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PROFESSOR: Welcome back.
00:00:22.280 --> 00:00:25.550
And today we will
continue doing some more
00:00:25.550 --> 00:00:27.570
problems to do with
standing waves.
00:00:27.570 --> 00:00:30.640
It's a very important topic,
so I thought it worthwhile
00:00:30.640 --> 00:00:34.380
considering some different
kinds of problems.
00:00:34.380 --> 00:00:36.660
So the first problem I
want to discuss with you
00:00:36.660 --> 00:00:38.270
is the following.
00:00:38.270 --> 00:00:43.210
Suppose you have two
equal-length strings,
00:00:43.210 --> 00:00:47.010
each of length L, but
they are very different.
00:00:47.010 --> 00:00:49.460
One is very massive.
00:00:49.460 --> 00:00:52.920
The other one is very light.
00:00:52.920 --> 00:00:57.620
And we tie the massive
one at one end.
00:00:57.620 --> 00:01:00.070
It's connected to the
other one in the middle.
00:01:00.070 --> 00:01:02.530
And the other one is tied here.
00:01:02.530 --> 00:01:05.190
The whole system is taut.
00:01:05.190 --> 00:01:13.120
And the question is, what
are two lowest normal modes
00:01:13.120 --> 00:01:14.900
of vibrations?
00:01:14.900 --> 00:01:17.910
Calculate the
frequencies of these
00:01:17.910 --> 00:01:20.770
and sketch what they look like.
00:01:20.770 --> 00:01:24.930
And I like this problem
because it's counter-intuitive.
00:01:24.930 --> 00:01:27.950
You could take a few
seconds for yourself
00:01:27.950 --> 00:01:30.290
and think whether you
could predict by guessing.
00:01:30.290 --> 00:01:34.920
And I'm almost certain
your guess will be wrong.
00:01:37.480 --> 00:01:41.210
Certainly, the first time
I tried to figure this out,
00:01:41.210 --> 00:01:45.810
my intuition gave
me the wrong answer.
00:01:45.810 --> 00:01:46.700
OK.
00:01:46.700 --> 00:01:50.570
Now, again, it's idealized.
00:01:50.570 --> 00:01:58.235
We'll assume that these strings
are ideal, lossless strings.
00:01:58.235 --> 00:02:01.120
All the displacements
are very small.
00:02:01.120 --> 00:02:04.860
So in the derivations of the
equations of motion, et cetera,
00:02:04.860 --> 00:02:07.210
the usual sine
theta equals theta,
00:02:07.210 --> 00:02:11.640
there is a constant
tension in this string T.
00:02:11.640 --> 00:02:15.730
Because they are connected,
of course, like this,
00:02:15.730 --> 00:02:19.260
the tension everywhere
must be the same.
00:02:19.260 --> 00:02:23.490
And all we know is
that the density
00:02:23.490 --> 00:02:27.760
per unit length of this
string, which I call mu sub H,
00:02:27.760 --> 00:02:33.470
this is the heavy
string, is much bigger
00:02:33.470 --> 00:02:39.860
than the density, the mass per
unit length, of the light one.
00:02:39.860 --> 00:02:41.550
Now, the fact that
they don't give me
00:02:41.550 --> 00:02:45.730
the values of these,
just say this,
00:02:45.730 --> 00:02:48.090
basically is
telling me they want
00:02:48.090 --> 00:02:50.490
almost a qualitative answer.
00:02:50.490 --> 00:02:55.865
I cannot calculate it exactly if
I don't know these two numbers.
00:02:55.865 --> 00:02:56.500
All right?
00:02:59.020 --> 00:03:03.480
Now, the other thing we
are told is, of course,
00:03:03.480 --> 00:03:07.420
a system like that can
oscillate in three dimensions.
00:03:07.420 --> 00:03:10.860
It could oscillate up and
down, out of the board,
00:03:10.860 --> 00:03:11.810
in, et cetera.
00:03:11.810 --> 00:03:16.400
They're telling
us limit ourselves
00:03:16.400 --> 00:03:20.955
to oscillations of the string
in the plane of the board.
00:03:20.955 --> 00:03:21.550
OK?
00:03:21.550 --> 00:03:26.200
So it's a 2D problem, not a
three-dimensional problem.
00:03:26.200 --> 00:03:27.200
OK.
00:03:27.200 --> 00:03:29.460
So how do we do this?
00:03:33.920 --> 00:03:37.060
We immediately
recognize that when
00:03:37.060 --> 00:03:43.560
you have a continuous line
of harmonic oscillators
00:03:43.560 --> 00:03:48.560
in the continuum limit, and
that's what the string is,
00:03:48.560 --> 00:03:55.150
the equation of motion
of the string is this.
00:03:55.150 --> 00:03:57.260
This is the standard
wave equation.
00:03:59.950 --> 00:04:03.980
It's the same equation
for the heavy string
00:04:03.980 --> 00:04:05.690
and the light string.
00:04:05.690 --> 00:04:12.580
And the only difference
is the phase velocity v,
00:04:12.580 --> 00:04:23.710
which depends on the tension and
the mass density of the string.
00:04:23.710 --> 00:04:27.570
So the heavy one we
will call v sub H,
00:04:27.570 --> 00:04:30.170
and it's the square
root of T over mu H.
00:04:30.170 --> 00:04:37.470
And the light one I call v sub
L, square root of T over mu L.
00:04:37.470 --> 00:04:38.380
OK.
00:04:38.380 --> 00:04:45.750
Since we know that mu H
is much bigger than mu L,
00:04:45.750 --> 00:04:50.860
we immediately know that
vH is much less than vL.
00:04:50.860 --> 00:04:53.390
That's the information we
have about those strings.
00:04:53.390 --> 00:04:56.130
OK.
00:04:56.130 --> 00:04:58.610
What else do we know
about this system?
00:04:58.610 --> 00:05:00.940
I mean, we are now
in the usual process
00:05:00.940 --> 00:05:03.520
of translating a
physical situation
00:05:03.520 --> 00:05:06.603
into the mathematical
description of the system.
00:05:06.603 --> 00:05:07.350
OK?
00:05:07.350 --> 00:05:09.920
In this particular
idealized case,
00:05:09.920 --> 00:05:11.710
this is almost a
mathematical description
00:05:11.710 --> 00:05:13.500
up there, but still.
00:05:13.500 --> 00:05:15.110
So this is the
equation of motion.
00:05:15.110 --> 00:05:17.520
And the other thing we
know about the system
00:05:17.520 --> 00:05:19.930
are the boundary conditions.
00:05:19.930 --> 00:05:30.500
We know that at both ends, at
x equals 0 and at x equals 2L,
00:05:30.500 --> 00:05:32.850
the string is tied down.
00:05:32.850 --> 00:05:36.660
So that for all time, therefore,
the displacement will be 0,
00:05:36.660 --> 00:05:39.890
y equals 0.
00:05:39.890 --> 00:05:45.520
We know that at the boundary,
the strings are connected.
00:05:45.520 --> 00:05:50.880
So at all times, y of
H must be equal to yL.
00:05:50.880 --> 00:05:53.300
Otherwise, it would be broken.
00:05:53.300 --> 00:05:54.890
So they're connected.
00:05:54.890 --> 00:05:58.680
The displacement of the
string in the middle,
00:05:58.680 --> 00:06:01.980
both the heavy one and the
light one, is the same,
00:06:01.980 --> 00:06:05.250
and this is true at all times.
00:06:05.250 --> 00:06:09.910
And the other thing
is that the slope
00:06:09.910 --> 00:06:14.590
of the string on both
sides must be the same.
00:06:14.590 --> 00:06:15.830
How do I know that?
00:06:15.830 --> 00:06:22.430
Well, imagine the junction
to be some object.
00:06:22.430 --> 00:06:25.440
Then that junction has 0 mass.
00:06:28.005 --> 00:06:33.620
If it has 0 mass, you cannot
have a net vertical force
00:06:33.620 --> 00:06:34.520
on that.
00:06:34.520 --> 00:06:37.190
You cannot have a net
force on it, period.
00:06:37.190 --> 00:06:41.800
Or because of Newton's laws,
force on the 0 mass object
00:06:41.800 --> 00:06:44.630
will give it an
infinite acceleration.
00:06:44.630 --> 00:06:46.630
It will just disappear there.
