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PROFESSOR YEN-JIE LEE:
Welcome back, everybody,
9
00:00:25,300 --> 00:00:28,810
to 8.03 Today, we
are going to continue
10
00:00:28,810 --> 00:00:31,870
the discussion of
symmetry matrix,
11
00:00:31,870 --> 00:00:33,970
which we started last time.
12
00:00:33,970 --> 00:00:37,460
And this is what
we have been doing.
13
00:00:37,460 --> 00:00:39,370
OK?
14
00:00:39,370 --> 00:00:41,050
So the thing which
we have been doing
15
00:00:41,050 --> 00:00:45,480
is to solve the normal mode
frequencies of component
16
00:00:45,480 --> 00:00:48,490
systems by looking at
equation of motion,
17
00:00:48,490 --> 00:00:51,850
MX double dot equal to minus KX.
18
00:00:51,850 --> 00:00:56,740
And in the end of the day,
what we are doing is really
19
00:00:56,740 --> 00:00:59,910
to solve Eigenvalue problem--
20
00:00:59,910 --> 00:01:02,830
M minus one K matrix
Eigenvalue problem.
21
00:01:02,830 --> 00:01:07,770
And we have been exercising
this several times
22
00:01:07,770 --> 00:01:10,570
the last few lectures.
23
00:01:10,570 --> 00:01:12,617
And the M minus
one K matrix says
24
00:01:12,617 --> 00:01:16,540
you're describing how each
component in the system
25
00:01:16,540 --> 00:01:18,110
interact with each other.
26
00:01:18,110 --> 00:01:18,610
OK?
27
00:01:18,610 --> 00:01:21,760
So that's actually what
we are trying to do,
28
00:01:21,760 --> 00:01:24,310
to solve the normal modes.
29
00:01:24,310 --> 00:01:28,420
So we have been making
progress, and we
30
00:01:28,420 --> 00:01:31,330
are increasing the number
of coupled oscillators
31
00:01:31,330 --> 00:01:32,630
as a function of time.
32
00:01:32,630 --> 00:01:35,230
And now we finally arrive
at our limit, which
33
00:01:35,230 --> 00:01:36,990
is infinite system, right?
34
00:01:36,990 --> 00:01:39,310
So basically, what we
have been discussing
35
00:01:39,310 --> 00:01:43,960
is a special kind of
infinite system, which
36
00:01:43,960 --> 00:01:47,470
actually satisfies
translation symmetry, right?
37
00:01:47,470 --> 00:01:51,730
So in general, we don't know
how to solve infinite system.
38
00:01:51,730 --> 00:01:53,490
If this system is
really complicated,
39
00:01:53,490 --> 00:01:57,730
no symmetry, then who knows
how to solve this, right?
40
00:01:57,730 --> 00:02:01,330
But very luckily,
in 8.03 we have
41
00:02:01,330 --> 00:02:06,340
started to get a highly
symmetrical infinite system,
42
00:02:06,340 --> 00:02:09,580
and in this case, it's
translation symmetry.
43
00:02:09,580 --> 00:02:13,390
And that is really pretty
nice, and we can actually
44
00:02:13,390 --> 00:02:18,490
use this example to learn
an interesting fact, which
45
00:02:18,490 --> 00:02:22,380
we can see from the physical
system we discuss here.
46
00:02:22,380 --> 00:02:26,440
So, one thing which we
have been discussing
47
00:02:26,440 --> 00:02:30,420
is the space translation
symmetry matrix.
48
00:02:30,420 --> 00:02:35,800
As a reminder, S
matrix, as defined here.
49
00:02:35,800 --> 00:02:39,460
Basically, if you
have a vector A, which
50
00:02:39,460 --> 00:02:43,910
describes the amplitude
of an individual component
51
00:02:43,910 --> 00:02:47,500
in a system, basically
these vectors
52
00:02:47,500 --> 00:02:51,700
have Aj, Aj plus one, Aj
plus two blah blah blah.
53
00:02:51,700 --> 00:02:58,520
All those amplitudes are
included in this vector.
54
00:02:58,520 --> 00:03:01,310
And what does this S matrix do?
55
00:03:01,310 --> 00:03:03,790
It's actually the following--
56
00:03:03,790 --> 00:03:08,200
so if A prime is
equal to S times A,
57
00:03:08,200 --> 00:03:12,910
then the base
component of A prime
58
00:03:12,910 --> 00:03:16,100
will be equal to Aj plus one.
59
00:03:16,100 --> 00:03:20,890
OK, so that's actually what
this S matrix does to A vector.
60
00:03:20,890 --> 00:03:21,550
OK?
61
00:03:21,550 --> 00:03:25,230
So you can think about S
metrics as an operator.
62
00:03:25,230 --> 00:03:28,840
It's actually picking
up the Aj plus
63
00:03:28,840 --> 00:03:33,080
one component, and moving
it to Aj in the new vector.
64
00:03:33,080 --> 00:03:33,580
OK?
65
00:03:33,580 --> 00:03:37,180
So that's actually what
this S matrix does.
66
00:03:37,180 --> 00:03:42,790
And we were talking about
the Eigenvector of S matrix.
67
00:03:42,790 --> 00:03:47,470
For example, if A is an
Eigenvector of S matrix,
68
00:03:47,470 --> 00:03:52,000
then we have this relation,
SA equal to beta A. Now,
69
00:03:52,000 --> 00:03:57,830
beta is actually the
Eigenvalue of S matrix.
70
00:03:57,830 --> 00:03:58,600
OK?
71
00:03:58,600 --> 00:04:02,380
And then we also discussed
last time, that means,
72
00:04:02,380 --> 00:04:05,680
based on this logical
extension, Aj prime
73
00:04:05,680 --> 00:04:08,170
would be equal to
Aj plus one based
74
00:04:08,170 --> 00:04:10,300
on the definition of S matrix.
75
00:04:10,300 --> 00:04:14,320
And now this is going to be
equal to beta times Aj, right?
76
00:04:14,320 --> 00:04:21,450
Because we assume that that A
vector is an Eigenvector of S
77
00:04:21,450 --> 00:04:21,950
matrix.
78
00:04:21,950 --> 00:04:23,050
OK?
79
00:04:23,050 --> 00:04:28,340
That means Aj will be equal
to beta to the j A zero.
80
00:04:28,340 --> 00:04:29,320
Right?
81
00:04:29,320 --> 00:04:33,550
According to this relation.
82
00:04:33,550 --> 00:04:36,460
Therefore, we can
conclude that Aj will be
83
00:04:36,460 --> 00:04:39,460
proportional to beta to the j.
84
00:04:39,460 --> 00:04:43,000
OK, beta is still
some coefficient,
85
00:04:43,000 --> 00:04:48,020
which we have not determined,
and it can be anything.
86
00:04:48,020 --> 00:04:49,180
OK?
87
00:04:49,180 --> 00:04:56,470
So, in order to consider a
system of little masses, which
88
00:04:56,470 --> 00:04:58,760
are oscillating up
and down instead
89
00:04:58,760 --> 00:05:05,580
of going in one direction
forever or the amplitude
90
00:05:05,580 --> 00:05:08,770
grows exponentially.
91
00:05:08,770 --> 00:05:11,740
And we also don't
want the system
92
00:05:11,740 --> 00:05:14,720
to have the amplitude
go to infinity
93
00:05:14,720 --> 00:05:19,360
when we go to very very large
j value or very small j value,
94
00:05:19,360 --> 00:05:22,633
therefore we limit
our discussion
95
00:05:22,633 --> 00:05:28,440
in the case of beta
equal to exponential ika.
96
00:05:28,440 --> 00:05:29,030
OK?
97
00:05:29,030 --> 00:05:35,570
In this case, the absolute
value of beta is one.
98
00:05:35,570 --> 00:05:36,660
Right?
99
00:05:36,660 --> 00:05:40,160
OK, and that means--
100
00:05:40,160 --> 00:05:44,990
OK, Aj is actually proportional
to beta to Aj, right?
101
00:05:44,990 --> 00:05:49,040
So if you take a ratio of
Aj plus one and the Aj,
102
00:05:49,040 --> 00:05:51,160
the ratio is beta.
103
00:05:51,160 --> 00:05:55,100
If beta Is not equal to
one or its absolute value
104
00:05:55,100 --> 00:05:59,180
is not equal to one,
then this Aj value
105
00:05:59,180 --> 00:06:02,540
is going to be
increasing according
106
00:06:02,540 --> 00:06:04,220
to a power law, right?
107
00:06:04,220 --> 00:06:06,350
So the amplitude is
going to be, whoa,
108
00:06:06,350 --> 00:06:08,930
going to a very, very
large value, right?
109
00:06:08,930 --> 00:06:11,810
And that corresponds to some
kind of physical system,
110
00:06:11,810 --> 00:06:14,060
but not corresponding
to oscillation.
111
00:06:14,060 --> 00:06:15,710
OK?
112
00:06:15,710 --> 00:06:19,010
And if we do this
and assume that beta
113
00:06:19,010 --> 00:06:23,120
is equal to exponential ika.
114
00:06:23,120 --> 00:06:28,220
OK, if we do this, what is going
to happen is the following.
115
00:06:28,220 --> 00:06:32,390
So again, the ratio
of Aj plus one and Aj
116
00:06:32,390 --> 00:06:39,410
is a fixed value beta,
equal to exponential ika.
117
00:06:39,410 --> 00:06:42,980
But what it does is, instead
of changing the amplitude,
118
00:06:42,980 --> 00:06:47,911
it's actually doing a
rotation in the complex plane.
119
00:06:47,911 --> 00:06:48,410
OK?
120
00:06:48,410 --> 00:06:54,080
So if this is Aj, J plus
one, J plus two, et cetera,
121
00:06:54,080 --> 00:06:57,850
as a function of say j, the
variable, then what it does
122
00:06:57,850 --> 00:07:01,810
is that this operation
multiplied by beta
123
00:07:01,810 --> 00:07:07,640
is really a rotation
in the complex plane.
124
00:07:07,640 --> 00:07:11,240
And while we actually see
in the physical system,
125
00:07:11,240 --> 00:07:15,620
it's actually a projection of
this complex imaginary plane
126
00:07:15,620 --> 00:07:17,990
to the real axis.
127
00:07:17,990 --> 00:07:21,610
And that would give you a sine
function and cosine function.
128
00:07:21,610 --> 00:07:25,760
So very interestingly,
if we choose wisely,
129
00:07:25,760 --> 00:07:29,030
the beta to have
absolute value of one,
130
00:07:29,030 --> 00:07:36,020
then that will give you a system
which is actually oscillating
131
00:07:36,020 --> 00:07:37,850
up and down and the
amplitude is actually
132
00:07:37,850 --> 00:07:40,730
confined within some value.
133
00:07:40,730 --> 00:07:43,340
OK, so that is actually
quite interesting.
134
00:07:43,340 --> 00:07:46,760
The other thing which I
would like to talk about is--
135
00:07:46,760 --> 00:07:52,690
OK, I choose to have
exponential ika as my beta.
136
00:07:52,690 --> 00:07:54,890
OK, why ka, right?
137
00:07:54,890 --> 00:07:56,620
It looks really strange here.
138
00:07:56,620 --> 00:08:00,140
Suddenly the k and and the
a are coming to play, right?
