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PROFESSOR YEN-JIE LEE:
Welcome back, everybody,
00:00:25.300 --> 00:00:28.810
to 8.03 Today, we
are going to continue
00:00:28.810 --> 00:00:31.870
the discussion of
symmetry matrix,
00:00:31.870 --> 00:00:33.970
which we started last time.
00:00:33.970 --> 00:00:37.460
And this is what
we have been doing.
00:00:37.460 --> 00:00:39.370
OK?
00:00:39.370 --> 00:00:41.050
So the thing which
we have been doing
00:00:41.050 --> 00:00:45.480
is to solve the normal mode
frequencies of component
00:00:45.480 --> 00:00:48.490
systems by looking at
equation of motion,
00:00:48.490 --> 00:00:51.850
MX double dot equal to minus KX.
00:00:51.850 --> 00:00:56.740
And in the end of the day,
what we are doing is really
00:00:56.740 --> 00:00:59.910
to solve Eigenvalue problem--
00:00:59.910 --> 00:01:02.830
M minus one K matrix
Eigenvalue problem.
00:01:02.830 --> 00:01:07.770
And we have been exercising
this several times
00:01:07.770 --> 00:01:10.570
the last few lectures.
00:01:10.570 --> 00:01:12.617
And the M minus
one K matrix says
00:01:12.617 --> 00:01:16.540
you're describing how each
component in the system
00:01:16.540 --> 00:01:18.110
interact with each other.
00:01:18.110 --> 00:01:18.610
OK?
00:01:18.610 --> 00:01:21.760
So that's actually what
we are trying to do,
00:01:21.760 --> 00:01:24.310
to solve the normal modes.
00:01:24.310 --> 00:01:28.420
So we have been making
progress, and we
00:01:28.420 --> 00:01:31.330
are increasing the number
of coupled oscillators
00:01:31.330 --> 00:01:32.630
as a function of time.
00:01:32.630 --> 00:01:35.230
And now we finally arrive
at our limit, which
00:01:35.230 --> 00:01:36.990
is infinite system, right?
00:01:36.990 --> 00:01:39.310
So basically, what we
have been discussing
00:01:39.310 --> 00:01:43.960
is a special kind of
infinite system, which
00:01:43.960 --> 00:01:47.470
actually satisfies
translation symmetry, right?
00:01:47.470 --> 00:01:51.730
So in general, we don't know
how to solve infinite system.
00:01:51.730 --> 00:01:53.490
If this system is
really complicated,
00:01:53.490 --> 00:01:57.730
no symmetry, then who knows
how to solve this, right?
00:01:57.730 --> 00:02:01.330
But very luckily,
in 8.03 we have
00:02:01.330 --> 00:02:06.340
started to get a highly
symmetrical infinite system,
00:02:06.340 --> 00:02:09.580
and in this case, it's
translation symmetry.
00:02:09.580 --> 00:02:13.390
And that is really pretty
nice, and we can actually
00:02:13.390 --> 00:02:18.490
use this example to learn
an interesting fact, which
00:02:18.490 --> 00:02:22.380
we can see from the physical
system we discuss here.
00:02:22.380 --> 00:02:26.440
So, one thing which we
have been discussing
00:02:26.440 --> 00:02:30.420
is the space translation
symmetry matrix.
00:02:30.420 --> 00:02:35.800
As a reminder, S
matrix, as defined here.
00:02:35.800 --> 00:02:39.460
Basically, if you
have a vector A, which
00:02:39.460 --> 00:02:43.910
describes the amplitude
of an individual component
00:02:43.910 --> 00:02:47.500
in a system, basically
these vectors
00:02:47.500 --> 00:02:51.700
have Aj, Aj plus one, Aj
plus two blah blah blah.
00:02:51.700 --> 00:02:58.520
All those amplitudes are
included in this vector.
00:02:58.520 --> 00:03:01.310
And what does this S matrix do?
00:03:01.310 --> 00:03:03.790
It's actually the following--
00:03:03.790 --> 00:03:08.200
so if A prime is
equal to S times A,
00:03:08.200 --> 00:03:12.910
then the base
component of A prime
00:03:12.910 --> 00:03:16.100
will be equal to Aj plus one.
00:03:16.100 --> 00:03:20.890
OK, so that's actually what
this S matrix does to A vector.
00:03:20.890 --> 00:03:21.550
OK?
00:03:21.550 --> 00:03:25.230
So you can think about S
metrics as an operator.
00:03:25.230 --> 00:03:28.840
It's actually picking
up the Aj plus
00:03:28.840 --> 00:03:33.080
one component, and moving
it to Aj in the new vector.
00:03:33.080 --> 00:03:33.580
OK?
00:03:33.580 --> 00:03:37.180
So that's actually what
this S matrix does.
00:03:37.180 --> 00:03:42.790
And we were talking about
the Eigenvector of S matrix.
00:03:42.790 --> 00:03:47.470
For example, if A is an
Eigenvector of S matrix,
00:03:47.470 --> 00:03:52.000
then we have this relation,
SA equal to beta A. Now,
00:03:52.000 --> 00:03:57.830
beta is actually the
Eigenvalue of S matrix.
00:03:57.830 --> 00:03:58.600
OK?
00:03:58.600 --> 00:04:02.380
And then we also discussed
last time, that means,
00:04:02.380 --> 00:04:05.680
based on this logical
extension, Aj prime
00:04:05.680 --> 00:04:08.170
would be equal to
Aj plus one based
00:04:08.170 --> 00:04:10.300
on the definition of S matrix.
00:04:10.300 --> 00:04:14.320
And now this is going to be
equal to beta times Aj, right?
00:04:14.320 --> 00:04:21.450
Because we assume that that A
vector is an Eigenvector of S
00:04:21.450 --> 00:04:21.950
matrix.
00:04:21.950 --> 00:04:23.050
OK?
00:04:23.050 --> 00:04:28.340
That means Aj will be equal
to beta to the j A zero.
00:04:28.340 --> 00:04:29.320
Right?
00:04:29.320 --> 00:04:33.550
According to this relation.
00:04:33.550 --> 00:04:36.460
Therefore, we can
conclude that Aj will be
00:04:36.460 --> 00:04:39.460
proportional to beta to the j.
00:04:39.460 --> 00:04:43.000
OK, beta is still
some coefficient,
00:04:43.000 --> 00:04:48.020
which we have not determined,
and it can be anything.
00:04:48.020 --> 00:04:49.180
OK?
00:04:49.180 --> 00:04:56.470
So, in order to consider a
system of little masses, which
00:04:56.470 --> 00:04:58.760
are oscillating up
and down instead
00:04:58.760 --> 00:05:05.580
of going in one direction
forever or the amplitude
00:05:05.580 --> 00:05:08.770
grows exponentially.
00:05:08.770 --> 00:05:11.740
And we also don't
want the system
00:05:11.740 --> 00:05:14.720
to have the amplitude
go to infinity
00:05:14.720 --> 00:05:19.360
when we go to very very large
j value or very small j value,
00:05:19.360 --> 00:05:22.633
therefore we limit
our discussion
00:05:22.633 --> 00:05:28.440
in the case of beta
equal to exponential ika.
00:05:28.440 --> 00:05:29.030
OK?
00:05:29.030 --> 00:05:35.570
In this case, the absolute
value of beta is one.
00:05:35.570 --> 00:05:36.660
Right?
00:05:36.660 --> 00:05:40.160
OK, and that means--
00:05:40.160 --> 00:05:44.990
OK, Aj is actually proportional
to beta to Aj, right?
00:05:44.990 --> 00:05:49.040
So if you take a ratio of
Aj plus one and the Aj,
00:05:49.040 --> 00:05:51.160
the ratio is beta.
00:05:51.160 --> 00:05:55.100
If beta Is not equal to
one or its absolute value
00:05:55.100 --> 00:05:59.180
is not equal to one,
then this Aj value
00:05:59.180 --> 00:06:02.540
is going to be
increasing according
00:06:02.540 --> 00:06:04.220
to a power law, right?
00:06:04.220 --> 00:06:06.350
So the amplitude is
going to be, whoa,
00:06:06.350 --> 00:06:08.930
going to a very, very
large value, right?
00:06:08.930 --> 00:06:11.810
And that corresponds to some
kind of physical system,
00:06:11.810 --> 00:06:14.060
but not corresponding
to oscillation.
00:06:14.060 --> 00:06:15.710
OK?
00:06:15.710 --> 00:06:19.010
And if we do this
and assume that beta
00:06:19.010 --> 00:06:23.120
is equal to exponential ika.
00:06:23.120 --> 00:06:28.220
OK, if we do this, what is going
to happen is the following.
00:06:28.220 --> 00:06:32.390
So again, the ratio
of Aj plus one and Aj
00:06:32.390 --> 00:06:39.410
is a fixed value beta,
equal to exponential ika.
00:06:39.410 --> 00:06:42.980
But what it does is, instead
of changing the amplitude,
00:06:42.980 --> 00:06:47.911
it's actually doing a
rotation in the complex plane.
00:06:47.911 --> 00:06:48.410
OK?
00:06:48.410 --> 00:06:54.080
So if this is Aj, J plus
one, J plus two, et cetera,
00:06:54.080 --> 00:06:57.850
as a function of say j, the
variable, then what it does
00:06:57.850 --> 00:07:01.810
is that this operation
multiplied by beta
00:07:01.810 --> 00:07:07.640
is really a rotation
in the complex plane.
00:07:07.640 --> 00:07:11.240
And while we actually see
in the physical system,
00:07:11.240 --> 00:07:15.620
it's actually a projection of
this complex imaginary plane
00:07:15.620 --> 00:07:17.990
to the real axis.
00:07:17.990 --> 00:07:21.610
And that would give you a sine
function and cosine function.
00:07:21.610 --> 00:07:25.760
So very interestingly,
if we choose wisely,
00:07:25.760 --> 00:07:29.030
the beta to have
absolute value of one,
00:07:29.030 --> 00:07:36.020
then that will give you a system
which is actually oscillating
00:07:36.020 --> 00:07:37.850
up and down and the
amplitude is actually
00:07:37.850 --> 00:07:40.730
confined within some value.
00:07:40.730 --> 00:07:43.340
OK, so that is actually
quite interesting.
00:07:43.340 --> 00:07:46.760
The other thing which I
would like to talk about is--
00:07:46.760 --> 00:07:52.690
OK, I choose to have
exponential ika as my beta.
00:07:52.690 --> 00:07:54.890
OK, why ka, right?
00:07:54.890 --> 00:07:56.620
It looks really strange here.
00:07:56.620 --> 00:08:00.140
Suddenly the k and and the
a are coming to play, right?
