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YEN-JIE LEE: So welcome
back, everybody, to 8.03.
00:00:26.540 --> 00:00:28.730
Happy to see you again.
00:00:28.730 --> 00:00:32.650
So here is the current
status of the 8.03.
00:00:32.650 --> 00:00:35.550
So right now, we have
finished the discussion
00:00:35.550 --> 00:00:37.520
of coupled oscillator.
00:00:37.520 --> 00:00:40.830
And then we go to infinite
number of coupled oscillators.
00:00:40.830 --> 00:00:46.040
And we found that there's a
wave equation coming out of it.
00:00:46.040 --> 00:00:52.990
And that means, in short, waves
are really a group effort.
00:00:52.990 --> 00:00:56.810
So many, many objects are
working really together,
00:00:56.810 --> 00:01:00.380
so that they create
a wave phenomena.
00:01:00.380 --> 00:01:03.290
And you can also see,
there's a close connection
00:01:03.290 --> 00:01:06.830
between vibration
of a single object
00:01:06.830 --> 00:01:10.310
and the formation of
the wave structure.
00:01:10.310 --> 00:01:12.350
So what we are
going to do today is
00:01:12.350 --> 00:01:16.250
to give you a short review of
what we have done last time.
00:01:16.250 --> 00:01:20.030
Then we actually will
continue to our understanding
00:01:20.030 --> 00:01:24.210
of wave equation today.
00:01:24.210 --> 00:01:26.150
So what we have
learned last time,
00:01:26.150 --> 00:01:30.950
we have learned how to solve
infinite system with space
00:01:30.950 --> 00:01:32.940
translation symmetry.
00:01:32.940 --> 00:01:35.750
And also, we learned
how to use it
00:01:35.750 --> 00:01:40.790
to solve finite systems
by imposing-- or adding
00:01:40.790 --> 00:01:42.460
boundary conditions.
00:01:42.460 --> 00:01:46.310
That would limit the infinite
number of normal modes
00:01:46.310 --> 00:01:49.367
to finite number of
normal modes based on--
00:01:49.367 --> 00:01:53.720
I mean, it's actually
closely related
00:01:53.720 --> 00:01:58.030
to how many objects
you have in the system.
00:01:58.030 --> 00:02:02.580
And also, we went ahead and
go to a continuum limit.
00:02:02.580 --> 00:02:05.240
And we found out, there is a
surprising result coming out
00:02:05.240 --> 00:02:06.020
of this.
00:02:06.020 --> 00:02:09.500
And this is actually
the wave equation.
00:02:09.500 --> 00:02:14.310
So what do we mean by
going to continuous limit?
00:02:14.310 --> 00:02:16.780
So the limit we
are talking about
00:02:16.780 --> 00:02:20.570
is that, before when we
discussed this closed
00:02:20.570 --> 00:02:24.160
system of infinite number
of objects holding together
00:02:24.160 --> 00:02:28.170
by strings, there is
a length scale, which
00:02:28.170 --> 00:02:30.970
is the separation
between objects, which
00:02:30.970 --> 00:02:34.390
is called a in my notation.
00:02:34.390 --> 00:02:38.440
And to make it continuous,
we are taking a limit such
00:02:38.440 --> 00:02:42.910
that the a, which is the
separation between objects,
00:02:42.910 --> 00:02:47.380
is so much smaller
than the wavelength.
00:02:47.380 --> 00:02:49.870
Basically, the
wavelength is actually
00:02:49.870 --> 00:02:54.280
the sinusoidal shape you
see when I perturb a system.
00:02:54.280 --> 00:03:00.220
And what I assuming is that
the distance between objects
00:03:00.220 --> 00:03:04.090
are so much smaller
than the wavelengths.
00:03:04.090 --> 00:03:07.650
So that's actually what I
called continuous limit.
00:03:07.650 --> 00:03:11.050
And that is actually true for
most of the example, which we
00:03:11.050 --> 00:03:14.080
see in the previous lectures.
00:03:14.080 --> 00:03:18.400
For example, I was
holding a giant spring.
00:03:18.400 --> 00:03:19.930
And I oscillate that.
00:03:19.930 --> 00:03:22.430
And all the little
components, or, say,
00:03:22.430 --> 00:03:28.240
all the little mass on that
spring, the space between all
00:03:28.240 --> 00:03:32.860
those little mass on the
spring are so much smaller
00:03:32.860 --> 00:03:34.730
than the length scale
we are talking about,
00:03:34.730 --> 00:03:38.530
which is at the
order of 1 meter.
00:03:38.530 --> 00:03:43.000
So that actually is
a sensible limit,
00:03:43.000 --> 00:03:46.490
which describes the
physics we are interested.
00:03:46.490 --> 00:03:51.830
When we go to a
continuous limit,
00:03:51.830 --> 00:03:55.790
we find that something
really interesting happens.
00:03:55.790 --> 00:04:00.470
So M minus 1 K matrix
originally is infinite times
00:04:00.470 --> 00:04:02.370
infinite dimension matrix.
00:04:02.370 --> 00:04:05.390
It becomes the operator,
which is actually
00:04:05.390 --> 00:04:10.400
minus T over rho L partial
square partial x squared.
00:04:10.400 --> 00:04:12.950
And also--
00:04:12.950 --> 00:04:15.030
OK, I changed the notation here.
00:04:15.030 --> 00:04:18.769
It was aj, and I
changed it to psi,
00:04:18.769 --> 00:04:23.700
because what we are going to
use later on, when we describe
00:04:23.700 --> 00:04:27.800
wave functions, et cetera,
especially in 8.04,
00:04:27.800 --> 00:04:29.390
we usually use psi.
00:04:29.390 --> 00:04:33.940
And the psi j, which were
discrete and evaluated
00:04:33.940 --> 00:04:38.660
in the individual discrete
position in the x direction,
00:04:38.660 --> 00:04:40.880
it's becoming a
continuous function,
00:04:40.880 --> 00:04:46.980
which is psi x and
is also a function t.
00:04:46.980 --> 00:04:50.750
Therefore, from
this exercise, we
00:04:50.750 --> 00:04:56.210
found out we see wave
equation, actually after we
00:04:56.210 --> 00:04:58.040
go to a continuous limit.
00:04:58.040 --> 00:04:59.750
And for more
information, you can also
00:04:59.750 --> 00:05:04.640
take a look at the textbook
in the relevant page.
00:05:04.640 --> 00:05:07.050
So what are we
going to do today?
00:05:07.050 --> 00:05:09.350
So today, what we
are going to do
00:05:09.350 --> 00:05:13.180
is to understand the wave
equation, the structure,
00:05:13.180 --> 00:05:15.380
and what does that
mean, and also
00:05:15.380 --> 00:05:20.210
what are the normal modes coming
out of this wave equation.
00:05:20.210 --> 00:05:22.730
And the next time,
in later lectures,
00:05:22.730 --> 00:05:26.570
we will also discuss another
special kind of motion, which
00:05:26.570 --> 00:05:29.960
is progressing wave solutions.
00:05:29.960 --> 00:05:32.780
So let's immediately
gets started
00:05:32.780 --> 00:05:37.800
by looking at a concrete
example and also
00:05:37.800 --> 00:05:43.970
to derive the normal modes
from this wave equation.
00:05:43.970 --> 00:05:48.380
Before we do that, let's take
a look at this wave equation.
00:05:48.380 --> 00:05:54.220
This wave equation is actually
equivalent to infinite
00:05:54.220 --> 00:05:58.620
number of equations of
motion, if you think about it.
00:05:58.620 --> 00:05:59.670
Why is that?
00:05:59.670 --> 00:06:02.930
That is because each x--
00:06:02.930 --> 00:06:07.640
each partition x you put in will
produce a equation of motion.
00:06:07.640 --> 00:06:13.100
So basically, originally, when
we were doing a discrete case,
00:06:13.100 --> 00:06:15.230
those are labeled by c.
00:06:15.230 --> 00:06:20.750
c is actually telling you
which mass I'm talking about.
00:06:20.750 --> 00:06:23.020
Now, it's actually
replaced by x.
00:06:23.020 --> 00:06:24.860
And what we are
actually doing is
00:06:24.860 --> 00:06:29.690
to solve infinite number
equation of motion in one go.
00:06:29.690 --> 00:06:34.230
And that is actually
the wave equation.
00:06:34.230 --> 00:06:40.730
So the first question we ask is,
what is actually normal modes
00:06:40.730 --> 00:06:46.790
based on this infinitely long
continuous system described
00:06:46.790 --> 00:06:48.900
by wave equations?
00:06:48.900 --> 00:06:52.690
So let's get
started immediately.
00:06:52.690 --> 00:06:55.750
So basically, we
can first assume
00:06:55.750 --> 00:06:59.270
what is actually the functional
form for normal modes.
00:06:59.270 --> 00:07:01.130
So what we can
actually do is we can
00:07:01.130 --> 00:07:06.680
assume that psi x, t
is actually equal to A,
00:07:06.680 --> 00:07:10.610
is actually a function
of x, times B,
00:07:10.610 --> 00:07:14.550
is actually a function of time.
00:07:14.550 --> 00:07:20.070
So what I am doing is actually
have a meaning actually.
00:07:20.070 --> 00:07:24.450
So A of x actually
give you a description
00:07:24.450 --> 00:07:27.250
of the functional form--
00:07:27.250 --> 00:07:33.960
the shape of the normal
mode as a function of x.
00:07:33.960 --> 00:07:40.640
So that's actually
giving you the shape
00:07:40.640 --> 00:07:42.000
as a function of x-axis.
00:07:44.770 --> 00:07:49.990
And B, which is a
function of t, is actually
00:07:49.990 --> 00:07:55.390
giving you information
how individual component
00:07:55.390 --> 00:08:00.370
goes up and down or move
as a function of time.
00:08:00.370 --> 00:08:04.060
So that actually control
the time evolution.
00:08:09.430 --> 00:08:13.250
And we were using
this wave equation
00:08:13.250 --> 00:08:18.445
to describe a continuous
system, like, for example,
00:08:18.445 --> 00:08:20.590
a string with tension t.
00:08:23.740 --> 00:08:25.670
So what we could do
is the following.
00:08:25.670 --> 00:08:29.500
So we are interested in
the solution of the wave
00:08:29.500 --> 00:08:31.600
equation, which is shown there.
00:08:31.600 --> 00:08:35.620
So what we could do is
that, OK, let's first assume
00:08:35.620 --> 00:08:39.280
this functional form, assuming
that every component is
00:08:39.280 --> 00:08:44.000
actually following the
same time-depend evolution.
00:08:44.000 --> 00:08:47.020
And then we can actually
plug in this functional form
00:08:47.020 --> 00:08:50.780
to the equation of motion
and see what we will get.
00:08:50.780 --> 00:09:02.170
So if we plug this in
into the wave equation,
00:09:02.170 --> 00:09:04.750
so what we are going to get is,
if you look at the left hand
00:09:04.750 --> 00:09:07.770
side, it's actually a partial
derivation with respect
00:09:07.770 --> 00:09:08.970
to time.