139
00:08:00,140 --> 00:08:02,591
So what is actually the small a?
140
00:08:02,591 --> 00:08:09,440
The small a is actually the
distance between little mass,
141
00:08:09,440 --> 00:08:11,360
just a reminder.
142
00:08:11,360 --> 00:08:13,520
So this is actually
the length scale
143
00:08:13,520 --> 00:08:16,410
between all those
little mass, right?
144
00:08:16,410 --> 00:08:21,950
Therefore, what I'm doing is to
factorize out the length scale,
145
00:08:21,950 --> 00:08:25,620
and then we suddenly found that.
146
00:08:25,620 --> 00:08:31,520
OK, after I do this, I define
beta equal to exponential ika
147
00:08:31,520 --> 00:08:33,980
Instead of exponential
i theta, right?
148
00:08:33,980 --> 00:08:36,980
So you can also do
exponential i theta, right?
149
00:08:36,980 --> 00:08:40,309
But instead, I gave
theta a fancy name,
150
00:08:40,309 --> 00:08:42,850
which is k times
a. a is actually
151
00:08:42,850 --> 00:08:45,940
the distance between mass.
152
00:08:45,940 --> 00:08:49,090
Then something interesting
happens, because k suddenly
153
00:08:49,090 --> 00:08:51,000
also will have a meaning.
154
00:08:51,000 --> 00:09:00,660
It's actually the wave number
of the resulting sine wave.
155
00:09:00,660 --> 00:09:01,160
OK?
156
00:09:01,160 --> 00:09:03,980
So that's actually
why we actually
157
00:09:03,980 --> 00:09:08,050
factor out this a factor.
158
00:09:08,050 --> 00:09:10,340
Then, after that, we
actually found out
159
00:09:10,340 --> 00:09:12,210
the amplitude would
be proportional
160
00:09:12,210 --> 00:09:16,580
to exponential ijka,
and j is actually
161
00:09:16,580 --> 00:09:20,480
just a label of a phase
component in the system.
162
00:09:20,480 --> 00:09:22,330
OK.
163
00:09:22,330 --> 00:09:27,890
So, once we have solved
the Eigenvalue problem
164
00:09:27,890 --> 00:09:32,360
for the symmetry matrix,
S, as we discussed before,
165
00:09:32,360 --> 00:09:35,720
if you look at the slides, OK?
166
00:09:35,720 --> 00:09:40,430
If S and the M minus one
K matrix, they commute.
167
00:09:40,430 --> 00:09:41,720
OK?
168
00:09:41,720 --> 00:09:45,860
Commutes means that you can
actually change S and and the M
169
00:09:45,860 --> 00:09:48,950
minus one matrix, you
can swap them, OK,
170
00:09:48,950 --> 00:09:50,735
when you multiply them together.
171
00:09:50,735 --> 00:09:51,740
OK?
172
00:09:51,740 --> 00:09:55,190
If you can swap S and
M minus one K matrix,
173
00:09:55,190 --> 00:09:57,680
that means they commute.
174
00:09:57,680 --> 00:10:00,740
And our conclusion
from last time
175
00:10:00,740 --> 00:10:06,230
is that they will share
the same Eigenvectors.
176
00:10:06,230 --> 00:10:07,130
OK?
177
00:10:07,130 --> 00:10:09,910
Of course, not necessarily
the same Eigenvalue,
178
00:10:09,910 --> 00:10:11,990
but they share the
same Eigenvector.
179
00:10:11,990 --> 00:10:13,400
So that's great news!
180
00:10:13,400 --> 00:10:17,330
Because instead of solving
M minus one K matrix, which
181
00:10:17,330 --> 00:10:19,820
can be really
complicated; depends
182
00:10:19,820 --> 00:10:23,490
on what kind of physical
system you are talking about.
183
00:10:23,490 --> 00:10:28,060
I can solve S matrix
Eigenvalue problem.
184
00:10:28,060 --> 00:10:29,310
OK?
185
00:10:29,310 --> 00:10:34,560
And then this Eigenvector,
which I just found here,
186
00:10:34,560 --> 00:10:41,240
is going to be the Eigenvector
of M minus one K matrix.
187
00:10:41,240 --> 00:10:45,110
And that's actually really
making things much easier,
188
00:10:45,110 --> 00:10:49,460
because now instead of solving
Eigenvalue problem of M
189
00:10:49,460 --> 00:10:52,210
minus one K matrix,
what I am doing
190
00:10:52,210 --> 00:10:55,550
is just multiplying M
minus one K times A,
191
00:10:55,550 --> 00:11:00,740
then you can actually obtain
the normal mode frequency omega.
192
00:11:00,740 --> 00:11:01,310
OK?
193
00:11:01,310 --> 00:11:03,750
So that's the issue of
the great excitement.
194
00:11:03,750 --> 00:11:05,210
What does that mean?
195
00:11:05,210 --> 00:11:09,370
That means, if you have
all kinds of systems, which
196
00:11:09,370 --> 00:11:13,200
are translation symmetric--
you can have a line,
197
00:11:13,200 --> 00:11:17,960
you have however
many people together,
198
00:11:17,960 --> 00:11:20,540
whatever a system
which is so on here.
199
00:11:20,540 --> 00:11:25,330
They are all going to
have the same Eigenvector.
200
00:11:25,330 --> 00:11:25,830
You see?
201
00:11:25,830 --> 00:11:27,290
You already know
how they are going
202
00:11:27,290 --> 00:11:30,020
to interact with each
other, and what is actually
203
00:11:30,020 --> 00:11:35,010
the amplitude as a function of
j, which is the location label.
204
00:11:35,010 --> 00:11:35,850
OK?
205
00:11:35,850 --> 00:11:39,160
So that's actually
really wonderful.
206
00:11:39,160 --> 00:11:43,040
So, in order to help
you with understanding
207
00:11:43,040 --> 00:11:46,490
of this system some
more, we are going
208
00:11:46,490 --> 00:11:51,290
to discuss another system
which is actually also
209
00:11:51,290 --> 00:11:52,870
very interesting.
210
00:11:52,870 --> 00:11:55,330
It's actually a spring--
211
00:11:55,330 --> 00:11:59,540
OK, last time we discussed a
spring and mass system, right?
212
00:11:59,540 --> 00:12:01,920
And then we solved it together.
213
00:12:01,920 --> 00:12:05,210
And this time we are going
to solve a system which
214
00:12:05,210 --> 00:12:08,060
is made of mass and strings.
215
00:12:08,060 --> 00:12:08,900
Ok?
216
00:12:08,900 --> 00:12:11,680
So that may actually copy--
217
00:12:11,680 --> 00:12:13,640
OK, let me actually
introduce you
218
00:12:13,640 --> 00:12:17,770
to this new system we are
going to talk about today.
219
00:12:17,770 --> 00:12:23,060
It has many little mass
here, from left hand side
220
00:12:23,060 --> 00:12:25,610
of the universe and right
hand side of the universe.
221
00:12:25,610 --> 00:12:28,740
Take forever to actually
construct this system.
222
00:12:28,740 --> 00:12:29,430
OK?
223
00:12:29,430 --> 00:12:33,860
Then my student actually
carefully link them by strings,
224
00:12:33,860 --> 00:12:38,750
and we make sure that the
string tension, OK, is actually
225
00:12:38,750 --> 00:12:41,630
a fixed value, which is T. OK.
226
00:12:41,630 --> 00:12:44,090
T is actually the
string tension.
227
00:12:44,090 --> 00:12:47,300
And of course, as what
we discussed before,
228
00:12:47,300 --> 00:12:50,570
the space or the distance
between little mass
229
00:12:50,570 --> 00:12:53,240
is actually A. OK?
230
00:12:53,240 --> 00:12:54,860
So that's, again,
the same length
231
00:12:54,860 --> 00:12:57,080
scale, which we were using.
232
00:12:57,080 --> 00:12:59,560
And finally, in
order to describe
233
00:12:59,560 --> 00:13:04,740
all those little
masses in the system,
234
00:13:04,740 --> 00:13:10,020
I label them as j, j plus one, j
plus two, et cetera, et cetera.
235
00:13:10,020 --> 00:13:11,270
OK.
236
00:13:11,270 --> 00:13:15,470
So, the question
we are asking is,
237
00:13:15,470 --> 00:13:20,190
what would be the resulting
motion of this system, right?
238
00:13:20,190 --> 00:13:24,650
So what we can do is what we
have done last time, right?
239
00:13:24,650 --> 00:13:28,820
We take jth object
in this system,
240
00:13:28,820 --> 00:13:34,520
and we look at the force
diagram and this M minus one--
241
00:13:34,520 --> 00:13:37,240
to write down the equational
motion and also the
242
00:13:37,240 --> 00:13:39,650
and M minus one K metrics.
243
00:13:39,650 --> 00:13:49,130
So what is actually the force
diagram of j's component?
244
00:13:52,010 --> 00:13:52,510
OK.
245
00:13:52,510 --> 00:13:54,495
If I take this--
246
00:13:54,495 --> 00:14:02,200
so first, I take my jth
mass, and it's connected
247
00:14:02,200 --> 00:14:05,560
to two strings, right?
248
00:14:05,560 --> 00:14:10,150
So there are two strings
connected to jth mass.
249
00:14:10,150 --> 00:14:10,960
OK?
250
00:14:10,960 --> 00:14:13,810
And of course,
left hand side you
251
00:14:13,810 --> 00:14:17,470
have another mass, right hand
side you have another option.
252
00:14:17,470 --> 00:14:19,660
OK?
253
00:14:19,660 --> 00:14:24,380
And I know the tension
of this string.
254
00:14:24,380 --> 00:14:27,160
Its actually a fixed
value, and the string
255
00:14:27,160 --> 00:14:31,510
tension is fixed at the T. OK?
256
00:14:31,510 --> 00:14:33,460
So, in order to
describe this system,
257
00:14:33,460 --> 00:14:35,960
I need to find my
coordinate system, right?
258
00:14:35,960 --> 00:14:36,810
As usual.
259
00:14:36,810 --> 00:14:38,560
So, what is actually
the coordinate system
260
00:14:38,560 --> 00:14:40,010
I'm going to use.?
261
00:14:40,010 --> 00:14:42,520
So I need to define
horizontal direction
262
00:14:42,520 --> 00:14:48,460
to be x, and the vertical
direction to be y.
263
00:14:48,460 --> 00:14:52,560
Therefore, I cannot express
this little mass to be--
264
00:14:52,560 --> 00:14:58,580
the position of jth little
mass as Xj and the Yj.
265
00:15:01,540 --> 00:15:03,040
OK?
266
00:15:03,040 --> 00:15:07,350
We can do the same thing
for the left-hand side mass
267
00:15:07,350 --> 00:15:16,567
is Xa minus one, Yj minus one,
and Xj plus one Yj plus one.
268
00:15:19,250 --> 00:15:22,870
So to simplify the discussion,
what I'm going to assume
269
00:15:22,870 --> 00:15:28,930
is that all those little masses
can only move up and down, OK,
270
00:15:28,930 --> 00:15:30,580
instead of back and forth.
271
00:15:30,580 --> 00:15:33,940
OK, so only one
direction is allowed,
272
00:15:33,940 --> 00:15:38,520
and also I assume that
the up and down motion
273
00:15:38,520 --> 00:15:42,640
of all those little masses
is really really very small.
