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PROFESSOR: Welcome back.
00:00:23.240 --> 00:00:27.660
Today, we will
consider problems where
00:00:27.660 --> 00:00:30.940
the solutions are
standing waves.
00:00:30.940 --> 00:00:36.550
So the first problem I'm going
to do is shown over here.
00:00:36.550 --> 00:00:47.298
What we have is a string
attached at both ends.
00:00:47.298 --> 00:00:50.990
The length of the
string is L. And it
00:00:50.990 --> 00:00:55.800
is distorted in
this strange shape.
00:00:55.800 --> 00:00:57.720
The dimensions are
given everywhere.
00:00:57.720 --> 00:01:02.210
It's symmetric about
the middle, OK?
00:01:02.210 --> 00:01:07.040
Somehow miraculously,
it's held fixed.
00:01:07.040 --> 00:01:10.120
You could imagine maybe sort
of nails nailed here and here
00:01:10.120 --> 00:01:11.890
and fingers holding there.
00:01:11.890 --> 00:01:17.020
And then miraculously,
we let go, all right?
00:01:17.020 --> 00:01:20.950
And the question
is after we let go,
00:01:20.950 --> 00:01:25.210
what is the wavelength of
the lowest mode that is not
00:01:25.210 --> 00:01:32.860
excited, the amplitude of the
lowest mode which is excited?
00:01:32.860 --> 00:01:36.110
And finally, the
question is, what
00:01:36.110 --> 00:01:41.640
is the shape of this string
at this strange time?
00:01:41.640 --> 00:01:44.712
Now, then what do they tell us?
00:01:44.712 --> 00:01:49.180
That this is an ideal
lossless string,
00:01:49.180 --> 00:01:53.800
which has a mass per
unit length of mu.
00:01:53.800 --> 00:01:59.460
It has a uniform tension T.
For all the equations to work
00:01:59.460 --> 00:02:01.760
and the derivation
of the wave equation,
00:02:01.760 --> 00:02:08.090
you have to assume that the
angle is sine angle, et cetera,
00:02:08.090 --> 00:02:11.720
is equal like this, which
is-- you look at this
00:02:11.720 --> 00:02:13.100
and say, holy smoke.
00:02:13.100 --> 00:02:14.930
How can this be the case?
00:02:14.930 --> 00:02:19.870
Well, of course, this is
an idealized situation.
00:02:19.870 --> 00:02:23.310
In reality, this would
be very, very small
00:02:23.310 --> 00:02:25.240
compared to that, all right?
00:02:25.240 --> 00:02:31.042
And one would have to have
some curvature at these points.
00:02:31.042 --> 00:02:35.450
The differences will
be, the real situation
00:02:35.450 --> 00:02:38.420
from this idealised
one, will only
00:02:38.420 --> 00:02:41.150
differ in some very
high frequencies.
00:02:41.150 --> 00:02:45.760
So the first approximation,
one can almost
00:02:45.760 --> 00:02:49.520
imagine that it's
possible to set this up.
00:02:49.520 --> 00:02:52.010
And this assumption, as
I told you, initially
00:02:52.010 --> 00:02:58.970
is the system, at t
equals 0, is stationary.
00:02:58.970 --> 00:02:59.470
OK.
00:03:03.220 --> 00:03:10.430
The above, as I was just saying,
is really highly idealized.
00:03:10.430 --> 00:03:14.500
It is almost a
mathematical representation
00:03:14.500 --> 00:03:18.620
already of a real situation.
00:03:18.620 --> 00:03:20.790
So you could imagine
this almost being
00:03:20.790 --> 00:03:24.700
a mathematical description
of this situation.
00:03:24.700 --> 00:03:30.470
And what makes it physics
rather than the mathematics
00:03:30.470 --> 00:03:35.040
is that string obeys
the wave equation.
00:03:35.040 --> 00:03:38.470
It obeys this wave equation.
00:03:38.470 --> 00:03:46.010
Once I've told you that
that picture on that diagram
00:03:46.010 --> 00:03:50.420
satisfies the wave equation
where V is a constant given
00:03:50.420 --> 00:03:53.590
by this quantity, we've
now converted this problem
00:03:53.590 --> 00:03:54.750
completely to mathematics.
00:03:58.610 --> 00:04:03.930
One thing to notice, this
is the phase velocity
00:04:03.930 --> 00:04:08.080
of propagation of a
progressive wave on a string.
00:04:08.080 --> 00:04:11.150
So here, that time,
they ask us in part
00:04:11.150 --> 00:04:19.290
C-- is the shape of the string
at a time L over V from time
00:04:19.290 --> 00:04:21.209
equals 0.
00:04:21.209 --> 00:04:25.350
So we're now solving this
problem, mathematics problem.