00:08:00.140 --> 00:08:02.591
So what is actually the small a?
00:08:02.591 --> 00:08:09.440
The small a is actually the
distance between little mass,
00:08:09.440 --> 00:08:11.360
just a reminder.
00:08:11.360 --> 00:08:13.520
So this is actually
the length scale
00:08:13.520 --> 00:08:16.410
between all those
little mass, right?
00:08:16.410 --> 00:08:21.950
Therefore, what I'm doing is to
factorize out the length scale,
00:08:21.950 --> 00:08:25.620
and then we suddenly found that.
00:08:25.620 --> 00:08:31.520
OK, after I do this, I define
beta equal to exponential ika
00:08:31.520 --> 00:08:33.980
Instead of exponential
i theta, right?
00:08:33.980 --> 00:08:36.980
So you can also do
exponential i theta, right?
00:08:36.980 --> 00:08:40.309
But instead, I gave
theta a fancy name,
00:08:40.309 --> 00:08:42.850
which is k times
a. a is actually
00:08:42.850 --> 00:08:45.940
the distance between mass.
00:08:45.940 --> 00:08:49.090
Then something interesting
happens, because k suddenly
00:08:49.090 --> 00:08:51.000
also will have a meaning.
00:08:51.000 --> 00:09:00.660
It's actually the wave number
of the resulting sine wave.
00:09:00.660 --> 00:09:01.160
OK?
00:09:01.160 --> 00:09:03.980
So that's actually
why we actually
00:09:03.980 --> 00:09:08.050
factor out this a factor.
00:09:08.050 --> 00:09:10.340
Then, after that, we
actually found out
00:09:10.340 --> 00:09:12.210
the amplitude would
be proportional
00:09:12.210 --> 00:09:16.580
to exponential ijka,
and j is actually
00:09:16.580 --> 00:09:20.480
just a label of a phase
component in the system.
00:09:20.480 --> 00:09:22.330
OK.
00:09:22.330 --> 00:09:27.890
So, once we have solved
the Eigenvalue problem
00:09:27.890 --> 00:09:32.360
for the symmetry matrix,
S, as we discussed before,
00:09:32.360 --> 00:09:35.720
if you look at the slides, OK?
00:09:35.720 --> 00:09:40.430
If S and the M minus one
K matrix, they commute.
00:09:40.430 --> 00:09:41.720
OK?
00:09:41.720 --> 00:09:45.860
Commutes means that you can
actually change S and and the M
00:09:45.860 --> 00:09:48.950
minus one matrix, you
can swap them, OK,
00:09:48.950 --> 00:09:50.735
when you multiply them together.
00:09:50.735 --> 00:09:51.740
OK?
00:09:51.740 --> 00:09:55.190
If you can swap S and
M minus one K matrix,
00:09:55.190 --> 00:09:57.680
that means they commute.
00:09:57.680 --> 00:10:00.740
And our conclusion
from last time
00:10:00.740 --> 00:10:06.230
is that they will share
the same Eigenvectors.
00:10:06.230 --> 00:10:07.130
OK?
00:10:07.130 --> 00:10:09.910
Of course, not necessarily
the same Eigenvalue,
00:10:09.910 --> 00:10:11.990
but they share the
same Eigenvector.
00:10:11.990 --> 00:10:13.400
So that's great news!
00:10:13.400 --> 00:10:17.330
Because instead of solving
M minus one K matrix, which
00:10:17.330 --> 00:10:19.820
can be really
complicated; depends
00:10:19.820 --> 00:10:23.490
on what kind of physical
system you are talking about.
00:10:23.490 --> 00:10:28.060
I can solve S matrix
Eigenvalue problem.
00:10:28.060 --> 00:10:29.310
OK?
00:10:29.310 --> 00:10:34.560
And then this Eigenvector,
which I just found here,
00:10:34.560 --> 00:10:41.240
is going to be the Eigenvector
of M minus one K matrix.
00:10:41.240 --> 00:10:45.110
And that's actually really
making things much easier,
00:10:45.110 --> 00:10:49.460
because now instead of solving
Eigenvalue problem of M
00:10:49.460 --> 00:10:52.210
minus one K matrix,
what I am doing
00:10:52.210 --> 00:10:55.550
is just multiplying M
minus one K times A,
00:10:55.550 --> 00:11:00.740
then you can actually obtain
the normal mode frequency omega.
00:11:00.740 --> 00:11:01.310
OK?
00:11:01.310 --> 00:11:03.750
So that's the issue of
the great excitement.
00:11:03.750 --> 00:11:05.210
What does that mean?
00:11:05.210 --> 00:11:09.370
That means, if you have
all kinds of systems, which
00:11:09.370 --> 00:11:13.200
are translation symmetric--
you can have a line,
00:11:13.200 --> 00:11:17.960
you have however
many people together,
00:11:17.960 --> 00:11:20.540
whatever a system
which is so on here.
00:11:20.540 --> 00:11:25.330
They are all going to
have the same Eigenvector.
00:11:25.330 --> 00:11:25.830
You see?
00:11:25.830 --> 00:11:27.290
You already know
how they are going
00:11:27.290 --> 00:11:30.020
to interact with each
other, and what is actually
00:11:30.020 --> 00:11:35.010
the amplitude as a function of
j, which is the location label.
00:11:35.010 --> 00:11:35.850
OK?
00:11:35.850 --> 00:11:39.160
So that's actually
really wonderful.
00:11:39.160 --> 00:11:43.040
So, in order to help
you with understanding
00:11:43.040 --> 00:11:46.490
of this system some
more, we are going
00:11:46.490 --> 00:11:51.290
to discuss another system
which is actually also
00:11:51.290 --> 00:11:52.870
very interesting.
00:11:52.870 --> 00:11:55.330
It's actually a spring--
00:11:55.330 --> 00:11:59.540
OK, last time we discussed a
spring and mass system, right?
00:11:59.540 --> 00:12:01.920
And then we solved it together.
00:12:01.920 --> 00:12:05.210
And this time we are going
to solve a system which
00:12:05.210 --> 00:12:08.060
is made of mass and strings.
00:12:08.060 --> 00:12:08.900
Ok?
00:12:08.900 --> 00:12:11.680
So that may actually copy--
00:12:11.680 --> 00:12:13.640
OK, let me actually
introduce you
00:12:13.640 --> 00:12:17.770
to this new system we are
going to talk about today.
00:12:17.770 --> 00:12:23.060
It has many little mass
here, from left hand side
00:12:23.060 --> 00:12:25.610
of the universe and right
hand side of the universe.
00:12:25.610 --> 00:12:28.740
Take forever to actually
construct this system.
00:12:28.740 --> 00:12:29.430
OK?
00:12:29.430 --> 00:12:33.860
Then my student actually
carefully link them by strings,
00:12:33.860 --> 00:12:38.750
and we make sure that the
string tension, OK, is actually
00:12:38.750 --> 00:12:41.630
a fixed value, which is T. OK.
00:12:41.630 --> 00:12:44.090
T is actually the
string tension.
00:12:44.090 --> 00:12:47.300
And of course, as what
we discussed before,
00:12:47.300 --> 00:12:50.570
the space or the distance
between little mass
00:12:50.570 --> 00:12:53.240
is actually A. OK?
00:12:53.240 --> 00:12:54.860
So that's, again,
the same length
00:12:54.860 --> 00:12:57.080
scale, which we were using.
00:12:57.080 --> 00:12:59.560
And finally, in
order to describe
00:12:59.560 --> 00:13:04.740
all those little
masses in the system,
00:13:04.740 --> 00:13:10.020
I label them as j, j plus one, j
plus two, et cetera, et cetera.
00:13:10.020 --> 00:13:11.270
OK.
00:13:11.270 --> 00:13:15.470
So, the question
we are asking is,
00:13:15.470 --> 00:13:20.190
what would be the resulting
motion of this system, right?
00:13:20.190 --> 00:13:24.650
So what we can do is what we
have done last time, right?
00:13:24.650 --> 00:13:28.820
We take jth object
in this system,
00:13:28.820 --> 00:13:34.520
and we look at the force
diagram and this M minus one--
00:13:34.520 --> 00:13:37.240
to write down the equational
motion and also the
00:13:37.240 --> 00:13:39.650
and M minus one K metrics.
00:13:39.650 --> 00:13:49.130
So what is actually the force
diagram of j's component?
00:13:52.010 --> 00:13:52.510
OK.
00:13:52.510 --> 00:13:54.495
If I take this--
00:13:54.495 --> 00:14:02.200
so first, I take my jth
mass, and it's connected
00:14:02.200 --> 00:14:05.560
to two strings, right?
00:14:05.560 --> 00:14:10.150
So there are two strings
connected to jth mass.
00:14:10.150 --> 00:14:10.960
OK?
00:14:10.960 --> 00:14:13.810
And of course,
left hand side you
00:14:13.810 --> 00:14:17.470
have another mass, right hand
side you have another option.
00:14:17.470 --> 00:14:19.660
OK?
00:14:19.660 --> 00:14:24.380
And I know the tension
of this string.
00:14:24.380 --> 00:14:27.160
Its actually a fixed
value, and the string
00:14:27.160 --> 00:14:31.510
tension is fixed at the T. OK?
00:14:31.510 --> 00:14:33.460
So, in order to
describe this system,
00:14:33.460 --> 00:14:35.960
I need to find my
coordinate system, right?
00:14:35.960 --> 00:14:36.810
As usual.
00:14:36.810 --> 00:14:38.560
So, what is actually
the coordinate system
00:14:38.560 --> 00:14:40.010
I'm going to use.?
00:14:40.010 --> 00:14:42.520
So I need to define
horizontal direction
00:14:42.520 --> 00:14:48.460
to be x, and the vertical
direction to be y.
00:14:48.460 --> 00:14:52.560
Therefore, I cannot express
this little mass to be--
00:14:52.560 --> 00:14:58.580
the position of jth little
mass as Xj and the Yj.
00:15:01.540 --> 00:15:03.040
OK?
00:15:03.040 --> 00:15:07.350
We can do the same thing
for the left-hand side mass
00:15:07.350 --> 00:15:16.567
is Xa minus one, Yj minus one,
and Xj plus one Yj plus one.
00:15:19.250 --> 00:15:22.870
So to simplify the discussion,
what I'm going to assume
00:15:22.870 --> 00:15:28.930
is that all those little masses
can only move up and down, OK,
00:15:28.930 --> 00:15:30.580
instead of back and forth.
00:15:30.580 --> 00:15:33.940
OK, so only one
direction is allowed,
00:15:33.940 --> 00:15:38.520
and also I assume that
the up and down motion
00:15:38.520 --> 00:15:42.640
of all those little masses
is really really very small.