00:06:46.630 --> 00:06:52.060
So at the junction, there
can be no net force.
00:06:52.060 --> 00:06:55.370
The tension on the string
on both sides is the same.
00:06:55.370 --> 00:07:00.640
And therefore, the slope of
the strings at all times,
00:07:00.640 --> 00:07:03.740
on the left it must
equal the right.
00:07:03.740 --> 00:07:07.010
So this is the complete
mathematical description.
00:07:07.010 --> 00:07:15.560
And we are asked to,
with this information,
00:07:15.560 --> 00:07:19.570
figure out, what are
the two lowest mode
00:07:19.570 --> 00:07:23.880
frequencies and what
are their shapes?
00:07:23.880 --> 00:07:24.780
OK?
00:07:24.780 --> 00:07:26.640
So that's what now I will do.
00:07:32.160 --> 00:07:34.150
OK.
00:07:34.150 --> 00:07:38.510
Now, we know that
what we're interested
00:07:38.510 --> 00:07:42.260
is in the normal mode.
00:07:42.260 --> 00:07:48.580
If you have a normal system
oscillating in the normal mode,
00:07:48.580 --> 00:07:52.340
you know that the
all parts oscillate
00:07:52.340 --> 00:07:57.070
with the same frequency
and same phase.
00:07:57.070 --> 00:08:00.070
That's the meaning of
being in the normal mode.
00:08:00.070 --> 00:08:01.250
OK?
00:08:01.250 --> 00:08:05.800
So we know that when this
system is oscillating
00:08:05.800 --> 00:08:09.940
in the normal mode,
both sides of the string
00:08:09.940 --> 00:08:13.685
will be oscillating with the
same omega, same frequency
00:08:13.685 --> 00:08:14.805
and phase.
00:08:14.805 --> 00:08:15.380
All right?
00:08:18.140 --> 00:08:25.020
And also, on both sides
we have a string, a row,
00:08:25.020 --> 00:08:28.270
a continuous row of
oscillators, and we
00:08:28.270 --> 00:08:32.510
know what the normal mode
oscillations of such a system
00:08:32.510 --> 00:08:33.059
is.
00:08:33.059 --> 00:08:35.270
We've seen it over
and over again.
00:08:35.270 --> 00:08:38.360
It is sinusoidal in fall.
00:08:38.360 --> 00:08:43.770
So the general expression of
the displacement of the string,
00:08:43.770 --> 00:08:46.760
whether it's on the left
side or on the right side,
00:08:46.760 --> 00:08:51.140
will be the displacement
of it from equilibrium
00:08:51.140 --> 00:08:58.250
is a sinusoidal function
in position multiplied
00:08:58.250 --> 00:09:03.930
by a sinusoidal
function in time.
00:09:03.930 --> 00:09:10.650
This function will be different
for the left and the right,
00:09:10.650 --> 00:09:16.901
the spatial distribution.
00:09:16.901 --> 00:09:18.710
This part will be the same.
00:09:22.040 --> 00:09:25.320
Why would this be different
for the two sides?
00:09:25.320 --> 00:09:33.290
Well, this equation
satisfies the wave equation.
00:09:35.980 --> 00:09:37.180
It is the solution.
00:09:37.180 --> 00:09:38.760
I am looking for
that wave equation.
00:09:42.370 --> 00:09:46.490
If this is to satisfy
the wave equation,
00:09:46.490 --> 00:09:50.520
you'll find that
omega/k, which you
00:09:50.520 --> 00:09:54.880
can write like lambda times f
where lambda is the wavelength
00:09:54.880 --> 00:09:57.160
and f is the frequency--
this is the angular
00:09:57.160 --> 00:10:00.930
frequency on the
wave number-- has
00:10:00.930 --> 00:10:03.440
to be equal to v.
Otherwise, this
00:10:03.440 --> 00:10:06.470
would not satisfy
the wave equation.
00:10:06.470 --> 00:10:07.240
OK?
00:10:07.240 --> 00:10:12.126
And v is different
for the two strings.
00:10:12.126 --> 00:10:13.050
All right?
00:10:13.050 --> 00:10:13.890
We showed that.
00:10:13.890 --> 00:10:15.810
We discussed it a second ago.
00:10:15.810 --> 00:10:19.110
v, the phase velocity
for the heavy string,
00:10:19.110 --> 00:10:23.670
is much smaller than
for the light one.
00:10:23.670 --> 00:10:27.150
And therefore, if
omega is the same,
00:10:27.150 --> 00:10:30.430
the k would be different
in the two cases.
00:10:30.430 --> 00:10:32.340
OK.
00:10:32.340 --> 00:10:33.240
Fine.
00:10:33.240 --> 00:10:42.910
Now, using this, we know that
the wavelength is given by v/f.
00:10:42.910 --> 00:10:45.420
Just I've copied it from there.
00:10:45.420 --> 00:10:51.420
And since the frequency
on both sides is the same
00:10:51.420 --> 00:10:53.900
but v is different,
I can immediately
00:10:53.900 --> 00:10:58.270
conclude that the wavelength
of the part of the string which
00:10:58.270 --> 00:11:03.530
is massive, the heavy one, over
the wavelength of the string
00:11:03.530 --> 00:11:08.755
on the light side is
simply equal to vH over vL.
00:11:08.755 --> 00:11:18.660
But we've been told that
the density of the heavy one
00:11:18.660 --> 00:11:21.250
is much greater
than the light one.
00:11:21.250 --> 00:11:23.470
Therefore, the
velocity, this one,
00:11:23.470 --> 00:11:26.100
is much smaller than this one.
00:11:26.100 --> 00:11:27.580
And therefore, we
immediately can
00:11:27.580 --> 00:11:32.490
conclude that the wavelength
in the normal mode
00:11:32.490 --> 00:11:35.250
oscillation of the
heavy one is much
00:11:35.250 --> 00:11:39.220
shorter than of the light one.
00:11:39.220 --> 00:11:40.660
All right.
00:11:40.660 --> 00:11:46.120
So now if we were
given more information
00:11:46.120 --> 00:11:49.780
exactly what these
quantities are, et cetera,
00:11:49.780 --> 00:11:56.980
we could write the equations
of the string on both sides
00:11:56.980 --> 00:11:58.170
and try to solve it.
00:11:58.170 --> 00:11:59.640
And in the next
problem, I'll try
00:11:59.640 --> 00:12:01.580
to do an example which is exact.
00:12:01.580 --> 00:12:05.140
Here I'm treating the problem
of it more qualitatively.
00:12:05.140 --> 00:12:09.730
And let's see what we can
figure out just from this fact
00:12:09.730 --> 00:12:13.240
that the wavelength
of the massive one
00:12:13.240 --> 00:12:16.900
is much smaller than
of the light one.
00:12:16.900 --> 00:12:23.900
We are now looking for a
solution to this problem
00:12:23.900 --> 00:12:28.680
where everything is moving with
the same phase and frequency.
00:12:28.680 --> 00:12:30.490
All right?
00:12:30.490 --> 00:12:34.980
And we are looking for
the lowest normal mode,
00:12:34.980 --> 00:12:37.730
in other words, when
everything is moving
00:12:37.730 --> 00:12:42.662
as slowly as possible subject
to the boundary conditions.
00:12:45.500 --> 00:12:50.360
Now, moving slowly
means big wavelengths.
00:12:50.360 --> 00:12:53.150
So what we have to try
to figure out-- here
00:12:53.150 --> 00:12:58.184
is a sketch of this string.
00:12:58.184 --> 00:12:59.100
This is the heavy one.
00:12:59.100 --> 00:13:01.440
This is the light one.
00:13:01.440 --> 00:13:08.150
And what we want, we want
to find, on each side,
00:13:08.150 --> 00:13:10.670
this string is just
like any old string.
00:13:10.670 --> 00:13:14.210
In the normal mode--
in a single mode,
00:13:14.210 --> 00:13:17.070
will have a sinusoidal
function, all right,
00:13:17.070 --> 00:13:20.390
of a single wavelength
on both sides.
00:13:20.390 --> 00:13:24.890
So this must be a part of a
sine curve or a cosine curve.
00:13:24.890 --> 00:13:25.925
So must this.