00:04:29.870 --> 00:04:42.110
We know that if you have
a coupled oscillator,
00:04:42.110 --> 00:04:46.700
there are normal mode
solutions, solutions
00:04:46.700 --> 00:04:49.010
where every oscillator
is oscillating
00:04:49.010 --> 00:04:53.170
with the same
frequency and phase.
00:04:53.170 --> 00:04:57.050
For a system, a
continuous system
00:04:57.050 --> 00:05:00.700
like this is, an infinite
number of identical oscillators
00:05:00.700 --> 00:05:06.910
in a straight line for a
string like that, the most
00:05:06.910 --> 00:05:12.540
general solution
will also be-- one
00:05:12.540 --> 00:05:16.610
will be able to represent
it as a superposition
00:05:16.610 --> 00:05:19.110
of normal modes.
00:05:19.110 --> 00:05:23.330
Of course, if you
look at such a system
00:05:23.330 --> 00:05:25.430
and you distort this
from equilibrium,
00:05:25.430 --> 00:05:28.950
you can set up an infinite
different varieties
00:05:28.950 --> 00:05:30.220
of solutions.
00:05:30.220 --> 00:05:33.330
But what I'm saying is that
any one of those solutions
00:05:33.330 --> 00:05:36.180
however complicated,
one could always
00:05:36.180 --> 00:05:41.520
reduce it to a superposition
of normal modes.
00:05:41.520 --> 00:05:45.180
In particular, if you
study the solutions
00:05:45.180 --> 00:05:50.620
of a system like this,
one dimensional system
00:05:50.620 --> 00:05:56.740
like that, you will find
that the normal modes are,
00:05:56.740 --> 00:06:00.830
in fact, standing waves
which are sinusoidal.
00:06:07.610 --> 00:06:16.210
So I've written for you here
a completely general solution
00:06:16.210 --> 00:06:17.405
to a string.
00:06:20.510 --> 00:06:26.390
And you can write it, as
I said, as a superposition
00:06:26.390 --> 00:06:31.855
of an infinite number of normal
modes, pretty complicated.
00:06:34.550 --> 00:06:38.270
It tells you at every
position on the string
00:06:38.270 --> 00:06:41.130
and at every time what
is the displacement.
00:06:41.130 --> 00:06:43.480
You can calculate the
transverse velocity
00:06:43.480 --> 00:06:45.710
by differentiating, et cetera.
00:06:45.710 --> 00:06:48.230
It just tells you everything
about the oscillations
00:06:48.230 --> 00:06:50.260
of a string.
00:06:50.260 --> 00:06:52.760
This is not any old string.
00:06:52.760 --> 00:06:55.520
Because we are told
some things about it.
00:06:55.520 --> 00:06:58.810
We are told it's
attached at both ends
00:06:58.810 --> 00:07:02.500
so that string has no
displacement in the Y
00:07:02.500 --> 00:07:08.770
direction at x equals 0 and
at x equals L. For all times
00:07:08.770 --> 00:07:18.280
t this equation will satisfy
that string-- doesn't matter
00:07:18.280 --> 00:07:21.760
even what distortion
it is-- provided
00:07:21.760 --> 00:07:27.060
we put some constraints
on these various constants
00:07:27.060 --> 00:07:28.510
such that this is true.
00:07:31.550 --> 00:07:33.850
Well, you can almost
do it in your head.
00:07:33.850 --> 00:07:41.370
If you look at this, all
these-- if for all times, at x
00:07:41.370 --> 00:07:49.980
equals 0, y is 0, this
term, B of N, has to be 0.
00:07:49.980 --> 00:07:51.790
So this, you could forget about.
00:07:51.790 --> 00:07:53.995
And you're down to this.
00:07:56.550 --> 00:07:58.310
Furthermore, you
know it has to be
00:07:58.310 --> 00:08:06.980
0 at x equals L. That puts
a constraint on k of n.
00:08:06.980 --> 00:08:14.410
So your solution
then is of this form.
00:08:14.410 --> 00:08:16.960
This is any arbitrary constant.
00:08:21.210 --> 00:08:24.320
For the wave equation,
you can show for a string
00:08:24.320 --> 00:08:29.880
that these two are related
through the phase velocity.
00:08:29.880 --> 00:08:32.440
And so once you've
established this,
00:08:32.440 --> 00:08:34.820
then we've got that constant.
00:08:34.820 --> 00:08:37.450
So you're still left with
quite a number of constants
00:08:37.450 --> 00:08:43.260
and an infinite series where
the An can be anything.
00:08:43.260 --> 00:08:44.710
This phase can be anything.
00:08:48.200 --> 00:08:50.470
It's always useful
to check yourself.
00:08:50.470 --> 00:08:51.980
There are some
things you can check.
00:08:51.980 --> 00:08:58.870
So for example, let's consider
the n equals 1 harmonic,
00:08:58.870 --> 00:09:03.660
that normal mode, the
first normal mode.