00:15:42.640 --> 00:15:46.450
So I can use small
angle approximation.
00:15:46.450 --> 00:15:49.160
OK, so therefore I can--
00:15:49.160 --> 00:15:57.040
this is essentially the zero
of Y axis, when I actually--
00:15:57.040 --> 00:15:59.350
before I actually move--
00:15:59.350 --> 00:16:02.720
wind up with the mass away
from the equilibrium position.
00:16:02.720 --> 00:16:07.720
So when the string mass
system is at rest, OK,
00:16:07.720 --> 00:16:14.800
not really moving, then all the
mass are at Y equal to zero.
00:16:14.800 --> 00:16:16.420
OK?
00:16:16.420 --> 00:16:23.880
So now, I actually move
this mass to Yj, OK?
00:16:23.880 --> 00:16:28.960
Then, apparently there are two
forces acting on this mass.
00:16:28.960 --> 00:16:31.060
The left hand side
is string force,
00:16:31.060 --> 00:16:32.740
and the right hand
side is string force.
00:16:32.740 --> 00:16:38.300
And the magnitude of the
force is actually T. OK?
00:16:38.300 --> 00:16:42.590
So, in order to help us
with solving this problem.
00:16:42.590 --> 00:16:44.360
I would define two angles--
00:16:44.360 --> 00:16:50.140
the left hand side angle to be
theta one, and the right hand
00:16:50.140 --> 00:16:52.540
side angle to be theta two.
00:16:52.540 --> 00:16:56.320
Then I can now write down
the equation of motion
00:16:56.320 --> 00:16:59.800
in the horizontal direction, and
then in the vertical direction.
00:16:59.800 --> 00:17:01.960
OK?
00:17:01.960 --> 00:17:02.710
All right.
00:17:02.710 --> 00:17:06.280
Since I have assumed
that all of those masses
00:17:06.280 --> 00:17:10.089
can only move up and down,
now I mean-- and also
00:17:10.089 --> 00:17:13.880
the displacement with respect
to the equilibrium position,
00:17:13.880 --> 00:17:18.400
which is Y equal to zero, is
really small, compared to,
00:17:18.400 --> 00:17:21.400
for example, the length scale a.
00:17:21.400 --> 00:17:21.970
OK?
00:17:21.970 --> 00:17:25.569
So therefore, I can write
that condition explicitly.
00:17:25.569 --> 00:17:27.310
So I have a condition.
00:17:27.310 --> 00:17:32.680
I assume that Yj is actually
much, much smaller than a,
00:17:32.680 --> 00:17:36.670
which is the distance
between those masses.
00:17:36.670 --> 00:17:41.530
And that means theta
one and theta two
00:17:41.530 --> 00:17:47.320
are going to be much,
much smaller than one.
00:17:47.320 --> 00:17:47.860
OK?
00:17:47.860 --> 00:17:49.930
So that's actually
given us a chance
00:17:49.930 --> 00:17:53.240
to use small angle
approximation.
00:17:53.240 --> 00:17:55.660
So based on this
force diagram, I
00:17:55.660 --> 00:17:59.070
can now write down the
two equations of motion;
00:17:59.070 --> 00:18:01.220
one in the horizontal
direction, the other one
00:18:01.220 --> 00:18:02.960
in the vertical direction.
00:18:02.960 --> 00:18:09.670
So what I'm going to get
in the horizontal direction
00:18:09.670 --> 00:18:14.530
is M X j double dot.
00:18:14.530 --> 00:18:16.000
These will be equal to--
00:18:16.000 --> 00:18:17.480
OK, there are two forces.
00:18:17.480 --> 00:18:17.980
Right?
00:18:17.980 --> 00:18:19.990
Horizontal direction,
I would need
00:18:19.990 --> 00:18:23.320
to calculate the projection
to the X direction,
00:18:23.320 --> 00:18:25.310
therefore the left
hand side force
00:18:25.310 --> 00:18:30.970
will give you minus
T cosine theta one,
00:18:30.970 --> 00:18:33.620
and the right hand side
of this string force
00:18:33.620 --> 00:18:40.500
is going to give you
plus T cosine theta two.
00:18:40.500 --> 00:18:42.160
OK?
00:18:42.160 --> 00:18:48.020
And then in the vertical
direction, what I'm
00:18:48.020 --> 00:18:51.750
going to get is Myj double dot.
00:18:51.750 --> 00:18:57.210
This is equal to
minus T sine theta y.
00:18:57.210 --> 00:19:01.220
Now I'm doing the
projection in the y
00:19:01.220 --> 00:19:10.030
axis in the vertical direction,
and minus T sine theta two.
00:19:10.030 --> 00:19:11.510
OK.
00:19:11.510 --> 00:19:14.960
And since I have this
condition, all the mass
00:19:14.960 --> 00:19:20.810
can only move up and down,
and also the displacement
00:19:20.810 --> 00:19:25.820
is much, much smaller
than a; therefore, I
00:19:25.820 --> 00:19:28.080
have the small
angle approximation.
00:19:28.080 --> 00:19:33.440
Cosine theta is
roughly equal to one,
00:19:33.440 --> 00:19:38.700
and the sine theta is
roughly equal to theta.
00:19:38.700 --> 00:19:39.410
OK?
00:19:39.410 --> 00:19:42.440
And I would call this
the equation number one,
00:19:42.440 --> 00:19:45.590
and the second equation
in the vertical direction
00:19:45.590 --> 00:19:48.920
to be equation number two.
00:19:48.920 --> 00:19:50.900
OK.
00:19:50.900 --> 00:19:54.710
So, up to here, everything
is essentially exact,
00:19:54.710 --> 00:19:58.040
and now I would like to make
a small angle approximation
00:19:58.040 --> 00:19:59.570
and see what will happen.
00:19:59.570 --> 00:20:01.860
And now equation
number one will be
00:20:01.860 --> 00:20:07.820
called MXa double dot
equal to minus T plus T,
00:20:07.820 --> 00:20:10.580
and this is equal to zero.
00:20:10.580 --> 00:20:15.710
OK, so that means we will
not have horizontal direction
00:20:15.710 --> 00:20:16.400
acceleration.
00:20:16.400 --> 00:20:18.980
Therefore, in the
horizontal direction,
00:20:18.980 --> 00:20:21.740
there will be no
acceleration and therefore
00:20:21.740 --> 00:20:26.900
no movement in the X direction,
or horizontal direction.
00:20:26.900 --> 00:20:31.190
OK, and now I can take a look
at the vertical direction.
00:20:31.190 --> 00:20:32.630
OK?
00:20:32.630 --> 00:20:38.180
So basically, what I am going
to get is MYj double dot,
00:20:38.180 --> 00:20:41.670
and this will be
equal to minus T. Now,
00:20:41.670 --> 00:20:47.450
sine theta one will be
roughly equal to theta one.
00:20:47.450 --> 00:20:49.460
So what is actually theta one?
00:20:49.460 --> 00:20:52.110
Theta one is going to be--
00:20:52.110 --> 00:21:00.010
OK, so this is actually
the Yj minus Yj minus one.
00:21:00.010 --> 00:21:01.169
Right?
00:21:01.169 --> 00:21:02.710
So now that's actually
the difference
00:21:02.710 --> 00:21:10.140
between the amplitude of the
displacement of the mass j,
00:21:10.140 --> 00:21:13.030
and the displacement
of mass j minus one.
00:21:13.030 --> 00:21:13.770
Right?
00:21:13.770 --> 00:21:16.980
Divided by a, I get theta one.
00:21:16.980 --> 00:21:17.940
Right?
00:21:17.940 --> 00:21:20.480
Therefore, the first
sine theta one,
00:21:20.480 --> 00:21:29.430
will become Yj minus Yj
minus one, divided by a.
00:21:29.430 --> 00:21:30.820
OK?
00:21:30.820 --> 00:21:34.250
And of course, you can do the
same thing for the sine theta
00:21:34.250 --> 00:21:35.320
two, right?
00:21:35.320 --> 00:21:38.735
Which is actually
just the theta two.
00:21:38.735 --> 00:21:40.360
Then, basically, what
I am going to get
00:21:40.360 --> 00:21:51.330
is minus Tyj minus Yj
plus one divided by a.
00:21:51.330 --> 00:21:53.110
OK.
00:21:53.110 --> 00:21:54.128
Any questions?
00:21:56.880 --> 00:21:57.500
OK.
00:21:57.500 --> 00:21:59.360
I hope everybody
is following it.
00:21:59.360 --> 00:22:03.740
OK, so now I can actually
simplify equation number two,
00:22:03.740 --> 00:22:07.850
and basically, what I
am going to get is MYj
00:22:07.850 --> 00:22:12.670
double dot will be
equal to minus T over a.
00:22:12.670 --> 00:22:17.700
Basically, I take T
over a out of it again.
00:22:17.700 --> 00:22:23.180
And I also collect all the
terms related to Yj minus one,
00:22:23.180 --> 00:22:30.850
minus two Yj, plus Yj plus one.
00:22:30.850 --> 00:22:34.690
OK, I just ask you to
rewrite equation number two
00:22:34.690 --> 00:22:40.800
in a form which we like more.
00:22:40.800 --> 00:22:45.110
OK, so that is actually the
equation of motion, so from now
00:22:45.110 --> 00:22:47.670
on, I am going to
ignore all the motion
00:22:47.670 --> 00:22:51.920
in the horizontal direction,
because in this small angle
00:22:51.920 --> 00:22:54.545
approximation we have
shown you that there
00:22:54.545 --> 00:22:58.700
will be no acceleration
in the x direction, right?
00:22:58.700 --> 00:23:04.520
So now it's actually getting
a step forward again.
00:23:04.520 --> 00:23:06.470
Basically, we have the
equation of motion,
00:23:06.470 --> 00:23:09.240
and what is usually
the next step?
00:23:09.240 --> 00:23:12.510
The next step is to
write down what matrix?
00:23:12.510 --> 00:23:14.384
Anybody can help me?
00:23:14.384 --> 00:23:15.960
STUDENT 1: M minus one k matrix.
00:23:15.960 --> 00:23:18.085
PROFESSOR YEN-JIE LEE: M
minus one k matrix, right?
00:23:18.085 --> 00:23:21.080
So actually, as usual, we
actually follow the procedure.
00:23:21.080 --> 00:23:25.020
Now I would like to write
down the m minus one k matrix.
00:23:25.020 --> 00:23:28.960
OK, so before I do that
what, I will define--
00:23:28.960 --> 00:23:35.600
I will actually assume my normal
mode has this functional form,
00:23:35.600 --> 00:23:43.850
yj is equal to the real part
of aj exponential i omega t
00:23:43.850 --> 00:23:45.500
plus phi.