00:13:29.180 --> 00:13:33.580
And we want to make it as
long wavelength as possible,
00:13:33.580 --> 00:13:37.060
and we want to satisfy all
the boundary conditions.
00:13:37.060 --> 00:13:41.360
So here it's located.
00:13:41.360 --> 00:13:43.570
Here this one is located.
00:13:43.570 --> 00:13:44.900
OK?
00:13:44.900 --> 00:13:49.220
The one on the right, you want
to make as long a wavelength as
00:13:49.220 --> 00:13:56.610
possible, and so this is
almost a straight line.
00:13:56.610 --> 00:13:58.710
This is part of a
sine curve, but it's
00:13:58.710 --> 00:14:01.430
almost a straight line.
00:14:01.430 --> 00:14:06.050
This one is part
of a sine curve.
00:14:06.050 --> 00:14:08.840
And at the boundary,
we must have
00:14:08.840 --> 00:14:11.910
the two touch, so the
string is not broken,
00:14:11.910 --> 00:14:13.980
and the slope to be the same.
00:14:13.980 --> 00:14:17.350
So the answer is obvious.
00:14:17.350 --> 00:14:21.270
The solution will be
something like a sine that
00:14:21.270 --> 00:14:25.540
just turns around
near the top here.
00:14:25.540 --> 00:14:29.100
And when the curvature
of this sine curve
00:14:29.100 --> 00:14:32.690
is such that it points
towards this point,
00:14:32.690 --> 00:14:36.750
then we have found the
lowest frequency solution.
00:14:36.750 --> 00:14:40.000
All right?
00:14:40.000 --> 00:14:41.040
This is approximate.
00:14:41.040 --> 00:14:44.590
This will have very slight
curvature but approximate.
00:14:44.590 --> 00:14:48.770
And so what we see,
when the wavelength
00:14:48.770 --> 00:14:51.710
on the left-hand
side is basically
00:14:51.710 --> 00:14:54.340
4 times this, which is 4L.
00:14:54.340 --> 00:14:57.530
So this will be the
lowest normal mode
00:14:57.530 --> 00:15:00.550
that we figured out without
solving any equations,
00:15:00.550 --> 00:15:05.140
from just our knowledge
of the boundary conditions
00:15:05.140 --> 00:15:10.740
and what normal mode solutions
are on the string like this.
00:15:10.740 --> 00:15:11.400
All right?
00:15:11.400 --> 00:15:14.920
Now, I can now,
knowing the wavelength,
00:15:14.920 --> 00:15:21.350
I can calculate the frequency
this corresponds to because we
00:15:21.350 --> 00:15:28.080
know that-- we've got it
here-- omega is simply
00:15:28.080 --> 00:15:30.440
k times v. All right?
00:15:30.440 --> 00:15:34.810
k is 2 pi over the wavelength.
00:15:34.810 --> 00:15:36.220
So we have omega 1.
00:15:36.220 --> 00:15:38.980
This is the lowest normal mode.
00:15:38.980 --> 00:15:43.370
Angular frequency, 2 pi
over lambda 1 times vH.
00:15:43.370 --> 00:15:45.520
We know what lambda 1 is.
00:15:45.520 --> 00:15:50.180
We just argued ourselves to
show it's approximately 4L.
00:15:50.180 --> 00:15:54.030
It's actually a little less
than 4L, but approximately 4L.
00:15:54.030 --> 00:15:58.420
And so this is the lowest
normal mode frequency.
00:15:58.420 --> 00:15:59.010
All right.
00:15:59.010 --> 00:16:00.470
Now, what's the next normal?
00:16:00.470 --> 00:16:03.710
They ask us for the two
lowest normal modes.
00:16:03.710 --> 00:16:06.010
We repeat the argument.
00:16:06.010 --> 00:16:11.930
We are now trying to figure
out-- each of these strings
00:16:11.930 --> 00:16:19.390
now we would like to oscillate
with slightly higher frequency.
00:16:19.390 --> 00:16:25.710
This, again, will be a
sine curve on both sides.
00:16:25.710 --> 00:16:28.770
And again, we've got to
find a situation where
00:16:28.770 --> 00:16:34.060
the curvature-- the sine
has shorter wavelength--
00:16:34.060 --> 00:16:38.610
but the minimum shorter
that I can make it
00:16:38.610 --> 00:16:43.200
so that I satisfy the
boundary conditions as before.
00:16:43.200 --> 00:16:46.820
The strings must not break,
and the slope must be the same.
00:16:46.820 --> 00:16:48.950
Well, if you play
around, you'll see
00:16:48.950 --> 00:16:53.150
that the next lowest normal
mode will look like this.
00:16:53.150 --> 00:16:56.150
The string goes like this.
00:16:56.150 --> 00:16:57.010
All right?
00:16:57.010 --> 00:17:02.480
And here, the massive
one is just curving.
00:17:02.480 --> 00:17:05.220
And when its curvature
is such that it
00:17:05.220 --> 00:17:10.119
points towards this origin,
where this is almost straight,
00:17:10.119 --> 00:17:15.240
that will be the sketch
of the normal mode.
00:17:15.240 --> 00:17:16.640
So that's what it
will look like.
00:17:16.640 --> 00:17:18.839
That will be the next
lowest normal mode.
00:17:18.839 --> 00:17:20.190
Now, what is the wavelength?
00:17:20.190 --> 00:17:22.550
You can calculate
that from this length.
00:17:22.550 --> 00:17:27.800
It's just a little
less than 4/3 L. OK?
00:17:27.800 --> 00:17:33.870
And again, like before, since we
know the vH and the wavelength,
00:17:33.870 --> 00:17:36.580
we can calculate the frequency.
00:17:36.580 --> 00:17:39.510
So these are the two
normal mode frequencies.
00:17:39.510 --> 00:17:41.970
And why did I say it's
counter-intuitive?
00:17:41.970 --> 00:17:44.240
For you, maybe it is intuitive.
00:17:44.240 --> 00:17:48.870
But I would have thought that
if you connect a heavy string
00:17:48.870 --> 00:17:51.370
to a light one, it's
the light one that
00:17:51.370 --> 00:17:55.090
would be wobbling up and
down and not the heavy one.
00:17:55.090 --> 00:17:58.880
And in the reality, it's
the opposite that happens.
00:17:58.880 --> 00:18:00.480
It's the little guy
that seems to be
00:18:00.480 --> 00:18:02.670
deciding what the
big guy is doing.
00:18:02.670 --> 00:18:05.540
And this is the
one which is-- also
00:18:05.540 --> 00:18:10.240
this one-- all the time
is nearly straight.
00:18:10.240 --> 00:18:13.960
The role of this one
is simply to keep
00:18:13.960 --> 00:18:18.920
this end of the heavy
string tied down,
00:18:18.920 --> 00:18:24.680
and it determines the
boundary condition here.
00:18:24.680 --> 00:18:25.450
OK?
00:18:25.450 --> 00:18:29.580
But it's an interesting
case to think about.
00:18:29.580 --> 00:18:30.140
OK.
00:18:30.140 --> 00:18:34.910
So that's the first
problem I wanted to do.
00:18:34.910 --> 00:18:38.260
Now let's go to another one.
00:18:38.260 --> 00:18:42.700
In some ways, it's similar
to the one we've done.
00:18:42.700 --> 00:18:48.500
But just in order to show you
how general these discussions
00:18:48.500 --> 00:18:54.930
are of these problems,
I'm taking a situation
00:18:54.930 --> 00:18:59.230
where the displacement
from equilibrium
00:18:59.230 --> 00:19:04.970
is not transverse to the
location of the oscillator.
00:19:04.970 --> 00:19:09.025
I'm going to consider an
example of longitudinal waves.
00:19:15.710 --> 00:19:17.670
Let me discuss the problem here.
00:19:28.230 --> 00:19:31.390
This is the problem
we're going to consider.
00:19:31.390 --> 00:19:37.520
We're going to consider a
solid rod attached to a wall.
00:19:37.520 --> 00:19:42.280
So this is just a round rod.
00:19:45.600 --> 00:19:48.650
It is connected to another rod.
00:19:51.305 --> 00:19:55.850
Let's say take some case,
maybe a tungsten, a lead alloy,
00:19:55.850 --> 00:19:59.200
a rod of tungsten
lead, which has
00:19:59.200 --> 00:20:04.010
a certain Young's modulus y.