00:09:08.970 --> 00:09:10.650
Therefore, what we
are going to get
00:09:10.650 --> 00:09:18.490
is A of x times partial square
of B, which is a function of t,
00:09:18.490 --> 00:09:21.050
partial t squared.
00:09:21.050 --> 00:09:23.050
The right hand side
of the equation,
00:09:23.050 --> 00:09:28.260
which is actually
equal to vp squared,
00:09:28.260 --> 00:09:34.330
is a partial derivative
with respect to x.
00:09:34.330 --> 00:09:39.870
Therefore, you have
B is a function of t
00:09:39.870 --> 00:09:47.100
only times partial square
A of x partial t squared.
00:09:49.810 --> 00:09:55.940
So actually, we can
just for convenience--
00:09:55.940 --> 00:09:57.880
oh, sorry, that's
supposed to be partial x.
00:09:57.880 --> 00:09:59.680
Thank you very much--
00:09:59.680 --> 00:10:01.780
partial x squared.
00:10:01.780 --> 00:10:04.600
So just for convenience,
we can actually
00:10:04.600 --> 00:10:11.790
divide the whole equation by A
times B times the vp squared.
00:10:11.790 --> 00:10:18.190
We can issue divide the
whole equation by A times B,
00:10:18.190 --> 00:10:20.650
for example, and the vp squared.
00:10:20.650 --> 00:10:23.010
If we do this,
then basically I'm
00:10:23.010 --> 00:10:25.750
moving this part to
the left hand side.
00:10:25.750 --> 00:10:35.285
So I get 1 over vp square B of
t partial square Bt partial t
00:10:35.285 --> 00:10:35.785
square.
00:10:38.510 --> 00:10:43.880
And the right hand side, because
I also divide AB vp square,
00:10:43.880 --> 00:10:53.720
therefore, I get 1 over A of x
partial square A of x partial
00:10:53.720 --> 00:10:54.720
x square.
00:11:00.350 --> 00:11:02.030
So far so good.
00:11:02.030 --> 00:11:06.290
And basically, what
I'm doing is just
00:11:06.290 --> 00:11:10.160
plug in the functional
form, which I assume here
00:11:10.160 --> 00:11:16.130
and then divide everything
by AB times vp square.
00:11:16.130 --> 00:11:19.080
And what I immediately
find is that left hand
00:11:19.080 --> 00:11:26.070
side is a function
which only depends on t.
00:11:26.070 --> 00:11:29.180
Left hand side
only depends on t.
00:11:29.180 --> 00:11:32.360
And right hand
side is a function
00:11:32.360 --> 00:11:36.430
which on the depends on x.
00:11:36.430 --> 00:11:41.250
So in short, I have
in this situation
00:11:41.250 --> 00:11:46.140
f of t, which is left hand
side, is equal to g of x.
00:11:49.640 --> 00:11:54.560
You will see this over and over
again in later lectures related
00:11:54.560 --> 00:11:56.830
to physics.
00:11:56.830 --> 00:12:01.450
This is actually the so-called
separation of variables.
00:12:01.450 --> 00:12:04.395
So basically, you are
facing a situation
00:12:04.395 --> 00:12:08.440
f of t equal to g of x.
00:12:08.440 --> 00:12:10.720
If you think about
this situation,
00:12:10.720 --> 00:12:12.650
that's actually
really, really helpful,
00:12:12.650 --> 00:12:21.420
because, OK, now what I can do
is I can stay at a specific x.
00:12:21.420 --> 00:12:24.210
For example, I
choose this point.
00:12:24.210 --> 00:12:28.890
Then I let the time go forward.
00:12:28.890 --> 00:12:33.450
Of course, I cannot stop
time, but if it goes forward,
00:12:33.450 --> 00:12:37.590
then left hand side
equivalent, if it's changing,
00:12:37.590 --> 00:12:42.420
you will change,
because I change t.
00:12:42.420 --> 00:12:45.210
If the left hand side
equation is changing,
00:12:45.210 --> 00:12:48.620
then that's actually
not going to--
00:12:48.620 --> 00:12:50.490
this equation is
not going to work,
00:12:50.490 --> 00:12:52.260
because I am not changing x.
00:12:52.260 --> 00:12:56.940
I am fixing myself at a specific
location, and a lot of time
00:12:56.940 --> 00:12:57.840
go on.
00:12:57.840 --> 00:13:01.590
Then if left hand
side is changing,
00:13:01.590 --> 00:13:05.470
then this equation cannot work.
00:13:05.470 --> 00:13:06.640
You see?
00:13:06.640 --> 00:13:10.610
Therefore, what is the
consequence of this equation?
00:13:10.610 --> 00:13:15.760
That means, left hand side,
f of t, must be a constant.
00:13:18.570 --> 00:13:22.810
Therefore, no matter what
I do, if I change time,
00:13:22.810 --> 00:13:25.220
it's not going to
change anything.
00:13:25.220 --> 00:13:28.430
I can put in whatever
time, like 1 billion years
00:13:28.430 --> 00:13:33.330
after this lecture or
now, it doesn't matter.
00:13:33.330 --> 00:13:34.680
It's a constant.
00:13:34.680 --> 00:13:35.970
So this must be a constant.
00:13:39.780 --> 00:13:42.290
I can do the same trick.
00:13:42.290 --> 00:13:45.710
I froze the time.
00:13:45.710 --> 00:13:50.300
I fix the time, and then
I compare this point
00:13:50.300 --> 00:13:54.650
to that point, or, say,
something billion billions
00:13:54.650 --> 00:13:59.250
of light years away
from this class room.
00:13:59.250 --> 00:14:03.980
I am changing the x, but
I'm not changing the t.
00:14:03.980 --> 00:14:06.590
The same argument also holds--
00:14:06.590 --> 00:14:09.950
if this function is
changing as a function of x,
00:14:09.950 --> 00:14:12.038
then I am screwed, because--
00:14:12.038 --> 00:14:13.280
[LAUGHTER]
00:14:13.280 --> 00:14:14.630
--it doesn't work, right?
00:14:14.630 --> 00:14:17.060
I mean, this--
00:14:17.060 --> 00:14:20.225
Therefore, it has to be
a constant also as well.
00:14:23.580 --> 00:14:26.230
Constant also equal
to a constant,
00:14:26.230 --> 00:14:29.204
that's really lovely, right?
00:14:29.204 --> 00:14:33.860
[LAUGHS] That means,
I can say this
00:14:33.860 --> 00:14:36.850
is equal to that is
equal to a constant.
00:14:39.780 --> 00:14:44.060
As usual, I call this constant
really, really strange
00:14:44.060 --> 00:14:49.430
fancy name-- minus Km square,
which you will not like it.
00:14:49.430 --> 00:14:51.590
But later, you would like it.
00:14:51.590 --> 00:14:54.650
[LAUGHTER]
00:14:54.650 --> 00:14:55.170
Very good.
00:14:55.170 --> 00:14:58.390
So we make tremendous
amount of progress.
00:14:58.390 --> 00:15:03.360
Originally, we saw
that we are in trouble.
00:15:03.360 --> 00:15:05.320
It's A times Bt--
00:15:05.320 --> 00:15:09.030
Ax times Bt, sounds
really horrible.
00:15:09.030 --> 00:15:12.350
Now, actually, you
see that this equation
00:15:12.350 --> 00:15:16.550
is really simple to solve.
00:15:16.550 --> 00:15:20.690
So let's actually take
a look at the solution
00:15:20.690 --> 00:15:24.380
to the f function
and the g function.
00:15:24.380 --> 00:15:30.200
So the first thing, if I
take the left hand side,
00:15:30.200 --> 00:15:35.590
which is a time-dependence
part, I can copy there--
00:15:35.590 --> 00:15:36.630
this is actually 1--
00:15:36.630 --> 00:15:39.350
copy the equivalent here--
00:15:39.350 --> 00:15:49.300
is 1 over vp square B of t
partial square Bt partial t
00:15:49.300 --> 00:15:50.090
square.
00:15:50.090 --> 00:15:52.370
And this is equal
to a fancy name
00:15:52.370 --> 00:15:55.765
of this constant--
minus Km square.
00:15:59.420 --> 00:16:04.260
And of course, I can multiply
everything by vp squared B.
00:16:04.260 --> 00:16:08.740
And I get partial square
B partial t square.
00:16:08.740 --> 00:16:14.690
And this is equal too
minus vp square Km square
00:16:14.690 --> 00:16:17.760
B. Wait a second.
00:16:17.760 --> 00:16:21.540
We have solved this
equation infinite number
00:16:21.540 --> 00:16:24.090
of times in this lecture.
00:16:26.850 --> 00:16:28.080
You remember the solution?
00:16:28.080 --> 00:16:30.550
What is the solution?
00:16:30.550 --> 00:16:31.480
Anybody can help me?
00:16:34.342 --> 00:16:35.296
AUDIENCE: [INAUDIBLE].
00:16:35.296 --> 00:16:36.727
YEN-JIE LEE: Anybody?
00:16:36.727 --> 00:16:39.640
It's sine or cosine
function, right?
00:16:39.640 --> 00:16:41.750
This is harmonic oscillation.
00:16:41.750 --> 00:16:46.160
It's almost like equation of
motion of a spring-mass system,
00:16:46.160 --> 00:16:47.390
right?
00:16:47.390 --> 00:16:49.730
I hope that you
are already bored.
00:16:49.730 --> 00:16:52.010
And that means I
very successful.
00:16:52.010 --> 00:16:54.710
[LAUGHTER]
00:16:54.710 --> 00:16:58.760
B of t will be equal to--
00:16:58.760 --> 00:17:10.760
[LAUGHS]---- B of m sine
omega m t plus beta m,
00:17:10.760 --> 00:17:18.210
where omega m is actually
equal to vp times Km.
00:17:18.210 --> 00:17:23.060
I define omega m
equal to vp time Km.
00:17:29.030 --> 00:17:36.820
So surprisingly, the solution
of B is really simple.
00:17:36.820 --> 00:17:42.630
It's actually Bm sine
omega m t plus beta m.
00:17:42.630 --> 00:17:43.970
So what does that mean?
00:17:43.970 --> 00:17:49.940
That means the overall motion
or overall time-dependent
00:17:49.940 --> 00:17:56.030
evolution of the system
is like harmonic motion.
00:17:56.030 --> 00:18:03.080
If you do get individual
component in this system.
00:18:03.080 --> 00:18:06.230
So that's really nice.
00:18:06.230 --> 00:18:08.590
So let's take a look
at the right hand side.
00:18:08.590 --> 00:18:12.370
The right hand side
what we have is
00:18:12.370 --> 00:18:22.800
1 over A A of x partial
square A x partial x squared.
00:18:22.800 --> 00:18:26.520
And now, this equal
to minus Km squared.
