1
00:00:24,000 --> 00:00:27,046
To avoid possible
misunderstandings,
2
00:00:27,046 --> 00:00:31,138
my lectures start at 9:30
Eastern Standard Time,
3
00:00:31,138 --> 00:00:36,274
which is different from lobby
seven time, which you may have
4
00:00:36,274 --> 00:00:39,321
noticed.
The clock in lobby seven is
5
00:00:39,321 --> 00:00:43,325
seven minutes slow.
Today, we're going to cover
6
00:00:43,325 --> 00:00:47,504
coupled oscillators,
which is a big part in 8.03,
7
00:00:47,504 --> 00:00:52,031
though we leave out damping in
order to avoid a major
8
00:00:52,031 --> 00:00:55,774
complication.
Imagine that I have a pendulum
9
00:00:55,774 --> 00:00:59,604
length L mass M,
and I have another pendulum,
10
00:00:59,604 --> 00:01:05,673
also mass M length L.
And I connect a spring between
11
00:01:05,673 --> 00:01:09,326
them, as you see there,
spring constant K.
12
00:01:09,326 --> 00:01:14,405
Imagine now at time T equals
zero that I give this object,
13
00:01:14,405 --> 00:01:19,841
which I call object number one,
and this is object number two,
14
00:01:19,841 --> 00:01:24,831
give it a certain position.
I give it a certain velocity,
15
00:01:24,831 --> 00:01:30,000
so I have four choices.
And, I let the system go.
16
00:01:30,000 --> 00:01:34,175
What you're going to see is
something extremely chaotic,
17
00:01:34,175 --> 00:01:38,123
and our task today is to
predict what the position of
18
00:01:38,123 --> 00:01:42,527
this one is at any moment in
time, and what the position of
19
00:01:42,527 --> 00:01:45,032
that one is at any moment in
time.
20
00:01:45,032 --> 00:01:48,297
And, to show you how chaotic
that motion is,
21
00:01:48,297 --> 00:01:52,396
I will just show you this.
So, I take this one and just
22
00:01:52,396 --> 00:01:56,420
displace it from equilibrium.
I displace this one from
23
00:01:56,420 --> 00:02:00,520
equilibrium, and at time T
equals zero, I will give the
24
00:02:00,520 --> 00:02:05,000
one, my left hand,
just a certain velocity.
25
00:02:05,000 --> 00:02:08,914
And, if you now look at the
position of the individual
26
00:02:08,914 --> 00:02:12,903
object, it would seem nearly
impossible to come with an
27
00:02:12,903 --> 00:02:16,153
analytic solution,
which tells you what these
28
00:02:16,153 --> 00:02:18,960
motions are.
You'll see that amplitudes
29
00:02:18,960 --> 00:02:20,806
build up.
Of certain ones,
30
00:02:20,806 --> 00:02:24,352
amplitude goes down.
This one is hardly moving at
31
00:02:24,352 --> 00:02:26,863
all now.
Now it's picking up again.
32
00:02:26,863 --> 00:02:31,000
And so, our task today is to
work on that.
33
00:02:31,000 --> 00:02:35,097
What is by no means obvious
that I will show that to you,
34
00:02:35,097 --> 00:02:39,487
that any motion no matter how
you start it off is going to be
35
00:02:39,487 --> 00:02:43,073
the superposition of two normal
[mode?] solutions.
36
00:02:43,073 --> 00:02:47,536
Anyway you start it can always
be written as the superposition
37
00:02:47,536 --> 00:02:51,268
of two normal mode solutions.
What is a normal mode?
38
00:02:51,268 --> 00:02:53,463
A normal mode is,
in this case,
39
00:02:53,463 --> 00:02:57,560
that both objects have exactly
the same frequency that is
40
00:02:57,560 --> 00:03:01,878
fundamental to normal mode and
that they are either in phase
41
00:03:01,878 --> 00:03:05,317
with each other or out of phase
with each other,
42
00:03:05,317 --> 00:03:10,000
nothing in between because
there's no damping.
43
00:03:10,000 --> 00:03:13,378
So, it's either in phase or
it's out of phase.
44
00:03:13,378 --> 00:03:16,307
That is the normal mode.
In other words,
45
00:03:16,307 --> 00:03:19,761
if I call one of those
frequencies omega minus,
46
00:03:19,761 --> 00:03:22,614
minus means it is the lowest
frequency.
47
00:03:22,614 --> 00:03:27,119
There are two frequencies in
the system because there are two
48
00:03:27,119 --> 00:03:28,621
objects.
Omega minus,
49
00:03:28,621 --> 00:03:32,000
I call that the lowest
frequency.