00:09:03.660 --> 00:09:08.390
If you look at this, it's
clear that the wavelength
00:09:08.390 --> 00:09:11.790
of that harmonic is 2L.
00:09:11.790 --> 00:09:18.790
If you look at this, the period
of that harmonic is 2L over v.
00:09:18.790 --> 00:09:23.900
Period is 1 over frequency,
or frequency is 1 over period.
00:09:23.900 --> 00:09:25.210
So the frequency is v/2L.
00:09:29.470 --> 00:09:36.080
We know that for harmonic
waves or standing waves
00:09:36.080 --> 00:09:40.090
that lambda times F
is the phase velocity.
00:09:40.090 --> 00:09:43.360
Let's check it here,
multiply lambda by F,
00:09:43.360 --> 00:09:45.980
and you get that
this is satisfied.
00:09:45.980 --> 00:09:47.730
So at least this
is a quick check
00:09:47.730 --> 00:09:51.070
that we haven't made
some stupid mistake.
00:09:51.070 --> 00:09:59.770
All right, so this
satisfies any string
00:09:59.770 --> 00:10:06.380
of length L which is
tied at both ends.
00:10:06.380 --> 00:10:09.830
But our string is
not any string.
00:10:09.830 --> 00:10:17.810
We've said that initially, at
t equals 0 for all positions,
00:10:17.810 --> 00:10:20.350
the string is stationary.
00:10:20.350 --> 00:10:26.020
So our description of it must
take into account the equation
00:10:26.020 --> 00:10:29.215
which describes this
shape of the string
00:10:29.215 --> 00:10:35.800
and it must have this
boundary condition, OK?
00:10:35.800 --> 00:10:40.510
So we take this
equation-- so far,
00:10:40.510 --> 00:10:42.880
we've reduced it to
that-- differentiate it
00:10:42.880 --> 00:10:45.980
with respect to time, get this.
00:10:45.980 --> 00:10:49.650
And that's an easy-- we
just have to differentiate--
00:10:49.650 --> 00:10:52.380
the cosine gives you minus sine.
00:10:52.380 --> 00:10:57.230
OK, so you end up with this.
00:10:57.230 --> 00:11:00.210
And this, we know is 0.
00:11:00.210 --> 00:11:05.705
It's 0 at-- I'm sorry,
this is a t equals 0.
00:11:05.705 --> 00:11:07.180
I'm sorry.
00:11:07.180 --> 00:11:13.490
We know that at t equals
0 at every position
00:11:13.490 --> 00:11:15.510
x-- we said that.
00:11:15.510 --> 00:11:19.060
We're holding this string
with nothing moving.
00:11:19.060 --> 00:11:20.930
At t equals 0, it's stationary.
00:11:20.930 --> 00:11:21.980
We then let go.
00:11:21.980 --> 00:11:23.780
So this is 0.
00:11:23.780 --> 00:11:26.310
And if you look
at this, this will
00:11:26.310 --> 00:11:32.710
be 0 at all values
of x at t equals 0,
00:11:32.710 --> 00:11:36.100
only if all values
of phi n is 0.
00:11:36.100 --> 00:11:40.370
So notice, gradually, by making
use of all the information we
00:11:40.370 --> 00:11:44.590
have about the situation--
what string it is, where it's
00:11:44.590 --> 00:11:50.540
attached, how long it is,
what is its motion at t
00:11:50.540 --> 00:11:55.740
equals 0-- we're gradually
getting rid of the constants
00:11:55.740 --> 00:11:59.400
or determining them.
00:11:59.400 --> 00:12:03.370
Now that phi n is
0, I go back to this
00:12:03.370 --> 00:12:05.050
and rewrite the equation.
00:12:05.050 --> 00:12:12.030
And now, we've ended up that
the general equation which
00:12:12.030 --> 00:12:18.530
describes a string
which has length L tied
00:12:18.530 --> 00:12:25.800
at both ends, stationary
at t equals 0 is this.
00:12:25.800 --> 00:12:29.810
An is still to be determined.
00:12:29.810 --> 00:12:34.200
We have not made use, so
far, of the information what
00:12:34.200 --> 00:12:36.900
is the shape of the
string at t equals 0.
00:12:36.900 --> 00:12:38.560
That's what we'll do next.
00:12:38.560 --> 00:12:42.230
But before I do this, I
just want to make a comment.
00:12:42.230 --> 00:12:45.360
In many books,
problems, et cetera,
00:12:45.360 --> 00:12:49.190
you'll find people starting
with this equation.
00:12:49.190 --> 00:12:53.500
And you'll be wondering, how
on Earth-- is this the most
00:12:53.500 --> 00:12:56.910
general equation
for string problems?
00:12:56.910 --> 00:12:58.480
No.