00:23:45.500 --> 00:23:48.200
So basically, that tells
you that all the components
00:23:48.200 --> 00:23:51.710
are oscillating at the
same frequency, omega,
00:23:51.710 --> 00:23:53.990
and the same phase, phi.
00:23:53.990 --> 00:23:55.338
Yes.
00:23:55.338 --> 00:23:58.124
STUDENT 2: When
there's a scenario--
00:23:58.124 --> 00:23:59.290
PROFESSOR YEN-JIE LEE: Yeah?
00:23:59.290 --> 00:24:01.266
STUDENT 2: [INAUDIBLE]
also [INAUDIBLE]..
00:24:01.266 --> 00:24:02.972
PROFESSOR YEN-JIE LEE: This one?
00:24:02.972 --> 00:24:07.578
STUDENT 2: No, there
should be one [INAUDIBLE]..
00:24:07.578 --> 00:24:08.744
PROFESSOR YEN-JIE LEE: Here?
00:24:08.744 --> 00:24:09.225
STUDENT 2: Yeah.
00:24:09.225 --> 00:24:10.308
PROFESSOR YEN-JIE LEE: Ah.
00:24:10.308 --> 00:24:12.524
OK, maybe I made a
mistake somewhere.
00:24:22.690 --> 00:24:26.880
Yeah I think it
should be plus, right?
00:24:26.880 --> 00:24:27.380
OK.
00:24:30.380 --> 00:24:31.117
All right.
00:24:31.117 --> 00:24:31.950
Thank you very much.
00:24:34.750 --> 00:24:36.850
So now we have all
the ingredients
00:24:36.850 --> 00:24:41.500
and we assume that it has
a normal mode of aj, yj
00:24:41.500 --> 00:24:43.130
in this functional form.
00:24:43.130 --> 00:24:44.350
And then now I can--
00:24:44.350 --> 00:24:47.858
the next step is to get
m minus one k matrix.
00:24:47.858 --> 00:24:49.100
All right.
00:24:49.100 --> 00:24:50.850
So what is actually m metrics?
00:24:50.850 --> 00:24:53.050
M matrix is really
really straightforward.
00:24:53.050 --> 00:25:00.940
It's m, m, m in
the diagonal terms,
00:25:00.940 --> 00:25:02.960
and all the rest of
the terms are zero.
00:25:02.960 --> 00:25:04.130
All right.
00:25:04.130 --> 00:25:05.980
And those, of
course, you can also
00:25:05.980 --> 00:25:09.940
write down k matrix, right?
00:25:09.940 --> 00:25:14.004
So the k matrix
will be equal to--
00:25:14.004 --> 00:25:17.020
there are many terms,
and in the middle
00:25:17.020 --> 00:25:25.180
you have minus t over a, two
t over a, minus t over a,
00:25:25.180 --> 00:25:27.780
and all the rest of
the terms are zero.
00:25:27.780 --> 00:25:31.030
And of course these
patterns go on and on.
00:25:31.030 --> 00:25:37.780
Minus t over a, two t over A,
minus t over a and zeros, et
00:25:37.780 --> 00:25:38.740
cetera et cetera.
00:25:38.740 --> 00:25:43.790
And this pattern is going to
go on forever, because this
00:25:43.790 --> 00:25:47.650
is actually infinitly long
matrix, infinite times
00:25:47.650 --> 00:25:49.540
infinitely long matrix.
00:25:53.200 --> 00:25:58.490
OK, so once we have this,
we can now write down
00:25:58.490 --> 00:26:01.916
the m minus one k matrix.
00:26:01.916 --> 00:26:04.845
What is actually the
m minus one k matrix?
00:26:04.845 --> 00:26:10.840
It has a similar structure
to k matrix, right?
00:26:10.840 --> 00:26:12.600
All those are zeros--
00:26:12.600 --> 00:26:14.520
OK, all those are
the other values,
00:26:14.520 --> 00:26:24.380
but it has a fixed structure
minus t over ma, two t over ma,
00:26:24.380 --> 00:26:29.660
minus t over ma, and then zeros.
00:26:29.660 --> 00:26:35.960
And this will go on forever
in the diagonal term and also
00:26:35.960 --> 00:26:37.640
the next two diagonal terms.
00:26:37.640 --> 00:26:41.520
And all the rest of
the terms. are zero.
00:26:41.520 --> 00:26:44.155
OK, Any questions?
00:26:48.770 --> 00:26:50.750
OK, so that's really nice.
00:26:50.750 --> 00:26:53.240
Now we have our m
minus one k matrix,
00:26:53.240 --> 00:26:54.980
and the good news
is that you don't
00:26:54.980 --> 00:26:58.790
have to solve m minus one k
matrix's Eigenvalue again,
00:26:58.790 --> 00:26:59.360
right?
00:26:59.360 --> 00:27:01.880
Because we have solved
the Eigenvalue problem
00:27:01.880 --> 00:27:05.540
of s matrix, therefore
what is our left over
00:27:05.540 --> 00:27:11.280
is to multiply m minus
one k by a, right?
00:27:11.280 --> 00:27:19.130
a is actually one of the
Eigenvectors of s matrix.
00:27:19.130 --> 00:27:23.900
OK, so I am going to
multiply that for you,
00:27:23.900 --> 00:27:30.010
and now we calculate m minus
one k equal to omega square a,
00:27:30.010 --> 00:27:34.480
then I can get omega square
out of this calculation.
00:27:34.480 --> 00:27:36.740
OK?
00:27:36.740 --> 00:27:39.920
If I again focus on this term.
00:27:42.830 --> 00:27:44.770
OK, so basically,
what I am going to get
00:27:44.770 --> 00:27:51.880
is, right hand side, I
have omega square aj, OK?
00:27:51.880 --> 00:27:55.420
Now that's actually from
the right hand side, OK?
00:27:55.420 --> 00:28:00.940
And left hand side m minus one
k times a, what I'm going to get
00:28:00.940 --> 00:28:16.641
is t over ma minus aj minus one
plus two aj minus aj plus one.
00:28:16.641 --> 00:28:17.140
All right?
00:28:17.140 --> 00:28:20.350
Because if you take this term--
00:28:20.350 --> 00:28:26.680
this term is actually
in the exact diagonal
00:28:26.680 --> 00:28:30.310
of this m minus one k matrix,
therefore this matches with j,
00:28:30.310 --> 00:28:36.270
and this will match with j minus
one; match with j plus one.
00:28:36.270 --> 00:28:40.750
OK, therefore, if you multiply
m minus one k and the a, you get
00:28:40.750 --> 00:28:46.510
this result. OK?
00:28:46.510 --> 00:28:53.810
And we also know that aj is
proportional to exponential
00:28:53.810 --> 00:28:55.870
ijka, right?
00:28:55.870 --> 00:28:59.430
Therefore I can
take aj out of this
00:28:59.430 --> 00:29:07.930
and basically I get t over
ma aj minus exponential minus
00:29:07.930 --> 00:29:11.640
ika plus two.
00:29:11.640 --> 00:29:19.350
Because I take aj out
of this bracket, OK.
00:29:19.350 --> 00:29:23.150
And minus exponential ika.
00:29:26.710 --> 00:29:27.620
OK.
00:29:27.620 --> 00:29:30.830
Now I actually can cancel aj.
00:29:30.830 --> 00:29:33.200
Basically, what I
get is, omega square
00:29:33.200 --> 00:29:43.200
will be equal to t over ma
two minus exponential ika
00:29:43.200 --> 00:29:45.884
plus exponential minus ika.
00:29:50.730 --> 00:29:57.161
And that will be equal
to two t over ma.
00:29:57.161 --> 00:29:57.660
OK.
00:30:00.330 --> 00:30:07.200
One minus-- OK, so exponential
ika plus exponential minus ika,
00:30:07.200 --> 00:30:12.960
you are going to get two
cosine ka, all right?
00:30:12.960 --> 00:30:16.594
Therefore, you get
one minus cosine ka.
00:30:20.797 --> 00:30:22.200
OK.
00:30:22.200 --> 00:30:31.770
I define omega zero to be
square root of t over ma,
00:30:31.770 --> 00:30:33.870
just to make my life easier.
00:30:33.870 --> 00:30:35.180
OK?
00:30:35.180 --> 00:30:39.270
Then, what is going
to happen is that I
00:30:39.270 --> 00:30:48.560
will have omega
square equal to two
00:30:48.560 --> 00:30:56.450
omega zero square
one minus cosine ka.
00:30:56.450 --> 00:30:57.650
Any questions?
00:31:02.510 --> 00:31:09.600
OK, of course if you like, you
can also rewrite this as four
00:31:09.600 --> 00:31:21.800
omega zero square sine
square ka divided by two.
00:31:21.800 --> 00:31:25.230
OK, so if you like.
00:31:25.230 --> 00:31:28.960
OK, so look at
what we have done.
00:31:28.960 --> 00:31:33.280
We studied a highly
symmetric system,
00:31:33.280 --> 00:31:35.800
which is as you were
shown in the slide.
00:31:35.800 --> 00:31:41.050
OK, basically you satisfy the
space translation symmetry.
00:31:41.050 --> 00:31:41.880
OK?
00:31:41.880 --> 00:31:44.080
Now what we have been
doing is to derive
00:31:44.080 --> 00:31:47.830
the equation of motion,
make use of the small angle
00:31:47.830 --> 00:31:51.670
approximation, then you
will be able to find that,
00:31:51.670 --> 00:31:55.390
OK, only the y
direction is actually
00:31:55.390 --> 00:31:57.850
moving as a function of time.
00:31:57.850 --> 00:32:02.110
Therefore, based on this
derivation of m minus one k
00:32:02.110 --> 00:32:06.920
matrix, I arrive, and also
based on the equation of motion,
00:32:06.920 --> 00:32:09.280
which I derived from
the first diagram,
00:32:09.280 --> 00:32:12.190
I get this m minus one k matrix.
00:32:12.190 --> 00:32:16.305
And since we know that m
minus one k matrix and s
00:32:16.305 --> 00:32:18.780
matrix will share
this Eigenvector,
00:32:18.780 --> 00:32:24.220
I can multiply m minus one k
matrix, and I come back to a.
00:32:24.220 --> 00:32:27.340
Then I will be able to
solve the functional
00:32:27.340 --> 00:32:33.970
form of omega square, and
that is actually given here.
00:32:33.970 --> 00:32:38.320
Omega square is equal to
two omega zero square y
00:32:38.320 --> 00:32:40.240
minus cosine ka.
00:32:40.240 --> 00:32:41.860
OK?