00:18:30.630 --> 00:18:32.760
I don't want to go
over this again.
00:18:32.760 --> 00:18:36.660
This is actually the
same thing as number one.
00:18:36.660 --> 00:18:38.040
The only thing
which is different
00:18:38.040 --> 00:18:44.550
is that now the partial
derivative is actually the x.
00:18:44.550 --> 00:18:47.190
It's partial square
A partial x square.
00:18:47.190 --> 00:18:50.750
Therefore, I can immediately
write down the solution.
00:18:50.750 --> 00:18:52.600
A of t-- oh, sorry--
00:18:52.600 --> 00:19:04.200
A of x will be equal to
Cm sine Km x plus alpha m.
00:19:08.520 --> 00:19:09.430
Any questions so far?
00:19:14.180 --> 00:19:23.610
So until now, you accept the
fact that f of t and the g of x
00:19:23.610 --> 00:19:27.620
has to be the same constant.
00:19:27.620 --> 00:19:32.160
And I call it minus Km square.
00:19:32.160 --> 00:19:38.540
And I didn't actually tell
you what Km I'm choosing.
00:19:38.540 --> 00:19:45.700
In reality, according to this
result, Km can the anything,
00:19:45.700 --> 00:19:50.530
can be any number, as
long as it's a constant.
00:19:50.530 --> 00:19:56.410
Therefore, I would like to write
the corresponding psi, which
00:19:56.410 --> 00:20:01.970
is the wave function, is
actually labeled by m.
00:20:01.970 --> 00:20:07.860
m is it just label which K
I was using, nothing fancy.
00:20:07.860 --> 00:20:09.920
It's just a label.
00:20:09.920 --> 00:20:16.160
Psi of m is a
function of x and t.
00:20:16.160 --> 00:20:27.520
And that will be equal to
Bm times Cm sine omega mt
00:20:27.520 --> 00:20:37.830
plus beta m sine
Km x plus alpha m.
00:20:37.830 --> 00:20:41.520
Bm and Cm are just
arbitrary constant.
00:20:41.520 --> 00:20:44.110
Therefore, I can merge them.
00:20:44.110 --> 00:20:46.650
And I will call it just Am.
00:20:53.059 --> 00:20:57.710
And, of course, don't forget
we have this condition.
00:20:57.710 --> 00:21:03.590
Omega m is actually
equal to vp times Km.
00:21:03.590 --> 00:21:05.130
So this is actually
defined here.
00:21:10.880 --> 00:21:14.840
So here, since this is actually
a second order differential
00:21:14.840 --> 00:21:20.200
equation, you have unknown
factor, which is beta m.
00:21:20.200 --> 00:21:24.360
You have also Bm is a unknown.
00:21:24.360 --> 00:21:26.570
And the right hand
side, you also
00:21:26.570 --> 00:21:33.834
have a Cm, which is unknown,
and alpha m, which is unknown.
00:21:36.500 --> 00:21:43.040
When we combine them, I
replace Bm times Cm by Am.
00:21:43.040 --> 00:21:46.820
Therefore, what we have
is that Am is actually
00:21:46.820 --> 00:21:50.690
some kind of amplitude,
which can be determined
00:21:50.690 --> 00:21:53.840
by initial conditions, which
I will talk about that later.
00:21:56.680 --> 00:22:00.930
Beta m is the unknown coming
from the left hand side
00:22:00.930 --> 00:22:02.710
derivation.
00:22:02.710 --> 00:22:05.650
Alpha m is also unknown,
which is actually
00:22:05.650 --> 00:22:09.430
coming from the right
side derivation related
00:22:09.430 --> 00:22:14.080
to the shape of the
normal mode of the system.
00:22:14.080 --> 00:22:18.580
And finally, there's one
additional unknown coefficient,
00:22:18.580 --> 00:22:19.940
which is Km--
00:22:19.940 --> 00:22:21.660
it's kind of arbitrary now--
00:22:21.660 --> 00:22:25.660
control actually
the wave number,
00:22:25.660 --> 00:22:32.490
or say the wavelength of the
shape of the normal mode.
00:22:32.490 --> 00:22:37.170
So when you see this,
doesn't this surprise you?
00:22:37.170 --> 00:22:40.110
May not surprise you
any more, because we
00:22:40.110 --> 00:22:44.390
have solved infinite number
of coupled oscillator.
00:22:44.390 --> 00:22:48.720
And you have learned
that, OK, the normal modes
00:22:48.720 --> 00:22:52.720
have a shape of sine function.
00:22:52.720 --> 00:22:55.250
It's like a sine function--
00:22:55.250 --> 00:22:58.920
before it was like
a sine as a Ka.
00:22:58.920 --> 00:23:01.140
And the Ka is performing x.
00:23:01.140 --> 00:23:03.030
So what we are
actually getting here
00:23:03.030 --> 00:23:09.540
is that doesn't surprise you
since this system also satisfy
00:23:09.540 --> 00:23:12.480
space translational symmetry.
00:23:12.480 --> 00:23:19.530
Therefore, the functional form
of the shape of the normal mode
00:23:19.530 --> 00:23:21.950
is also a sine function.
00:23:21.950 --> 00:23:25.050
So that's actually
pretty satisfactory
00:23:25.050 --> 00:23:30.690
and also come out as what we
would expect based on what
00:23:30.690 --> 00:23:32.050
we actually have learned.
00:23:35.030 --> 00:23:39.330
So let's actually take a look at
the structure of this function.
00:23:39.330 --> 00:23:42.740
So basically, as I
mentioned before,
00:23:42.740 --> 00:23:48.760
everything is oscillating
at the frequency omega m
00:23:48.760 --> 00:23:51.200
with a phase beta m.
00:23:51.200 --> 00:23:55.040
So that satisfy the
condition of normal mode,
00:23:55.040 --> 00:23:56.690
what is the condition.
00:23:56.690 --> 00:23:58.580
All the components
in the system are
00:23:58.580 --> 00:24:03.790
oscillating at the same
frequency and the same phase.
00:24:03.790 --> 00:24:08.140
Indeed, yes, that's correct.
00:24:08.140 --> 00:24:11.870
Also as a function of
time, it's actually
00:24:11.870 --> 00:24:16.040
going up and down harmonically
as we already discussed.
00:24:16.040 --> 00:24:19.670
And the relative amplitude,
as I said, is a sine function.
00:24:19.670 --> 00:24:25.030
And of course, I already
demonstrated this before,
00:24:25.030 --> 00:24:28.130
that you can see that
in this is system,
00:24:28.130 --> 00:24:32.610
you can see a sine function
when I start to drive it.
00:24:32.610 --> 00:24:36.650
So what I am doing is actually
to convert the kinetic energy
00:24:36.650 --> 00:24:43.780
from my hand to energy stored
as potential or kinetic energy
00:24:43.780 --> 00:24:50.480
in this string-rod system and
in this bell wave machine.
00:24:50.480 --> 00:24:54.500
So you can see that beautifully
those are sine function.
00:24:54.500 --> 00:24:57.800
And, of course, if I do
a higher frequency one,
00:24:57.800 --> 00:25:01.115
you can see the hand
oscillation frequency changed.
00:25:03.920 --> 00:25:09.330
And that is actually
controlled by this equation,
00:25:09.330 --> 00:25:12.800
this dispersion relation.
00:25:12.800 --> 00:25:15.470
And this dispersion
relation is actually
00:25:15.470 --> 00:25:18.770
relating two physical quantity.
00:25:18.770 --> 00:25:22.180
One is actually the wave number.
00:25:22.180 --> 00:25:26.065
Of course, if you are more
familiar with wavelength,
00:25:26.065 --> 00:25:29.330
it's actually 2 pi over Km.
00:25:29.330 --> 00:25:31.130
Lambda m is actually
the wavelength
00:25:31.130 --> 00:25:36.080
of the shape of the normal mode.
00:25:36.080 --> 00:25:38.000
And the oscillation
frequency is actually
00:25:38.000 --> 00:25:45.440
controlled by this dispersion
relation, this function.
00:25:45.440 --> 00:25:47.600
And you can say
that, Professor Lee,
00:25:47.600 --> 00:25:51.950
I have been so
tired of this demo.
00:25:51.950 --> 00:25:54.320
I've seen this
1,000 times, right?
00:25:54.320 --> 00:25:56.730
And basically,
you are showing me
00:25:56.730 --> 00:25:59.645
that, OK, you can actually
oscillate this system
00:25:59.645 --> 00:26:02.480
and excite this
system so that it's
00:26:02.480 --> 00:26:05.870
oscillating at some natural
frequency the system
00:26:05.870 --> 00:26:06.630
like, right?
00:26:06.630 --> 00:26:09.430
It's actually some kind
of resonance behavior.
00:26:09.430 --> 00:26:11.090
So I can actually excite--
00:26:11.090 --> 00:26:12.140
I can-- no.
00:26:12.140 --> 00:26:15.150
I can randomly
shake this system.
00:26:15.150 --> 00:26:18.290
And then it's going to be
a linear combination of all
00:26:18.290 --> 00:26:20.390
the excited normal modes, right?
00:26:20.390 --> 00:26:22.880
We have seen this
many, many times.
00:26:22.880 --> 00:26:28.490
What I am going to show you is
that there's another machine
00:26:28.490 --> 00:26:31.895
here, which is
actually demonstrating
00:26:31.895 --> 00:26:36.680
the resonance of some wave.
00:26:36.680 --> 00:26:47.360
So here is actually
so-called the Rijke tube.
00:26:47.360 --> 00:26:49.010
So the structure is like this.
00:26:49.010 --> 00:26:54.710
So basically, you have a metal
tube, which is red thing there.
00:26:54.710 --> 00:26:56.300
And inside the tube--
00:26:56.300 --> 00:26:59.030
you cannot see it now--
but inside the tube,
00:26:59.030 --> 00:27:06.710
there's a wire mesh, which the
air can flow freely up and down
00:27:06.710 --> 00:27:09.080
in this mesh.
00:27:09.080 --> 00:27:14.450
And what I'm going to do
now is to heat up the mesh
00:27:14.450 --> 00:27:16.760
and see what is going to happen.
00:27:16.760 --> 00:27:20.480
I will just heat it
up by like six second
00:27:20.480 --> 00:27:23.400
and see what is going to happen.
00:27:23.400 --> 00:27:25.840
So now, I'm going to
do this very carefully.
00:27:35.640 --> 00:27:38.580
[RESONATING SOUND]
00:27:38.580 --> 00:27:40.410
Can you hear that?
00:27:47.284 --> 00:27:56.760
[LAUGHS] OK, very good.
00:27:56.760 --> 00:28:00.290
So listen, what is happening?
00:28:00.290 --> 00:28:07.740
So you hear a mono frequency
sound generated from what?
00:28:07.740 --> 00:28:13.860
Generated from the heat
I gave to the wire mesh.