00:12:58.480 --> 00:13:02.090
The answer why they
started with this
00:13:02.090 --> 00:13:06.510
is, often without telling
you, doing all this analysis
00:13:06.510 --> 00:13:08.580
in their head.
00:13:08.580 --> 00:13:14.420
This is only true
for a situation
00:13:14.420 --> 00:13:18.550
where this string
is tied at both ends
00:13:18.550 --> 00:13:20.720
and initially stationary.
00:13:20.720 --> 00:13:23.260
That forces you to that.
00:13:23.260 --> 00:13:28.410
And An now is now
not determined.
00:13:28.410 --> 00:13:31.750
And in order to find out the
value of An-- in other words,
00:13:31.750 --> 00:13:35.340
the amplitude of the
different normal modes--
00:13:35.340 --> 00:13:41.340
we have to make use of the shape
of the string at t equals 0.
00:13:41.340 --> 00:13:42.710
And that's what we'll do now.
00:13:51.390 --> 00:13:55.080
All right, so now,
I'm going to make
00:13:55.080 --> 00:13:57.800
use of the last bit
of information I have.
00:13:57.800 --> 00:14:02.000
And since that defines the
physical situation completely,
00:14:02.000 --> 00:14:04.890
there'd better be enough
to find all the constants.
00:14:04.890 --> 00:14:06.550
There should be nothing left.
00:14:06.550 --> 00:14:09.720
They should be then
completely determined.
00:14:09.720 --> 00:14:14.270
So first of all, I said we
didn't make use of the shape
00:14:14.270 --> 00:14:15.630
at t equals 0.
00:14:15.630 --> 00:14:19.790
At t equals 0, cosine of 0 is 1.
00:14:19.790 --> 00:14:28.790
So what we know is that the
string shape is given by,
00:14:28.790 --> 00:14:32.420
from here, by this equation.
00:14:32.420 --> 00:14:39.820
So this tells us that--
this is a description
00:14:39.820 --> 00:14:43.320
of the shape of
the string in terms
00:14:43.320 --> 00:14:46.380
of an infinite series
of sinusoidal functions.
00:14:49.290 --> 00:15:00.360
OK, we notice this tells us the
shape of the string at every X
00:15:00.360 --> 00:15:03.280
position at t equals 0.
00:15:03.280 --> 00:15:07.540
So I look at that,
the original shape.
00:15:07.540 --> 00:15:10.870
And I can describe
it mathematically.
00:15:10.870 --> 00:15:17.170
y of x at time equals 0 is
0 at all points from one
00:15:17.170 --> 00:15:20.190
end of the string to where
the distortion starts.
00:15:20.190 --> 00:15:26.760
And it starts at 3/8 L if
you look at the diagram.
00:15:26.760 --> 00:15:35.530
Then from position x
of 3/8 L to x is 5/8 L,
00:15:35.530 --> 00:15:40.470
the string is straight again,
and its displaced by a distance
00:15:40.470 --> 00:15:46.530
H. And afterwards, past
the 5/8 L, it's back to 0.
00:15:46.530 --> 00:15:51.170
For a second, let's go
back and just look at that.
00:15:51.170 --> 00:15:54.690
So we're talking about
this shape, right?
00:15:54.690 --> 00:16:03.890
It's 0, 0 in Y, and then jumps
to H between 3/8 L and 5/8 L.
00:16:03.890 --> 00:16:11.040
This shape, when
you let go, that
00:16:11.040 --> 00:16:17.260
will determine the amplitude
of the different normal modes
00:16:17.260 --> 00:16:24.370
that are important in the
oscillation of that system.
00:16:24.370 --> 00:16:31.150
OK, so we must determine these
infinite number of constants.
00:16:31.150 --> 00:16:35.970
At first sight, it looks
like an impossible task
00:16:35.970 --> 00:16:39.550
if it wasn't for a
brilliant mathematician
00:16:39.550 --> 00:16:42.850
in the 18th century,
Fourier, who
00:16:42.850 --> 00:16:46.860
made a relatively simple
but incredibly important
00:16:46.860 --> 00:16:48.730
observation.
00:16:48.730 --> 00:16:53.960
When applied to this
problem, it is the following.
00:16:58.410 --> 00:17:03.850
He realized that one can
do the following trick.
00:17:08.230 --> 00:17:12.770
If you take this
function, sine n pi
00:17:12.770 --> 00:17:17.290
over L x to 1 which is
describing our shape
00:17:17.290 --> 00:17:22.220
over there, if you multiply
it-- and by the way,
00:17:22.220 --> 00:17:25.780
n-- I'm just reminding you--
goes from 1 to infinity.
00:17:25.780 --> 00:17:29.200
It's 1, 2, 3, 4,
5, 6, et cetera.