00:20:04.010 --> 00:20:06.860
And Young's modulus,
let me remind you,
00:20:06.860 --> 00:20:09.730
it is the extension
per unit length
00:20:09.730 --> 00:20:14.190
if you apply to it a force,
certain force per unit area.
00:20:14.190 --> 00:20:16.220
So it's extension
per unit length
00:20:16.220 --> 00:20:18.100
divided by force
per unit length.
00:20:18.100 --> 00:20:24.020
It tells you how,
under the system,
00:20:24.020 --> 00:20:28.400
what strain you get
into the system.
00:20:28.400 --> 00:20:32.060
This is analogous to
like, in a spring,
00:20:32.060 --> 00:20:35.960
we talk of the constant
k for the spring.
00:20:35.960 --> 00:20:38.070
Tells you how much
the spring [INAUDIBLE]
00:20:38.070 --> 00:20:39.510
when you apply a force.
00:20:39.510 --> 00:20:42.090
So the y will tell
you how much this
00:20:42.090 --> 00:20:46.330
extends under a
force per unit area.
00:20:46.330 --> 00:20:52.940
And I'm going to assume this rod
has a density rho, mass density
00:20:52.940 --> 00:20:54.390
rho.
00:20:54.390 --> 00:20:57.620
It's connected to another,
but this time much
00:20:57.620 --> 00:21:01.790
lighter rod, a rod
which has three
00:21:01.790 --> 00:21:06.690
lengths, 3L length, same area.
00:21:06.690 --> 00:21:09.320
They are welded together
here, connected together,
00:21:09.320 --> 00:21:10.640
glued together.
00:21:10.640 --> 00:21:13.210
OK?
00:21:13.210 --> 00:21:15.050
I've chosen two materials.
00:21:15.050 --> 00:21:16.970
And if you look
in the books, you
00:21:16.970 --> 00:21:23.490
can find different
materials having same y.
00:21:23.490 --> 00:21:25.950
This is a carbon
fiber composite.
00:21:25.950 --> 00:21:29.630
This is a tungsten-lead alloy.
00:21:29.630 --> 00:21:32.920
And they both have the
same Young's modulus.
00:21:32.920 --> 00:21:35.930
But this is much, much lighter.
00:21:35.930 --> 00:21:40.610
The density of this is 1/9 of
the density of this material.
00:21:40.610 --> 00:21:44.890
And I chose these-- I wanted
to come up with numbers where
00:21:44.890 --> 00:21:48.440
one can solve the
problem completely.
00:21:48.440 --> 00:21:54.790
In the previous problem, I
want to do this qualitatively
00:21:54.790 --> 00:21:58.330
so I didn't have to pick
up numbers which gave you
00:21:58.330 --> 00:22:02.755
a final mathematical description
which has an analytic solution.
00:22:07.360 --> 00:22:08.600
OK.
00:22:08.600 --> 00:22:12.490
So we have these two
rods connected like this.
00:22:12.490 --> 00:22:14.225
This one is attached to a wall.
00:22:14.225 --> 00:22:16.830
It cannot move.
00:22:16.830 --> 00:22:19.306
And this is free.
00:22:19.306 --> 00:22:21.605
I purposely wanted
a problem where
00:22:21.605 --> 00:22:23.530
there are new
things coming in so
00:22:23.530 --> 00:22:25.960
that you can see
how one tackles it.
00:22:25.960 --> 00:22:27.570
OK.
00:22:27.570 --> 00:22:29.180
As usual, there's
some assumption.
00:22:29.180 --> 00:22:32.310
We're going to assume that
this is a lossless system,
00:22:32.310 --> 00:22:37.480
while the rods are oscillating
that they're not losing energy,
00:22:37.480 --> 00:22:40.640
that the oscillations
are small so
00:22:40.640 --> 00:22:44.290
that we can do all the
usual approximations.
00:22:44.290 --> 00:22:48.400
We'll make the assumption
that this radius is small
00:22:48.400 --> 00:22:50.740
compared to this
length so that we
00:22:50.740 --> 00:22:53.590
don't have to worry about
transverse oscillations
00:22:53.590 --> 00:22:54.590
of the system.
00:22:54.590 --> 00:22:58.080
We're only talking about
longitudinal oscillations.
00:22:58.080 --> 00:23:01.870
So we're talking about
small oscillations
00:23:01.870 --> 00:23:05.310
in the longitudinal direction.
00:23:05.310 --> 00:23:08.860
And we are told that
for such a system--
00:23:08.860 --> 00:23:12.830
and you could derive
it for yourselves--
00:23:12.830 --> 00:23:17.270
that the phase velocity is
the square root of Young's
00:23:17.270 --> 00:23:20.070
modulus divided by the density.
00:23:20.070 --> 00:23:22.500
So it's different
for the two sides.
00:23:22.500 --> 00:23:23.240
OK?
00:23:23.240 --> 00:23:27.570
And the question
is, can we calculate
00:23:27.570 --> 00:23:30.270
for this-- knowing
all these quantities,
00:23:30.270 --> 00:23:32.950
these are now all
given quantities--
00:23:32.950 --> 00:23:37.030
can we calculate what is
the angular frequency omega
00:23:37.030 --> 00:23:43.200
1 of the first normal
mode as this oscillates?
00:23:43.200 --> 00:23:46.610
The problem, in some ways, is
similar to the previous one.
00:23:46.610 --> 00:23:51.160
But as I say, normally
we more often do things
00:23:51.160 --> 00:23:55.560
like strings oscillating
because it's easier
00:23:55.560 --> 00:24:03.740
to plot on the board the
displacement of the string,
00:24:03.740 --> 00:24:06.750
which is in the y
direction, and the position
00:24:06.750 --> 00:24:09.540
in the x direction.
00:24:09.540 --> 00:24:13.030
Here we have the
displacement from equilibrium
00:24:13.030 --> 00:24:17.000
at any point in
the same direction,
00:24:17.000 --> 00:24:21.160
in the same as the position, and
that makes it harder to plot.
00:24:21.160 --> 00:24:24.240
That's why one normally
doesn't take this example.
00:24:27.170 --> 00:24:28.990
We tend to take
examples which are
00:24:28.990 --> 00:24:33.290
easier to illustrate
on the board.
00:24:33.290 --> 00:24:34.140
OK.
00:24:34.140 --> 00:24:39.560
So let's now, as usual-- this
is the physical situation--
00:24:39.560 --> 00:24:43.170
what does it correspond to
as a description in terms
00:24:43.170 --> 00:24:44.980
of mathematics?
00:24:44.980 --> 00:24:50.020
Well, let's define, at
any point x and time
00:24:50.020 --> 00:24:54.320
t, the displacement
of the material
00:24:54.320 --> 00:24:58.450
of the rod from the
position of equilibrium
00:24:58.450 --> 00:25:00.610
by the Greek letter psi.
00:25:00.610 --> 00:25:06.120
So it's psi at x of t
00:25:06.120 --> 00:25:13.150
Now, this system is,
again, an ideal system
00:25:13.150 --> 00:25:19.930
of a continuous distribution
of harmonic oscillators.
00:25:19.930 --> 00:25:23.630
Each piece of mass oscillates
backwards and forwards
00:25:23.630 --> 00:25:24.620
longitudinally.
00:25:24.620 --> 00:25:26.290
That's the oscillator.
00:25:26.290 --> 00:25:31.090
It's continuous, that is,
an oscillator at every x.
00:25:31.090 --> 00:25:31.590
OK?
00:25:34.420 --> 00:25:38.060
And each oscillator is
coupled to its neighbors.
00:25:38.060 --> 00:25:43.360
So this is mathematically
exactly the same situation
00:25:43.360 --> 00:25:46.600
as a string, where the
oscillators are there
00:25:46.600 --> 00:25:50.850
oscillating up and down
instead of longitudinally.
00:25:50.850 --> 00:25:54.470
So the equation of
motion of this system
00:25:54.470 --> 00:25:56.348
will be, as always,
the wave equation.
00:25:58.976 --> 00:26:01.970
This is the longitudinal
displacement,
00:26:01.970 --> 00:26:03.190
how it's oscillating.
00:26:03.190 --> 00:26:06.150
v is the phase velocity.