00:28:13.860 --> 00:28:15.540
So what is actually happening?
00:28:15.540 --> 00:28:22.380
So when I heat up the mesh,
what is going to happen
00:28:22.380 --> 00:28:28.000
is that the air
around this screen
00:28:28.000 --> 00:28:30.410
is going to be heated up.
00:28:30.410 --> 00:28:33.350
Therefore, because
the air is heated up,
00:28:33.350 --> 00:28:35.520
it goes in the upward direction.
00:28:35.520 --> 00:28:38.630
And also, the
volume of the air is
00:28:38.630 --> 00:28:43.510
expanding, because of the
increased temperature.
00:28:43.510 --> 00:28:47.020
And that actually goes
through this system.
00:28:47.020 --> 00:28:52.860
And why it does is really like
what I'm doing to the bell lab
00:28:52.860 --> 00:28:59.180
wave machine, is actually
tyring to oscillate-- or excite
00:28:59.180 --> 00:29:06.070
any possible normal modes,
which this system actually like.
00:29:06.070 --> 00:29:10.870
So you can see that, after
a while, once the pressure
00:29:10.870 --> 00:29:18.520
and also the air inside the
tube get then self-organized,
00:29:18.520 --> 00:29:25.090
then you hear a very loud sound.
00:29:25.090 --> 00:29:28.045
So that means there are
energy flowing from the tube
00:29:28.045 --> 00:29:29.380
to your ear.
00:29:29.380 --> 00:29:31.490
And that is actually
coming from what?
00:29:31.490 --> 00:29:35.810
Coming from the heat
I put into the system.
00:29:35.810 --> 00:29:41.410
So it's actually a heat
sound wave conversion.
00:29:41.410 --> 00:29:43.860
So I hope you enjoyed this demo.
00:29:43.860 --> 00:29:46.750
And we will take a
five-minute break
00:29:46.750 --> 00:29:50.380
to take questions
before we move on.
00:29:50.380 --> 00:29:52.540
And of course, you are
welcome to come here
00:29:52.540 --> 00:29:57.370
and to play with the
demo if you want.
00:29:57.370 --> 00:30:03.440
[LAUGHS]
00:30:03.440 --> 00:30:07.010
So welcome back from the break.
00:30:07.010 --> 00:30:10.460
So what we are
going to do next is
00:30:10.460 --> 00:30:14.720
to understand how to determine
all those unknown coefficients.
00:30:14.720 --> 00:30:21.080
So you get to see here, there
are Am, which is the amplitude.
00:30:21.080 --> 00:30:25.846
There are beta m, which
is basically the phase.
00:30:25.846 --> 00:30:30.170
There are Km, which is
actually the wave number,
00:30:30.170 --> 00:30:34.830
and alpha m, which is
the phase for the shape.
00:30:34.830 --> 00:30:42.710
So what I'm going to show you
is that Am and the beta m,
00:30:42.710 --> 00:30:45.770
these two quantity
will be determined
00:30:45.770 --> 00:30:48.497
by initial conditions.
00:30:55.960 --> 00:31:03.460
Well, Km and alpha
m, as you may guess,
00:31:03.460 --> 00:31:13.170
those can be determined
by boundary condition
00:31:13.170 --> 00:31:17.560
for the Km and alpha m.
00:31:17.560 --> 00:31:20.620
So why don't we
just immediately get
00:31:20.620 --> 00:31:26.290
started with a concrete example.
00:31:26.290 --> 00:31:30.760
So let's take a look
at this situation.
00:31:30.760 --> 00:31:34.420
So this equation and
those all the possible Km
00:31:34.420 --> 00:31:40.120
are allowed when we talk
about infinitely long system.
00:31:40.120 --> 00:31:44.620
So far, we have not imposed
any boundary conditions.
00:31:44.620 --> 00:31:50.140
And what I'm going to do now is
to show you a example boundary
00:31:50.140 --> 00:31:55.550
condition and see how we can
actually fix Km and alpha m.
00:31:55.550 --> 00:32:00.510
So suppose we are
interested in this system.
00:32:00.510 --> 00:32:05.000
So I have a wall in
the left hand side.
00:32:05.000 --> 00:32:10.880
And I have a string
with length L.
00:32:10.880 --> 00:32:13.550
And it's actually
connected to massless ring.
00:32:23.470 --> 00:32:30.762
And this ring can actually
move up and down a long rod
00:32:30.762 --> 00:32:31.720
in the right hand side.
00:32:35.070 --> 00:32:44.550
And I also assume that this
string have a constant tension
00:32:44.550 --> 00:32:51.930
T. And also the
density is rho L.
00:32:51.930 --> 00:32:55.800
So basically, it's a
mass per unit length.
00:32:55.800 --> 00:32:58.290
So that's the system
which I am interested.
00:32:58.290 --> 00:33:03.590
And, of course, I need to define
my coordinate system as usual.
00:33:03.590 --> 00:33:08.040
I define horizontal
direction to be x direction.
00:33:08.040 --> 00:33:15.240
And I define the vertical
direction to be y direction.
00:33:15.240 --> 00:33:20.070
And I define y equal
to 0 is the equilibrium
00:33:20.070 --> 00:33:23.340
portion of the string.
00:33:23.340 --> 00:33:28.650
When psi is equal to 0, that
means this string is actually
00:33:28.650 --> 00:33:30.420
at rest.
00:33:30.420 --> 00:33:36.240
And not moving-- is actually
in the equilibrium position,
00:33:36.240 --> 00:33:41.310
it's not displaced at all
with respect to y equal to 0.
00:33:41.310 --> 00:33:45.450
And I can also define
that x equal to 0
00:33:45.450 --> 00:33:50.460
is the position of the
left hand side wall.
00:33:50.460 --> 00:33:53.170
So this is actually
the physical situation.
00:33:53.170 --> 00:33:56.940
And I would like to
actually find out what
00:33:56.940 --> 00:33:59.320
are the boundary conditions.
00:33:59.320 --> 00:34:01.510
So what are the
boundary condition?
00:34:01.510 --> 00:34:04.550
So from what we actually
discussed last time,
00:34:04.550 --> 00:34:08.670
left hand side, since this
string is actually fixed
00:34:08.670 --> 00:34:13.280
on the wall, I nailed it
there, it cannot move.
00:34:13.280 --> 00:34:17.927
Therefore, what is actually
the first boundary condition?
00:34:22.600 --> 00:34:25.870
Why is actually the
first boundary condition?
00:34:25.870 --> 00:34:28.350
Anybody can tell me?
00:34:28.350 --> 00:34:31.259
Which describes the situation,
the physical situation
00:34:31.259 --> 00:34:33.231
on your left hand side?
00:34:33.231 --> 00:34:35.219
AUDIENCE: y0 is 0.
00:34:35.219 --> 00:34:37.290
YEN-JIE LEE: y0 is 0.
00:34:37.290 --> 00:34:38.250
Very good.
00:34:38.250 --> 00:34:44.699
So when x is equal
to 0, y0 is 0.
00:34:44.699 --> 00:34:48.060
So on my note, I was using
a different notation.
00:34:48.060 --> 00:34:50.550
So I would just use psi.
00:34:50.550 --> 00:34:55.800
So psi 0 is equal to 0.
00:34:55.800 --> 00:35:00.900
Apparently, there's another
boundary of this system.
00:35:00.900 --> 00:35:03.390
The other boundary
condition is actually
00:35:03.390 --> 00:35:08.600
happening at x equal to L.
What is actually the boundary
00:35:08.600 --> 00:35:09.210
condition?
00:35:09.210 --> 00:35:10.250
Can somebody help me?
00:35:10.250 --> 00:35:10.750
Yes.
00:35:10.750 --> 00:35:13.654
AUDIENCE: Is it the
derivative of psi is 0?
00:35:13.654 --> 00:35:17.380
YEN-JIE LEE: The derivative
of the psi is equal to 0.
00:35:17.380 --> 00:35:22.270
So we will explain to
everybody why is that the case.
00:35:22.270 --> 00:35:31.560
The answer proposed is that
partial psi partial x L
00:35:31.560 --> 00:35:33.120
t is equal to 0.
00:35:33.120 --> 00:35:35.580
And this is 0, t,
because that has
00:35:35.580 --> 00:35:39.880
to be true no matter when I
actually invented this boundary
00:35:39.880 --> 00:35:42.240
condition.
00:35:42.240 --> 00:35:48.360
So what is actually giving
us this strange boundary
00:35:48.360 --> 00:35:50.820
condition?
00:35:50.820 --> 00:35:58.540
So suppose I focus on the
force diagram on this ring.
00:35:58.540 --> 00:36:00.700
So this ring is
actually connected
00:36:00.700 --> 00:36:07.490
to a string with
string tension T. Also,
00:36:07.490 --> 00:36:09.340
there's another force
which is actually
00:36:09.340 --> 00:36:13.300
trying to balance the
string tension, which
00:36:13.300 --> 00:36:15.460
is a normal force--
00:36:15.460 --> 00:36:19.440
normal force coming from
the rod, which is actually
00:36:19.440 --> 00:36:22.210
trying to stop the
ring from moving
00:36:22.210 --> 00:36:23.990
in the horizontal direction.
00:36:23.990 --> 00:36:26.156
So there's normal force then.
00:36:28.710 --> 00:36:34.110
And we also know that this
ring is actually massless.
00:36:34.110 --> 00:36:38.340
So m is equal to 0.
00:36:38.340 --> 00:36:46.325
Suppose that this partial psi
partial x, the slope is not 0.
00:36:46.325 --> 00:36:52.800
If slope is not 0, that means
the string may be pulling
00:36:52.800 --> 00:36:57.650
this ring to some direction.
00:36:57.650 --> 00:37:00.050
What is going to happen?
00:37:00.050 --> 00:37:05.960
So if this happens, it is
actually clear that the normal
00:37:05.960 --> 00:37:09.340
force cannot balance
the string for us.
00:37:12.270 --> 00:37:14.500
Everybody get it?
00:37:14.500 --> 00:37:16.480
What will happen?
00:37:16.480 --> 00:37:20.920
If this happened, then
this massless ring
00:37:20.920 --> 00:37:26.860
will suffer from infinitely
large acceleration.
00:37:26.860 --> 00:37:30.550
Because F is equal to ma.
00:37:30.550 --> 00:37:34.270
And m is 0, so a
goes to infinity.
00:37:34.270 --> 00:37:37.900
So that means this ring
will- peeew- disappear, go
00:37:37.900 --> 00:37:40.200
to the edge of the universe.
00:37:40.200 --> 00:37:41.950
Did that happen?
00:37:41.950 --> 00:37:44.735
No, it didn't happen.
00:37:44.735 --> 00:37:49.800
Therefore, this condition
must be satisfied.
00:37:49.800 --> 00:37:50.300
You see?
00:37:50.300 --> 00:37:54.140
So the slope of the
string cannot be nonzero.