00:17:29.200 --> 00:17:33.350
If you take this function
and you multiply it
00:17:33.350 --> 00:17:39.330
by a similar function, but
with different value of m,
00:17:39.330 --> 00:17:46.120
then this integral from one
end of the string to the other,
00:17:46.120 --> 00:17:53.390
is equal to 0 if this,
the m you've chosen here,
00:17:53.390 --> 00:17:55.010
is different than n.
00:17:58.300 --> 00:17:58.970
Try it.
00:17:58.970 --> 00:18:02.600
I don't want-- a good
exercise for yourself.
00:18:02.600 --> 00:18:05.630
Take this function.
00:18:05.630 --> 00:18:10.970
And do the integration, and
you'll find it comes out to 0.
00:18:10.970 --> 00:18:13.320
It's not hard to do.
00:18:13.320 --> 00:18:17.740
So that is 0 if n
is not equal to m.
00:18:17.740 --> 00:18:23.740
On the other hand,
if m equals n--
00:18:23.740 --> 00:18:29.200
so this is the integral of
sine squared n pi over L--
00:18:29.200 --> 00:18:30.885
it comes out to be L/2.
00:18:30.885 --> 00:18:32.010
You can do it for yourself.
00:18:35.350 --> 00:18:43.400
This trick allows us to
solve for all values of An.
00:18:43.400 --> 00:18:45.880
Now, let me tell
you, I did not do
00:18:45.880 --> 00:18:53.320
this in-- the Fourier
trick-- in general.
00:18:53.320 --> 00:18:59.330
If you go into books,
you'll find the formulation
00:18:59.330 --> 00:19:04.250
of this trick known
as Fourier's theorem
00:19:04.250 --> 00:19:06.090
for any periodic function.
00:19:09.800 --> 00:19:13.850
The advantage of learning it,
how to do it for any function,
00:19:13.850 --> 00:19:17.710
you don't have to do
this integral every time.
00:19:17.710 --> 00:19:19.840
Because suppose our
function turned out
00:19:19.840 --> 00:19:23.710
there to be a cosine, we
would have to then figure out,
00:19:23.710 --> 00:19:28.225
what do I multiply it by to
make this come out so simply?
00:19:35.505 --> 00:19:37.240
As I say, the
advantage is then you
00:19:37.240 --> 00:19:41.670
can use formulate
in books, et cetera.
00:19:41.670 --> 00:19:43.310
There's nothing wrong with that.
00:19:43.310 --> 00:19:44.590
It is good.
00:19:44.590 --> 00:19:49.940
And the way one changes a
physical situation into one
00:19:49.940 --> 00:19:53.840
where you can apply Fourier's
theorem, which only applies
00:19:53.840 --> 00:20:00.650
to periodic functions, is
to take your original shape,
00:20:00.650 --> 00:20:05.260
and out of it, create
a periodic function.
00:20:05.260 --> 00:20:09.450
Once you've done that, you
can use Fourier's theorem.
00:20:09.450 --> 00:20:15.140
I've, in this particular
case because I find it easier
00:20:15.140 --> 00:20:19.230
to explain it, simply
essentially derived
00:20:19.230 --> 00:20:24.380
the Fourier theorem for
this particular problem.
00:20:24.380 --> 00:20:29.960
For this particular
problem, the function
00:20:29.960 --> 00:20:34.940
by which you multiply
this is sine m pi.
00:20:34.940 --> 00:20:39.860
OK, once you've realized
that this is true,
00:20:39.860 --> 00:20:50.420
I can go back and use it
to determine every An.
00:20:57.890 --> 00:20:59.160
I go to this function.
00:21:03.510 --> 00:21:12.390
I multiply both sides in
turn by sine m pi over L
00:21:12.390 --> 00:21:21.670
for every n where I want to find
the value of An, for each one.
00:21:21.670 --> 00:21:24.070
So I have to do
an infinite number
00:21:24.070 --> 00:21:27.010
of integrals in principle.
00:21:27.010 --> 00:21:31.100
But often you don't need
to know every value of n.
00:21:31.100 --> 00:21:34.920
You may want to
know each one of An
00:21:34.920 --> 00:21:39.030
is the amplitude
of the harmonic.
00:21:39.030 --> 00:21:41.830
You may be interested
in what is the amplitude
00:21:41.830 --> 00:21:45.320
of the first harmonic solution
or the second or third.
00:21:45.320 --> 00:21:49.600
However many you want, so
many integrals you have to do.
00:21:49.600 --> 00:21:51.730
No magic.
00:21:51.730 --> 00:21:54.545
You're basically solving
infinite number of equations.
00:21:58.260 --> 00:22:02.890
And if I apply this
to our problem,
00:22:02.890 --> 00:22:11.410
we find that An is equal
to 2/L. It's from here.
00:22:11.410 --> 00:22:19.400
The integral of 0 of
2L of y sine at dx.