00:26:09.040 --> 00:26:14.500
And we know that
the phase velocity
00:26:14.500 --> 00:26:18.560
is the square root of
Young's modulus over density.
00:26:18.560 --> 00:26:21.590
Since we know the
ratio of the densities,
00:26:21.590 --> 00:26:25.520
and the y is the same
for both materials,
00:26:25.520 --> 00:26:29.560
you know that the ratio of
the phase velocity in the two
00:26:29.560 --> 00:26:34.710
materials, v2 over v1, will
be 3, the square root of 9.
00:26:34.710 --> 00:26:36.320
OK?
00:26:36.320 --> 00:26:43.420
And as always, we know that
omega/k is the phase velocity.
00:26:43.420 --> 00:26:48.560
Otherwise, it wouldn't
satisfy this equation.
00:26:48.560 --> 00:26:53.720
So we know that the k2/k1--
and I'm reminding you,
00:26:53.720 --> 00:26:56.380
k is 2 pi over the
wavelength, right?
00:26:56.380 --> 00:26:57.910
So k2/k1 is 1/3.
00:27:00.820 --> 00:27:01.320
OK.
00:27:01.320 --> 00:27:05.630
So this is the equations of
motion and the constants of it.
00:27:05.630 --> 00:27:08.020
What are the
boundary conditions?
00:27:08.020 --> 00:27:10.410
Well, they're slightly
different to the one
00:27:10.410 --> 00:27:13.840
we've had of strings
attached at both ends.
00:27:17.830 --> 00:27:23.140
We know that, in the
middle, the two rods
00:27:23.140 --> 00:27:25.840
are attached to each other.
00:27:25.840 --> 00:27:28.790
So the displacement
from equilibrium,
00:27:28.790 --> 00:27:31.860
whether you're on the left
rod or on the right rod,
00:27:31.860 --> 00:27:33.262
will be the same.
00:27:33.262 --> 00:27:37.590
So the one on the
left, which I call
00:27:37.590 --> 00:27:40.880
psi 1, at position
L [INAUDIBLE] t,
00:27:40.880 --> 00:27:43.120
will be equal to
psi 2 at L of t.
00:27:43.120 --> 00:27:45.690
So that's one
boundary condition.
00:27:45.690 --> 00:27:48.460
Another boundary
condition is the one
00:27:48.460 --> 00:27:51.530
analogous to when you
have two strings joined,
00:27:51.530 --> 00:27:55.350
that the slope of the strings
must be the same on both sides
00:27:55.350 --> 00:27:57.360
because the junction
has no mass.
00:27:57.360 --> 00:28:00.540
Similarly here, the junction
between the two rods has no
00:28:00.540 --> 00:28:04.920
mass, so there cannot be a
net force on that junction.
00:28:04.920 --> 00:28:10.720
And since the y, the Young's
modulus, is the same,
00:28:10.720 --> 00:28:14.070
the tension is the
same inside, then
00:28:14.070 --> 00:28:21.940
the slope of d psi 1
dx at that junction
00:28:21.940 --> 00:28:24.920
must equal to d
psi dx at the ther.
00:28:24.920 --> 00:28:29.730
As I repeat, this is analogous
to a junction between two
00:28:29.730 --> 00:28:32.960
strings when you're talking
about transverse oscillations
00:28:32.960 --> 00:28:33.800
of strings.
00:28:33.800 --> 00:28:38.670
The slope at the junction must
be the same on both sides,
00:28:38.670 --> 00:28:42.990
unless there is a massive
object at that junction.
00:28:42.990 --> 00:28:44.360
And we don't have such a thing.
00:28:44.360 --> 00:28:47.715
If there was a heavy bead or
something at that junction,
00:28:47.715 --> 00:28:50.250
this would no longer be true.
00:28:50.250 --> 00:28:52.600
The other thing is, what
are the boundary conditions
00:28:52.600 --> 00:28:54.360
of the ends?
00:28:54.360 --> 00:28:56.690
On the left end, it's easier.
00:28:56.690 --> 00:28:59.890
We said the left end is
attached to the wall,
00:28:59.890 --> 00:29:02.080
therefore it cannot move.
00:29:02.080 --> 00:29:02.780
OK?
00:29:02.780 --> 00:29:08.120
Therefore, the displacement from
equilibrium of this left end
00:29:08.120 --> 00:29:11.610
is equal to 0 at all times.
00:29:11.610 --> 00:29:13.930
The right end is different.
00:29:13.930 --> 00:29:17.470
It's free, and so
it's certainly not 0.
00:29:17.470 --> 00:29:19.180
What do we know about that?
00:29:19.180 --> 00:29:25.390
Well, at the end, the
rod ends, and there
00:29:25.390 --> 00:29:27.430
is no mass or anything there.
00:29:27.430 --> 00:29:31.710
So the end of the rod
is not pushing anything.
00:29:31.710 --> 00:29:36.680
There's effective, it's pushing
a 0 mass at the end of it.
00:29:36.680 --> 00:29:42.100
Therefore, there cannot
be a net force on that.
00:29:42.100 --> 00:29:47.070
And so the rod cannot
be compressed at the end
00:29:47.070 --> 00:29:51.590
or elongated because then it
would be exerting like a spring
00:29:51.590 --> 00:29:55.790
attached to nothing with
exerted force on it,
00:29:55.790 --> 00:29:57.270
which is not possible.
00:29:57.270 --> 00:30:02.410
And so the rate of
change of psi 1 with x
00:30:02.410 --> 00:30:04.345
of the displacement
from equilibrium
00:30:04.345 --> 00:30:11.950
from x at the-- sorry,
I'm here, I'm here.
00:30:11.950 --> 00:30:19.980
The slope of psi 2 with respect
to x at the end, which is-- I
00:30:19.980 --> 00:30:20.895
am sorry.
00:30:20.895 --> 00:30:24.240
This will be 4L.
00:30:24.240 --> 00:30:25.190
This is my mistake.
00:30:25.190 --> 00:30:26.260
This is 4L here.
00:30:30.850 --> 00:30:31.570
OK.
00:30:31.570 --> 00:30:36.290
At the end of the
rod, it has to be 0.
00:30:36.290 --> 00:30:36.940
OK.
00:30:36.940 --> 00:30:39.950
So this is now the
mathematical description.
00:30:39.950 --> 00:30:45.840
And all we have to do is solve
the equation of motion subject
00:30:45.840 --> 00:30:49.150
to these boundary
conditions and find out
00:30:49.150 --> 00:30:54.480
what is the lowest frequency
that satisfies the boundary
00:30:54.480 --> 00:30:58.280
conditions and
the wave equation.
00:30:58.280 --> 00:30:59.110
OK?
00:30:59.110 --> 00:31:05.430
And again, I have written
out the solution for you.
00:31:05.430 --> 00:31:07.610
And let me go a
little faster now.
00:31:10.500 --> 00:31:12.760
We now have experience.
00:31:12.760 --> 00:31:18.420
We know that the wave equation
has standing wave solutions,
00:31:18.420 --> 00:31:24.510
which are some amplitude times
a sinusoidal function in x
00:31:24.510 --> 00:31:28.980
and a sinusoidal
function of time.
00:31:28.980 --> 00:31:37.290
And now to save time, I am
going to inspect immediately
00:31:37.290 --> 00:31:40.610
the boundary
conditions, and using,
00:31:40.610 --> 00:31:44.010
in particular, these
boundary conditions,
00:31:44.010 --> 00:31:49.640
the ones at x equals
0 and 4L for all time.
00:31:49.640 --> 00:31:56.100
And I'm going to figure out
sinusoidal functions of x
00:31:56.100 --> 00:32:03.250
and t, which satisfy
these boundary conditions.
00:32:03.250 --> 00:32:06.380
Now, we know that we are
talking about standing waves.
00:32:06.380 --> 00:32:16.360
So for both rods, the
angular frequency,
00:32:16.360 --> 00:32:20.120
the [INAUDIBLE] oscillations
in time, will be the same.
00:32:20.120 --> 00:32:23.550
So this part will be the same.
00:32:23.550 --> 00:32:27.965
This part must satisfy
the boundary condition
00:32:27.965 --> 00:32:30.880
at the extreme left
and right, which
00:32:30.880 --> 00:32:36.840
forces me to write here
a sine k and here cosine
00:32:36.840 --> 00:32:41.790
k2 of what happens
at the boundaries.