00:37:54.140 --> 00:37:56.900
Otherwise, some
crisis will happen.
00:37:59.810 --> 00:38:00.450
Very good.
00:38:00.450 --> 00:38:03.850
So we have the two conditions.
00:38:03.850 --> 00:38:09.410
And the second thing, which we
are going to demonstrate you,
00:38:09.410 --> 00:38:13.220
is that, OK, I promise you
that boundary condition can
00:38:13.220 --> 00:38:17.120
fix these two constants.
00:38:17.120 --> 00:38:19.510
So therefore, we are
going to demonstrate that.
00:38:23.740 --> 00:38:29.250
So let's use the first condition
we have in the right hand side
00:38:29.250 --> 00:38:30.460
board.
00:38:30.460 --> 00:38:37.070
And basically, from 1, you
can actually get psi m 0, t.
00:38:37.070 --> 00:38:40.780
I am plugging in
this condition--
00:38:40.780 --> 00:38:47.440
plugging in this solution to
boundary condition number 1.
00:38:47.440 --> 00:38:49.060
And basically, what
I am going to get
00:38:49.060 --> 00:38:58.840
is this is equal to Am
sine alpha m sine omega m t
00:38:58.840 --> 00:39:02.550
plus beta m.
00:39:02.550 --> 00:39:05.910
And this is actually equal to 0.
00:39:05.910 --> 00:39:10.770
So you only have a
alpha m here because I
00:39:10.770 --> 00:39:15.000
am setting x to be equal to 0.
00:39:15.000 --> 00:39:17.370
I'm setting x to be equal to 0.
00:39:17.370 --> 00:39:21.690
Therefore, you already
have that functional form.
00:39:21.690 --> 00:39:24.000
So now, we are facing a choice.
00:39:24.000 --> 00:39:27.750
So you can set Am to be equal
to 0 is arbitrary number.
00:39:27.750 --> 00:39:31.440
But if you set m equal
to 0, everything is 0.
00:39:31.440 --> 00:39:33.930
And it's not fun,
it's not moving.
00:39:33.930 --> 00:39:39.360
Therefore, I don't want to
set Am to be equal to 0.
00:39:39.360 --> 00:39:43.080
And you can say, huh,
maybe this is equal to 0--
00:39:43.080 --> 00:39:46.170
sine omega m t plus
beta m is equal to 0.
00:39:46.170 --> 00:39:48.780
But this is really
a sine function.
00:39:48.780 --> 00:39:55.350
And this condition has
to be satisfied no matter
00:39:55.350 --> 00:39:59.610
at which time you are revisiting
this boundary condition.
00:39:59.610 --> 00:40:03.450
At all times, this boundary
condition has to be satisfied.
00:40:03.450 --> 00:40:08.520
Therefore, this
cannot be equal to 0.
00:40:08.520 --> 00:40:11.520
Therefore, I conclude
that this is the 0.
00:40:14.372 --> 00:40:15.330
So what does that mean?
00:40:15.330 --> 00:40:19.350
That means I can choose
alpha m is equal to 0.
00:40:21.930 --> 00:40:26.031
So that's actually given by
the first boundary condition.
00:40:28.860 --> 00:40:32.470
So let's actually take a look at
the second boundary condition--
00:40:32.470 --> 00:40:39.660
partial psi partial x evaluated
at x equal to L and any time t
00:40:39.660 --> 00:40:41.610
is equal to 0.
00:40:41.610 --> 00:40:44.310
So now, I can plug in,
again, the solution
00:40:44.310 --> 00:40:53.100
in the middle board partial
psi m L, t, partial x.
00:40:53.100 --> 00:41:05.300
And that will be equal to Am
Km sine omega m t plus beta m
00:41:05.300 --> 00:41:10.990
cosine Km x.
00:41:10.990 --> 00:41:14.590
And this is equal to 0.
00:41:14.590 --> 00:41:17.950
So I am taking a partial
derivative partial psi
00:41:17.950 --> 00:41:19.090
partial x.
00:41:19.090 --> 00:41:21.190
Therefore, the
sine become cosine.
00:41:21.190 --> 00:41:27.660
The sine Km x plus
alpha m becoming cosine.
00:41:27.660 --> 00:41:30.460
And also, I know already from
the first boundary condition,
00:41:30.460 --> 00:41:33.180
alpha m is equal to 0.
00:41:33.180 --> 00:41:36.720
Therefore, I get
cosine Km x here.
00:41:41.070 --> 00:41:45.790
And this is evaluated
at x equal to L.
00:41:45.790 --> 00:41:50.410
So that means this
thing must be equal to 0
00:41:50.410 --> 00:41:53.330
based on the second
boundary condition.
00:41:53.330 --> 00:41:56.020
Of course, we can have
a losing argument--
00:41:56.020 --> 00:41:57.730
Am should not be equal to 0.
00:41:57.730 --> 00:42:00.730
Otherwise, you will be
equal to 0 all the time,
00:42:00.730 --> 00:42:04.500
the whole wave function is 0.
00:42:04.500 --> 00:42:06.840
And this is actually
changing as a function
00:42:06.840 --> 00:42:09.960
of time, the same argument,
because this boundary condition
00:42:09.960 --> 00:42:13.040
has to be satisfied
at all times.
00:42:13.040 --> 00:42:14.730
From the beginning
of the Universe
00:42:14.730 --> 00:42:20.100
to the end of the Universe, this
condition has to be satisfied.
00:42:20.100 --> 00:42:22.410
Therefore, these
cannot be equal to 0.
00:42:22.410 --> 00:42:29.150
And what is actually left over
is cosine Km x evaluated at L
00:42:29.150 --> 00:42:30.250
equal to 0.
00:42:30.250 --> 00:42:37.230
So cosine Km L is equal to 0.
00:42:37.230 --> 00:42:44.040
So that means you cannot
arbitrarily choose Km anymore.
00:42:44.040 --> 00:42:47.900
Before we introduced boundary
conditions, we were saying, ah,
00:42:47.900 --> 00:42:51.480
Km is actually some
arbitrary constant.
00:42:51.480 --> 00:42:54.720
And now, it's not
arbitrary any more.
00:42:54.720 --> 00:42:58.380
It has to satisfy
this condition.
00:42:58.380 --> 00:42:59.890
What does that mean?
00:42:59.890 --> 00:43:05.210
This means that Km
has to be equal to 2m
00:43:05.210 --> 00:43:09.150
minus 1 divided by 2L times pi.
00:43:11.860 --> 00:43:13.730
You can actually check this.
00:43:13.730 --> 00:43:21.724
And this small m is equal to
1, 2, 3, et cetera, et cetera.
00:43:26.180 --> 00:43:29.370
And then you can
see that there are
00:43:29.370 --> 00:43:32.200
many, many different solutions.
00:43:35.560 --> 00:43:40.270
So you can see that,
as I mentioned before,
00:43:40.270 --> 00:43:45.820
the boundary conditions
determine Km and alpha m.
00:43:45.820 --> 00:43:49.510
So you can see that the first
condition at x equal to 0
00:43:49.510 --> 00:43:51.460
determine alpha m.
00:43:51.460 --> 00:43:53.620
The second boundary
condition also
00:43:53.620 --> 00:43:58.600
help us to determine what
are the possible Km value.
00:43:58.600 --> 00:44:04.210
And that is actually
listed here.
00:44:04.210 --> 00:44:07.430
Any questions?
00:44:07.430 --> 00:44:07.930
No?
00:44:12.490 --> 00:44:16.130
So in order to help
you to visualize
00:44:16.130 --> 00:44:18.370
what we have
learned from here, I
00:44:18.370 --> 00:44:22.450
can now choose m is equal to 1.
00:44:22.450 --> 00:44:26.650
So you can see that I
carefully choose my notation
00:44:26.650 --> 00:44:27.910
from the beginning.
00:44:27.910 --> 00:44:34.480
So therefore, m is now the
index of the normal mode
00:44:34.480 --> 00:44:36.400
I am referring to.
00:44:36.400 --> 00:44:40.930
So now, if I choose
m equal to 1,
00:44:40.930 --> 00:44:43.300
then I can actually
evaluate what
00:44:43.300 --> 00:44:49.280
would be the resulting K. So
K1, according to last formula 2
00:44:49.280 --> 00:44:51.770
minus 1 is giving you 1.
00:44:51.770 --> 00:44:53.750
Therefore, you'll
get pi over 2L.
00:44:56.530 --> 00:44:58.810
And, of course, you
can also calculate
00:44:58.810 --> 00:45:02.460
based on the wave number
what will be the wavelength.
00:45:02.460 --> 00:45:08.010
So the wavelength lambda 1
will be equal to 2 pi over K1.
00:45:08.010 --> 00:45:11.170
Wine And that will give you 4L.
00:45:14.430 --> 00:45:22.730
Don't forget, once you
actually already decide K,
00:45:22.730 --> 00:45:27.980
the omega is also
determined, because omega,
00:45:27.980 --> 00:45:31.920
which is the angular
frequency of this normal mode,
00:45:31.920 --> 00:45:34.980
is determined by that
dispersion relation omega
00:45:34.980 --> 00:45:38.520
m equal to vp times Km.
00:45:38.520 --> 00:45:41.640
So therefore, I can
now calculate omega 1.
00:45:41.640 --> 00:45:46.100
That will be equal
to vp times K1.
00:45:46.100 --> 00:45:56.310
And that will you square root
of T over rho L pi over 2L.
00:45:56.310 --> 00:45:59.730
So this is actually coming
from the last lecture,
00:45:59.730 --> 00:46:00.750
the formula of vp.
00:46:03.840 --> 00:46:10.230
So that means, if you fix the
shape of you are normal mode,
00:46:10.230 --> 00:46:13.200
then the angular
frequency is also fixed,
00:46:13.200 --> 00:46:17.880
according to this
dispersion relation.
00:46:17.880 --> 00:46:21.690
So of course, I can now
visualize this situation.
00:46:21.690 --> 00:46:28.490
And basically, I can plot this
system as a function of time--
00:46:28.490 --> 00:46:31.200
as a function of x, not
as a function of time.
00:46:31.200 --> 00:46:36.380
So when this system reach
the maxima amplitude,
00:46:36.380 --> 00:46:38.640
it would look like this.
00:46:38.640 --> 00:46:43.940
And this is actually
amplitude A1.
00:46:43.940 --> 00:46:48.470
Because I am talking about the
first normal mode labeled by m
00:46:48.470 --> 00:46:52.730
equal to 1, and there is
an unknown amplitude A1.
00:46:52.730 --> 00:46:54.400
And that is actually
showing here.
00:46:57.350 --> 00:47:02.186
Of course, I can go
ahead and calculate
00:47:02.186 --> 00:47:06.380
if m is equal to 2,
what is going to happen?
00:47:06.380 --> 00:47:09.430
If I increase the m,
what is going to happen
00:47:09.430 --> 00:47:14.120
is that K is also increased.
