
then here these algebraic
equations must be satisfied.
00:43:16.390 --> 00:43:18.125
But you'll notice
something interesting.
00:43:20.930 --> 00:43:24.010
If I take the first
of these equations,
00:43:24.010 --> 00:43:30.000
it boils down to A1 times
this quantity plus A2
00:43:30.000 --> 00:43:33.160
to [? then ?] this
quantity equal to 0.
00:43:33.160 --> 00:43:36.320
And the second equation
boils down to this.
00:43:39.580 --> 00:43:44.450
If this is true, then
this must be true.
00:43:44.450 --> 00:43:49.850
It's A1 divided by A2
has to be equal to that.
00:43:49.850 --> 00:43:54.940
If this is true,
then A1 divided by A2
00:43:54.940 --> 00:43:56.590
has got to be equal to this.
00:43:59.300 --> 00:44:02.960
That means that this
has to be equal to that,
00:44:02.960 --> 00:44:06.460
or else I have an inconsistency.
00:44:06.460 --> 00:44:13.660
So what I have found is
that, in general, my guess
00:44:13.660 --> 00:44:15.020
is not a good one.
00:44:15.020 --> 00:44:19.010
In general, these
equations would not
00:44:19.010 --> 00:44:24.580
satisfy my second-order
differential equations.
00:44:24.580 --> 00:44:30.660
There is only under very
special circumstances
00:44:30.660 --> 00:44:33.470
that they do
satisfy it, and that
00:44:33.470 --> 00:44:37.470
is if this is equal to that.
00:44:40.440 --> 00:44:42.824
In general, they
will not be equal.
00:44:46.460 --> 00:44:50.410
So under what conditions
will these be equal?
00:44:50.410 --> 00:44:51.850
Well, let me force them.
00:44:51.850 --> 00:44:55.760
Let me say this is
equal to that and see
00:44:55.760 --> 00:44:59.680
what it means for omega.
00:44:59.680 --> 00:45:03.890
I originally took omega
to be an unknown quantity.
00:45:03.890 --> 00:45:08.060
What I am now seeing,
that those functions
00:45:08.060 --> 00:45:13.430
would work only for
specific values of omega.
00:45:13.430 --> 00:45:18.050
So I'm going to solve
this equation for omega
00:45:18.050 --> 00:45:24.090
and see for what values of omega
do I get a consistent solution.
00:45:26.620 --> 00:45:29.820
And this is a hard
grind, unfortunately.
00:45:29.820 --> 00:45:34.880
If I take this,
multiply it out, I
00:45:34.880 --> 00:45:39.730
get a quadratic equation
in omega squared.
00:45:39.730 --> 00:45:43.200
If this quadratic
equation is satisfied,
00:45:43.200 --> 00:45:49.790
then all these are
self-consistent.
00:45:49.790 --> 00:45:50.290
All right?
00:45:50.290 --> 00:45:56.170
You know how to solve
a quadratic equation,
00:45:56.170 --> 00:46:05.430
and this is a quadratic
equation in omega squared.
00:46:05.430 --> 00:46:08.530
Omega square will have
to be equal to minus
00:46:08.530 --> 00:46:13.350
this plus or minus the
square root of this,
00:46:13.350 --> 00:46:16.840
4 times this times that,
divided by 2a, right?
00:46:16.840 --> 00:46:19.960
This is the standard
formula for the solution
00:46:19.960 --> 00:46:21.840
of a quadratic equation.
00:46:26.090 --> 00:46:28.280
And there's a plus or minus.
00:46:28.280 --> 00:46:32.970
If I calculate this out,
I get that omega squared
00:46:32.970 --> 00:46:36.470
has to be equal
to 2 plus or minus
00:46:36.470 --> 00:46:40.680
the square root of 2
times omega 0 squared.
00:46:40.680 --> 00:46:41.690
All right?
00:46:41.690 --> 00:46:49.200
Now, omega 0 squared, earlier
on, I defined to be g/l.
00:46:49.200 --> 00:46:53.830
So omega squared has to be this.
00:46:53.830 --> 00:46:57.030
So let's stop for
a second and think
00:46:57.030 --> 00:47:00.560
and review what we've done.
00:47:00.560 --> 00:47:06.370
We found that our
physical situation
00:47:06.370 --> 00:47:13.690
can be described by these two
coupled differential equations.
00:47:13.690 --> 00:47:20.666
We looked for solutions, which
are normal modes where both x1
00:47:20.666 --> 00:47:25.990
and x2 is oscillating with
same frequency and phase.
00:47:25.990 --> 00:47:33.700
And we found that it was
possible to find solution
00:47:33.700 --> 00:47:42.730
of this form to these equations
if and only if omega squared
00:47:42.730 --> 00:47:46.210
has one of two possible values.
00:47:46.210 --> 00:47:48.050
I've rewritten them here.
00:47:48.050 --> 00:47:53.020
One is 2 plus root 2
times g/l, and the other's
00:47:53.020 --> 00:47:54.890
2 minus root 2 g/l.
00:47:57.420 --> 00:48:04.900
With these values, these two
equations satisfies that.
00:48:04.900 --> 00:48:10.850
So what we found is
the two normal modes.
00:48:10.850 --> 00:48:17.420
Since this system consists
of the coupled oscillators,
00:48:17.420 --> 00:48:21.290
in other words, two oscillators
coupled with each other,
00:48:21.290 --> 00:48:23.960
we know there are
two normal modes.
00:48:23.960 --> 00:48:30.180
So these are the normal mode
frequencies in this situation.