00:32:41.860 --> 00:32:49.710
And is this actually telling you
that omega is a function of k.
00:32:49.710 --> 00:32:51.540
OK.
00:32:51.540 --> 00:32:52.884
What is k?
00:32:52.884 --> 00:33:01.730
k is actually the wave
number and omega is actually
00:33:01.730 --> 00:33:06.610
the angular frequency of
the normal modes, right?
00:33:06.610 --> 00:33:10.665
So that means what we
were talking about--
00:33:10.665 --> 00:33:20.960
that means if we fix the
wavelength or the wave number,
00:33:20.960 --> 00:33:27.601
k, then there will be
a corresponding omega.
00:33:27.601 --> 00:33:28.100
OK?
00:33:28.100 --> 00:33:31.040
If you fix the wavelengths
you are talking about,
00:33:31.040 --> 00:33:37.460
then the omega is also fixed
by this omega of k function.
00:33:37.460 --> 00:33:41.080
OK, now we actually call
it dispersion relation.
00:33:50.700 --> 00:33:54.300
This term may not
mean much to you now,
00:33:54.300 --> 00:33:57.650
but later in the discussion,
you will find, aha!
00:33:57.650 --> 00:34:00.960
It really makes sense, and that
we will talk about dispersion
00:34:00.960 --> 00:34:03.315
in the later lectures.
00:34:06.160 --> 00:34:11.010
OK, so the conclusion from
here is that, basically,
00:34:11.010 --> 00:34:17.000
if I have this distance that
satisfies this translation
00:34:17.000 --> 00:34:25.010
symmetry, then what it tells us
is that the normal modes, what
00:34:25.010 --> 00:34:31.260
looks like some kind
of sinusoidal function,
00:34:31.260 --> 00:34:32.550
as we discussed last time.
00:34:35.139 --> 00:34:39.219
And also this--
00:34:39.219 --> 00:34:44.260
OK, so this is actually
the amplitude, what
00:34:44.260 --> 00:34:47.530
I am drawing here, this curve.
00:34:47.530 --> 00:34:53.800
And all those masses are
only moving up and down, OK?
00:34:53.800 --> 00:34:55.250
As a function of time.
00:34:55.250 --> 00:35:00.040
And this is aj, and that is the
oscillation frequency, which
00:35:00.040 --> 00:35:03.190
is actually the frequency
of moving up and down,
00:35:03.190 --> 00:35:05.790
this kind of motion,
is actually omega.
00:35:08.510 --> 00:35:15.140
And also, we learned that
omega is equal to omega of k.
00:35:15.140 --> 00:35:19.030
And that is actually
decided by the length,
00:35:19.030 --> 00:35:25.190
and how distorted is
this normal mode--
00:35:25.190 --> 00:35:27.110
the shape of the normal mode?
00:35:27.110 --> 00:35:31.190
And this is actually determined
by the k, which is actually
00:35:31.190 --> 00:35:36.770
the wave number, and of course
you can also get the wavelength
00:35:36.770 --> 00:35:38.810
from two pi over k.
00:35:38.810 --> 00:35:41.180
OK?
00:35:41.180 --> 00:35:44.930
In short, if you give
it a specific wave
00:35:44.930 --> 00:35:49.370
number or wavelength, than
the oscillation frequency
00:35:49.370 --> 00:35:53.830
is already fixed because
of the equation of motion,
00:35:53.830 --> 00:35:57.200
which we did, right,
from the first diagram.
00:35:57.200 --> 00:35:58.388
Any questions?
00:36:04.250 --> 00:36:05.980
OK.
00:36:05.980 --> 00:36:08.080
The last point which I
would like to remind you
00:36:08.080 --> 00:36:12.730
is that, at this point,
since we are talking
00:36:12.730 --> 00:36:17.110
about infinitely long
systems, therefore
00:36:17.110 --> 00:36:22.226
all possible k are allowed.
00:36:22.226 --> 00:36:23.020
Right?
00:36:23.020 --> 00:36:25.400
Because, basically, you
have an infinite number
00:36:25.400 --> 00:36:28.370
of coupled oscillators,
and therefore you
00:36:28.370 --> 00:36:31.280
have an infinite
number of normal modes.
00:36:31.280 --> 00:36:38.030
So all possible cases are
allowed, and that actually
00:36:38.030 --> 00:36:41.300
because we have even an
infinitely long system.
00:36:41.300 --> 00:36:45.080
After the break, which we
will take a five minute break,
00:36:45.080 --> 00:36:48.950
we will discuss how to use
infinitely long systems
00:36:48.950 --> 00:36:53.340
to actually understand
a finite system.
00:36:53.340 --> 00:36:56.840
So you will see that,
actually I can use, now,
00:36:56.840 --> 00:36:59.750
this space translation
symmetry, and to solve,
00:36:59.750 --> 00:37:02.690
in general, infinitely
long systems.
00:37:02.690 --> 00:37:05.815
And I can actually even go
back to find to a finite system
00:37:05.815 --> 00:37:07.860
and see what we
can get from there.
00:37:07.860 --> 00:37:08.360
OK?
00:37:08.360 --> 00:37:12.920
So we will be back at 12:20.
00:37:12.920 --> 00:37:18.370
If you have any
questions, I will be here
00:37:18.370 --> 00:37:20.530
OK, welcome back, everybody.
00:37:20.530 --> 00:37:23.590
So we will continue
the discussion.
00:37:23.590 --> 00:37:27.680
So there were a few questions
asked during the break.
00:37:27.680 --> 00:37:30.730
So, the first question is
related to how we actually
00:37:30.730 --> 00:37:32.870
arrive at this equation.
00:37:32.870 --> 00:37:35.620
And that is actually because--
00:37:35.620 --> 00:37:39.180
OK, two t over ma is
actually really happening
00:37:39.180 --> 00:37:40.570
in a diagonal term.
00:37:40.570 --> 00:37:43.370
Therefore, if you
multiply m minus one k
00:37:43.370 --> 00:37:48.200
matrix and A matrix, which
is actually shown there,
00:37:48.200 --> 00:37:51.750
then you will get
this term, minus t
00:37:51.750 --> 00:37:54.770
over ma multiplied
by aj minus one
00:37:54.770 --> 00:37:59.860
plus two t over ma
times aj price minus t
00:37:59.860 --> 00:38:02.560
over ma times aj plus one.
00:38:02.560 --> 00:38:06.070
And that is actually why we
can arrive at this expression.
00:38:06.070 --> 00:38:06.610
OK?
00:38:06.610 --> 00:38:08.990
Then what happens
afterward is that we
00:38:08.990 --> 00:38:13.180
found that aj can be factorized
out, and they cancel.
00:38:13.180 --> 00:38:17.560
And then now, my solution
depends now upon the amplitude,
00:38:17.560 --> 00:38:23.560
and still omega is actually
dependent on the k value, which
00:38:23.560 --> 00:38:24.540
we actually choose.
00:38:24.540 --> 00:38:25.780
OK.
00:38:25.780 --> 00:38:32.370
The second question is, why
do I say k is the wave number?
00:38:32.370 --> 00:38:34.390
OK, where is that coming from?
00:38:34.390 --> 00:38:36.220
So that is because--
00:38:36.220 --> 00:38:40.250
OK, so aj is proportional
to exponential ijka.
00:38:40.250 --> 00:38:40.750
OK?
00:38:40.750 --> 00:38:43.060
It has a fancy name.
00:38:43.060 --> 00:38:48.900
If I take the real
part, OK, as we
00:38:48.900 --> 00:38:51.370
did when we went
to the description
00:38:51.370 --> 00:38:55.840
of physical systems,
then you get cosine jka.
00:38:55.840 --> 00:38:57.070
OK?
00:38:57.070 --> 00:39:00.730
And j times a is actually--
00:39:00.730 --> 00:39:02.470
j is actually a label, right?
00:39:02.470 --> 00:39:05.260
Labeling which mass
I am talking about.
00:39:05.260 --> 00:39:07.960
A is actually the distance
between all those masses.
00:39:07.960 --> 00:39:14.440
j times a will give you
the x location of the mass.
00:39:14.440 --> 00:39:19.970
So j times a is actually the
the x position of the mass.
00:39:19.970 --> 00:39:20.470
OK?
00:39:20.470 --> 00:39:26.800
Therefore, if you accept
that, this becomes cosine kx,
00:39:26.800 --> 00:39:28.690
and from there you
will see immediately
00:39:28.690 --> 00:39:32.580
that k has a meaning, which
is actually the wave number.
00:39:32.580 --> 00:39:34.630
OK.
00:39:34.630 --> 00:39:37.240
All right, is that?
00:39:37.240 --> 00:39:42.430
OK, so that was the
questions raised,
00:39:42.430 --> 00:39:45.130
which I can quickly explain.
00:39:45.130 --> 00:39:47.960
So, what I am going
to do now is that--
00:39:47.960 --> 00:39:52.120
OK, we have solved, in general,
an infinitely long system.
00:39:52.120 --> 00:39:56.000
What are actually the
resulting normal modes
00:39:56.000 --> 00:39:57.470
of infinitely long systems?
00:39:57.470 --> 00:40:02.830
It has an infinite
number of normal modes,
00:40:02.830 --> 00:40:06.970
and we will wonder
if I can actually
00:40:06.970 --> 00:40:08.980
borrow this
infinitely long system
00:40:08.980 --> 00:40:14.680
and solve finite systems to see
if I can arrive at the solution
00:40:14.680 --> 00:40:15.890
really quickly.
00:40:15.890 --> 00:40:16.620
OK?
00:40:16.620 --> 00:40:20.320
So the answer is actually yes.
00:40:20.320 --> 00:40:25.210
So if I consider a finite
system that looks like this;
00:40:25.210 --> 00:40:32.510
so I have many, many little
masses on this system
00:40:32.510 --> 00:40:35.620
and they are connected
to each other
00:40:35.620 --> 00:40:41.290
by the center strings,
which I prepared before, OK?
00:40:41.290 --> 00:40:44.830
And I call this the
position in the y
00:40:44.830 --> 00:40:51.310
direction of this object y1,
and then the next object y2, y3,
00:40:51.310 --> 00:40:52.330
etc.
00:40:52.330 --> 00:40:56.670
And I have an object
in this system
00:40:56.670 --> 00:41:03.310
and both ends of the string
are fixed on the wall.
00:41:03.310 --> 00:41:05.380
OK?
00:41:05.380 --> 00:41:10.270
So I can actually now argue
that the infinitely long system
00:41:10.270 --> 00:41:13.540
can help us with
the understanding
00:41:13.540 --> 00:41:15.460
of this finite system.
00:41:15.460 --> 00:41:17.150
Why is that?