00:47:14.120 --> 00:47:15.420
So K is increased.
00:47:15.420 --> 00:47:19.010
Then that means the
wavelength is decreased.
00:47:19.010 --> 00:47:21.550
I have calculated the K for you.
00:47:21.550 --> 00:47:26.630
And that is equal
to 3 pi over 2L.
00:47:26.630 --> 00:47:33.300
And those are the lambda 2
will be equal to 4L over 3.
00:47:33.300 --> 00:47:38.560
You can actually double
check this at home.
00:47:38.560 --> 00:47:41.900
And of course, I
can now demonstrate
00:47:41.900 --> 00:47:47.090
what would be the resulting
shape of the massless mode.
00:47:47.090 --> 00:47:50.550
It would look like this.
00:47:50.550 --> 00:47:57.930
And this is essentially
telling you the amplitude A2.
00:47:57.930 --> 00:48:01.320
You can also do m equal to 3.
00:48:01.320 --> 00:48:03.855
If you doing that,
basically what you get
00:48:03.855 --> 00:48:10.020
is something like this,
et cetera, et cetera.
00:48:12.640 --> 00:48:14.790
Any questions?
00:48:14.790 --> 00:48:19.880
And the motion of this system
that is a function of time
00:48:19.880 --> 00:48:22.890
is that this whole
shape, this shape,
00:48:22.890 --> 00:48:29.350
is multiplied by sine
omega m t plus beta m.
00:48:29.350 --> 00:48:36.180
So the whole shape is going to
scale up and down harmonically.
00:48:36.180 --> 00:48:40.830
And so if you focus on
one of the point here,
00:48:40.830 --> 00:48:45.330
it's going to be going
up and down harmonically.
00:48:45.330 --> 00:48:49.140
Very important, there's no
back and forth movement.
00:48:49.140 --> 00:48:51.540
Everything is only
moving up and down.
00:48:51.540 --> 00:48:56.660
If you focus on only one of
the particle in this string,
00:48:56.660 --> 00:48:58.900
everything is
moving up and down.
00:48:58.900 --> 00:48:59.640
Like here, right?
00:48:59.640 --> 00:49:02.530
So when I create a curve--
00:49:02.530 --> 00:49:08.140
when I create some kind of wave,
all the components are always
00:49:08.140 --> 00:49:11.730
moving only up and down,
instead of back and forth,
00:49:11.730 --> 00:49:12.810
because they can't.
00:49:12.810 --> 00:49:14.230
They can't move back and forth.
00:49:14.230 --> 00:49:18.530
But you maybe cheated by
the shape-- the evolution
00:49:18.530 --> 00:49:20.081
as a boundary of
time, it seems to me
00:49:20.081 --> 00:49:22.330
that, ah, something is
actually moving back and forth.
00:49:22.330 --> 00:49:26.930
But never-- all the particles
are moving up and down--
00:49:26.930 --> 00:49:29.340
very important.
00:49:29.340 --> 00:49:36.640
Finally, we have shown you
the first three normal modes.
00:49:36.640 --> 00:49:41.035
And what is actually the
most general solution?
00:49:41.035 --> 00:49:42.160
What is a general solution?
00:49:48.700 --> 00:49:53.020
Of course, as we had
before, general solution
00:49:53.020 --> 00:49:58.690
would be a linear combination of
all the possible normal modes.
00:49:58.690 --> 00:50:02.320
So now, I would like
to write psi x, t,
00:50:02.320 --> 00:50:07.050
as the general solution is
going to be the sum of all
00:50:07.050 --> 00:50:10.450
the allowed normal mode.
00:50:10.450 --> 00:50:13.960
In this specific
case, it's going
00:50:13.960 --> 00:50:17.140
to be a summation from 1--
00:50:17.140 --> 00:50:34.630
m equal to 1 to infinity Am sine
omega m t plus beta m sine Km x
00:50:34.630 --> 00:50:36.420
plus alpha m.
00:50:36.420 --> 00:50:40.260
And in this specific
case, it become cosine--
00:50:40.260 --> 00:50:42.600
the Km is there--
00:50:42.600 --> 00:50:46.020
cosine 2m minus 1--
00:50:46.020 --> 00:50:48.220
sorry, it should be sine.
00:50:48.220 --> 00:50:54.660
It should be sine 2m
minus 1 over 2L pi.
00:50:54.660 --> 00:50:59.220
And the alpha m in this
case is equal to 0.
00:50:59.220 --> 00:51:02.820
So the upper formula is
the most general case.
00:51:02.820 --> 00:51:06.380
You're summing over
the possible m's.
00:51:06.380 --> 00:51:08.910
And the Km and alpha
m can be determined
00:51:08.910 --> 00:51:11.250
by boundary conditions.
00:51:11.250 --> 00:51:16.080
And in this specific case,
the right hand side expression
00:51:16.080 --> 00:51:21.861
is reading like sine 2m
minus 1 over 2L times pi.
00:51:25.370 --> 00:51:27.650
So that's very nice.
00:51:27.650 --> 00:51:31.490
And then you can see
another sets of example
00:51:31.490 --> 00:51:33.500
here in the slide.
00:51:33.500 --> 00:51:37.520
So this is another
set of normal modes
00:51:37.520 --> 00:51:41.900
from m equal to 1
to m equal to 6.
00:51:41.900 --> 00:51:45.560
And you can see
that in this example
00:51:45.560 --> 00:51:51.710
both ends are fixed, instead
of one end is actually
00:51:51.710 --> 00:51:54.190
attached to a massless ring.
00:51:54.190 --> 00:51:57.710
If both ends are fixed, then
you get these normal modes.
00:51:57.710 --> 00:52:00.290
And of course, in the later--
00:52:00.290 --> 00:52:02.840
in your p set, you
will be exercising
00:52:02.840 --> 00:52:06.170
this kind of normal
modes and solve
00:52:06.170 --> 00:52:08.720
the corresponding
Km and alpha m.
00:52:08.720 --> 00:52:14.180
And you can see that, if you
focus on the upper left corner,
00:52:14.180 --> 00:52:20.600
you will see that the
oscillation frequency is low.
00:52:20.600 --> 00:52:21.650
Why is that?
00:52:21.650 --> 00:52:27.290
That is because the wave
number Km is small, therefore,
00:52:27.290 --> 00:52:28.970
wavelength is long.
00:52:28.970 --> 00:52:31.760
According to that
formula, omega m
00:52:31.760 --> 00:52:35.040
is proportional to wave number.
00:52:35.040 --> 00:52:38.270
Therefore, you can see that
the oscillation frequency
00:52:38.270 --> 00:52:43.120
is actually two
times slower compared
00:52:43.120 --> 00:52:47.510
to m equal to 2 case, which
is the upper right corner
00:52:47.510 --> 00:52:52.310
result And you can see that, if
you increase m more and more,
00:52:52.310 --> 00:52:55.520
you get larger and larger
K. And therefore, you
00:52:55.520 --> 00:52:59.581
see that the oscillation
frequency is getting larger
00:52:59.581 --> 00:53:00.080
and larger.
00:53:06.320 --> 00:53:11.240
So now, we are actually
facing an issue here.
00:53:11.240 --> 00:53:17.960
Wait a second, so now we
have solved the functional
00:53:17.960 --> 00:53:19.490
form of the normal mode.
00:53:19.490 --> 00:53:23.750
We have learned how to
determine Km and alpha m.
00:53:23.750 --> 00:53:25.520
But we are facing
a difficulty here,
00:53:25.520 --> 00:53:28.610
because Am is very
difficult to solve,
00:53:28.610 --> 00:53:31.910
because you have infinite
number of terms here.
00:53:31.910 --> 00:53:35.480
And beta m, how
do we solve this?
00:53:35.480 --> 00:53:39.870
So it's getting really,
really difficult.
00:53:39.870 --> 00:53:41.990
So what I am going
to tell you is
00:53:41.990 --> 00:53:45.560
that we can actually, again,
use the help from the math
00:53:45.560 --> 00:53:46.610
department.
00:53:46.610 --> 00:53:49.920
They have actually
proposed the solution.
00:53:49.920 --> 00:53:52.770
They actually say that,
huh, this is actually
00:53:52.770 --> 00:53:55.910
identical problem that we
solved in the math department,
00:53:55.910 --> 00:54:01.430
is just for the decomposition
and for the series.
00:54:01.430 --> 00:54:03.200
So what is actually
for the series?
00:54:03.200 --> 00:54:07.220
So you can see, from
here, there's a triangular
00:54:07.220 --> 00:54:10.160
shape between 0 and 1.
00:54:10.160 --> 00:54:13.720
It's a function-- probably
is a function of x.
00:54:13.720 --> 00:54:16.670
And between 0 and 1, it
looks like a triangle.
00:54:16.670 --> 00:54:19.850
And if you do for
the decomposition,
00:54:19.850 --> 00:54:26.390
it can be decomposed
as small k sine
00:54:26.390 --> 00:54:31.940
function plus the
second normal mode
00:54:31.940 --> 00:54:33.590
and plus a second massless mode.
00:54:33.590 --> 00:54:39.780
And you can see that, if you
increase the number of terms
00:54:39.780 --> 00:54:43.790
included in this
Fourier series, then
00:54:43.790 --> 00:54:45.800
you will see that
the shape is actually
00:54:45.800 --> 00:54:51.290
getting closer and closer
to the triangular shape.
00:54:51.290 --> 00:54:55.410
In order to help you
with the visualization,
00:54:55.410 --> 00:54:58.790
here is actually
what I prepared.
00:54:58.790 --> 00:55:04.820
So this actually extracted from
essentially a real example,
00:55:04.820 --> 00:55:07.940
which I really used a
computer to calculate.
00:55:07.940 --> 00:55:12.140
And I tracked the contribution
from m equal to 1.
00:55:12.140 --> 00:55:16.690
This means that the first
term in this summation--
00:55:16.690 --> 00:55:18.830
infinite number of
term summation--
00:55:18.830 --> 00:55:21.000
the first term looks like this.
00:55:21.000 --> 00:55:25.340
And if you include the first
and second and third term,
00:55:25.340 --> 00:55:28.310
it becomes something
like a plateau.
00:55:28.310 --> 00:55:31.260
And then if you
increase 1 to 5, it's
00:55:31.260 --> 00:55:36.880
evolving as a function of m,
becoming more and more-- hm,
00:55:36.880 --> 00:55:38.840
strange shape.
00:55:38.840 --> 00:55:42.920
And that is actually including
the summation from first term
00:55:42.920 --> 00:55:50.090
to 11 terms and,
finally, 11 to 19.
00:55:50.090 --> 00:55:51.990
Huh, what this is--
00:55:51.990 --> 00:55:53.700
what is actually the
function I put in?
00:55:53.700 --> 00:55:55.860
It's actually a MIT function!
00:55:55.860 --> 00:55:58.100
[LAUGHTER]
00:55:58.100 --> 00:56:02.040
I put in a MIT function
into this again.