274
00:15:42,640 --> 00:15:46,450
So I can use small
angle approximation.
275
00:15:46,450 --> 00:15:49,160
OK, so therefore I can--
276
00:15:49,160 --> 00:15:57,040
this is essentially the zero
of Y axis, when I actually--
277
00:15:57,040 --> 00:15:59,350
before I actually move--
278
00:15:59,350 --> 00:16:02,720
wind up with the mass away
from the equilibrium position.
279
00:16:02,720 --> 00:16:07,720
So when the string mass
system is at rest, OK,
280
00:16:07,720 --> 00:16:14,800
not really moving, then all the
mass are at Y equal to zero.
281
00:16:14,800 --> 00:16:16,420
OK?
282
00:16:16,420 --> 00:16:23,880
So now, I actually move
this mass to Yj, OK?
283
00:16:23,880 --> 00:16:28,960
Then, apparently there are two
forces acting on this mass.
284
00:16:28,960 --> 00:16:31,060
The left hand side
is string force,
285
00:16:31,060 --> 00:16:32,740
and the right hand
side is string force.
286
00:16:32,740 --> 00:16:38,300
And the magnitude of the
force is actually T. OK?
287
00:16:38,300 --> 00:16:42,590
So, in order to help us
with solving this problem.
288
00:16:42,590 --> 00:16:44,360
I would define two angles--
289
00:16:44,360 --> 00:16:50,140
the left hand side angle to be
theta one, and the right hand
290
00:16:50,140 --> 00:16:52,540
side angle to be theta two.
291
00:16:52,540 --> 00:16:56,320
Then I can now write down
the equation of motion
292
00:16:56,320 --> 00:16:59,800
in the horizontal direction, and
then in the vertical direction.
293
00:16:59,800 --> 00:17:01,960
OK?
294
00:17:01,960 --> 00:17:02,710
All right.
295
00:17:02,710 --> 00:17:06,280
Since I have assumed
that all of those masses
296
00:17:06,280 --> 00:17:10,089
can only move up and down,
now I mean-- and also
297
00:17:10,089 --> 00:17:13,880
the displacement with respect
to the equilibrium position,
298
00:17:13,880 --> 00:17:18,400
which is Y equal to zero, is
really small, compared to,
299
00:17:18,400 --> 00:17:21,400
for example, the length scale a.
300
00:17:21,400 --> 00:17:21,970
OK?
301
00:17:21,970 --> 00:17:25,569
So therefore, I can write
that condition explicitly.
302
00:17:25,569 --> 00:17:27,310
So I have a condition.
303
00:17:27,310 --> 00:17:32,680
I assume that Yj is actually
much, much smaller than a,
304
00:17:32,680 --> 00:17:36,670
which is the distance
between those masses.
305
00:17:36,670 --> 00:17:41,530
And that means theta
one and theta two
306
00:17:41,530 --> 00:17:47,320
are going to be much,
much smaller than one.
307
00:17:47,320 --> 00:17:47,860
OK?
308
00:17:47,860 --> 00:17:49,930
So that's actually
given us a chance
309
00:17:49,930 --> 00:17:53,240
to use small angle
approximation.
310
00:17:53,240 --> 00:17:55,660
So based on this
force diagram, I
311
00:17:55,660 --> 00:17:59,070
can now write down the
two equations of motion;
312
00:17:59,070 --> 00:18:01,220
one in the horizontal
direction, the other one
313
00:18:01,220 --> 00:18:02,960
in the vertical direction.
314
00:18:02,960 --> 00:18:09,670
So what I'm going to get
in the horizontal direction
315
00:18:09,670 --> 00:18:14,530
is M X j double dot.
316
00:18:14,530 --> 00:18:16,000
These will be equal to--
317
00:18:16,000 --> 00:18:17,480
OK, there are two forces.
318
00:18:17,480 --> 00:18:17,980
Right?
319
00:18:17,980 --> 00:18:19,990
Horizontal direction,
I would need
320
00:18:19,990 --> 00:18:23,320
to calculate the projection
to the X direction,
321
00:18:23,320 --> 00:18:25,310
therefore the left
hand side force
322
00:18:25,310 --> 00:18:30,970
will give you minus
T cosine theta one,
323
00:18:30,970 --> 00:18:33,620
and the right hand side
of this string force
324
00:18:33,620 --> 00:18:40,500
is going to give you
plus T cosine theta two.
325
00:18:40,500 --> 00:18:42,160
OK?
326
00:18:42,160 --> 00:18:48,020
And then in the vertical
direction, what I'm
327
00:18:48,020 --> 00:18:51,750
going to get is Myj double dot.
328
00:18:51,750 --> 00:18:57,210
This is equal to
minus T sine theta y.
329
00:18:57,210 --> 00:19:01,220
Now I'm doing the
projection in the y
330
00:19:01,220 --> 00:19:10,030
axis in the vertical direction,
and minus T sine theta two.
331
00:19:10,030 --> 00:19:11,510
OK.
332
00:19:11,510 --> 00:19:14,960
And since I have this
condition, all the mass
333
00:19:14,960 --> 00:19:20,810
can only move up and down,
and also the displacement
334
00:19:20,810 --> 00:19:25,820
is much, much smaller
than a; therefore, I
335
00:19:25,820 --> 00:19:28,080
have the small
angle approximation.
336
00:19:28,080 --> 00:19:33,440
Cosine theta is
roughly equal to one,
337
00:19:33,440 --> 00:19:38,700
and the sine theta is
roughly equal to theta.
338
00:19:38,700 --> 00:19:39,410
OK?
339
00:19:39,410 --> 00:19:42,440
And I would call this
the equation number one,
340
00:19:42,440 --> 00:19:45,590
and the second equation
in the vertical direction
341
00:19:45,590 --> 00:19:48,920
to be equation number two.
342
00:19:48,920 --> 00:19:50,900
OK.
343
00:19:50,900 --> 00:19:54,710
So, up to here, everything
is essentially exact,
344
00:19:54,710 --> 00:19:58,040
and now I would like to make
a small angle approximation
345
00:19:58,040 --> 00:19:59,570
and see what will happen.
346
00:19:59,570 --> 00:20:01,860
And now equation
number one will be
347
00:20:01,860 --> 00:20:07,820
called MXa double dot
equal to minus T plus T,
348
00:20:07,820 --> 00:20:10,580
and this is equal to zero.
349
00:20:10,580 --> 00:20:15,710
OK, so that means we will
not have horizontal direction
350
00:20:15,710 --> 00:20:16,400
acceleration.
351
00:20:16,400 --> 00:20:18,980
Therefore, in the
horizontal direction,
352
00:20:18,980 --> 00:20:21,740
there will be no
acceleration and therefore
353
00:20:21,740 --> 00:20:26,900
no movement in the X direction,
or horizontal direction.
354
00:20:26,900 --> 00:20:31,190
OK, and now I can take a look
at the vertical direction.
355
00:20:31,190 --> 00:20:32,630
OK?
356
00:20:32,630 --> 00:20:38,180
So basically, what I am going
to get is MYj double dot,
357
00:20:38,180 --> 00:20:41,670
and this will be
equal to minus T. Now,
358
00:20:41,670 --> 00:20:47,450
sine theta one will be
roughly equal to theta one.
359
00:20:47,450 --> 00:20:49,460
So what is actually theta one?
360
00:20:49,460 --> 00:20:52,110
Theta one is going to be--
361
00:20:52,110 --> 00:21:00,010
OK, so this is actually
the Yj minus Yj minus one.
362
00:21:00,010 --> 00:21:01,169
Right?
363
00:21:01,169 --> 00:21:02,710
So now that's actually
the difference
364
00:21:02,710 --> 00:21:10,140
between the amplitude of the
displacement of the mass j,
365
00:21:10,140 --> 00:21:13,030
and the displacement
of mass j minus one.
366
00:21:13,030 --> 00:21:13,770
Right?
367
00:21:13,770 --> 00:21:16,980
Divided by a, I get theta one.
368
00:21:16,980 --> 00:21:17,940
Right?
369
00:21:17,940 --> 00:21:20,480
Therefore, the first
sine theta one,
370
00:21:20,480 --> 00:21:29,430
will become Yj minus Yj
minus one, divided by a.
371
00:21:29,430 --> 00:21:30,820
OK?
372
00:21:30,820 --> 00:21:34,250
And of course, you can do the
same thing for the sine theta
373
00:21:34,250 --> 00:21:35,320
two, right?
374
00:21:35,320 --> 00:21:38,735
Which is actually
just the theta two.
375
00:21:38,735 --> 00:21:40,360
Then, basically, what
I am going to get
376
00:21:40,360 --> 00:21:51,330
is minus Tyj minus Yj
plus one divided by a.
377
00:21:51,330 --> 00:21:53,110
OK.
378
00:21:53,110 --> 00:21:54,128
Any questions?
379
00:21:56,880 --> 00:21:57,500
OK.
380
00:21:57,500 --> 00:21:59,360
I hope everybody
is following it.
381
00:21:59,360 --> 00:22:03,740
OK, so now I can actually
simplify equation number two,
382
00:22:03,740 --> 00:22:07,850
and basically, what I
am going to get is MYj
383
00:22:07,850 --> 00:22:12,670
double dot will be
equal to minus T over a.
384
00:22:12,670 --> 00:22:17,700
Basically, I take T
over a out of it again.
385
00:22:17,700 --> 00:22:23,180
And I also collect all the
terms related to Yj minus one,
386
00:22:23,180 --> 00:22:30,850
minus two Yj, plus Yj plus one.
387
00:22:30,850 --> 00:22:34,690
OK, I just ask you to
rewrite equation number two
388
00:22:34,690 --> 00:22:40,800
in a form which we like more.
389
00:22:40,800 --> 00:22:45,110
OK, so that is actually the
equation of motion, so from now
390
00:22:45,110 --> 00:22:47,670
on, I am going to
ignore all the motion
391
00:22:47,670 --> 00:22:51,920
in the horizontal direction,
because in this small angle
392
00:22:51,920 --> 00:22:54,545
approximation we have
shown you that there
393
00:22:54,545 --> 00:22:58,700
will be no acceleration
in the x direction, right?
394
00:22:58,700 --> 00:23:04,520
So now it's actually getting
a step forward again.
395
00:23:04,520 --> 00:23:06,470
Basically, we have the
equation of motion,
396
00:23:06,470 --> 00:23:09,240
and what is usually
the next step?
397
00:23:09,240 --> 00:23:12,510
The next step is to
write down what matrix?
398
00:23:12,510 --> 00:23:14,384
Anybody can help me?
399
00:23:14,384 --> 00:23:15,960
STUDENT 1: M minus one k matrix.
400
00:23:15,960 --> 00:23:18,085
PROFESSOR YEN-JIE LEE: M
minus one k matrix, right?
401
00:23:18,085 --> 00:23:21,080
So actually, as usual, we
actually follow the procedure.
402
00:23:21,080 --> 00:23:25,020
Now I would like to write
down the m minus one k matrix.
403
00:23:25,020 --> 00:23:28,960
OK, so before I do that
what, I will define--
404
00:23:28,960 --> 00:23:35,600
I will actually assume my normal
mode has this functional form,
405
00:23:35,600 --> 00:23:43,850
yj is equal to the real part
of aj exponential i omega t
406
00:23:43,850 --> 00:23:45,500
plus phi.