50
00:03:32,000 --> 00:03:35,500
Then it would mean,
if they are in phase with each
51
00:03:35,500 --> 00:03:39,642
other, that they come to a halt
at the same moment in time.
52
00:03:39,642 --> 00:03:42,857
That means in phase and in the
same direction.
53
00:03:42,857 --> 00:03:47,214
So, they come to a halt at the
same moment in time in the same
54
00:03:47,214 --> 00:03:49,785
direction, and at the same
frequency.
55
00:03:49,785 --> 00:03:53,714
If, then, I have another
frequency which is higher there
56
00:03:53,714 --> 00:03:57,142
are two normal modes because we
have two objects.
57
00:03:57,142 --> 00:04:02,000
If we have three objects there
are three normal modes.
58
00:04:02,000 --> 00:04:06,599
If we go to the higher normal
mode, the higher frequency,
59
00:04:06,599 --> 00:04:11,033
they have the same frequency
that there are 180° out of
60
00:04:11,033 --> 00:04:13,826
phase.
So, when one comes to a halt
61
00:04:13,826 --> 00:04:17,357
here at the other one comes to a
halt there.
62
00:04:17,357 --> 00:04:20,478
That's what it means 180° out
of phase.
63
00:04:20,478 --> 00:04:23,024
I can excite,
and I will excite,
64
00:04:23,024 --> 00:04:26,144
this system into its normal
modes only.
65
00:04:26,144 --> 00:04:31,154
I can excite this normal mode
alone and this normal mode alone
66
00:04:31,154 --> 00:04:36,000
if I choose the correct initial
conditions.
67
00:04:36,000 --> 00:04:39,613
So, for any randomly chosen
initial condition,
68
00:04:39,613 --> 00:04:43,709
the motion of each object can
be written as a linear
69
00:04:43,709 --> 00:04:48,448
combination of these two modes.
If you take my word for that
70
00:04:48,448 --> 00:04:53,106
for now, but of course I will
demonstrate to you and I will
71
00:04:53,106 --> 00:04:55,676
prove that to you,
it would mean,
72
00:04:55,676 --> 00:05:00,575
then, that [SOUND OFF/THEN ON]
function of time can be written
73
00:05:00,575 --> 00:05:04,028
as having some kind of an
amplitude, X zero,
74
00:05:04,028 --> 00:05:10,111
I give it a minus sign.
Of course, it's related to that
75
00:05:10,111 --> 00:05:16,034
normal mode frequency times the
cosine of omega minus T plus
76
00:05:16,034 --> 00:05:20,451
some [phi?] minus,
plus some other amplitude,
77
00:05:20,451 --> 00:05:26,073
which I call X zero plus,
which is going to be related to
78
00:05:26,073 --> 00:05:33,000
this frequency times the cosine
of omega plus T plus phi plus.
79
00:05:33,000 --> 00:05:37,695
Now, let's look at this.
Let's try to see through this,
80
00:05:37,695 --> 00:05:41,347
what this means.
It means that if I know my
81
00:05:41,347 --> 00:05:45,173
initial conditions,
that I can determine this
82
00:05:45,173 --> 00:05:48,391
amplitude.
I can determine this phase.
83
00:05:48,391 --> 00:05:53,782
I can determine this amplitude,
and I can determine this phase.
84
00:05:53,782 --> 00:05:58,217
There have to be four
adjustable constants because I
85
00:05:58,217 --> 00:06:02,304
have the choice between the
positions at T zero,
86
00:06:02,304 --> 00:06:06,875
and the velocity.
So, there must be four.
87
00:06:06,875 --> 00:06:10,708
What you see is that omega
minus and omega plus,
88
00:06:10,708 --> 00:06:15,437
which are these normal mode
frequencies, are independent of
89
00:06:15,437 --> 00:06:19,759
the initial condition.
So, now I'm going to write down
90
00:06:19,759 --> 00:06:23,184
the position in time for object
number two.
91
00:06:23,184 --> 00:06:27,261
And I know that in this mode,
it must have the same
92
00:06:27,261 --> 00:06:30,523
frequency.
So there must be here a cosine
93
00:06:30,523 --> 00:06:34,763
minus T plus the same,
exactly the same phase because
94
00:06:34,763 --> 00:06:39,085
I told you, normal mode means
same frequency in phase,
95
00:06:39,085 --> 00:06:45,478
or same frequency out of phase.
And, it is the next one which
96
00:06:45,478 --> 00:06:50,521
is going to be out of phase.
But, you will see that shortly.
97
00:06:50,521 --> 00:06:53,769
So, this term must become omega
plus T.