00:48:30.180 --> 00:48:33.440
And by the way, I'm
delighted to see
00:48:33.440 --> 00:48:38.430
that, if you remember Professor
Walter Lewin's lectures,
00:48:38.430 --> 00:48:41.830
he actually quoted
these numbers.
00:48:41.830 --> 00:48:42.580
So he was right.
00:48:42.580 --> 00:48:45.270
He did not make a mistake,
as we've just seen.
00:48:45.270 --> 00:48:46.280
OK?
00:48:46.280 --> 00:48:48.960
These are the two
normal mode frequencies.
00:48:48.960 --> 00:48:51.830
Now, how about if
we want to predict
00:48:51.830 --> 00:48:54.250
everything, the
amplitude, et cetera?
00:48:57.330 --> 00:49:00.380
We've got to be careful here
with the logic we've applied.
00:49:03.500 --> 00:49:10.780
For any one of these normal
frequencies, say this one,
00:49:10.780 --> 00:49:23.390
earlier on we found that A1 to
A2 is given by this equation
00:49:23.390 --> 00:49:25.220
or by this equation.
00:49:25.220 --> 00:49:27.820
But for the given
values of omega,
00:49:27.820 --> 00:49:30.940
we found these are
now equivalent.
00:49:30.940 --> 00:49:37.890
And so if you know what
omega is, say omega A,
00:49:37.890 --> 00:49:43.400
one of the normal
modes, this we know.
00:49:43.400 --> 00:49:45.930
It's g/l.
00:49:45.930 --> 00:49:47.550
And this is known.
00:49:47.550 --> 00:49:56.170
So A1/A2, once we fix
omega A, is fixed.
00:49:56.170 --> 00:50:01.370
So these are not two
independent amplitudes.
00:50:01.370 --> 00:50:07.620
So if I go back to here when
we said let's guess a solution,
00:50:07.620 --> 00:50:10.150
and this was an arbitrary
number and this was an arbitrary
00:50:10.150 --> 00:50:15.550
number, now we see
that in a normal mode
00:50:15.550 --> 00:50:18.720
the ratio between them is fixed.
00:50:18.720 --> 00:50:22.840
It depends on the
normal mode frequency.
00:50:22.840 --> 00:50:25.340
But the [? overall ?]
normalization
00:50:25.340 --> 00:50:26.530
is still arbitrary.
00:50:26.530 --> 00:50:29.230
I could make this 10 times
this and 10 times that.
00:50:29.230 --> 00:50:31.460
[? It would still ?] work.
00:50:31.460 --> 00:50:39.670
So back to here, we found one of
the normal mode's frequencies.
00:50:39.670 --> 00:50:43.630
Using the equation
above, it determines
00:50:43.630 --> 00:50:48.560
the value of A1 to A2.
00:50:48.560 --> 00:50:52.400
Taking the other
normal mode frequency
00:50:52.400 --> 00:50:57.000
determines the
ratio of those two
00:50:57.000 --> 00:51:00.380
but not the overall amplitude.
00:51:00.380 --> 00:51:04.780
Not to confuse the two, I
will call this B1 and B2.
00:51:04.780 --> 00:51:07.180
This is the amplitude
of the first mass,
00:51:07.180 --> 00:51:08.870
and this is of the other one.
00:51:08.870 --> 00:51:14.840
The ratio is fixed, determined
by the value of omega B.
00:51:14.840 --> 00:51:18.570
But the overall
amplitude is not.
00:51:18.570 --> 00:51:24.020
So now we have ended up
in the same situation
00:51:24.020 --> 00:51:29.110
we were in our three masses,
where, if you remember,
00:51:29.110 --> 00:51:33.020
I started by guessing
the normal modes
00:51:33.020 --> 00:51:37.690
and guessing the
relative amplitudes.
00:51:37.690 --> 00:51:42.490
And with that information alone
and the initial conditions,
00:51:42.490 --> 00:51:44.500
I could predict
what will happen.
00:51:44.500 --> 00:51:46.650
I could now repeat that here.
00:51:46.650 --> 00:51:51.510
But to save time, I think
you can do that for yourself.
00:51:51.510 --> 00:51:55.970
For each, you can write
now a single equation
00:51:55.970 --> 00:52:03.830
saying what a given mass will do
as the sum of two normal modes,
00:52:03.830 --> 00:52:11.280
where you know the frequency
of them and the amplitudes.
00:52:11.280 --> 00:52:14.040
And you have to remember
that for the two masses,
00:52:14.040 --> 00:52:16.570
the ratio of the amplitudes
in each normal mode
00:52:16.570 --> 00:52:20.330
have to be fixed given
by what we've done here.
00:52:20.330 --> 00:52:24.810
And therefore, you can
get the single equation
00:52:24.810 --> 00:52:28.530
for each mass, one for one
mass, one for the other.
00:52:28.530 --> 00:52:31.690
Which will tell you what
they will do in the future
00:52:31.690 --> 00:52:36.500
if you know what they're
doing initially, for example,
00:52:36.500 --> 00:52:40.700
the initial conditions,
that you can solve
00:52:40.700 --> 00:52:45.710
to get the final predicted
position of each mass.
00:52:45.710 --> 00:52:47.350
So that's as much
as I was hoping
00:52:47.350 --> 00:52:51.700
to do today on
coupled oscillators.
00:52:51.700 --> 00:52:56.010
And we'll continue
next time on situations
00:52:56.010 --> 00:52:58.920
with many, many more
degrees of freedom.
00:52:58.920 --> 00:53:00.770
Thank you.