407
00:23:45,500 --> 00:23:48,200
So basically, that tells
you that all the components
408
00:23:48,200 --> 00:23:51,710
are oscillating at the
same frequency, omega,
409
00:23:51,710 --> 00:23:53,990
and the same phase, phi.
410
00:23:53,990 --> 00:23:55,338
Yes.
411
00:23:55,338 --> 00:23:58,124
STUDENT 2: When
there's a scenario--
412
00:23:58,124 --> 00:23:59,290
PROFESSOR YEN-JIE LEE: Yeah?
413
00:23:59,290 --> 00:24:01,266
STUDENT 2: [INAUDIBLE]
also [INAUDIBLE]..
414
00:24:01,266 --> 00:24:02,972
PROFESSOR YEN-JIE LEE: This one?
415
00:24:02,972 --> 00:24:07,578
STUDENT 2: No, there
should be one [INAUDIBLE]..
416
00:24:07,578 --> 00:24:08,744
PROFESSOR YEN-JIE LEE: Here?
417
00:24:08,744 --> 00:24:09,225
STUDENT 2: Yeah.
418
00:24:09,225 --> 00:24:10,308
PROFESSOR YEN-JIE LEE: Ah.
419
00:24:10,308 --> 00:24:12,524
OK, maybe I made a
mistake somewhere.
420
00:24:22,690 --> 00:24:26,880
Yeah I think it
should be plus, right?
421
00:24:26,880 --> 00:24:27,380
OK.
422
00:24:30,380 --> 00:24:31,117
All right.
423
00:24:31,117 --> 00:24:31,950
Thank you very much.
424
00:24:34,750 --> 00:24:36,850
So now we have all
the ingredients
425
00:24:36,850 --> 00:24:41,500
and we assume that it has
a normal mode of aj, yj
426
00:24:41,500 --> 00:24:43,130
in this functional form.
427
00:24:43,130 --> 00:24:44,350
And then now I can--
428
00:24:44,350 --> 00:24:47,858
the next step is to get
m minus one k matrix.
429
00:24:47,858 --> 00:24:49,100
All right.
430
00:24:49,100 --> 00:24:50,850
So what is actually m metrics?
431
00:24:50,850 --> 00:24:53,050
M matrix is really
really straightforward.
432
00:24:53,050 --> 00:25:00,940
It's m, m, m in
the diagonal terms,
433
00:25:00,940 --> 00:25:02,960
and all the rest of
the terms are zero.
434
00:25:02,960 --> 00:25:04,130
All right.
435
00:25:04,130 --> 00:25:05,980
And those, of
course, you can also
436
00:25:05,980 --> 00:25:09,940
write down k matrix, right?
437
00:25:09,940 --> 00:25:14,004
So the k matrix
will be equal to--
438
00:25:14,004 --> 00:25:17,020
there are many terms,
and in the middle
439
00:25:17,020 --> 00:25:25,180
you have minus t over a, two
t over a, minus t over a,
440
00:25:25,180 --> 00:25:27,780
and all the rest of
the terms are zero.
441
00:25:27,780 --> 00:25:31,030
And of course these
patterns go on and on.
442
00:25:31,030 --> 00:25:37,780
Minus t over a, two t over A,
minus t over a and zeros, et
443
00:25:37,780 --> 00:25:38,740
cetera et cetera.
444
00:25:38,740 --> 00:25:43,790
And this pattern is going to
go on forever, because this
445
00:25:43,790 --> 00:25:47,650
is actually infinitly long
matrix, infinite times
446
00:25:47,650 --> 00:25:49,540
infinitely long matrix.
447
00:25:53,200 --> 00:25:58,490
OK, so once we have this,
we can now write down
448
00:25:58,490 --> 00:26:01,916
the m minus one k matrix.
449
00:26:01,916 --> 00:26:04,845
What is actually the
m minus one k matrix?
450
00:26:04,845 --> 00:26:10,840
It has a similar structure
to k matrix, right?
451
00:26:10,840 --> 00:26:12,600
All those are zeros--
452
00:26:12,600 --> 00:26:14,520
OK, all those are
the other values,
453
00:26:14,520 --> 00:26:24,380
but it has a fixed structure
minus t over ma, two t over ma,
454
00:26:24,380 --> 00:26:29,660
minus t over ma, and then zeros.
455
00:26:29,660 --> 00:26:35,960
And this will go on forever
in the diagonal term and also
456
00:26:35,960 --> 00:26:37,640
the next two diagonal terms.
457
00:26:37,640 --> 00:26:41,520
And all the rest of
the terms. are zero.
458
00:26:41,520 --> 00:26:44,155
OK, Any questions?
459
00:26:48,770 --> 00:26:50,750
OK, so that's really nice.
460
00:26:50,750 --> 00:26:53,240
Now we have our m
minus one k matrix,
461
00:26:53,240 --> 00:26:54,980
and the good news
is that you don't
462
00:26:54,980 --> 00:26:58,790
have to solve m minus one k
matrix's Eigenvalue again,
463
00:26:58,790 --> 00:26:59,360
right?
464
00:26:59,360 --> 00:27:01,880
Because we have solved
the Eigenvalue problem
465
00:27:01,880 --> 00:27:05,540
of s matrix, therefore
what is our left over
466
00:27:05,540 --> 00:27:11,280
is to multiply m minus
one k by a, right?
467
00:27:11,280 --> 00:27:19,130
a is actually one of the
Eigenvectors of s matrix.
468
00:27:19,130 --> 00:27:23,900
OK, so I am going to
multiply that for you,
469
00:27:23,900 --> 00:27:30,010
and now we calculate m minus
one k equal to omega square a,
470
00:27:30,010 --> 00:27:34,480
then I can get omega square
out of this calculation.
471
00:27:34,480 --> 00:27:36,740
OK?
472
00:27:36,740 --> 00:27:39,920
If I again focus on this term.
473
00:27:42,830 --> 00:27:44,770
OK, so basically,
what I am going to get
474
00:27:44,770 --> 00:27:51,880
is, right hand side, I
have omega square aj, OK?
475
00:27:51,880 --> 00:27:55,420
Now that's actually from
the right hand side, OK?
476
00:27:55,420 --> 00:28:00,940
And left hand side m minus one
k times a, what I'm going to get
477
00:28:00,940 --> 00:28:16,641
is t over ma minus aj minus one
plus two aj minus aj plus one.
478
00:28:16,641 --> 00:28:17,140
All right?
479
00:28:17,140 --> 00:28:20,350
Because if you take this term--
480
00:28:20,350 --> 00:28:26,680
this term is actually
in the exact diagonal
481
00:28:26,680 --> 00:28:30,310
of this m minus one k matrix,
therefore this matches with j,
482
00:28:30,310 --> 00:28:36,270
and this will match with j minus
one; match with j plus one.
483
00:28:36,270 --> 00:28:40,750
OK, therefore, if you multiply
m minus one k and the a, you get
484
00:28:40,750 --> 00:28:46,510
this result. OK?
485
00:28:46,510 --> 00:28:53,810
And we also know that aj is
proportional to exponential
486
00:28:53,810 --> 00:28:55,870
ijka, right?
487
00:28:55,870 --> 00:28:59,430
Therefore I can
take aj out of this
488
00:28:59,430 --> 00:29:07,930
and basically I get t over
ma aj minus exponential minus
489
00:29:07,930 --> 00:29:11,640
ika plus two.
490
00:29:11,640 --> 00:29:19,350
Because I take aj out
of this bracket, OK.
491
00:29:19,350 --> 00:29:23,150
And minus exponential ika.
492
00:29:26,710 --> 00:29:27,620
OK.
493
00:29:27,620 --> 00:29:30,830
Now I actually can cancel aj.
494
00:29:30,830 --> 00:29:33,200
Basically, what I
get is, omega square
495
00:29:33,200 --> 00:29:43,200
will be equal to t over ma
two minus exponential ika
496
00:29:43,200 --> 00:29:45,884
plus exponential minus ika.
497
00:29:50,730 --> 00:29:57,161
And that will be equal
to two t over ma.
498
00:29:57,161 --> 00:29:57,660
OK.
499
00:30:00,330 --> 00:30:07,200
One minus-- OK, so exponential
ika plus exponential minus ika,
500
00:30:07,200 --> 00:30:12,960
you are going to get two
cosine ka, all right?
501
00:30:12,960 --> 00:30:16,594
Therefore, you get
one minus cosine ka.
502
00:30:20,797 --> 00:30:22,200
OK.
503
00:30:22,200 --> 00:30:31,770
I define omega zero to be
square root of t over ma,
504
00:30:31,770 --> 00:30:33,870
just to make my life easier.
505
00:30:33,870 --> 00:30:35,180
OK?
506
00:30:35,180 --> 00:30:39,270
Then, what is going
to happen is that I
507
00:30:39,270 --> 00:30:48,560
will have omega
square equal to two
508
00:30:48,560 --> 00:30:56,450
omega zero square
one minus cosine ka.
509
00:30:56,450 --> 00:30:57,650
Any questions?
510
00:31:02,510 --> 00:31:09,600
OK, of course if you like, you
can also rewrite this as four
511
00:31:09,600 --> 00:31:21,800
omega zero square sine
square ka divided by two.
512
00:31:21,800 --> 00:31:25,230
OK, so if you like.
513
00:31:25,230 --> 00:31:28,960
OK, so look at
what we have done.
514
00:31:28,960 --> 00:31:33,280
We studied a highly
symmetric system,
515
00:31:33,280 --> 00:31:35,800
which is as you were
shown in the slide.
516
00:31:35,800 --> 00:31:41,050
OK, basically you satisfy the
space translation symmetry.
517
00:31:41,050 --> 00:31:41,880
OK?
518
00:31:41,880 --> 00:31:44,080
Now what we have been
doing is to derive
519
00:31:44,080 --> 00:31:47,830
the equation of motion,
make use of the small angle
520
00:31:47,830 --> 00:31:51,670
approximation, then you
will be able to find that,
521
00:31:51,670 --> 00:31:55,390
OK, only the y
direction is actually
522
00:31:55,390 --> 00:31:57,850
moving as a function of time.
523
00:31:57,850 --> 00:32:02,110
Therefore, based on this
derivation of m minus one k
524
00:32:02,110 --> 00:32:06,920
matrix, I arrive, and also
based on the equation of motion,
525
00:32:06,920 --> 00:32:09,280
which I derived from
the first diagram,
526
00:32:09,280 --> 00:32:12,190
I get this m minus one k matrix.
527
00:32:12,190 --> 00:32:16,305
And since we know that m
minus one k matrix and s
528
00:32:16,305 --> 00:32:18,780
matrix will share
this Eigenvector,
529
00:32:18,780 --> 00:32:24,220
I can multiply m minus one k
matrix, and I come back to a.
530
00:32:24,220 --> 00:32:27,340
Then I will be able to
solve the functional
531
00:32:27,340 --> 00:32:33,970
form of omega square, and
that is actually given here.
532
00:32:33,970 --> 00:32:38,320
Omega square is equal to
two omega zero square y
533
00:32:38,320 --> 00:32:40,240
minus cosine ka.
534
00:32:40,240 --> 00:32:41,860
OK?
535
00:32:41,860 --> 00:32:49,710
And is this actually telling you
that omega is a function of k.