00:41:17.150 --> 00:41:20.860
That is because, now, I
can assume that, huh, this
00:41:20.860 --> 00:41:24.870
is actually just part of
an infinitely long system.
00:41:24.870 --> 00:41:26.440
All right.
00:41:26.440 --> 00:41:29.710
So I construct my
infinitely long system,
00:41:29.710 --> 00:41:40.270
and now I nail the yth mass,
I nail the y n plus one mass,
00:41:40.270 --> 00:41:43.030
and I fix that so that
it cannot move, OK?
00:41:43.030 --> 00:41:46.720
So it's still an
infinitely long system,
00:41:46.720 --> 00:41:52.090
but there are two interesting
boundary conditions
00:41:52.090 --> 00:41:57.481
at j equal to zero and
j equal to n plus one.
00:41:57.481 --> 00:41:57.980
OK.
00:41:57.980 --> 00:42:00.340
What are the two
boundary conditions?
00:42:08.880 --> 00:42:15.260
The first one is y
zero equal to zero.
00:42:15.260 --> 00:42:16.610
OK?
00:42:16.610 --> 00:42:24.060
And the second condition is
y n plus one equal to zero.
00:42:24.060 --> 00:42:25.520
OK?
00:42:25.520 --> 00:42:27.280
So there are two
boundary conditions,
00:42:27.280 --> 00:42:31.180
so basically what I'm looking
at is still an infinitely long
00:42:31.180 --> 00:42:36.800
system, but I require
y zero and y n plus one
00:42:36.800 --> 00:42:39.610
to satisfy these two conditions.
00:42:39.610 --> 00:42:40.550
OK?
00:42:40.550 --> 00:42:44.090
And we will find that,
huh, with this procedure,
00:42:44.090 --> 00:42:48.165
we can also solve this finite
number of couple oscillators.
00:42:48.165 --> 00:42:52.610
The problem, in this case
we, have coupled oscillators.
00:42:52.610 --> 00:42:54.317
OK?
00:42:54.317 --> 00:42:56.150
So the first thing which
I would like to say
00:42:56.150 --> 00:43:01.190
is, based on the functional
form, the functional
00:43:01.190 --> 00:43:05.660
form of omega square, now
this is equal to four omega
00:43:05.660 --> 00:43:09.380
zero square sine
square ka over two.
00:43:09.380 --> 00:43:10.220
OK?
00:43:10.220 --> 00:43:15.260
What we actually have
is that omega k is
00:43:15.260 --> 00:43:19.670
equal to omega minus k.
00:43:19.670 --> 00:43:20.720
OK?
00:43:20.720 --> 00:43:24.920
So both of them will give you
the same angular frequency.
00:43:24.920 --> 00:43:29.250
OK, therefore, what
does that mean?
00:43:29.250 --> 00:43:32.080
This means that
linear combination
00:43:32.080 --> 00:43:39.260
of exponential ijka and the
exponential minus ijka--
00:43:42.250 --> 00:43:46.545
OK, linear combination
of these two vectors--
00:43:49.240 --> 00:43:54.980
is also an Eigenvector
of m minus one k metrics.
00:43:54.980 --> 00:43:58.290
OK, so you can do
linear combination
00:43:58.290 --> 00:44:00.995
of these two vectors.
00:44:06.390 --> 00:44:13.380
OK, so if we do that, now
I can guess my solution
00:44:13.380 --> 00:44:23.340
will be like yj equal to real
part of exponential i omega t
00:44:23.340 --> 00:44:26.100
plus phi.
00:44:26.100 --> 00:44:30.500
I can now have a linear
combination of exponential ijka
00:44:30.500 --> 00:44:33.240
and the exponential minus ijka.
00:44:33.240 --> 00:44:39.670
Basically, I have
alpha exponential ijka
00:44:39.670 --> 00:44:43.600
plus beta exponential
minus ijka.
00:44:48.400 --> 00:44:50.120
OK?
00:44:50.120 --> 00:44:52.580
And I would like to
determine why that's actually
00:44:52.580 --> 00:44:57.800
alpha and beta which actually
satisfy these boundary
00:44:57.800 --> 00:45:01.010
conditions, one and two.
00:45:01.010 --> 00:45:06.290
OK, so now I can use the
first boundary condition,
00:45:06.290 --> 00:45:09.740
y zero equal to zero, right?
00:45:09.740 --> 00:45:11.810
So j equal to zero.
00:45:11.810 --> 00:45:16.510
Therefore, basically, what I
get is, when j is equal to zero,
00:45:16.510 --> 00:45:21.060
then this is actually one and
this is actually one, right?
00:45:21.060 --> 00:45:27.530
And this actually gives
you y zero equal to zero.
00:45:27.530 --> 00:45:31.160
If y zero is equal
to 0 at all times,
00:45:31.160 --> 00:45:36.320
no matter what t as you give it
to this system, then basically
00:45:36.320 --> 00:45:41.060
you have alpha plus beta
equal to zero, right?
00:45:41.060 --> 00:45:43.580
Because j is equal to zero.
00:45:43.580 --> 00:45:49.085
So you have alpha plus beta, and
that has to be equal to zero.
00:45:49.085 --> 00:45:52.850
Therefore, you can conclude that
alpha is equal to minus beta.
00:45:55.980 --> 00:46:00.000
And I've reused the
second boundary condition,
00:46:00.000 --> 00:46:05.270
y n plus one equal to zero,
because I nailed this mass
00:46:05.270 --> 00:46:08.510
and then fixed that so
that it cannot move.
00:46:08.510 --> 00:46:13.810
Then basically, what you get is
y n plus one is equal to zero,
00:46:13.810 --> 00:46:21.510
then basically you have
alpha exponential i n
00:46:21.510 --> 00:46:24.735
plus one ka plus--
00:46:28.350 --> 00:46:32.460
okay, so beta is your equal
to minus alpha, right?
00:46:32.460 --> 00:46:44.530
So basically, you can get
minus exponential minus i
00:46:44.530 --> 00:46:49.000
n plus one ka, right?
00:46:49.000 --> 00:46:51.640
Multiplied by alpha.
00:46:51.640 --> 00:46:56.720
And now this is
actually equal to zero.
00:46:56.720 --> 00:46:59.150
Now we have the choice.
00:46:59.150 --> 00:47:02.690
We can actually set alpha
to be equal to zero,
00:47:02.690 --> 00:47:05.300
but if I set alpha
to be equal to zero,
00:47:05.300 --> 00:47:08.030
then beta is also zero.
00:47:08.030 --> 00:47:10.040
Then I have zero
everywhere, right?
00:47:10.040 --> 00:47:11.810
Then there's no oscillation.
00:47:11.810 --> 00:47:15.450
And that's not fun, right?
00:47:15.450 --> 00:47:16.260
OK.
00:47:16.260 --> 00:47:18.790
Therefore, what I'm
going to set is actually
00:47:18.790 --> 00:47:21.030
the second turn equal to zero.
00:47:21.030 --> 00:47:21.530
OK.
00:47:21.530 --> 00:47:24.020
The second turn, I
can actually simplify
00:47:24.020 --> 00:47:33.980
that to be two i
sine n plus one ka.
00:47:33.980 --> 00:47:37.380
And now this is
actually equal to zero.
00:47:37.380 --> 00:47:38.290
OK.
00:47:38.290 --> 00:47:42.150
What is actually the condition
of this thing equal to zero?
00:47:42.150 --> 00:47:46.640
Basically, n plus 1 is
actually a given number,
00:47:46.640 --> 00:47:52.040
a is actually the distance
between those masses,
00:47:52.040 --> 00:47:56.690
therefore, what I can actually
change is the k value.
00:47:56.690 --> 00:47:57.530
Right?
00:47:57.530 --> 00:48:00.500
So I can now solve
this condition,
00:48:00.500 --> 00:48:03.080
and I will conclude
that k will have
00:48:03.080 --> 00:48:09.800
to be equal to n times
pi divided by N plus one.
00:48:09.800 --> 00:48:12.500
I hope you can see it.
00:48:12.500 --> 00:48:24.550
Where small n is equal to one,
two, three, until capital N.
00:48:24.550 --> 00:48:27.500
So what does that mean?
00:48:27.500 --> 00:48:29.450
This means that--
00:48:29.450 --> 00:48:33.730
OK, originally, before I
introduced the boundary
00:48:33.730 --> 00:48:38.080
condition, this system
is infinitely long, OK,
00:48:38.080 --> 00:48:45.000
and it has an infinite number
of normal modes, right?
00:48:45.000 --> 00:48:47.880
But once I introduced
these boundary conditions,
00:48:47.880 --> 00:48:53.760
which I actually require
y zero equal to zero,
00:48:53.760 --> 00:48:57.080
because I fixed this
point on the wall.
00:48:57.080 --> 00:49:00.660
Y m plus one equal to zero
because I fixed, also,
00:49:00.660 --> 00:49:02.910
that point on the wall.
00:49:02.910 --> 00:49:05.220
Something really happened.
00:49:05.220 --> 00:49:14.960
Now it actually gives us, first,
the shape of the system when it
00:49:14.960 --> 00:49:16.470
is actually in the normal mode.
00:49:16.470 --> 00:49:18.840
Basically, the shape--
what I mean here
00:49:18.840 --> 00:49:22.350
is the amplitude as
a function of j, OK?
00:49:22.350 --> 00:49:24.600
That's actually what
I mean by shape, OK?
00:49:24.600 --> 00:49:29.430
The shape is now
like a sine function.
00:49:29.430 --> 00:49:32.670
That's the first thing
which we get from here.
00:49:32.670 --> 00:49:35.540
The second thing
which we get here
00:49:35.540 --> 00:49:41.750
is that now, the k values
are not arbitrary anymore.
00:49:41.750 --> 00:49:47.290
The k values are equal
to n pi over n plus one.
00:49:47.290 --> 00:49:51.910
And the the small n is actually
equal for one, two, three,
00:49:51.910 --> 00:49:57.280
until N. So now, once you
actually fix this two point,
00:49:57.280 --> 00:50:05.370
you actually have only
how many normal modes?
00:50:05.370 --> 00:50:07.990
N normal modes!
00:50:07.990 --> 00:50:08.830
Right?
00:50:08.830 --> 00:50:12.880
So what I want to tell
you is that, in general,
00:50:12.880 --> 00:50:16.690
the sinusoidal shape is
actually fixed already
00:50:16.690 --> 00:50:20.440
by this translation
symmetry argument.
00:50:20.440 --> 00:50:21.240
OK.