00:22:19.400 --> 00:22:19.910
Right?
00:22:19.910 --> 00:22:25.200
All I'm taking, I'm multiplying
both sides of this equation
00:22:25.200 --> 00:22:32.540
by sine of m pi L over
x and integrating them
00:22:32.540 --> 00:22:37.680
both from 0 to L.
On one side, you
00:22:37.680 --> 00:22:42.870
end up with An and on the
other, 2/L times this integral.
00:22:42.870 --> 00:22:47.890
Then once you've done this
integral, you know the answer.
00:22:47.890 --> 00:22:55.680
In our case, y of x of
0 is particularly easy.
00:22:55.680 --> 00:22:59.500
It's 0 in this range,
0 in this range,
00:22:59.500 --> 00:23:02.960
and a constant for the rest.
00:23:02.960 --> 00:23:07.760
Well, over the part of
the range where it's 0,
00:23:07.760 --> 00:23:10.490
if I multiply
something by 0, it's 0.
00:23:10.490 --> 00:23:15.240
And so I only have to do
this integral over the range
00:23:15.240 --> 00:23:24.780
where y of x for 0 time is H.
So I have to do this integral.
00:23:24.780 --> 00:23:31.350
And now I have reduced
the situation that every n
00:23:31.350 --> 00:23:33.840
that we are
interested in, we can
00:23:33.840 --> 00:23:36.460
calculate by doing
this integral.
00:23:36.460 --> 00:23:42.476
So in principle, we've solved
it for all values of n.
00:23:46.650 --> 00:23:50.410
So finally, I can
answer the question.
00:23:50.410 --> 00:23:54.990
So the first question was
what lowest value of n
00:23:54.990 --> 00:23:59.460
for which An is 0.
00:23:59.460 --> 00:24:00.220
Well, let's start.
00:24:00.220 --> 00:24:03.390
Take number 1, 2, 3.
00:24:03.390 --> 00:24:13.870
And for n equals 1, this
integral is not equal to 0,
00:24:13.870 --> 00:24:17.360
because what you have
to do is you basically
00:24:17.360 --> 00:24:21.260
are integrating
this-- for n equals
00:24:21.260 --> 00:24:26.390
1, that's what the sine
pi over L x looks like.
00:24:26.390 --> 00:24:30.930
You're integrating that
and multiply it by 0.
00:24:30.930 --> 00:24:33.610
And here, you're multiplying
by a constant and 0.
00:24:33.610 --> 00:24:37.210
Clearly, this is not
going to give you a 0.
00:24:37.210 --> 00:24:40.840
Let's take the second
one, the second harmonic
00:24:40.840 --> 00:24:42.090
in this calculation.
00:24:42.090 --> 00:24:48.260
You're multiplying
this y by this function
00:24:48.260 --> 00:24:50.910
and integrating it, OK?
00:24:50.910 --> 00:24:55.830
Notice the sine function
is symmetric about 0
00:24:55.830 --> 00:24:57.940
and so is this symmetric.
00:24:57.940 --> 00:25:01.710
Except here, this is an
anti-symmetric function.
00:25:01.710 --> 00:25:06.370
So the sine is negative
to the right of the middle
00:25:06.370 --> 00:25:08.110
and positive to the left.
00:25:08.110 --> 00:25:12.320
If I multiply this positive by
that, I get a positive number.
00:25:12.320 --> 00:25:15.370
If I multiply this by that,
I get a negative number.
00:25:15.370 --> 00:25:17.030
The two are equal.
00:25:17.030 --> 00:25:20.900
And so the integral
across there is 0.
00:25:20.900 --> 00:25:28.200
So n equals 2 is the lowest
harmonic for which An is 0.
00:25:28.200 --> 00:25:37.940
When I let that string
go, the second harmonic
00:25:37.940 --> 00:25:40.150
will not be excited.
00:25:40.150 --> 00:25:44.750
Which is the lowest n for which
it is excited-- in other words
00:25:44.750 --> 00:25:46.010
that An is not equal 0.
00:25:46.010 --> 00:25:52.370
Well, I told you when n
equals 1, clearly An is not 0.
00:25:52.370 --> 00:25:54.300
So that will be excited.
00:25:54.300 --> 00:25:58.920
So the first harmonic
will be excited.
00:25:58.920 --> 00:26:00.250
What is the amplitude?
00:26:00.250 --> 00:26:02.630
Well, we have to calculate it.
00:26:02.630 --> 00:26:05.130
I told you, every An--
I told you what it is.
00:26:05.130 --> 00:26:06.880
I have to do some work.
00:26:06.880 --> 00:26:12.210
So you take A of 1 is 2/L. The
integral from that to the H
00:26:12.210 --> 00:26:17.560
times this function
where n is 1, all right?
00:26:17.560 --> 00:26:21.090
And that, fortunately, is
an easy integral to do.