536
00:32:49,710 --> 00:32:51,540
OK.
537
00:32:51,540 --> 00:32:52,884
What is k?
538
00:32:52,884 --> 00:33:01,730
k is actually the wave
number and omega is actually
539
00:33:01,730 --> 00:33:06,610
the angular frequency of
the normal modes, right?
540
00:33:06,610 --> 00:33:10,665
So that means what we
were talking about--
541
00:33:10,665 --> 00:33:20,960
that means if we fix the
wavelength or the wave number,
542
00:33:20,960 --> 00:33:27,601
k, then there will be
a corresponding omega.
543
00:33:27,601 --> 00:33:28,100
OK?
544
00:33:28,100 --> 00:33:31,040
If you fix the wavelengths
you are talking about,
545
00:33:31,040 --> 00:33:37,460
then the omega is also fixed
by this omega of k function.
546
00:33:37,460 --> 00:33:41,080
OK, now we actually call
it dispersion relation.
547
00:33:50,700 --> 00:33:54,300
This term may not
mean much to you now,
548
00:33:54,300 --> 00:33:57,650
but later in the discussion,
you will find, aha!
549
00:33:57,650 --> 00:34:00,960
It really makes sense, and that
we will talk about dispersion
550
00:34:00,960 --> 00:34:03,315
in the later lectures.
551
00:34:06,160 --> 00:34:11,010
OK, so the conclusion from
here is that, basically,
552
00:34:11,010 --> 00:34:17,000
if I have this distance that
satisfies this translation
553
00:34:17,000 --> 00:34:25,010
symmetry, then what it tells us
is that the normal modes, what
554
00:34:25,010 --> 00:34:31,260
looks like some kind
of sinusoidal function,
555
00:34:31,260 --> 00:34:32,550
as we discussed last time.
556
00:34:35,139 --> 00:34:39,219
And also this--
557
00:34:39,219 --> 00:34:44,260
OK, so this is actually
the amplitude, what
558
00:34:44,260 --> 00:34:47,530
I am drawing here, this curve.
559
00:34:47,530 --> 00:34:53,800
And all those masses are
only moving up and down, OK?
560
00:34:53,800 --> 00:34:55,250
As a function of time.
561
00:34:55,250 --> 00:35:00,040
And this is aj, and that is the
oscillation frequency, which
562
00:35:00,040 --> 00:35:03,190
is actually the frequency
of moving up and down,
563
00:35:03,190 --> 00:35:05,790
this kind of motion,
is actually omega.
564
00:35:08,510 --> 00:35:15,140
And also, we learned that
omega is equal to omega of k.
565
00:35:15,140 --> 00:35:19,030
And that is actually
decided by the length,
566
00:35:19,030 --> 00:35:25,190
and how distorted is
this normal mode--
567
00:35:25,190 --> 00:35:27,110
the shape of the normal mode?
568
00:35:27,110 --> 00:35:31,190
And this is actually determined
by the k, which is actually
569
00:35:31,190 --> 00:35:36,770
the wave number, and of course
you can also get the wavelength
570
00:35:36,770 --> 00:35:38,810
from two pi over k.
571
00:35:38,810 --> 00:35:41,180
OK?
572
00:35:41,180 --> 00:35:44,930
In short, if you give
it a specific wave
573
00:35:44,930 --> 00:35:49,370
number or wavelength, than
the oscillation frequency
574
00:35:49,370 --> 00:35:53,830
is already fixed because
of the equation of motion,
575
00:35:53,830 --> 00:35:57,200
which we did, right,
from the first diagram.
576
00:35:57,200 --> 00:35:58,388
Any questions?
577
00:36:04,250 --> 00:36:05,980
OK.
578
00:36:05,980 --> 00:36:08,080
The last point which I
would like to remind you
579
00:36:08,080 --> 00:36:12,730
is that, at this point,
since we are talking
580
00:36:12,730 --> 00:36:17,110
about infinitely long
systems, therefore
581
00:36:17,110 --> 00:36:22,226
all possible k are allowed.
582
00:36:22,226 --> 00:36:23,020
Right?
583
00:36:23,020 --> 00:36:25,400
Because, basically, you
have an infinite number
584
00:36:25,400 --> 00:36:28,370
of coupled oscillators,
and therefore you
585
00:36:28,370 --> 00:36:31,280
have an infinite
number of normal modes.
586
00:36:31,280 --> 00:36:38,030
So all possible cases are
allowed, and that actually
587
00:36:38,030 --> 00:36:41,300
because we have even an
infinitely long system.
588
00:36:41,300 --> 00:36:45,080
After the break, which we
will take a five minute break,
589
00:36:45,080 --> 00:36:48,950
we will discuss how to use
infinitely long systems
590
00:36:48,950 --> 00:36:53,340
to actually understand
a finite system.
591
00:36:53,340 --> 00:36:56,840
So you will see that,
actually I can use, now,
592
00:36:56,840 --> 00:36:59,750
this space translation
symmetry, and to solve,
593
00:36:59,750 --> 00:37:02,690
in general, infinitely
long systems.
594
00:37:02,690 --> 00:37:05,815
And I can actually even go
back to find to a finite system
595
00:37:05,815 --> 00:37:07,860
and see what we
can get from there.
596
00:37:07,860 --> 00:37:08,360
OK?
597
00:37:08,360 --> 00:37:12,920
So we will be back at 12:20.
598
00:37:12,920 --> 00:37:18,370
If you have any
questions, I will be here
599
00:37:18,370 --> 00:37:20,530
OK, welcome back, everybody.
600
00:37:20,530 --> 00:37:23,590
So we will continue
the discussion.
601
00:37:23,590 --> 00:37:27,680
So there were a few questions
asked during the break.
602
00:37:27,680 --> 00:37:30,730
So, the first question is
related to how we actually
603
00:37:30,730 --> 00:37:32,870
arrive at this equation.
604
00:37:32,870 --> 00:37:35,620
And that is actually because--
605
00:37:35,620 --> 00:37:39,180
OK, two t over ma is
actually really happening
606
00:37:39,180 --> 00:37:40,570
in a diagonal term.
607
00:37:40,570 --> 00:37:43,370
Therefore, if you
multiply m minus one k
608
00:37:43,370 --> 00:37:48,200
matrix and A matrix, which
is actually shown there,
609
00:37:48,200 --> 00:37:51,750
then you will get
this term, minus t
610
00:37:51,750 --> 00:37:54,770
over ma multiplied
by aj minus one
611
00:37:54,770 --> 00:37:59,860
plus two t over ma
times aj price minus t
612
00:37:59,860 --> 00:38:02,560
over ma times aj plus one.
613
00:38:02,560 --> 00:38:06,070
And that is actually why we
can arrive at this expression.
614
00:38:06,070 --> 00:38:06,610
OK?
615
00:38:06,610 --> 00:38:08,990
Then what happens
afterward is that we
616
00:38:08,990 --> 00:38:13,180
found that aj can be factorized
out, and they cancel.
617
00:38:13,180 --> 00:38:17,560
And then now, my solution
depends now upon the amplitude,
618
00:38:17,560 --> 00:38:23,560
and still omega is actually
dependent on the k value, which
619
00:38:23,560 --> 00:38:24,540
we actually choose.
620
00:38:24,540 --> 00:38:25,780
OK.
621
00:38:25,780 --> 00:38:32,370
The second question is, why
do I say k is the wave number?
622
00:38:32,370 --> 00:38:34,390
OK, where is that coming from?
623
00:38:34,390 --> 00:38:36,220
So that is because--
624
00:38:36,220 --> 00:38:40,250
OK, so aj is proportional
to exponential ijka.
625
00:38:40,250 --> 00:38:40,750
OK?
626
00:38:40,750 --> 00:38:43,060
It has a fancy name.
627
00:38:43,060 --> 00:38:48,900
If I take the real
part, OK, as we
628
00:38:48,900 --> 00:38:51,370
did when we went
to the description
629
00:38:51,370 --> 00:38:55,840
of physical systems,
then you get cosine jka.
630
00:38:55,840 --> 00:38:57,070
OK?
631
00:38:57,070 --> 00:39:00,730
And j times a is actually--
632
00:39:00,730 --> 00:39:02,470
j is actually a label, right?
633
00:39:02,470 --> 00:39:05,260
Labeling which mass
I am talking about.
634
00:39:05,260 --> 00:39:07,960
A is actually the distance
between all those masses.
635
00:39:07,960 --> 00:39:14,440
j times a will give you
the x location of the mass.
636
00:39:14,440 --> 00:39:19,970
So j times a is actually the
the x position of the mass.
637
00:39:19,970 --> 00:39:20,470
OK?
638
00:39:20,470 --> 00:39:26,800
Therefore, if you accept
that, this becomes cosine kx,
639
00:39:26,800 --> 00:39:28,690
and from there you
will see immediately
640
00:39:28,690 --> 00:39:32,580
that k has a meaning, which
is actually the wave number.
641
00:39:32,580 --> 00:39:34,630
OK.
642
00:39:34,630 --> 00:39:37,240
All right, is that?
643
00:39:37,240 --> 00:39:42,430
OK, so that was the
questions raised,
644
00:39:42,430 --> 00:39:45,130
which I can quickly explain.
645
00:39:45,130 --> 00:39:47,960
So, what I am going
to do now is that--
646
00:39:47,960 --> 00:39:52,120
OK, we have solved, in general,
an infinitely long system.
647
00:39:52,120 --> 00:39:56,000
What are actually the
resulting normal modes
648
00:39:56,000 --> 00:39:57,470
of infinitely long systems?
649
00:39:57,470 --> 00:40:02,830
It has an infinite
number of normal modes,
650
00:40:02,830 --> 00:40:06,970
and we will wonder
if I can actually
651
00:40:06,970 --> 00:40:08,980
borrow this
infinitely long system
652
00:40:08,980 --> 00:40:14,680
and solve finite systems to see
if I can arrive at the solution
653
00:40:14,680 --> 00:40:15,890
really quickly.
654
00:40:15,890 --> 00:40:16,620
OK?
655
00:40:16,620 --> 00:40:20,320
So the answer is actually yes.
656
00:40:20,320 --> 00:40:25,210
So if I consider a finite
system that looks like this;
657
00:40:25,210 --> 00:40:32,510
so I have many, many little
masses on this system
658
00:40:32,510 --> 00:40:35,620
and they are connected
to each other
659
00:40:35,620 --> 00:40:41,290
by the center strings,
which I prepared before, OK?
660
00:40:41,290 --> 00:40:44,830
And I call this the
position in the y
661
00:40:44,830 --> 00:40:51,310
direction of this object y1,
and then the next object y2, y3,
662
00:40:51,310 --> 00:40:52,330
etc.
663
00:40:52,330 --> 00:40:56,670
And I have an object
in this system
664
00:40:56,670 --> 00:41:03,310
and both ends of the string
are fixed on the wall.
665
00:41:03,310 --> 00:41:05,380
OK?
666
00:41:05,380 --> 00:41:10,270
So I can actually now argue
that the infinitely long system
667
00:41:10,270 --> 00:41:13,540
can help us with
the understanding
668
00:41:13,540 --> 00:41:15,460
of this finite system.
669
00:41:15,460 --> 00:41:17,150
Why is that?