00:50:21.240 --> 00:50:26.560
And once we nail both sides--
00:50:26.560 --> 00:50:29.680
actually, we also
restrict ourselves
00:50:29.680 --> 00:50:37.360
to the discussion of only a
few k which actually satisfy
00:50:37.360 --> 00:50:38.730
the boundary conditions.
00:50:38.730 --> 00:50:43.150
And if I plot all those normal
modes as a function of i,
00:50:43.150 --> 00:50:46.780
basically what you
can see from here,
00:50:46.780 --> 00:50:50.800
you can see if I have n
equals to one and capital
00:50:50.800 --> 00:50:54.560
N equal to four in this case.
00:50:54.560 --> 00:50:56.620
OK, so I have four--
00:50:56.620 --> 00:51:00.200
basically, I'm going to
have four normal modes.
00:51:00.200 --> 00:51:06.510
The first one will be like
a really long wavelength
00:51:06.510 --> 00:51:09.730
one, when n is equal to one.
00:51:09.730 --> 00:51:14.350
And if I increase
the small n value
00:51:14.350 --> 00:51:17.110
so that the k becomes
bigger, then you
00:51:17.110 --> 00:51:23.330
can see that there is more
distortion when this system is
00:51:23.330 --> 00:51:24.910
in one of the normal modes.
00:51:24.910 --> 00:51:30.440
And this shape is actually going
to be oscillating up and down,
00:51:30.440 --> 00:51:31.170
instead of--
00:51:31.170 --> 00:51:34.230
OK, so all those points are
only moving up and down, right?
00:51:34.230 --> 00:51:35.140
Just a reminder.
00:51:35.140 --> 00:51:35.790
OK?
00:51:35.790 --> 00:51:41.170
And why do we have
all those cases?
00:51:41.170 --> 00:51:45.300
Because of the
boundary conditions.
00:51:45.300 --> 00:51:46.871
OK, any questions?
00:51:50.720 --> 00:51:54.150
OK, so I have
several other cases,
00:51:54.150 --> 00:51:57.720
which is open end and
the closed end, and also
00:51:57.720 --> 00:52:01.320
the driven and the coupled
oscillator examples.
00:52:01.320 --> 00:52:04.190
Also in the lecture notes,
but unfortunately, we
00:52:04.190 --> 00:52:06.050
are will not be able
to go over them,
00:52:06.050 --> 00:52:08.240
but I think they are
very, very detailed,
00:52:08.240 --> 00:52:10.400
the notes in the lecture notes.
00:52:10.400 --> 00:52:11.360
OK.
00:52:11.360 --> 00:52:14.980
So, let me-- before I
move on to the discussion
00:52:14.980 --> 00:52:17.900
of continuous
systems, OK, I would
00:52:17.900 --> 00:52:23.400
like to discuss with you
what we have learned so far.
00:52:23.400 --> 00:52:27.200
So what we have learned is
that, if I have a symmetry which
00:52:27.200 --> 00:52:29.810
is a translation
symmetry, and plus,
00:52:29.810 --> 00:52:34.730
we only limit ourselves in
the discussion of oscillation.
00:52:34.730 --> 00:52:39.530
OK, in other words,
we limit the amplitude
00:52:39.530 --> 00:52:42.680
so that it doesn't explode
at the edge of the universe.
00:52:42.680 --> 00:52:43.640
OK.
00:52:43.640 --> 00:52:46.760
And I will give you
a beta value which
00:52:46.760 --> 00:52:52.160
is the functional form
of exponential ika,
00:52:52.160 --> 00:52:57.290
and equation of motion can be
derived from the first diagram.
00:52:57.290 --> 00:53:01.070
Once we entered the
equation of motion,
00:53:01.070 --> 00:53:03.320
we can get m minus
one k matrix, then
00:53:03.320 --> 00:53:07.630
we can derive omega square
from this expression.
00:53:07.630 --> 00:53:12.860
And finally, we actually can
simplify everything and then
00:53:12.860 --> 00:53:15.470
get the dispersion
relation omega
00:53:15.470 --> 00:53:20.050
equal to omega k, which is a
function of k, the wave number.
00:53:22.640 --> 00:53:25.850
Before we actually introduce
boundary conditions
00:53:25.850 --> 00:53:30.420
to go from an infinitely long
system to a finite system,
00:53:30.420 --> 00:53:33.610
all the k values are allowed.
00:53:33.610 --> 00:53:36.290
Once you introduce
boundary conditions,
00:53:36.290 --> 00:53:42.050
you find that you only have a
limited number of normal modes.
00:53:42.050 --> 00:53:46.460
Second, the k value not
continuous at any value
00:53:46.460 --> 00:53:48.410
any more, it becomes discrete.
00:53:48.410 --> 00:53:53.710
And only n values allowed
from this exercise.
00:53:53.710 --> 00:53:57.410
And finally, what is actually
the most general solution
00:53:57.410 --> 00:53:59.690
is actually the
linear combination
00:53:59.690 --> 00:54:03.760
of all those normal
modes, which we show here.
00:54:03.760 --> 00:54:08.760
And what is actually the ratio
between all those normal modes?
00:54:08.760 --> 00:54:11.480
All those free
coefficients are determined
00:54:11.480 --> 00:54:14.220
by initial conditions
if you are given.
00:54:14.220 --> 00:54:15.650
OK?
00:54:15.650 --> 00:54:18.485
So that's actually what
we have learned so far.
00:54:22.160 --> 00:54:29.240
And now I would like
to make a leap of faith
00:54:29.240 --> 00:54:30.340
to see what happens.
00:54:30.340 --> 00:54:34.460
OK, what we are going to
do is to introduce you
00:54:34.460 --> 00:54:39.950
to a continuous infinite
number of coupled oscillators.
00:54:39.950 --> 00:54:40.850
OK?
00:54:40.850 --> 00:54:41.970
So what does that mean?
00:54:41.970 --> 00:54:44.900
What I am going to do
is to go from this--
00:54:44.900 --> 00:54:55.080
so I have t and then a t, a
lot of string and mass system,
00:54:55.080 --> 00:54:56.700
et cetera, et
cetera, but I would
00:54:56.700 --> 00:55:06.510
like to go from there to just
an infinitely long string
00:55:06.510 --> 00:55:15.110
with string tension t and
some kind of density or mass.
00:55:15.110 --> 00:55:16.240
OK?
00:55:16.240 --> 00:55:19.830
We like to make it continuous
to see what will happen.
00:55:19.830 --> 00:55:21.690
OK?
00:55:21.690 --> 00:55:30.710
So, just a reminder of the
jth term of the m minus one
00:55:30.710 --> 00:55:32.250
k matrix operation.
00:55:32.250 --> 00:55:37.890
Basically, we have m minus
one k times a, the j's term
00:55:37.890 --> 00:55:40.180
of m minus one k a.
00:55:40.180 --> 00:55:45.540
That is actually given
by omega square aj.
00:55:45.540 --> 00:55:54.480
This is equal to t over
ma minus aj minus one
00:55:54.480 --> 00:56:01.060
plus two aj minus aj plus one?
00:56:01.060 --> 00:56:01.680
OK.
00:56:01.680 --> 00:56:05.510
So this is actually just a
copy of that formula here.
00:56:05.510 --> 00:56:10.570
OK, so if I make it continuous--
00:56:16.360 --> 00:56:19.400
OK, so that means
what I'm going to get
00:56:19.400 --> 00:56:25.060
is omega square a but
evaluated at position
00:56:25.060 --> 00:56:30.340
x, where x is actually
equal to j times a.
00:56:30.340 --> 00:56:32.200
OK?
00:56:32.200 --> 00:56:37.750
And this will be
equal to t over ma
00:56:37.750 --> 00:56:50.067
minus a, evaluated at x minus a
plus two a x minus a x plus a.
00:56:53.730 --> 00:56:59.100
Now I am going to make this
a very, very small, right?
00:56:59.100 --> 00:57:02.350
So that, when I make
a very, very small,
00:57:02.350 --> 00:57:06.240
then it becomes a very,
very continuous system.
00:57:06.240 --> 00:57:07.530
OK?
00:57:07.530 --> 00:57:09.900
So what I'm going to do is--
00:57:09.900 --> 00:57:14.030
I can now make a go to zero.
00:57:16.680 --> 00:57:25.470
I can now use Taylor's
series fx plus delta x--
00:57:25.470 --> 00:57:29.880
and just a reminder, if you
do a Taylor expansion of this,
00:57:29.880 --> 00:57:39.640
you are going to get f of x
plus delta x f prime x plus one
00:57:39.640 --> 00:57:46.080
over two factorial delta
x square f double prime x.
00:57:49.349 --> 00:57:53.400
OK, so that means
now, I can actually
00:57:53.400 --> 00:57:59.670
do a Taylor expansion of a x
minus a and then a x plus a.
00:57:59.670 --> 00:58:05.230
So what I'm going
to get is like this.
00:58:05.230 --> 00:58:11.770
So a x minus a
will become a of x
00:58:11.770 --> 00:58:22.240
minus a a prime x plus one over
two a square a double prime x.
00:58:26.048 --> 00:58:29.500
I can also do the same thing
to do a Taylor expansion
00:58:29.500 --> 00:58:33.380
for a x plus a.
00:58:33.380 --> 00:58:44.960
a x plus a will be equal to a
of x plus a prime x plus one
00:58:44.960 --> 00:58:51.355
over two a square
a double prime x.
00:58:51.355 --> 00:58:51.855
OK.
00:58:59.620 --> 00:59:02.545
Once I have done this,
basically, then, I
00:59:02.545 --> 00:59:11.315
can calculate minus a
x minus a plus two a
00:59:11.315 --> 00:59:17.360
of x minus a x plus a.
00:59:17.360 --> 00:59:22.630
OK, I'm just taking the middle
term and copying it there.
00:59:22.630 --> 00:59:24.170
OK?
00:59:24.170 --> 00:59:30.320
If I use, now, this expression,
OK, then basically I
00:59:30.320 --> 00:59:36.200
will see that, OK, a of x
terms actually cancel, right?
00:59:36.200 --> 00:59:40.790
Because I have two a of
x from these two terms,
00:59:40.790 --> 00:59:44.670
and they cancel
with this two a x.
00:59:44.670 --> 00:59:45.951
OK?
00:59:45.951 --> 00:59:50.600
And also, a prime terms,
also cancel, right?
00:59:50.600 --> 00:59:59.780
Because you have a x minus
and a x plus a, therefore,
00:59:59.780 --> 01:00:01.790
the x prime turns cancel here.
01:00:01.790 --> 01:00:05.010
This is a minus sign,
this is a plus sign.
01:00:05.010 --> 01:00:05.790
You see?
01:00:05.790 --> 01:00:11.550
So therefore, what is actually
left over is all those terms.