00:26:21.090 --> 00:26:23.140
Integrate the sine,
you get the cosine.
00:26:23.140 --> 00:26:26.480
And putting in the limits,
you get this answer.
00:26:26.480 --> 00:26:31.490
So that is our answer to
that part of the problem.
00:26:31.490 --> 00:26:34.860
The final part of
the problem was
00:26:34.860 --> 00:26:39.350
they said, OK, I
have this shape.
00:26:39.350 --> 00:26:40.800
I let go.
00:26:40.800 --> 00:26:46.160
What will that shape
look like at a time
00:26:46.160 --> 00:26:52.360
was L-- in the problem,
the time was L times
00:26:52.360 --> 00:26:57.550
the square root of mu over T.
T's the tension in the string.
00:26:57.550 --> 00:27:00.570
But I know that square
root of mu over t
00:27:00.570 --> 00:27:03.490
is 1/d, the phase velocity.
00:27:03.490 --> 00:27:08.900
So they ask us to calculate
the shape of that string
00:27:08.900 --> 00:27:12.560
at the time which is the
length of the string divided
00:27:12.560 --> 00:27:15.470
by the phase velocity
of progressive waves
00:27:15.470 --> 00:27:17.410
on the string.
00:27:17.410 --> 00:27:27.720
OK, now we know completely what
is the shape of the string.
00:27:27.720 --> 00:27:28.980
It is this formula.
00:27:28.980 --> 00:27:33.330
This is the formula
that tells us
00:27:33.330 --> 00:27:37.545
what the string looks like
at all places, all times.
00:27:41.130 --> 00:27:46.650
From our knowledge
of this integral,
00:27:46.650 --> 00:27:52.520
depending where it's
symmetric or anti-symmetric
00:27:52.520 --> 00:27:56.290
sinusoidal function
about the center,
00:27:56.290 --> 00:28:00.050
we know that every
second one will be 0.
00:28:00.050 --> 00:28:05.330
So this sum now,
I've only summed
00:28:05.330 --> 00:28:09.830
for n equals 1, 3,
5, 7, et cetera.
00:28:09.830 --> 00:28:17.880
The terms n equals 2, 4, 6, et
cetera, will have 0 amplitude.
00:28:17.880 --> 00:28:25.540
But this now describes our
situation in its entirety--
00:28:25.540 --> 00:28:29.950
because I've now not only know
the spatial shape, but also
00:28:29.950 --> 00:28:32.380
as a function of time.
00:28:32.380 --> 00:28:38.240
So this describes
what it was at time t.
00:28:38.240 --> 00:28:45.130
What will it be if I increase
the time-- let's start from 0,
00:28:45.130 --> 00:28:48.690
and I go to a time L/v?
00:28:48.690 --> 00:28:56.280
All right, well, if this
changes from 0 to L/v,
00:28:56.280 --> 00:29:01.150
this changes to cosine n pi.
00:29:01.150 --> 00:29:05.550
But you know that for n
equals 1, 3, 5, et cetera,
00:29:05.550 --> 00:29:09.540
cosine of n pi-- in other
words, at pi, at 3 pi,
00:29:09.540 --> 00:29:12.710
et cetera-- is minus 1.
00:29:12.710 --> 00:29:21.690
So at this later time,
every term in this expansion
00:29:21.690 --> 00:29:23.500
has changed signs.
00:29:23.500 --> 00:29:30.250
But it's exactly the same
series, except it's minus 1
00:29:30.250 --> 00:29:31.220
here.
00:29:31.220 --> 00:29:33.190
So what has happened
to the string?
00:29:33.190 --> 00:29:38.210
It is exactly the same
shape with an opposite sign.
00:29:38.210 --> 00:29:42.660
So with the string, instead
of being up here like that
00:29:42.660 --> 00:29:47.800
like it was over here, it's
flipped over like that.
00:29:47.800 --> 00:29:50.900
Without doing any more work,
I can conclude that simply
00:29:50.900 --> 00:29:54.460
because I have the same series
as before with a negative sign.
00:29:54.460 --> 00:29:56.360
So that is the answer.
00:29:56.360 --> 00:29:57.550
So I've solved the problem.
00:29:57.550 --> 00:30:00.960
But at this instance, I just
want to digress for a second
00:30:00.960 --> 00:30:04.900
and tell you this
actually is a problem
00:30:04.900 --> 00:30:10.000
that we could have almost
done all by thinking alone.
00:30:10.000 --> 00:30:15.160
It wasn't necessary to do big
fraction of the work we did.
00:30:15.160 --> 00:30:18.680
The only part which
there was no choice
00:30:18.680 --> 00:30:21.330
is this amplitude [INAUDIBLE].
00:30:21.330 --> 00:30:28.140
Because I could have gone
back, scratched my memory
00:30:28.140 --> 00:30:30.310
and knowledge, and
said, look, hold on.