00:32:41.790 --> 00:32:42.290
OK?
00:32:51.470 --> 00:32:54.660
A good exercise
for you is slowly
00:32:54.660 --> 00:32:57.410
think through the
boundary conditions
00:32:57.410 --> 00:33:01.530
and the general solutions,
the normal mode solutions,
00:33:01.530 --> 00:33:02.890
of the wave equation.
00:33:02.890 --> 00:33:04.350
And you'll come
to the conclusion
00:33:04.350 --> 00:33:14.010
that this describes the
standing waves in the two rods.
00:33:16.670 --> 00:33:17.640
OK.
00:33:17.640 --> 00:33:27.130
We know further that omega is
equal to k1 v1 in the left rod,
00:33:27.130 --> 00:33:30.230
and equal to k2 v2
in the right rod.
00:33:30.230 --> 00:33:33.160
This has to be the same
in both, as I said,
00:33:33.160 --> 00:33:35.860
because we are in
the normal mode.
00:33:35.860 --> 00:33:39.610
We're in a normal mode, where
the whole system is oscillating
00:33:39.610 --> 00:33:42.430
with the same
frequency and phase.
00:33:42.430 --> 00:33:42.930
OK.
00:33:42.930 --> 00:33:48.320
Now, we will use the other bit
of information, the boundary
00:33:48.320 --> 00:33:54.280
conditions at the junction,
which I've written here.
00:33:54.280 --> 00:34:07.810
And from that, try to find a
solution that satisfies these
00:34:07.810 --> 00:34:11.489
and see what is the angular
frequency of oscillation
00:34:11.489 --> 00:34:13.719
for that.
00:34:13.719 --> 00:34:31.179
So we can now write
at x equals L--
00:34:31.179 --> 00:34:34.159
these are the two equations.
00:34:34.159 --> 00:34:38.010
We'll take the displacement
of the rod on the left
00:34:38.010 --> 00:34:41.179
and make it equal to the right
and the slope on the left
00:34:41.179 --> 00:34:43.020
equal to the right.
00:34:43.020 --> 00:34:44.920
On the left, this
is the equation.
00:34:44.920 --> 00:34:46.639
On the right is that.
00:34:46.639 --> 00:34:50.139
At that boundary to
make the two equal,
00:34:50.139 --> 00:34:54.440
I get that A1 sine
k1 L at x equals
00:34:54.440 --> 00:35:01.230
L is equal to A2 cosine
times this, all right, at x.
00:35:01.230 --> 00:35:02.800
When we're talking
about x2 equals
00:35:02.800 --> 00:35:06.750
L, L minus 4L is
minus 3, et cetera.
00:35:06.750 --> 00:35:07.990
OK?
00:35:07.990 --> 00:35:13.580
The cosine terms cancel
because the time part
00:35:13.580 --> 00:35:16.410
cancels because they're
the same on both sides.
00:35:16.410 --> 00:35:19.010
So this is the equation
we have to satisfy
00:35:19.010 --> 00:35:20.680
so that the string
is not broken.
00:35:20.680 --> 00:35:23.160
And I'm just
writing it like this
00:35:23.160 --> 00:35:29.110
because we know that-- we
did that earlier-- that k2/k1
00:35:29.110 --> 00:35:30.340
is 1/3.
00:35:30.340 --> 00:35:34.640
And the cosine of a plus and
the minus angle is the same,
00:35:34.640 --> 00:35:38.890
so I've just replaced
that with plus k1 L.
00:35:38.890 --> 00:35:46.860
This is now looking
at the slopes
00:35:46.860 --> 00:35:52.355
of the displacement
psi at that boundary.
00:35:52.355 --> 00:35:55.350
So I differentiate
these with respect
00:35:55.350 --> 00:35:58.540
to x, all right, this one and
that one, and equate them.
00:35:58.540 --> 00:36:00.930
And I get this
equation, all right?
00:36:00.930 --> 00:36:05.110
And so in order to satisfy
the boundary conditions
00:36:05.110 --> 00:36:10.330
and the wave equation, I end
up with these two equations.
00:36:10.330 --> 00:36:13.260
I've chosen things
such that these
00:36:13.260 --> 00:36:16.330
can be solved analytically.
00:36:16.330 --> 00:36:17.240
It's not hard.
00:36:17.240 --> 00:36:18.620
You can do it for yourself.
00:36:18.620 --> 00:36:21.840
You have two equations,
two unknowns.
00:36:21.840 --> 00:36:25.240
The unknown is the
ratio of A2 to A1.
00:36:25.240 --> 00:36:26.440
That's one unknown.
00:36:26.440 --> 00:36:28.790
And the other unknown is k1.
00:36:28.790 --> 00:36:31.780
You can solve these
equations trigonometrically
00:36:31.780 --> 00:36:34.720
only because I've
chosen numbers of rho,
00:36:34.720 --> 00:36:37.830
et cetera, which make
the answer come out.
00:36:37.830 --> 00:36:42.940
And you end up that
you get k1 from it.
00:36:42.940 --> 00:36:46.130
Knowing k1, you can
calculate omega.
00:36:46.130 --> 00:36:48.080
And this is the
answer you will get.
00:36:48.080 --> 00:36:50.680
I won't waste your time here.
00:36:50.680 --> 00:36:54.405
You can do this
algebra for yourselves.
00:36:58.580 --> 00:37:02.730
So here I've got
the final answer
00:37:02.730 --> 00:37:04.490
because I had all the input.
00:37:04.490 --> 00:37:09.490
If I had tried to do the
same with the string one,
00:37:09.490 --> 00:37:13.350
if I chose some random ratio
of the density of the string
00:37:13.350 --> 00:37:17.350
on both sides, everything
else would have been similar.
00:37:17.350 --> 00:37:21.750
But here I would have ended up
with a transcendental equation,
00:37:21.750 --> 00:37:25.180
in general, which I
cannot solve analytically.
00:37:25.180 --> 00:37:29.170
And so if you take any
random case for yourself
00:37:29.170 --> 00:37:33.320
and try to solve it, you would
find that, at this point,
00:37:33.320 --> 00:37:34.487
you would be stuck.
00:37:34.487 --> 00:37:36.570
You would be stuck with a
transcendental equation,
00:37:36.570 --> 00:37:39.160
which, of course, you
could solve numerically
00:37:39.160 --> 00:37:40.990
over the computer, et cetera.
00:37:40.990 --> 00:37:41.500
OK.
00:37:41.500 --> 00:37:42.280
So that's the end.
00:37:42.280 --> 00:37:46.290
And now I'll come to the
third, yet another example
00:37:46.290 --> 00:37:49.760
of standing waves, this
time in three dimensions.
00:37:49.760 --> 00:37:54.260
So now we come to the third
standing wave problem.
00:37:54.260 --> 00:37:57.127
And just for variety,
I decided to take one
00:37:57.127 --> 00:37:57.960
in three dimensions.
00:38:00.474 --> 00:38:02.820
So the problem is the following.
00:38:02.820 --> 00:38:06.900
Suppose you have a room,
which is 2 meters in size
00:38:06.900 --> 00:38:08.290
by 3 meters by 4.
00:38:11.710 --> 00:38:17.990
What is the lowest
standing wave frequency
00:38:17.990 --> 00:38:20.170
that you can get in such a room?
00:38:20.170 --> 00:38:21.870
OK?
00:38:21.870 --> 00:38:29.360
So it's basically a problem
to do with pressure waves,
00:38:29.360 --> 00:38:33.200
because that's what sound
is, standing pressure
00:38:33.200 --> 00:38:36.610
waves in three dimensions.
00:38:36.610 --> 00:38:40.250
We are told that, in
this particular room,
00:38:40.250 --> 00:38:43.440
what the density of the air is.
00:38:43.440 --> 00:38:47.410
We're also told what the
bulk modulus, in other words,
00:38:47.410 --> 00:38:50.250
how compressed the
air can get and what's
00:38:50.250 --> 00:38:55.630
the resultant pressure--
what kind of compression
00:38:55.630 --> 00:38:57.880
gives rise to air.
00:38:57.880 --> 00:39:01.340
And we are asked to
calculate lowest normal mode
00:39:01.340 --> 00:39:03.770
frequency of this system.