670
00:41:17,150 --> 00:41:20,860
That is because, now, I
can assume that, huh, this
671
00:41:20,860 --> 00:41:24,870
is actually just part of
an infinitely long system.
672
00:41:24,870 --> 00:41:26,440
All right.
673
00:41:26,440 --> 00:41:29,710
So I construct my
infinitely long system,
674
00:41:29,710 --> 00:41:40,270
and now I nail the yth mass,
I nail the y n plus one mass,
675
00:41:40,270 --> 00:41:43,030
and I fix that so that
it cannot move, OK?
676
00:41:43,030 --> 00:41:46,720
So it's still an
infinitely long system,
677
00:41:46,720 --> 00:41:52,090
but there are two interesting
boundary conditions
678
00:41:52,090 --> 00:41:57,481
at j equal to zero and
j equal to n plus one.
679
00:41:57,481 --> 00:41:57,980
OK.
680
00:41:57,980 --> 00:42:00,340
What are the two
boundary conditions?
681
00:42:08,880 --> 00:42:15,260
The first one is y
zero equal to zero.
682
00:42:15,260 --> 00:42:16,610
OK?
683
00:42:16,610 --> 00:42:24,060
And the second condition is
y n plus one equal to zero.
684
00:42:24,060 --> 00:42:25,520
OK?
685
00:42:25,520 --> 00:42:27,280
So there are two
boundary conditions,
686
00:42:27,280 --> 00:42:31,180
so basically what I'm looking
at is still an infinitely long
687
00:42:31,180 --> 00:42:36,800
system, but I require
y zero and y n plus one
688
00:42:36,800 --> 00:42:39,610
to satisfy these two conditions.
689
00:42:39,610 --> 00:42:40,550
OK?
690
00:42:40,550 --> 00:42:44,090
And we will find that,
huh, with this procedure,
691
00:42:44,090 --> 00:42:48,165
we can also solve this finite
number of couple oscillators.
692
00:42:48,165 --> 00:42:52,610
The problem, in this case
we, have coupled oscillators.
693
00:42:52,610 --> 00:42:54,317
OK?
694
00:42:54,317 --> 00:42:56,150
So the first thing which
I would like to say
695
00:42:56,150 --> 00:43:01,190
is, based on the functional
form, the functional
696
00:43:01,190 --> 00:43:05,660
form of omega square, now
this is equal to four omega
697
00:43:05,660 --> 00:43:09,380
zero square sine
square ka over two.
698
00:43:09,380 --> 00:43:10,220
OK?
699
00:43:10,220 --> 00:43:15,260
What we actually have
is that omega k is
700
00:43:15,260 --> 00:43:19,670
equal to omega minus k.
701
00:43:19,670 --> 00:43:20,720
OK?
702
00:43:20,720 --> 00:43:24,920
So both of them will give you
the same angular frequency.
703
00:43:24,920 --> 00:43:29,250
OK, therefore, what
does that mean?
704
00:43:29,250 --> 00:43:32,080
This means that
linear combination
705
00:43:32,080 --> 00:43:39,260
of exponential ijka and the
exponential minus ijka--
706
00:43:42,250 --> 00:43:46,545
OK, linear combination
of these two vectors--
707
00:43:49,240 --> 00:43:54,980
is also an Eigenvector
of m minus one k metrics.
708
00:43:54,980 --> 00:43:58,290
OK, so you can do
linear combination
709
00:43:58,290 --> 00:44:00,995
of these two vectors.
710
00:44:06,390 --> 00:44:13,380
OK, so if we do that, now
I can guess my solution
711
00:44:13,380 --> 00:44:23,340
will be like yj equal to real
part of exponential i omega t
712
00:44:23,340 --> 00:44:26,100
plus phi.
713
00:44:26,100 --> 00:44:30,500
I can now have a linear
combination of exponential ijka
714
00:44:30,500 --> 00:44:33,240
and the exponential minus ijka.
715
00:44:33,240 --> 00:44:39,670
Basically, I have
alpha exponential ijka
716
00:44:39,670 --> 00:44:43,600
plus beta exponential
minus ijka.
717
00:44:48,400 --> 00:44:50,120
OK?
718
00:44:50,120 --> 00:44:52,580
And I would like to
determine why that's actually
719
00:44:52,580 --> 00:44:57,800
alpha and beta which actually
satisfy these boundary
720
00:44:57,800 --> 00:45:01,010
conditions, one and two.
721
00:45:01,010 --> 00:45:06,290
OK, so now I can use the
first boundary condition,
722
00:45:06,290 --> 00:45:09,740
y zero equal to zero, right?
723
00:45:09,740 --> 00:45:11,810
So j equal to zero.
724
00:45:11,810 --> 00:45:16,510
Therefore, basically, what I
get is, when j is equal to zero,
725
00:45:16,510 --> 00:45:21,060
then this is actually one and
this is actually one, right?
726
00:45:21,060 --> 00:45:27,530
And this actually gives
you y zero equal to zero.
727
00:45:27,530 --> 00:45:31,160
If y zero is equal
to 0 at all times,
728
00:45:31,160 --> 00:45:36,320
no matter what t as you give it
to this system, then basically
729
00:45:36,320 --> 00:45:41,060
you have alpha plus beta
equal to zero, right?
730
00:45:41,060 --> 00:45:43,580
Because j is equal to zero.
731
00:45:43,580 --> 00:45:49,085
So you have alpha plus beta, and
that has to be equal to zero.
732
00:45:49,085 --> 00:45:52,850
Therefore, you can conclude that
alpha is equal to minus beta.
733
00:45:55,980 --> 00:46:00,000
And I've reused the
second boundary condition,
734
00:46:00,000 --> 00:46:05,270
y n plus one equal to zero,
because I nailed this mass
735
00:46:05,270 --> 00:46:08,510
and then fixed that so
that it cannot move.
736
00:46:08,510 --> 00:46:13,810
Then basically, what you get is
y n plus one is equal to zero,
737
00:46:13,810 --> 00:46:21,510
then basically you have
alpha exponential i n
738
00:46:21,510 --> 00:46:24,735
plus one ka plus--
739
00:46:28,350 --> 00:46:32,460
okay, so beta is your equal
to minus alpha, right?
740
00:46:32,460 --> 00:46:44,530
So basically, you can get
minus exponential minus i
741
00:46:44,530 --> 00:46:49,000
n plus one ka, right?
742
00:46:49,000 --> 00:46:51,640
Multiplied by alpha.
743
00:46:51,640 --> 00:46:56,720
And now this is
actually equal to zero.
744
00:46:56,720 --> 00:46:59,150
Now we have the choice.
745
00:46:59,150 --> 00:47:02,690
We can actually set alpha
to be equal to zero,
746
00:47:02,690 --> 00:47:05,300
but if I set alpha
to be equal to zero,
747
00:47:05,300 --> 00:47:08,030
then beta is also zero.
748
00:47:08,030 --> 00:47:10,040
Then I have zero
everywhere, right?
749
00:47:10,040 --> 00:47:11,810
Then there's no oscillation.
750
00:47:11,810 --> 00:47:15,450
And that's not fun, right?
751
00:47:15,450 --> 00:47:16,260
OK.
752
00:47:16,260 --> 00:47:18,790
Therefore, what I'm
going to set is actually
753
00:47:18,790 --> 00:47:21,030
the second turn equal to zero.
754
00:47:21,030 --> 00:47:21,530
OK.
755
00:47:21,530 --> 00:47:24,020
The second turn, I
can actually simplify
756
00:47:24,020 --> 00:47:33,980
that to be two i
sine n plus one ka.
757
00:47:33,980 --> 00:47:37,380
And now this is
actually equal to zero.
758
00:47:37,380 --> 00:47:38,290
OK.
759
00:47:38,290 --> 00:47:42,150
What is actually the condition
of this thing equal to zero?
760
00:47:42,150 --> 00:47:46,640
Basically, n plus 1 is
actually a given number,
761
00:47:46,640 --> 00:47:52,040
a is actually the distance
between those masses,
762
00:47:52,040 --> 00:47:56,690
therefore, what I can actually
change is the k value.
763
00:47:56,690 --> 00:47:57,530
Right?
764
00:47:57,530 --> 00:48:00,500
So I can now solve
this condition,
765
00:48:00,500 --> 00:48:03,080
and I will conclude
that k will have
766
00:48:03,080 --> 00:48:09,800
to be equal to n times
pi divided by N plus one.
767
00:48:09,800 --> 00:48:12,500
I hope you can see it.
768
00:48:12,500 --> 00:48:24,550
Where small n is equal to one,
two, three, until capital N.
769
00:48:24,550 --> 00:48:27,500
So what does that mean?
770
00:48:27,500 --> 00:48:29,450
This means that--
771
00:48:29,450 --> 00:48:33,730
OK, originally, before I
introduced the boundary
772
00:48:33,730 --> 00:48:38,080
condition, this system
is infinitely long, OK,
773
00:48:38,080 --> 00:48:45,000
and it has an infinite number
of normal modes, right?
774
00:48:45,000 --> 00:48:47,880
But once I introduced
these boundary conditions,
775
00:48:47,880 --> 00:48:53,760
which I actually require
y zero equal to zero,
776
00:48:53,760 --> 00:48:57,080
because I fixed this
point on the wall.
777
00:48:57,080 --> 00:49:00,660
Y m plus one equal to zero
because I fixed, also,
778
00:49:00,660 --> 00:49:02,910
that point on the wall.
779
00:49:02,910 --> 00:49:05,220
Something really happened.
780
00:49:05,220 --> 00:49:14,960
Now it actually gives us, first,
the shape of the system when it
781
00:49:14,960 --> 00:49:16,470
is actually in the normal mode.
782
00:49:16,470 --> 00:49:18,840
Basically, the shape--
what I mean here
783
00:49:18,840 --> 00:49:22,350
is the amplitude as
a function of j, OK?
784
00:49:22,350 --> 00:49:24,600
That's actually what
I mean by shape, OK?
785
00:49:24,600 --> 00:49:29,430
The shape is now
like a sine function.
786
00:49:29,430 --> 00:49:32,670
That's the first thing
which we get from here.
787
00:49:32,670 --> 00:49:35,540
The second thing
which we get here
788
00:49:35,540 --> 00:49:41,750
is that now, the k values
are not arbitrary anymore.
789
00:49:41,750 --> 00:49:47,290
The k values are equal
to n pi over n plus one.
790
00:49:47,290 --> 00:49:51,910
And the the small n is actually
equal for one, two, three,
791
00:49:51,910 --> 00:49:57,280
until N. So now, once you
actually fix this two point,
792
00:49:57,280 --> 00:50:05,370
you actually have only
how many normal modes?
793
00:50:05,370 --> 00:50:07,990
N normal modes!
794
00:50:07,990 --> 00:50:08,830
Right?
795
00:50:08,830 --> 00:50:12,880
So what I want to tell
you is that, in general,
796
00:50:12,880 --> 00:50:16,690
the sinusoidal shape is
actually fixed already
797
00:50:16,690 --> 00:50:20,440
by this translation
symmetry argument.
798
00:50:20,440 --> 00:50:21,240
OK.
799
00:50:21,240 --> 00:50:26,560
And once we nail both sides--
800
00:50:26,560 --> 00:50:29,680
actually, we also
restrict ourselves
801
00:50:29,680 --> 00:50:37,360
to the discussion of only a
few k which actually satisfy
802
00:50:37,360 --> 00:50:38,730
the boundary conditions.