00:30:30.310 --> 00:30:33.440
When I have been a
string, I could always
00:30:33.440 --> 00:30:37.560
analyze it in terms
of normal modes.
00:30:37.560 --> 00:30:45.380
But also, we know that we
can describe the solutions
00:30:45.380 --> 00:30:48.370
in terms of progressive waves.
00:30:48.370 --> 00:30:50.890
The two are equivalent
to each other.
00:30:50.890 --> 00:30:52.550
It's not extra solutions.
00:30:52.550 --> 00:30:56.400
But by adding
appropriately normal modes,
00:30:56.400 --> 00:31:02.220
I can get a superposition
of progressive waves.
00:31:02.220 --> 00:31:06.080
I can use the uniqueness
theorem to see
00:31:06.080 --> 00:31:13.110
if I've got a solution
which represents
00:31:13.110 --> 00:31:16.760
the situation at hand.
00:31:16.760 --> 00:31:24.580
If you look at the
original problem,
00:31:24.580 --> 00:31:31.920
if you look at this problem,
this is stationary, all right?
00:31:31.920 --> 00:31:39.530
So I could imagine this
solution being the superposition
00:31:39.530 --> 00:31:45.020
of two progressive waves
that look like this but half
00:31:45.020 --> 00:31:49.080
the amplitude, one moving
to the left and one
00:31:49.080 --> 00:31:51.090
moving to the right.
00:31:51.090 --> 00:31:57.190
As they overlap--
in other words,
00:31:57.190 --> 00:32:02.750
I have two waves, one like
this moving to the right,
00:32:02.750 --> 00:32:06.250
and one like this
moving to the left.
00:32:06.250 --> 00:32:11.610
If I add them, I'll
get this distortion,
00:32:11.610 --> 00:32:14.610
which is stationary.
00:32:14.610 --> 00:32:18.370
Therefore, if I
solved the problem
00:32:18.370 --> 00:32:25.860
for each one of these
progressive waves and add them,
00:32:25.860 --> 00:32:28.340
I'll get the answer to this.
00:32:28.340 --> 00:32:31.300
It'll satisfy all the
boundary conditions.
00:32:31.300 --> 00:32:32.840
By the uniqueness
theorem, it will
00:32:32.840 --> 00:32:38.720
be the correct description
of what happened.
00:32:38.720 --> 00:32:44.320
And then with
that, I can predict
00:32:44.320 --> 00:32:49.000
what will be the
shape at any time.
00:32:49.000 --> 00:32:54.220
So for example, in
the time they ask L/v,
00:32:54.220 --> 00:33:00.220
a progressive wave
will move a distance L.
00:33:00.220 --> 00:33:05.010
So the wave over
here, which was going
00:33:05.010 --> 00:33:10.310
to do the right would hit this
boundary-- that pulse-- would
00:33:10.310 --> 00:33:12.180
flip over.
00:33:12.180 --> 00:33:16.940
And you know that this
string is fixed there.
00:33:16.940 --> 00:33:21.250
What happens if a pulse
comes to a fixed end?
00:33:21.250 --> 00:33:26.440
Imagine you are
causing the pulse.
00:33:26.440 --> 00:33:29.340
You are the forcewave
moving along
00:33:29.340 --> 00:33:31.300
which causes the distortion.
00:33:31.300 --> 00:33:34.810
When you come to the
rigid end, you're pulling.
00:33:34.810 --> 00:33:36.300
It won't give.
00:33:36.300 --> 00:33:39.550
So the only way to happen,
the string will pull you down.
00:33:39.550 --> 00:33:41.450
So you'll flip over.
00:33:41.450 --> 00:33:45.040
So this pulse, when
it hits this end,
00:33:45.040 --> 00:33:47.340
will flip over
upside down, and then
00:33:47.340 --> 00:33:49.600
progress this way upside down.
00:33:49.600 --> 00:33:54.710
This one will go to this end,
flip over, and progress back.
00:33:54.710 --> 00:34:02.370
By the time t has changed
by L over the velocity,
00:34:02.370 --> 00:34:05.670
those flipped pulses would have
come back here and overlapped,
00:34:05.670 --> 00:34:07.130
but upside down.
00:34:07.130 --> 00:34:08.909
And so this would have flipped.
00:34:08.909 --> 00:34:12.730
By this kind of analysis, if
you use your wits about it,
00:34:12.730 --> 00:34:20.060
sometimes you can solve
more difficult problems
00:34:20.060 --> 00:34:25.159
quickly by using all the
knowledge you have about waves,
00:34:25.159 --> 00:34:26.912
the propagation, et cetera.
00:34:26.912 --> 00:34:28.120
Probably a good time to stop.
00:34:28.120 --> 00:34:29.970
Thank you.