01:00:11.550 --> 01:00:24.060
This is going to give you
a double prime x a squared,
01:00:24.060 --> 01:00:28.390
plus many other higher
order terms, right?
01:00:28.390 --> 01:00:31.560
Because those are
actually not completed,
01:00:31.560 --> 01:00:34.030
we have many, many
higher order terms.
01:00:34.030 --> 01:00:35.230
OK?
01:00:35.230 --> 01:00:39.640
In the limit of x go to zero--
01:00:39.640 --> 01:00:40.398
Yes.
01:00:40.398 --> 01:00:43.624
STUDENT: Isn't the
xa prime [INAUDIBLE]??
01:00:43.624 --> 01:00:44.790
PROFESSOR YEN-JIE LEE: Yeah.
01:00:44.790 --> 01:00:46.860
Oh yeah, you are right.
01:00:46.860 --> 01:00:49.280
Thank you very much.
01:00:49.280 --> 01:00:50.920
OK, thank you for that.
01:00:50.920 --> 01:00:54.370
So there should be a minus
sign in front of it, OK.
01:00:54.370 --> 01:00:59.470
So basically what we have is a
double prime x times a square
01:00:59.470 --> 01:01:10.570
plus some higher order turn of
a cubed, et cetera, et cetera.
01:01:10.570 --> 01:01:16.540
OK, since we are talking about
the limit of a goes to zero,
01:01:16.540 --> 01:01:20.670
we can safely ignore all
the higher order terms.
01:01:20.670 --> 01:01:22.120
OK?
01:01:22.120 --> 01:01:32.550
So now, if we go back to
this equation, omega square a
01:01:32.550 --> 01:01:40.840
will become, basically,
t over m minus t over ma
01:01:40.840 --> 01:01:46.234
a double prime x
times a squared.
01:01:51.180 --> 01:01:54.900
OK, and then plus some
higher order terms.
01:01:54.900 --> 01:02:02.850
OK, now I can define rho l to
be equal to m divided by a.
01:02:02.850 --> 01:02:03.460
OK.
01:02:03.460 --> 01:02:07.500
And I will I will have
to be very careful when
01:02:07.500 --> 01:02:11.490
I go to the continuous
limit, OK, I also
01:02:11.490 --> 01:02:17.670
do not want to make this system
infinitely massive, right?
01:02:17.670 --> 01:02:20.700
Therefore, I would like
to fix the rho l when
01:02:20.700 --> 01:02:24.030
I go to the continuous limit.
01:02:24.030 --> 01:02:28.980
So basically what I am doing is
that I cut this system in half
01:02:28.980 --> 01:02:32.880
so that a becomes smaller and
the mass also because smaller,
01:02:32.880 --> 01:02:37.320
so that rho l stays
as a constant, OK,
01:02:37.320 --> 01:02:39.350
when I go to a continuous limit.
01:02:39.350 --> 01:02:43.640
OK, so what I'm going
to get is omega square a
01:02:43.640 --> 01:02:54.010
will be equal to minus t
over rho l a double prime x.
01:02:54.010 --> 01:02:56.730
And to write it explicitly,
this is actually
01:02:56.730 --> 01:03:01.620
equal to minus t over rho l.
01:03:01.620 --> 01:03:06.730
I'll just square a
partial x squared.
01:03:06.730 --> 01:03:07.230
OK?
01:03:07.230 --> 01:03:11.160
Because each prime is
actually the differentiation
01:03:11.160 --> 01:03:14.580
which is spread to x, right?
01:03:14.580 --> 01:03:15.270
OK.
01:03:15.270 --> 01:03:17.220
Don't forget-- what
is actually this?
01:03:17.220 --> 01:03:21.530
This is actually m
minus one ka, right?
01:03:21.530 --> 01:03:23.788
Equal to omega square a.
01:03:27.180 --> 01:03:32.330
And that this is
actually just partial
01:03:32.330 --> 01:03:39.860
square a partial t square.
01:03:39.860 --> 01:03:40.360
Right?
01:03:40.360 --> 01:03:42.850
Because this is actually
m minus one k matrix.
01:03:42.850 --> 01:03:49.256
It's actually originally-- going
back to the original equation,
01:03:49.256 --> 01:03:55.330
we are actually solving
m x double dot equal
01:03:55.330 --> 01:03:57.328
to minus kx problem, right?
01:04:00.610 --> 01:04:02.970
So there should be a
minus sign there as well.
01:04:02.970 --> 01:04:03.830
Right?
01:04:03.830 --> 01:04:08.400
So mx double dot
equal to minus kx,
01:04:08.400 --> 01:04:14.490
therefore x double dot will be
equal to minus m minus one kx.
01:04:14.490 --> 01:04:15.330
Right?
01:04:15.330 --> 01:04:21.176
So minus m minus one k matrix
will be equal to x double dot,
01:04:21.176 --> 01:04:21.950
right?
01:04:21.950 --> 01:04:25.460
Therefore, what this is
actually x double dot?
01:04:25.460 --> 01:04:29.600
It's basically-- in the
current presentation,
01:04:29.600 --> 01:04:33.570
it's actually just partial
square a, partial t square.
01:04:33.570 --> 01:04:34.650
OK?
01:04:34.650 --> 01:04:37.470
Can everybody accept this?
01:04:37.470 --> 01:04:40.440
OK, so now we have
some sensibility
01:04:40.440 --> 01:04:43.400
going from a discrete
system, which
01:04:43.400 --> 01:04:46.560
you have a length scale of
a, to a continuous system,
01:04:46.560 --> 01:04:48.120
because a goes to zero.
01:04:48.120 --> 01:04:52.400
By a certain time, I fix
the ratio of m and a so
01:04:52.400 --> 01:04:56.910
that the system doesn't
grow too infinitely massive.
01:04:56.910 --> 01:05:01.530
OK, so if I do this, then in
short, you get this equation.
01:05:01.530 --> 01:05:05.220
Partial square a
partial t square,
01:05:05.220 --> 01:05:09.660
and that is equal to t
over rho l partial square
01:05:09.660 --> 01:05:13.740
a partial x square.
01:05:13.740 --> 01:05:14.550
OK?
01:05:14.550 --> 01:05:24.190
I can now define vp as equal
to square root of t over rho l.
01:05:24.190 --> 01:05:24.930
OK?
01:05:24.930 --> 01:05:29.750
And this will become the
p square partial square
01:05:29.750 --> 01:05:31.800
a partial x square.
01:05:34.850 --> 01:05:37.310
What is this?
01:05:37.310 --> 01:05:40.460
This is what equation?
01:05:40.460 --> 01:05:43.580
Wave equation!
01:05:43.580 --> 01:05:45.440
OK, you see that now?
01:05:45.440 --> 01:05:49.250
After all the hard
work, OK, going
01:05:49.250 --> 01:05:52.940
from infinity long
systems, discrete systems,
01:05:52.940 --> 01:05:54.810
then go to a
continuous limit, we
01:05:54.810 --> 01:05:59.060
discovered the wave equation.
01:05:59.060 --> 01:06:02.990
This is probably the
most important equation
01:06:02.990 --> 01:06:07.070
you actually learn, until now.
01:06:07.070 --> 01:06:11.120
More important than
f equal to ma, right?
01:06:11.120 --> 01:06:13.390
Because you can
listen to my lecture,
01:06:13.390 --> 01:06:16.850
even, if this equation
didn't exist, right?
01:06:16.850 --> 01:06:20.220
Then I can not propagate
a sound wave to your ear.
01:06:20.220 --> 01:06:22.100
Right?
01:06:22.100 --> 01:06:24.690
And you cannot even
see the black board,
01:06:24.690 --> 01:06:26.810
because the
electromagnetic wave,
01:06:26.810 --> 01:06:30.050
which I will show
in a later lecture,
01:06:30.050 --> 01:06:34.670
also kind of satisfies
the same function or form.
01:06:34.670 --> 01:06:37.280
So I think this is an
achievement and a highlight
01:06:37.280 --> 01:06:38.980
of today's class.
01:06:38.980 --> 01:06:42.110
We actually realize
now, what we have
01:06:42.110 --> 01:06:45.980
been doing is really
solving something
01:06:45.980 --> 01:06:48.920
related to the wave equation.
01:06:48.920 --> 01:06:53.180
And next time, what I would like
to actually discuss with you
01:06:53.180 --> 01:06:56.070
is the solution of
this wave equation.
01:06:56.070 --> 01:06:58.070
So here, I have--
01:06:58.070 --> 01:06:59.930
again, last time
you have seen this.
01:06:59.930 --> 01:07:02.765
This is a coupled
oscillator system.
01:07:02.765 --> 01:07:05.420
It has 72 components.
01:07:05.420 --> 01:07:08.810
As a physicist, that's actually
equal to an infinite number
01:07:08.810 --> 01:07:11.090
of coupled oscillators, OK.
01:07:11.090 --> 01:07:12.560
That's good enough.
01:07:12.560 --> 01:07:18.590
And you can see
that, if I do this,
01:07:18.590 --> 01:07:21.210
it does something
really strange, right?
01:07:21.210 --> 01:07:27.380
You see a progressing
wave going back and forth,
01:07:27.380 --> 01:07:30.290
and it disappears
because of friction.
01:07:30.290 --> 01:07:31.850
OK?
01:07:31.850 --> 01:07:35.270
If I can construct
something closer
01:07:35.270 --> 01:07:39.290
to a perfect coupled
system without friction,
01:07:39.290 --> 01:07:42.470
what is going to happen
is that this wave
01:07:42.470 --> 01:07:47.800
is going to be there bouncing
back and forth forever.
01:07:47.800 --> 01:07:49.790
And that actually
can be understood
01:07:49.790 --> 01:07:52.260
by the wave equation.
01:07:52.260 --> 01:07:58.810
And also, if I oscillate this
system at a fixed frequency,
01:07:58.810 --> 01:08:02.380
you will see that these
become standing waves.
01:08:04.950 --> 01:08:07.780
And of course I can
I do crazy things,
01:08:07.780 --> 01:08:11.070
I can oscillate and stop it
and it becomes really, really
01:08:11.070 --> 01:08:13.050
complicated motion.
01:08:13.050 --> 01:08:15.360
And of course you are
welcome to come here and play
01:08:15.360 --> 01:08:16.500
after the class.
01:08:16.500 --> 01:08:20.370
And next time, on
Thursday, we are
01:08:20.370 --> 01:08:24.540
going to talk about the
solution to this equation
01:08:24.540 --> 01:08:28.710
and how to understand
all kinds of fancy motion
01:08:28.710 --> 01:08:32.220
this system can do,
given by the nature.
01:08:32.220 --> 01:08:34.210
Thank you very much.