803
00:50:38,730 --> 00:50:43,150
And if I plot all those normal
modes as a function of i,
804
00:50:43,150 --> 00:50:46,780
basically what you
can see from here,
805
00:50:46,780 --> 00:50:50,800
you can see if I have n
equals to one and capital
806
00:50:50,800 --> 00:50:54,560
N equal to four in this case.
807
00:50:54,560 --> 00:50:56,620
OK, so I have four--
808
00:50:56,620 --> 00:51:00,200
basically, I'm going to
have four normal modes.
809
00:51:00,200 --> 00:51:06,510
The first one will be like
a really long wavelength
810
00:51:06,510 --> 00:51:09,730
one, when n is equal to one.
811
00:51:09,730 --> 00:51:14,350
And if I increase
the small n value
812
00:51:14,350 --> 00:51:17,110
so that the k becomes
bigger, then you
813
00:51:17,110 --> 00:51:23,330
can see that there is more
distortion when this system is
814
00:51:23,330 --> 00:51:24,910
in one of the normal modes.
815
00:51:24,910 --> 00:51:30,440
And this shape is actually going
to be oscillating up and down,
816
00:51:30,440 --> 00:51:31,170
instead of--
817
00:51:31,170 --> 00:51:34,230
OK, so all those points are
only moving up and down, right?
818
00:51:34,230 --> 00:51:35,140
Just a reminder.
819
00:51:35,140 --> 00:51:35,790
OK?
820
00:51:35,790 --> 00:51:41,170
And why do we have
all those cases?
821
00:51:41,170 --> 00:51:45,300
Because of the
boundary conditions.
822
00:51:45,300 --> 00:51:46,871
OK, any questions?
823
00:51:50,720 --> 00:51:54,150
OK, so I have
several other cases,
824
00:51:54,150 --> 00:51:57,720
which is open end and
the closed end, and also
825
00:51:57,720 --> 00:52:01,320
the driven and the coupled
oscillator examples.
826
00:52:01,320 --> 00:52:04,190
Also in the lecture notes,
but unfortunately, we
827
00:52:04,190 --> 00:52:06,050
are will not be able
to go over them,
828
00:52:06,050 --> 00:52:08,240
but I think they are
very, very detailed,
829
00:52:08,240 --> 00:52:10,400
the notes in the lecture notes.
830
00:52:10,400 --> 00:52:11,360
OK.
831
00:52:11,360 --> 00:52:14,980
So, let me-- before I
move on to the discussion
832
00:52:14,980 --> 00:52:17,900
of continuous
systems, OK, I would
833
00:52:17,900 --> 00:52:23,400
like to discuss with you
what we have learned so far.
834
00:52:23,400 --> 00:52:27,200
So what we have learned is
that, if I have a symmetry which
835
00:52:27,200 --> 00:52:29,810
is a translation
symmetry, and plus,
836
00:52:29,810 --> 00:52:34,730
we only limit ourselves in
the discussion of oscillation.
837
00:52:34,730 --> 00:52:39,530
OK, in other words,
we limit the amplitude
838
00:52:39,530 --> 00:52:42,680
so that it doesn't explode
at the edge of the universe.
839
00:52:42,680 --> 00:52:43,640
OK.
840
00:52:43,640 --> 00:52:46,760
And I will give you
a beta value which
841
00:52:46,760 --> 00:52:52,160
is the functional form
of exponential ika,
842
00:52:52,160 --> 00:52:57,290
and equation of motion can be
derived from the first diagram.
843
00:52:57,290 --> 00:53:01,070
Once we entered the
equation of motion,
844
00:53:01,070 --> 00:53:03,320
we can get m minus
one k matrix, then
845
00:53:03,320 --> 00:53:07,630
we can derive omega square
from this expression.
846
00:53:07,630 --> 00:53:12,860
And finally, we actually can
simplify everything and then
847
00:53:12,860 --> 00:53:15,470
get the dispersion
relation omega
848
00:53:15,470 --> 00:53:20,050
equal to omega k, which is a
function of k, the wave number.
849
00:53:22,640 --> 00:53:25,850
Before we actually introduce
boundary conditions
850
00:53:25,850 --> 00:53:30,420
to go from an infinitely long
system to a finite system,
851
00:53:30,420 --> 00:53:33,610
all the k values are allowed.
852
00:53:33,610 --> 00:53:36,290
Once you introduce
boundary conditions,
853
00:53:36,290 --> 00:53:42,050
you find that you only have a
limited number of normal modes.
854
00:53:42,050 --> 00:53:46,460
Second, the k value not
continuous at any value
855
00:53:46,460 --> 00:53:48,410
any more, it becomes discrete.
856
00:53:48,410 --> 00:53:53,710
And only n values allowed
from this exercise.
857
00:53:53,710 --> 00:53:57,410
And finally, what is actually
the most general solution
858
00:53:57,410 --> 00:53:59,690
is actually the
linear combination
859
00:53:59,690 --> 00:54:03,760
of all those normal
modes, which we show here.
860
00:54:03,760 --> 00:54:08,760
And what is actually the ratio
between all those normal modes?
861
00:54:08,760 --> 00:54:11,480
All those free
coefficients are determined
862
00:54:11,480 --> 00:54:14,220
by initial conditions
if you are given.
863
00:54:14,220 --> 00:54:15,650
OK?
864
00:54:15,650 --> 00:54:18,485
So that's actually what
we have learned so far.
865
00:54:22,160 --> 00:54:29,240
And now I would like
to make a leap of faith
866
00:54:29,240 --> 00:54:30,340
to see what happens.
867
00:54:30,340 --> 00:54:34,460
OK, what we are going to
do is to introduce you
868
00:54:34,460 --> 00:54:39,950
to a continuous infinite
number of coupled oscillators.
869
00:54:39,950 --> 00:54:40,850
OK?
870
00:54:40,850 --> 00:54:41,970
So what does that mean?
871
00:54:41,970 --> 00:54:44,900
What I am going to do
is to go from this--
872
00:54:44,900 --> 00:54:55,080
so I have t and then a t, a
lot of string and mass system,
873
00:54:55,080 --> 00:54:56,700
et cetera, et
cetera, but I would
874
00:54:56,700 --> 00:55:06,510
like to go from there to just
an infinitely long string
875
00:55:06,510 --> 00:55:15,110
with string tension t and
some kind of density or mass.
876
00:55:15,110 --> 00:55:16,240
OK?
877
00:55:16,240 --> 00:55:19,830
We like to make it continuous
to see what will happen.
878
00:55:19,830 --> 00:55:21,690
OK?
879
00:55:21,690 --> 00:55:30,710
So, just a reminder of the
jth term of the m minus one
880
00:55:30,710 --> 00:55:32,250
k matrix operation.
881
00:55:32,250 --> 00:55:37,890
Basically, we have m minus
one k times a, the j's term
882
00:55:37,890 --> 00:55:40,180
of m minus one k a.
883
00:55:40,180 --> 00:55:45,540
That is actually given
by omega square aj.
884
00:55:45,540 --> 00:55:54,480
This is equal to t over
ma minus aj minus one
885
00:55:54,480 --> 00:56:01,060
plus two aj minus aj plus one?
886
00:56:01,060 --> 00:56:01,680
OK.
887
00:56:01,680 --> 00:56:05,510
So this is actually just a
copy of that formula here.
888
00:56:05,510 --> 00:56:10,570
OK, so if I make it continuous--
889
00:56:16,360 --> 00:56:19,400
OK, so that means
what I'm going to get
890
00:56:19,400 --> 00:56:25,060
is omega square a but
evaluated at position
891
00:56:25,060 --> 00:56:30,340
x, where x is actually
equal to j times a.
892
00:56:30,340 --> 00:56:32,200
OK?
893
00:56:32,200 --> 00:56:37,750
And this will be
equal to t over ma
894
00:56:37,750 --> 00:56:50,067
minus a, evaluated at x minus a
plus two a x minus a x plus a.
895
00:56:53,730 --> 00:56:59,100
Now I am going to make this
a very, very small, right?
896
00:56:59,100 --> 00:57:02,350
So that, when I make
a very, very small,
897
00:57:02,350 --> 00:57:06,240
then it becomes a very,
very continuous system.
898
00:57:06,240 --> 00:57:07,530
OK?
899
00:57:07,530 --> 00:57:09,900
So what I'm going to do is--
900
00:57:09,900 --> 00:57:14,030
I can now make a go to zero.
901
00:57:16,680 --> 00:57:25,470
I can now use Taylor's
series fx plus delta x--
902
00:57:25,470 --> 00:57:29,880
and just a reminder, if you
do a Taylor expansion of this,
903
00:57:29,880 --> 00:57:39,640
you are going to get f of x
plus delta x f prime x plus one
904
00:57:39,640 --> 00:57:46,080
over two factorial delta
x square f double prime x.
905
00:57:49,349 --> 00:57:53,400
OK, so that means
now, I can actually
906
00:57:53,400 --> 00:57:59,670
do a Taylor expansion of a x
minus a and then a x plus a.
907
00:57:59,670 --> 00:58:05,230
So what I'm going
to get is like this.
908
00:58:05,230 --> 00:58:11,770
So a x minus a
will become a of x
909
00:58:11,770 --> 00:58:22,240
minus a a prime x plus one over
two a square a double prime x.
910
00:58:26,048 --> 00:58:29,500
I can also do the same thing
to do a Taylor expansion
911
00:58:29,500 --> 00:58:33,380
for a x plus a.
912
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
913
00:58:44,960 --> 00:58:51,355
over two a square
a double prime x.
914
00:58:51,355 --> 00:58:51,855
OK.
915
00:58:59,620 --> 00:59:02,545
Once I have done this,
basically, then, I
916
00:59:02,545 --> 00:59:11,315
can calculate minus a
x minus a plus two a
917
00:59:11,315 --> 00:59:17,360
of x minus a x plus a.
918
00:59:17,360 --> 00:59:22,630
OK, I'm just taking the middle
term and copying it there.
919
00:59:22,630 --> 00:59:24,170
OK?
920
00:59:24,170 --> 00:59:30,320
If I use, now, this expression,
OK, then basically I
921
00:59:30,320 --> 00:59:36,200
will see that, OK, a of x
terms actually cancel, right?
922
00:59:36,200 --> 00:59:40,790
Because I have two a of
x from these two terms,
923
00:59:40,790 --> 00:59:44,670
and they cancel
with this two a x.
924
00:59:44,670 --> 00:59:45,951
OK?
925
00:59:45,951 --> 00:59:50,600
And also, a prime terms,
also cancel, right?
926
00:59:50,600 --> 00:59:59,780
Because you have a x minus
and a x plus a, therefore,
927
00:59:59,780 --> 01:00:01,790
the x prime turns cancel here.
928
01:00:01,790 --> 01:00:05,010
This is a minus sign,
this is a plus sign.
929
01:00:05,010 --> 01:00:05,790
You see?
930
01:00:05,790 --> 01:00:11,550
So therefore, what is actually
left over is all those terms.
931
01:00:11,550 --> 01:00:24,060
This is going to give you
a double prime x a squared,
932
01:00:24,060 --> 01:00:28,390
plus many other higher
order terms, right?