00:56:02.040 --> 00:56:04.460
And you can see that--
00:56:04.460 --> 00:56:07.940
wow, 1 to 59 term.
00:56:07.940 --> 00:56:11.050
I need to use 59
terms to describe
00:56:11.050 --> 00:56:14.510
this really wonderful shape.
00:56:14.510 --> 00:56:22.490
[LAUGHS] So in order to help
you with the visualization,
00:56:22.490 --> 00:56:26.810
listen you can see I prepared a
little program, which actually
00:56:26.810 --> 00:56:34.588
can show you the evolutions as I
increase more and more m terms.
00:56:34.588 --> 00:56:36.020
Let's take a look.
00:56:36.020 --> 00:56:38.840
You can see that,
originally it looks--
00:56:38.840 --> 00:56:42.195
doesn't look-- oh, you
cannot see anything.
00:56:42.195 --> 00:56:43.990
Wait a second.
00:56:43.990 --> 00:56:46.422
What is going on?
00:56:46.422 --> 00:56:48.776
Let me see if I can--
00:56:48.776 --> 00:56:50.650
I hope I don't screw this up.
00:56:56.840 --> 00:56:57.430
Sorry.
00:56:57.430 --> 00:56:58.320
I need to restart.
00:57:06.550 --> 00:57:08.600
So let's get started.
00:57:08.600 --> 00:57:10.980
So you can see that from
the first few terms,
00:57:10.980 --> 00:57:12.760
it doesn't look anything.
00:57:12.760 --> 00:57:17.760
But very soon, when you have
20 terms added to each other,
00:57:17.760 --> 00:57:22.630
it looks really pretty
much like a MIT dome.
00:57:22.630 --> 00:57:26.530
And you can see that
this program is really
00:57:26.530 --> 00:57:31.230
trying really hard to
describe the sharp edge
00:57:31.230 --> 00:57:33.620
in the left-hand side
and the right-hand side.
00:57:33.620 --> 00:57:39.160
You can see that those kind of
really infinitely sharp edge
00:57:39.160 --> 00:57:42.260
will need infinite
number of terms,
00:57:42.260 --> 00:57:47.230
so that if your
m is really huge,
00:57:47.230 --> 00:57:50.630
then the K, the wave number,
is going to infinity.
00:57:50.630 --> 00:57:55.100
Then you can actually
produce infinitely sharp edge
00:57:55.100 --> 00:57:56.170
in this function.
00:57:56.170 --> 00:57:59.920
And that is actually, you can
see from this demonstration
00:57:59.920 --> 00:58:04.180
the program is really
struggling with this really
00:58:04.180 --> 00:58:06.500
super sharp edge.
00:58:06.500 --> 00:58:09.100
So look at the left hand side
and right hand side corner,
00:58:09.100 --> 00:58:15.520
originally the slope is
clearly not high enough.
00:58:15.520 --> 00:58:21.010
And thus, we include higher,
higher m value terms.
00:58:21.010 --> 00:58:24.930
And you can see the description
becoming much, much better
00:58:24.930 --> 00:58:26.400
at the edge but,
of course, still
00:58:26.400 --> 00:58:29.650
are not perfect, because you
need infinite number of terms
00:58:29.650 --> 00:58:33.940
to describe the
shape of MIT dome.
00:58:33.940 --> 00:58:39.040
Of course, we can also take
a look at other example,
00:58:39.040 --> 00:58:44.950
just testing my eyesight
I'm not sure if I--
00:58:44.950 --> 00:58:49.420
OK, so I can increase the
speed to save on time.
00:58:49.420 --> 00:58:52.540
So this is actually
a square pulse,
00:58:52.540 --> 00:58:57.400
which you can see from the
scope pretty often when
00:58:57.400 --> 00:59:00.010
you do experiment.
00:59:00.010 --> 00:59:03.430
And you can see that a square
pulse is really difficult
00:59:03.430 --> 00:59:10.070
to reproduce, as I said before,
due to these sharply rising h.
00:59:10.070 --> 00:59:12.160
And of course, I can
also demonstrate you
00:59:12.160 --> 00:59:16.090
another example, which
is a triangular shape.
00:59:16.090 --> 00:59:18.523
And you can see that--
00:59:18.523 --> 00:59:23.360
ah, still, you can see
it works pretty nicely.
00:59:23.360 --> 00:59:27.560
And the function doesn't like
at all the right hand side
00:59:27.560 --> 00:59:34.870
edge, because of
exactly the same reason.
00:59:34.870 --> 00:59:35.630
OK, very good.
00:59:35.630 --> 00:59:38.500
So let's come back
to the presentation.
00:59:38.500 --> 00:59:40.910
Can you see-- OK, very good.
00:59:40.910 --> 00:59:47.720
So the question is, how do we
actually extract Am and beta m?
00:59:47.720 --> 00:59:51.010
OK, I have done that
with a computer program.
00:59:51.010 --> 00:59:55.080
And what I'm going to do now is
to show you a concrete example.
00:59:55.080 --> 00:59:58.220
And we are going to
go through it together
00:59:58.220 --> 01:00:01.250
to see how we actually
can extract Am and beta m.
01:00:01.250 --> 01:00:06.630
So suppose I give you
an initial condition.
01:00:06.630 --> 01:00:10.910
It's exactly the same
system I am talking about.
01:00:10.910 --> 01:00:13.475
But now, I prepare
this system at t
01:00:13.475 --> 01:00:17.320
equal to 0 some
specific kind of shape.
01:00:17.320 --> 01:00:22.460
This L/2 is actually the
first half of the system,
01:00:22.460 --> 01:00:26.210
is actually untouched.
01:00:26.210 --> 01:00:28.920
The first part of the
string is actually
01:00:28.920 --> 01:00:32.100
at the equilibrium position.
01:00:32.100 --> 01:00:35.970
And this is actually
x equal to L/2.
01:00:35.970 --> 01:00:41.020
Suddenly, I actually move
the string sharply up.
01:00:41.020 --> 01:00:44.970
And the rest half of
the string is actually
01:00:44.970 --> 01:00:49.540
at the height of h in this case.
01:00:49.540 --> 01:00:56.990
And of course, the right hand
side edge is x equal to L.
01:00:56.990 --> 01:01:01.880
And this is actually a
snapshot, which I actually took
01:01:01.880 --> 01:01:06.680
with my camera at t equal to 0.
01:01:06.680 --> 01:01:12.350
And also, I assume that at
t equal to 0, the string--
01:01:12.350 --> 01:01:17.780
all the components of
the string is at rest.
01:01:23.740 --> 01:01:27.460
So based on this information,
which I give you,
01:01:27.460 --> 01:01:31.900
I can now translate this
information into mathematics.
01:01:31.900 --> 01:01:35.410
So that corresponds to
two initial conditions.
01:01:42.260 --> 01:01:47.810
The first one is
that, since the string
01:01:47.810 --> 01:02:00.710
is at rest, that means psi dot x
evaluated at t equal to 0 is 0,
01:02:00.710 --> 01:02:03.950
because the string is at rest.
01:02:03.950 --> 01:02:15.230
The second initial condition
is that psi x 0 is known and is
01:02:15.230 --> 01:02:17.060
actually shown in this graph.
01:02:21.340 --> 01:02:24.850
So from this, we
would like to see
01:02:24.850 --> 01:02:31.800
if we can actually extract
information about capital
01:02:31.800 --> 01:02:36.690
Am and the beta m.
01:02:36.690 --> 01:02:40.750
So let's immediately get
started to see how we can
01:02:40.750 --> 01:02:44.610
use those initial conditions.
01:02:44.610 --> 01:02:47.790
So from the first
initial condition, a,
01:02:47.790 --> 01:02:52.530
related to the initial velocity
of the string, basically,
01:02:52.530 --> 01:02:59.520
we can get psi dot x, t.
01:02:59.520 --> 01:03:05.040
And this will be equal to, let's
see, the sum over m equal to 1
01:03:05.040 --> 01:03:06.090
to infinity.
01:03:06.090 --> 01:03:11.250
So basically, I'm
taking this equation
01:03:11.250 --> 01:03:14.550
when I plug in
that equation into
01:03:14.550 --> 01:03:17.380
the first initial condition.
01:03:17.380 --> 01:03:27.070
So basically, what I have
is Am omega m sine omega m
01:03:27.070 --> 01:03:31.944
t plus beta m.
01:03:31.944 --> 01:03:33.640
Oh, this will become cosine--
01:03:33.640 --> 01:03:38.600
sorry-- because I'm doing
a derivative, psi dot.
01:03:38.600 --> 01:03:39.880
So this will become cosine.
01:03:42.850 --> 01:03:49.518
And sine Km x plus alpha m.
01:03:52.930 --> 01:03:59.250
And this is actually
equal to 0 when
01:03:59.250 --> 01:04:08.480
psi dot x, t, is actually
evaluated at t equal to 0.
01:04:08.480 --> 01:04:11.330
And this is equal to 0.
01:04:11.330 --> 01:04:18.920
So if I plug in t equal to 0 to
this equation, this becomes 0.
01:04:18.920 --> 01:04:21.630
And then we know that the
shape of the normal mode
01:04:21.630 --> 01:04:26.520
is some kind of sine function
from the previous discussion.
01:04:26.520 --> 01:04:33.170
And I am now requiring this
thing to be equal to 0.
01:04:33.170 --> 01:04:37.790
Of course, I cannot make
Am omega m equal to 0.
01:04:37.790 --> 01:04:39.830
That's what we discussed before.
01:04:39.830 --> 01:04:43.250
And this sine function
can be evaluated
01:04:43.250 --> 01:04:46.340
at any place, any x value.
01:04:46.340 --> 01:04:50.180
Therefore, this
cannot be equal to 0.
01:04:50.180 --> 01:04:53.000
Therefore, what is
actually the result?
01:04:53.000 --> 01:04:57.610
The resulting condition
is that cosine beta m
01:04:57.610 --> 01:05:01.220
will be equal to 0.
01:05:01.220 --> 01:05:05.030
Therefore-- huh, from
this initial condition,
01:05:05.030 --> 01:05:09.590
actually I can conclude
that beta m is actually
01:05:09.590 --> 01:05:11.390
equal to pi/2, for example.
01:05:20.420 --> 01:05:23.540
So therefore, you can
see that very clearly
01:05:23.540 --> 01:05:26.600
from the first initial
condition, the string
01:05:26.600 --> 01:05:28.610
is not moving at
the beginning, I
01:05:28.610 --> 01:05:34.490
can conclude that beta
m is equal to pi over 2.
01:05:34.490 --> 01:05:38.620
And just as reminder, alpha
m is actually equal to 0
01:05:38.620 --> 01:05:40.750
from the previous discussion.
01:05:40.750 --> 01:05:47.950
And Km is actually equal to
2m minus 1 pi divided by 2L,
01:05:47.950 --> 01:05:50.110
because I just
want to copy here,
01:05:50.110 --> 01:05:55.990
because somehow the board
is covered by another board.