00:39:03.770 --> 00:39:04.390
OK.
00:39:04.390 --> 00:39:05.430
So how do we do this?
00:39:08.290 --> 00:39:10.315
Not surprisingly,
you'll remember already
00:39:10.315 --> 00:39:14.980
when we were talking about
one-dimensional system, way
00:39:14.980 --> 00:39:20.510
the standing waves of strings
or on a rod, et cetera,
00:39:20.510 --> 00:39:22.670
it's quite complex.
00:39:22.670 --> 00:39:24.870
If you have three
dimensions, the situation
00:39:24.870 --> 00:39:26.660
is getting more and
more complicated.
00:39:29.410 --> 00:39:32.260
In this course,
we don't actually
00:39:32.260 --> 00:39:37.560
derive the three-dimensional
wave equation.
00:39:37.560 --> 00:39:42.730
But by analogy with
the one-dimensional one
00:39:42.730 --> 00:39:47.930
and two-dimensional
one, we sort of indicate
00:39:47.930 --> 00:39:50.290
what form you would
expect this to be.
00:39:50.290 --> 00:39:52.360
And it actually is the case.
00:39:52.360 --> 00:39:56.310
The three-dimensional
wave equation
00:39:56.310 --> 00:39:59.730
for a scalar quantity
like pressure
00:39:59.730 --> 00:40:04.150
that has no direction, all
right, is written here.
00:40:04.150 --> 00:40:09.590
It's the d2p dt squared,
like for the one dimension,
00:40:09.590 --> 00:40:13.650
is equal to v squared
times the gradient squared
00:40:13.650 --> 00:40:15.970
of the pressure,
which, if I write it
00:40:15.970 --> 00:40:21.770
in Cartesian coordinates,
it's written here.
00:40:21.770 --> 00:40:27.140
Clearly, if I take a
three-dimensional system
00:40:27.140 --> 00:40:31.080
but reduced two dimensions
with some small distances
00:40:31.080 --> 00:40:34.450
so that, in essence, it's
a one-dimensional problem,
00:40:34.450 --> 00:40:36.770
this gives you the wave
equation you would expect.
00:40:36.770 --> 00:40:39.700
If I remove two of
these terms, what I
00:40:39.700 --> 00:40:45.160
have is the wave equation
in one dimension.
00:40:45.160 --> 00:40:46.010
OK.
00:40:46.010 --> 00:40:53.165
In this equation, p
is the excess pressure
00:40:53.165 --> 00:40:58.420
of the air above its
equilibrium Imagine the room.
00:40:58.420 --> 00:40:59.960
Everything is at equilibrium.
00:40:59.960 --> 00:41:01.260
The air is not moving.
00:41:01.260 --> 00:41:03.470
The pressure is not
changing, et cetera.
00:41:03.470 --> 00:41:05.940
There will be some pressure.
00:41:05.940 --> 00:41:08.740
What we are
interested in is what
00:41:08.740 --> 00:41:13.290
happens if we displace
that air from equilibrium,
00:41:13.290 --> 00:41:18.650
change its pressure, and
let go, what will happen?
00:41:18.650 --> 00:41:21.390
And so it's that excess
pressure we are interested.
00:41:21.390 --> 00:41:23.050
That's the p.
00:41:23.050 --> 00:41:27.600
And we are interested
to see if there is a way
00:41:27.600 --> 00:41:31.070
to do that such that
the air in the room
00:41:31.070 --> 00:41:34.080
everywhere, the
pressure, oscillates
00:41:34.080 --> 00:41:36.750
with the same
frequency and phase.
00:41:36.750 --> 00:41:41.260
If it does, then in
that room the air
00:41:41.260 --> 00:41:45.530
will be in its normal mode,
oscillating in its normal mode.
00:41:48.130 --> 00:41:51.850
One can show when one derives
this-- and I did not derive it,
00:41:51.850 --> 00:41:55.890
of course-- is that
this v is the phase
00:41:55.890 --> 00:42:00.610
velocity of propagation
of pressure waves in air,
00:42:00.610 --> 00:42:06.790
the sound velocity, which is the
square root of the bulk modulus
00:42:06.790 --> 00:42:08.892
divided by the density.
00:42:08.892 --> 00:42:10.860
OK?
00:42:10.860 --> 00:42:11.520
OK.
00:42:11.520 --> 00:42:15.540
So the translation
of this problem
00:42:15.540 --> 00:42:19.565
is this is the equation
of motion of the system.
00:42:23.540 --> 00:42:26.400
And what are the
boundary conditions?
00:42:26.400 --> 00:42:32.510
The boundary conditions are
that at the edges of the room,
00:42:32.510 --> 00:42:37.940
at the walls, the
pressure does not
00:42:37.940 --> 00:42:43.050
change perpendicular
to the wall.
00:42:43.050 --> 00:42:50.240
So for example, the two walls,
at the end in the x direction,
00:42:50.240 --> 00:42:57.640
at the end there, at one side
of the wall, the rate of change
00:42:57.640 --> 00:43:01.570
of pressure with position and at
the other, it will not change.
00:43:01.570 --> 00:43:05.380
Top and bottom, perpendicular,
it will not change this way.
00:43:05.380 --> 00:43:06.455
It will not change.
00:43:09.700 --> 00:43:11.865
Why is this the
boundary condition?
00:43:16.020 --> 00:43:20.580
The reason for that is that
if you imagine at the wall,
00:43:20.580 --> 00:43:26.885
the molecules cannot move,
right, because of the wall.
00:43:26.885 --> 00:43:30.490
So therefore, you cannot
there, near the end,
00:43:30.490 --> 00:43:36.870
increase the number
of molecules.
00:43:36.870 --> 00:43:41.290
So perpendicular to
it, that the number
00:43:41.290 --> 00:43:45.530
will be independent of the
position, which is dp dx,
00:43:45.530 --> 00:43:47.840
is equal to 0.
00:43:47.840 --> 00:43:48.880
OK?
00:43:48.880 --> 00:43:57.330
And this is analogous to
the string situation, which
00:43:57.330 --> 00:44:00.420
is attached to a
massless ring where
00:44:00.420 --> 00:44:08.640
the slope is constant
at the boundary.
00:44:08.640 --> 00:44:09.300
OK.
00:44:09.300 --> 00:44:12.670
So with these
boundary conditions,
00:44:12.670 --> 00:44:16.186
what are the solutions
of this equation?
00:44:16.186 --> 00:44:16.685
OK?
00:44:24.180 --> 00:44:27.080
To find the most general
solution of this equation,
00:44:27.080 --> 00:44:29.440
it's obviously very,
very complicated.
00:44:29.440 --> 00:44:31.345
There are infinite
possibilities.
00:44:34.080 --> 00:44:40.910
But from my experience of
what happens in one dimension,
00:44:40.910 --> 00:44:44.130
we can do a pretty
good, intelligent guess.
00:44:44.130 --> 00:44:46.740
And what I've
written here, we can
00:44:46.740 --> 00:44:50.810
guess that the solution
will look like that.
00:44:50.810 --> 00:44:52.840
Why is it like this?
00:44:52.840 --> 00:44:56.460
Well, first of all,
we want a solution
00:44:56.460 --> 00:45:02.910
of this equation, which is a
normal mode, meaning everything
00:45:02.910 --> 00:45:06.790
oscillating with the
same frequency and phase.
00:45:06.790 --> 00:45:10.500
And so independent
of where you are,
00:45:10.500 --> 00:45:14.060
the time dependence must
be the same for everything
00:45:14.060 --> 00:45:16.990
or else it would not
be a normal mode.
00:45:16.990 --> 00:45:19.280
So that's why that.
00:45:19.280 --> 00:45:23.380
There is some amplitude,
so that's p maximum.
00:45:23.380 --> 00:45:28.630
You would expect symmetry
between the three directions.
00:45:28.630 --> 00:45:30.980
There's nothing
special between them.
00:45:30.980 --> 00:45:35.250
And you want some
function here which
00:45:35.250 --> 00:45:39.000
satisfies the
boundary conditions.
00:45:39.000 --> 00:45:46.430
And what you will find is
that this, a cosine kx,
00:45:46.430 --> 00:45:49.630
the slope of this with
respect to x gives you a sine.