933
01:00:28,390 --> 01:00:31,560
Because those are
actually not completed,
934
01:00:31,560 --> 01:00:34,030
we have many, many
higher order terms.
935
01:00:34,030 --> 01:00:35,230
OK?
936
01:00:35,230 --> 01:00:39,640
In the limit of x go to zero--
937
01:00:39,640 --> 01:00:40,398
Yes.
938
01:00:40,398 --> 01:00:43,624
STUDENT: Isn't the
xa prime [INAUDIBLE]??
939
01:00:43,624 --> 01:00:44,790
PROFESSOR YEN-JIE LEE: Yeah.
940
01:00:44,790 --> 01:00:46,860
Oh yeah, you are right.
941
01:00:46,860 --> 01:00:49,280
Thank you very much.
942
01:00:49,280 --> 01:00:50,920
OK, thank you for that.
943
01:00:50,920 --> 01:00:54,370
So there should be a minus
sign in front of it, OK.
944
01:00:54,370 --> 01:00:59,470
So basically what we have is a
double prime x times a square
945
01:00:59,470 --> 01:01:10,570
plus some higher order turn of
a cubed, et cetera, et cetera.
946
01:01:10,570 --> 01:01:16,540
OK, since we are talking about
the limit of a goes to zero,
947
01:01:16,540 --> 01:01:20,670
we can safely ignore all
the higher order terms.
948
01:01:20,670 --> 01:01:22,120
OK?
949
01:01:22,120 --> 01:01:32,550
So now, if we go back to
this equation, omega square a
950
01:01:32,550 --> 01:01:40,840
will become, basically,
t over m minus t over ma
951
01:01:40,840 --> 01:01:46,234
a double prime x
times a squared.
952
01:01:51,180 --> 01:01:54,900
OK, and then plus some
higher order terms.
953
01:01:54,900 --> 01:02:02,850
OK, now I can define rho l to
be equal to m divided by a.
954
01:02:02,850 --> 01:02:03,460
OK.
955
01:02:03,460 --> 01:02:07,500
And I will I will have
to be very careful when
956
01:02:07,500 --> 01:02:11,490
I go to the continuous
limit, OK, I also
957
01:02:11,490 --> 01:02:17,670
do not want to make this system
infinitely massive, right?
958
01:02:17,670 --> 01:02:20,700
Therefore, I would like
to fix the rho l when
959
01:02:20,700 --> 01:02:24,030
I go to the continuous limit.
960
01:02:24,030 --> 01:02:28,980
So basically what I am doing is
that I cut this system in half
961
01:02:28,980 --> 01:02:32,880
so that a becomes smaller and
the mass also because smaller,
962
01:02:32,880 --> 01:02:37,320
so that rho l stays
as a constant, OK,
963
01:02:37,320 --> 01:02:39,350
when I go to a continuous limit.
964
01:02:39,350 --> 01:02:43,640
OK, so what I'm going
to get is omega square a
965
01:02:43,640 --> 01:02:54,010
will be equal to minus t
over rho l a double prime x.
966
01:02:54,010 --> 01:02:56,730
And to write it explicitly,
this is actually
967
01:02:56,730 --> 01:03:01,620
equal to minus t over rho l.
968
01:03:01,620 --> 01:03:06,730
I'll just square a
partial x squared.
969
01:03:06,730 --> 01:03:07,230
OK?
970
01:03:07,230 --> 01:03:11,160
Because each prime is
actually the differentiation
971
01:03:11,160 --> 01:03:14,580
which is spread to x, right?
972
01:03:14,580 --> 01:03:15,270
OK.
973
01:03:15,270 --> 01:03:17,220
Don't forget-- what
is actually this?
974
01:03:17,220 --> 01:03:21,530
This is actually m
minus one ka, right?
975
01:03:21,530 --> 01:03:23,788
Equal to omega square a.
976
01:03:27,180 --> 01:03:32,330
And that this is
actually just partial
977
01:03:32,330 --> 01:03:39,860
square a partial t square.
978
01:03:39,860 --> 01:03:40,360
Right?
979
01:03:40,360 --> 01:03:42,850
Because this is actually
m minus one k matrix.
980
01:03:42,850 --> 01:03:49,256
It's actually originally-- going
back to the original equation,
981
01:03:49,256 --> 01:03:55,330
we are actually solving
m x double dot equal
982
01:03:55,330 --> 01:03:57,328
to minus kx problem, right?
983
01:04:00,610 --> 01:04:02,970
So there should be a
minus sign there as well.
984
01:04:02,970 --> 01:04:03,830
Right?
985
01:04:03,830 --> 01:04:08,400
So mx double dot
equal to minus kx,
986
01:04:08,400 --> 01:04:14,490
therefore x double dot will be
equal to minus m minus one kx.
987
01:04:14,490 --> 01:04:15,330
Right?
988
01:04:15,330 --> 01:04:21,176
So minus m minus one k matrix
will be equal to x double dot,
989
01:04:21,176 --> 01:04:21,950
right?
990
01:04:21,950 --> 01:04:25,460
Therefore, what this is
actually x double dot?
991
01:04:25,460 --> 01:04:29,600
It's basically-- in the
current presentation,
992
01:04:29,600 --> 01:04:33,570
it's actually just partial
square a, partial t square.
993
01:04:33,570 --> 01:04:34,650
OK?
994
01:04:34,650 --> 01:04:37,470
Can everybody accept this?
995
01:04:37,470 --> 01:04:40,440
OK, so now we have
some sensibility
996
01:04:40,440 --> 01:04:43,400
going from a discrete
system, which
997
01:04:43,400 --> 01:04:46,560
you have a length scale of
a, to a continuous system,
998
01:04:46,560 --> 01:04:48,120
because a goes to zero.
999
01:04:48,120 --> 01:04:52,400
By a certain time, I fix
the ratio of m and a so
1000
01:04:52,400 --> 01:04:56,910
that the system doesn't
grow too infinitely massive.
1001
01:04:56,910 --> 01:05:01,530
OK, so if I do this, then in
short, you get this equation.
1002
01:05:01,530 --> 01:05:05,220
Partial square a
partial t square,
1003
01:05:05,220 --> 01:05:09,660
and that is equal to t
over rho l partial square
1004
01:05:09,660 --> 01:05:13,740
a partial x square.
1005
01:05:13,740 --> 01:05:14,550
OK?
1006
01:05:14,550 --> 01:05:24,190
I can now define vp as equal
to square root of t over rho l.
1007
01:05:24,190 --> 01:05:24,930
OK?
1008
01:05:24,930 --> 01:05:29,750
And this will become the
p square partial square
1009
01:05:29,750 --> 01:05:31,800
a partial x square.
1010
01:05:34,850 --> 01:05:37,310
What is this?
1011
01:05:37,310 --> 01:05:40,460
This is what equation?
1012
01:05:40,460 --> 01:05:43,580
Wave equation!
1013
01:05:43,580 --> 01:05:45,440
OK, you see that now?
1014
01:05:45,440 --> 01:05:49,250
After all the hard
work, OK, going
1015
01:05:49,250 --> 01:05:52,940
from infinity long
systems, discrete systems,
1016
01:05:52,940 --> 01:05:54,810
then go to a
continuous limit, we
1017
01:05:54,810 --> 01:05:59,060
discovered the wave equation.
1018
01:05:59,060 --> 01:06:02,990
This is probably the
most important equation
1019
01:06:02,990 --> 01:06:07,070
you actually learn, until now.
1020
01:06:07,070 --> 01:06:11,120
More important than
f equal to ma, right?
1021
01:06:11,120 --> 01:06:13,390
Because you can
listen to my lecture,
1022
01:06:13,390 --> 01:06:16,850
even, if this equation
didn't exist, right?
1023
01:06:16,850 --> 01:06:20,220
Then I can not propagate
a sound wave to your ear.
1024
01:06:20,220 --> 01:06:22,100
Right?
1025
01:06:22,100 --> 01:06:24,690
And you cannot even
see the black board,
1026
01:06:24,690 --> 01:06:26,810
because the
electromagnetic wave,
1027
01:06:26,810 --> 01:06:30,050
which I will show
in a later lecture,
1028
01:06:30,050 --> 01:06:34,670
also kind of satisfies
the same function or form.
1029
01:06:34,670 --> 01:06:37,280
So I think this is an
achievement and a highlight
1030
01:06:37,280 --> 01:06:38,980
of today's class.
1031
01:06:38,980 --> 01:06:42,110
We actually realize
now, what we have
1032
01:06:42,110 --> 01:06:45,980
been doing is really
solving something
1033
01:06:45,980 --> 01:06:48,920
related to the wave equation.
1034
01:06:48,920 --> 01:06:53,180
And next time, what I would like
to actually discuss with you
1035
01:06:53,180 --> 01:06:56,070
is the solution of
this wave equation.
1036
01:06:56,070 --> 01:06:58,070
So here, I have--
1037
01:06:58,070 --> 01:06:59,930
again, last time
you have seen this.
1038
01:06:59,930 --> 01:07:02,765
This is a coupled
oscillator system.
1039
01:07:02,765 --> 01:07:05,420
It has 72 components.
1040
01:07:05,420 --> 01:07:08,810
As a physicist, that's actually
equal to an infinite number
1041
01:07:08,810 --> 01:07:11,090
of coupled oscillators, OK.
1042
01:07:11,090 --> 01:07:12,560
That's good enough.
1043
01:07:12,560 --> 01:07:18,590
And you can see
that, if I do this,
1044
01:07:18,590 --> 01:07:21,210
it does something
really strange, right?
1045
01:07:21,210 --> 01:07:27,380
You see a progressing
wave going back and forth,
1046
01:07:27,380 --> 01:07:30,290
and it disappears
because of friction.
1047
01:07:30,290 --> 01:07:31,850
OK?
1048
01:07:31,850 --> 01:07:35,270
If I can construct
something closer
1049
01:07:35,270 --> 01:07:39,290
to a perfect coupled
system without friction,
1050
01:07:39,290 --> 01:07:42,470
what is going to happen
is that this wave
1051
01:07:42,470 --> 01:07:47,800
is going to be there bouncing
back and forth forever.
1052
01:07:47,800 --> 01:07:49,790
And that actually
can be understood
1053
01:07:49,790 --> 01:07:52,260
by the wave equation.
1054
01:07:52,260 --> 01:07:58,810
And also, if I oscillate this
system at a fixed frequency,
1055
01:07:58,810 --> 01:08:02,380
you will see that these
become standing waves.
1056
01:08:04,950 --> 01:08:07,780
And of course I can
I do crazy things,
1057
01:08:07,780 --> 01:08:11,070
I can oscillate and stop it
and it becomes really, really
1058
01:08:11,070 --> 01:08:13,050
complicated motion.
1059
01:08:13,050 --> 01:08:15,360
And of course you are
welcome to come here and play
1060
01:08:15,360 --> 01:08:16,500
after the class.
1061
01:08:16,500 --> 01:08:20,370
And next time, on
Thursday, we are
1062
01:08:20,370 --> 01:08:24,540
going to talk about the
solution to this equation
1063
01:08:24,540 --> 01:08:28,710
and how to understand
all kinds of fancy motion
1064
01:08:28,710 --> 01:08:32,220
this system can do,
given by the nature.
1065
01:08:32,220 --> 01:08:34,210
Thank you very much.