01:05:55.990 --> 01:06:00.280
So now, I have done with
the first initial condition.
01:06:00.280 --> 01:06:03.880
And the other initial
condition I have
01:06:03.880 --> 01:06:07.960
is that, OK, I provided
you the picture
01:06:07.960 --> 01:06:10.720
I took at the beginning
of the experiment.
01:06:10.720 --> 01:06:13.560
Therefore, psi x, 0--
01:06:13.560 --> 01:06:15.010
at t equal to 0--
01:06:15.010 --> 01:06:17.910
is known.
01:06:17.910 --> 01:06:20.190
So very good.
01:06:20.190 --> 01:06:22.500
So I have this condition.
01:06:22.500 --> 01:06:25.290
But now, I am
facing a difficulty,
01:06:25.290 --> 01:06:28.670
because all those terms--
01:06:28.670 --> 01:06:32.610
all the terms, m equal
to 1 to infinity--
01:06:32.610 --> 01:06:37.200
contribute, as we demonstrated
before, to this shape.
01:06:37.200 --> 01:06:43.530
It's very difficult to
actually evaluate Am.
01:06:43.530 --> 01:06:52.140
So the trick is to make friends
from the math department.
01:06:52.140 --> 01:06:56.020
So what we could
do is that we can
01:06:56.020 --> 01:07:00.490
use the orthogonality
of the sine function
01:07:00.490 --> 01:07:04.290
to overcome this difficulty.
01:07:04.290 --> 01:07:06.760
So let me immediately
write down what do I
01:07:06.760 --> 01:07:10.550
mean by orthogonality
of the sine function.
01:07:10.550 --> 01:07:27.090
So if I do a integration from 0
to L on dx sine Km x sine Kn x,
01:07:27.090 --> 01:07:32.490
if I do this integral,
integrating from 0 to L,
01:07:32.490 --> 01:07:38.370
so what I am going to get
is that basically you either
01:07:38.370 --> 01:07:45.480
get L equal to 2
if m is equal to n,
01:07:45.480 --> 01:07:51.840
or you get 0, if m
is not equal to n.
01:07:51.840 --> 01:07:55.390
So basically, I have
two sine functions
01:07:55.390 --> 01:07:57.660
multiplied to each other.
01:07:57.660 --> 01:08:04.620
And I do integration from 0
to L multiplied by delta x.
01:08:04.620 --> 01:08:05.880
And this is Km.
01:08:05.880 --> 01:08:07.664
This is Kn.
01:08:07.664 --> 01:08:13.190
If Km and Kn are
different, you can actually
01:08:13.190 --> 01:08:15.720
go ahead and do this exercise.
01:08:15.720 --> 01:08:19.620
And you will find that,
indeed, if they are the same,
01:08:19.620 --> 01:08:22.200
then you will get L/2.
01:08:22.200 --> 01:08:25.470
On the other hand, if
they are not the same--
01:08:25.470 --> 01:08:29.100
the K value are not the same
for the first and second sine
01:08:29.100 --> 01:08:30.200
function--
01:08:30.200 --> 01:08:33.510
you are going to get 0.
01:08:33.510 --> 01:08:40.450
So that's very good news,
because if I do this--
01:08:40.450 --> 01:08:48.359
if I do this calculation,
I do 2/L integration from 0
01:08:48.359 --> 01:09:02.100
to L psi x, 0,
sine Km x dx, what
01:09:02.100 --> 01:09:06.220
is going to happen if
I do this integration?
01:09:06.220 --> 01:09:11.529
Remember, psi m is
a linear combination
01:09:11.529 --> 01:09:21.660
of infinite number of massless
modes with different sine Km x.
01:09:21.660 --> 01:09:27.840
If I multiplied
that by sine Km x,
01:09:27.840 --> 01:09:29.850
this is a very
crazy thing to do,
01:09:29.850 --> 01:09:34.170
because all the other
terms will become 0.
01:09:34.170 --> 01:09:38.220
If the K value of
one of the terms
01:09:38.220 --> 01:09:44.279
is not equal to the
dictator's value Km, it's 0.
01:09:46.790 --> 01:09:51.470
Otherwise, it's L/2,
and it is designed here
01:09:51.470 --> 01:09:54.380
to cancel that factor.
01:09:54.380 --> 01:09:59.340
So you can see that this
is like a mode picker.
01:09:59.340 --> 01:10:02.270
I'm picking up a
mode with this tool.
01:10:02.270 --> 01:10:04.780
This is like, this
tool, yeah, I'm
01:10:04.780 --> 01:10:08.415
picking this mode, which
is actually matching my Km.
01:10:11.570 --> 01:10:17.120
It's a miracle that
this become Am.
01:10:17.120 --> 01:10:21.200
I hope you get this idea,
even if probably you
01:10:21.200 --> 01:10:25.760
haven't heard about with
your decomposition before.
01:10:25.760 --> 01:10:28.150
But essentially,
what we are doing
01:10:28.150 --> 01:10:33.830
is that I'm going to evaluate
infinite number of integrals.
01:10:33.830 --> 01:10:39.960
And you are going to do that
in the exam, hopefully easy.
01:10:39.960 --> 01:10:41.660
[LAUGHS]
01:10:41.660 --> 01:10:44.600
What is going to happen is
that, if you do this integral,
01:10:44.600 --> 01:10:47.420
you are going to pick
only one mode out of it.
01:10:47.420 --> 01:10:50.870
And you are going
to be able to know
01:10:50.870 --> 01:10:53.240
the amplitude of that mode.
01:10:53.240 --> 01:10:56.215
So let's do this
immediately in this example.
01:11:02.395 --> 01:11:13.285
So Am is actually equal to
2/L. Since the amplitude
01:11:13.285 --> 01:11:17.190
is actually 0 between 0 and L/2.
01:11:17.190 --> 01:11:23.130
So I can safely just
integrate from L/2 to L.
01:11:23.130 --> 01:11:27.130
So I do a integration
from L/2 to L,
01:11:27.130 --> 01:11:35.920
because between 0 and L/2,
the initial position is 0.
01:11:35.920 --> 01:11:44.180
So what I'm going to
get is h sine Km x dx.
01:11:44.180 --> 01:11:47.570
And of course, everybody
know how to do this integral.
01:11:47.570 --> 01:11:49.780
It doesn't look that horrible.
01:11:49.780 --> 01:11:58.630
And this would become
2/L minus h over Km.
01:11:58.630 --> 01:12:05.050
And basically, this will
become cosine Km evaluated
01:12:05.050 --> 01:12:14.480
at L minus cosine
Km evaluated at L/2,
01:12:14.480 --> 01:12:18.400
and where this Km,
as just a reminder,
01:12:18.400 --> 01:12:25.450
is basically equal to 2m
minus 1 pi divided by 2/L.
01:12:25.450 --> 01:12:33.340
So I hope this actually help
you to understand the procedure
01:12:33.340 --> 01:12:37.420
to determining all those
unknown coefficients,
01:12:37.420 --> 01:12:40.620
starting from this equation.
01:12:40.620 --> 01:12:43.510
Am and the beta m
can be determined
01:12:43.510 --> 01:12:45.640
by initial conditions.
01:12:45.640 --> 01:12:48.250
As we actually
show here, you can
01:12:48.250 --> 01:12:50.710
use the initial
condition of velocity
01:12:50.710 --> 01:12:53.450
and the initial
condition of the shape
01:12:53.450 --> 01:12:56.850
and the help of a
mode picker to pick up
01:12:56.850 --> 01:13:01.240
the amplitude from
that tool function.
01:13:01.240 --> 01:13:06.040
And also, you can see that
Km and alpha m related
01:13:06.040 --> 01:13:11.200
to the shape of the normal
mode can be determined
01:13:11.200 --> 01:13:13.540
by boundary conditions--
01:13:13.540 --> 01:13:18.370
boundary conditions, how this
system is actually connected
01:13:18.370 --> 01:13:20.620
to the nearby systems.
01:13:20.620 --> 01:13:25.780
The nearby systems are
the rod and the wall.
01:13:25.780 --> 01:13:27.930
So that is actually the
two boundary conditions,
01:13:27.930 --> 01:13:32.020
which determine the
shape of the normal mode.
01:13:32.020 --> 01:13:35.470
And finally, very
important, as usual,
01:13:35.470 --> 01:13:39.980
the most general
solution is, of course,
01:13:39.980 --> 01:13:44.790
a linear combination of all of
those possible massless modes
01:13:44.790 --> 01:13:47.860
from m equal to 1 to infinity.
01:13:47.860 --> 01:13:53.050
And omega m, don't forget, is
determined by the dispersion
01:13:53.050 --> 01:13:56.560
relation, vp times Kl.
01:13:56.560 --> 01:13:59.680
Before the end, I would like
to mention to you something
01:13:59.680 --> 01:14:03.310
which you might
actually not notice
01:14:03.310 --> 01:14:05.110
when we were discussing this.
01:14:05.110 --> 01:14:11.710
So you can see that omega
is now proportional to Km.
01:14:11.710 --> 01:14:16.090
So if you plot omega
as a function of k,
01:14:16.090 --> 01:14:19.780
actually you can see that
it's becoming a straight line
01:14:19.780 --> 01:14:22.870
in this graph, which is
very straightforward.
01:14:22.870 --> 01:14:26.080
And on the other
hand, if you remember
01:14:26.080 --> 01:14:30.560
what we got last time
with discrete system,
01:14:30.560 --> 01:14:38.620
with length scale between
little mass is actually a,
01:14:38.620 --> 01:14:43.000
you get omega square
is equal to 4T
01:14:43.000 --> 01:14:47.770
over m sine squared ka over 2.
01:14:47.770 --> 01:14:51.250
So if you plot this
omega as a function of k,
01:14:51.250 --> 01:14:54.110
you will get the black curve.
01:14:54.110 --> 01:14:57.050
What does this
actually tell you?
01:14:57.050 --> 01:15:01.710
That is actually telling
you that, if you prepare
01:15:01.710 --> 01:15:07.750
a system at a specific normal
mode based on the oscillation
01:15:07.750 --> 01:15:12.670
frequency, you can actually
know the internal length
01:15:12.670 --> 01:15:17.820
scale of individual
mass, just in case you
01:15:17.820 --> 01:15:21.100
didn't notice this
interesting fact.
01:15:21.100 --> 01:15:22.820
So thank you very much.
01:15:22.820 --> 01:15:24.820
I hope you enjoyed the lecture.
01:15:24.820 --> 01:15:28.600
And I will see you
next Thursday--
01:15:28.600 --> 01:15:32.620
not here in the Walker room--
01:15:32.620 --> 01:15:34.770
Walker Memorial.
01:15:34.770 --> 01:15:36.340
So good luck, everybody.
01:15:36.340 --> 01:15:40.960
Maybe see some of you in
the office hour tomorrow.