00:45:49.630 --> 00:45:55.120
And at x equals 0, you will
get the boundary condition
00:45:55.120 --> 00:45:57.750
satisfied.
00:45:57.750 --> 00:46:01.140
And we'll see in a
second what constraint
00:46:01.140 --> 00:46:06.830
the boundary puts on
this at the other end.
00:46:06.830 --> 00:46:11.990
For this to be a solution
of our wave equation,
00:46:11.990 --> 00:46:18.790
we must satisfy this and that.
00:46:18.790 --> 00:46:21.920
And so in some ways, you could
say, look, I've guessed it.
00:46:21.920 --> 00:46:25.100
But then I'll use the uniqueness
there, say, OK, I've guessed it
00:46:25.100 --> 00:46:26.440
or I knew it.
00:46:26.440 --> 00:46:28.510
I guessed this was the solution.
00:46:28.510 --> 00:46:33.850
I will now check, does it
satisfy everything I know,
00:46:33.850 --> 00:46:36.660
the wave equation and
the boundary conditions?
00:46:36.660 --> 00:46:38.922
And the answer is yes, it does.
00:46:38.922 --> 00:46:40.380
And then I use the
uniqueness there
00:46:40.380 --> 00:46:44.500
that says therefore that is the
only solution that satisfies
00:46:44.500 --> 00:46:47.170
everything I know
about the situation.
00:46:47.170 --> 00:46:48.300
OK.
00:46:48.300 --> 00:47:00.200
Having done that, to get this
I used the boundary condition
00:47:00.200 --> 00:47:02.220
as x equals 0.
00:47:02.220 --> 00:47:09.110
But I want this slope also to
be 0 at the edges of the room.
00:47:09.110 --> 00:47:15.830
And that gives me a constraint
on what each one of these cases
00:47:15.830 --> 00:47:16.800
can be.
00:47:16.800 --> 00:47:29.395
And you find that kx has to
be some constant pi over Lx,
00:47:29.395 --> 00:47:36.060
where Lx is the length of
the room in the x direction,
00:47:36.060 --> 00:47:43.330
so that the slope of this at
the edge of the room is 0.
00:47:43.330 --> 00:47:46.800
Similarly for ky and kz.
00:47:46.800 --> 00:47:53.550
And so you find that these
L's, m's, and n's, have to be
00:47:53.550 --> 00:47:56.120
0, 1, 2, or 3.
00:47:56.120 --> 00:47:58.770
They cannot all be 0.
00:47:58.770 --> 00:48:01.980
If they are all 0,
then nothing goes on.
00:48:01.980 --> 00:48:08.826
This means that k is
0, which makes omega 0.
00:48:08.826 --> 00:48:12.230
It's a trivial, uninteresting
solution to our wave equation.
00:48:12.230 --> 00:48:14.995
So at least one of these
has to be non-zero.
00:48:17.780 --> 00:48:23.190
So which one will give
us the lowest frequency?
00:48:23.190 --> 00:48:24.580
All right?
00:48:24.580 --> 00:48:33.670
And so it will be the one
where the k, all right,
00:48:33.670 --> 00:48:43.630
is as small as possible, which
means that the wavelength is
00:48:43.630 --> 00:48:46.060
as long as possible.
00:48:46.060 --> 00:48:47.510
OK?
00:48:47.510 --> 00:48:56.750
So if we make the
L's and m, which
00:48:56.750 --> 00:49:02.030
correspond to the room,
the 2 meter times 3 meters,
00:49:02.030 --> 00:49:06.520
those to be 0, the only
direction we accept
00:49:06.520 --> 00:49:11.840
is in the direction of
z, which is the longest
00:49:11.840 --> 00:49:16.980
dimension of the room,
then that will give us
00:49:16.980 --> 00:49:22.686
the longest wavelength and
therefore lowest frequency.
00:49:22.686 --> 00:49:23.185
OK?
00:49:34.280 --> 00:49:44.310
So k, the largest value
of the wavelength,
00:49:44.310 --> 00:49:50.510
will be when kz is pi over 4m,
in other words, the wavelength
00:49:50.510 --> 00:49:55.420
in that direction
being 8 meters.
00:49:55.420 --> 00:50:02.790
And since we know that v is the
square root of the bulk modulus
00:50:02.790 --> 00:50:05.840
divided by the density, I
could take the numbers that
00:50:05.840 --> 00:50:09.535
are given, I get that the
phase velocity, or the velocity
00:50:09.535 --> 00:50:13.940
of sound in this
room, is 342 meters.
00:50:13.940 --> 00:50:17.160
The frequency, of course, is
equal to the phase velocity
00:50:17.160 --> 00:50:17.685
over lambda.
00:50:17.685 --> 00:50:20.380
Or the way I normally
remember, it's frequency times
00:50:20.380 --> 00:50:22.180
lambda is the phase velocity.
00:50:22.180 --> 00:50:23.660
All right?
00:50:23.660 --> 00:50:29.390
And I found the longest
wavelength is 8 meters.
00:50:29.390 --> 00:50:35.190
So from that, I get that
the lowest normal mode
00:50:35.190 --> 00:50:40.100
will oscillate at
almost 43 hertz.
00:50:40.100 --> 00:50:45.560
I'll just sketch for you here
what we've essentially-- we've
00:50:45.560 --> 00:50:55.370
got this room,
and we are looking
00:50:55.370 --> 00:51:00.270
for solutions which have
the longest wavelength.
00:51:00.270 --> 00:51:06.950
See, in principle,
one could have
00:51:06.950 --> 00:51:12.650
a wave like this
in this direction.
00:51:12.650 --> 00:51:15.930
This, the slope, has
to be perpendicular
00:51:15.930 --> 00:51:17.640
at the boundaries.
00:51:17.640 --> 00:51:19.550
You could have,
in this direction,
00:51:19.550 --> 00:51:25.230
for example, a wave like that.
00:51:25.230 --> 00:51:30.310
And in this direction, you could
have [INAUDIBLE] or a higher
00:51:30.310 --> 00:51:31.170
moment.
00:51:31.170 --> 00:51:32.800
So it's hard for me to draw it.
00:51:32.800 --> 00:51:36.020
But you can imagine
this pressure
00:51:36.020 --> 00:51:38.890
changing in all directions.
00:51:38.890 --> 00:51:42.310
But it has to be
represented, as I say,
00:51:42.310 --> 00:51:48.020
by a sinusoidal function
whose slope at the boundaries
00:51:48.020 --> 00:51:52.600
is always perpendicular
like this.
00:51:52.600 --> 00:51:57.730
And the wave number
is, of course,
00:51:57.730 --> 00:51:59.940
the sum of the wave
number of this squared
00:51:59.940 --> 00:52:01.790
plus this squared
plus this squared,
00:52:01.790 --> 00:52:04.830
taken the square root over.
00:52:04.830 --> 00:52:08.620
The lowest frequency
will be when,
00:52:08.620 --> 00:52:12.060
in this direction and this
direction, it's straight.
00:52:12.060 --> 00:52:15.680
There's no change of pressure,
no change in this direction.
00:52:15.680 --> 00:52:19.990
And the only direction
is in this direction,
00:52:19.990 --> 00:52:22.290
and this is the
4-meter direction.
00:52:22.290 --> 00:52:24.740
And that corresponds
to a wavelength
00:52:24.740 --> 00:52:28.130
of 8 meter, which
I calculated there.
00:52:28.130 --> 00:52:31.200
And so this will
behave in the same way
00:52:31.200 --> 00:52:33.450
as a one-dimensional system.
00:52:33.450 --> 00:52:35.920
Nothing changing
in two dimensions.
00:52:35.920 --> 00:52:38.150
And only in one
dimension it's changing.
00:52:38.150 --> 00:52:42.620
That will give you the
lowest normal mode.
00:52:42.620 --> 00:52:47.500
And the other normal modes,
you can calculate the frequency
00:52:47.500 --> 00:52:53.270
as you change the little
l, m, and n, play around.
00:52:53.270 --> 00:52:57.060
Always remember, at least one
of them has to be non-zero.
00:52:57.060 --> 00:53:00.300
And you can calculate all
the frequencies like this.
00:53:00.300 --> 00:53:00.840
OK.
00:53:00.840 --> 00:53:04.910
That's enough for today
on standing waves.
00:53:04.910 --> 00:53:06.760
Thank you.