98
00:06:53,769 --> 00:06:58,726
For now, I will say plus phi,
plus what you're going to see
99
00:06:58,726 --> 00:07:03,000
how the out of phase comes in
very shortly.
100
00:07:03,000 --> 00:07:06,241
Notice this omega minus must be
that omega minus.
101
00:07:06,241 --> 00:07:08,807
Otherwise it wouldn't be a
normal mode.
102
00:07:08,807 --> 00:07:11,441
This omega plus must be this
omega plus.
103
00:07:11,441 --> 00:07:14,007
Otherwise it wouldn't be a
normal mode.
104
00:07:14,007 --> 00:07:17,587
These phi's are the same to
allow them to be in phase.
105
00:07:17,587 --> 00:07:21,098
These phi's are the same,
even though they are out of
106
00:07:21,098 --> 00:07:23,462
phase.
But you will see very shortly
107
00:07:23,462 --> 00:07:27,311
that I will get a minus sign
here, which will take care of
108
00:07:27,311 --> 00:07:29,000
the 180°.
109
00:07:29,000 --> 00:07:34,000
110
00:07:34,000 --> 00:07:37,516
I can, with this system,
excite the lowest mode,
111
00:07:37,516 --> 00:07:40,958
so that it only oscillates in
that lowest mode.
112
00:07:40,958 --> 00:07:44,848
So, this term is not there,
and that this term is not
113
00:07:44,848 --> 00:07:46,344
there.
I can do that.
114
00:07:46,344 --> 00:07:50,310
I can always oscillates
something in one of the normal
115
00:07:50,310 --> 00:07:53,078
modes.
But, if I set it off at random,
116
00:07:53,078 --> 00:07:55,921
then it is going to be a
superposition.
117
00:07:55,921 --> 00:08:00,260
Even though it is early for
you, can you show me using your
118
00:08:00,260 --> 00:08:03,328
hands and your legs,
whatever you want to,
119
00:08:03,328 --> 00:08:07,518
when I set this off at just the
right initial conditions,
120
00:08:07,518 --> 00:08:10,735
what will that lowest normal
mode look like?
121
00:08:10,735 --> 00:08:15,000
How will these pendulums
oscillate?
122
00:08:15,000 --> 00:08:19,430
Very good, very good.
So you have a good instinct.
123
00:08:19,430 --> 00:08:21,601
Very good.
Let's do that.
124
00:08:21,601 --> 00:08:27,026
So, let's set them off in the
same direction and let them go.
125
00:08:27,026 --> 00:08:32,000
Those are ideal initial
conditions for this mode.
126
00:08:32,000 --> 00:08:35,500
And you will see that it is a
normal mode, namely,
127
00:08:35,500 --> 00:08:39,785
they have the same frequency,
and they are in phase with each
128
00:08:39,785 --> 00:08:42,428
other.
They come to a halt at the same
129
00:08:42,428 --> 00:08:45,428
moment in time.
So, what you see already is
130
00:08:45,428 --> 00:08:48,857
that this value here must,
also, be X zero minus.
131
00:08:48,857 --> 00:08:51,428
That's a must.
You have just seen it.
132
00:08:51,428 --> 00:08:54,571
In this case,
what it means is that you might
133
00:08:54,571 --> 00:08:58,071
as well remove the spring.
The spring is not doing
134
00:08:58,071 --> 00:09:00,500
anything.
The spring never gets any
135
00:09:00,500 --> 00:09:06,741
longer, never gets any shorter.
What do you think the pendulums
136
00:09:06,741 --> 00:09:11,605
will do when I excite the other
mode, which has a higher
137
00:09:11,605 --> 00:09:15,231
frequency?
Very good, can I see some other
138
00:09:15,231 --> 00:09:17,707
hands or heads or legs?
Good.
139
00:09:17,707 --> 00:09:19,564
That's this,
symmetry.
140
00:09:19,564 --> 00:09:22,482
There is a symmetry in the
system.
141
00:09:22,482 --> 00:09:27,877
And that's why you people are
so smart and see immediately the
142
00:09:27,877 --> 00:09:33,008
connection.
So, let me make here a drawing
143
00:09:33,008 --> 00:09:38,022
of the omega minus,
and then the amplitudes are the
144
00:09:38,022 --> 00:09:41,431
same.
And here, I make a drawing of
145
00:09:41,431 --> 00:09:44,640
the omega plus.
Higher frequency,
146
00:09:44,640 --> 00:09:50,456
and the amplitudes are now the
same, but 180° out of phase.
147
00:09:50,456 --> 00:09:54,568
So, it's easy to excite that.
There we go.
148
00:09:54,568 --> 00:09:59,080
Notice they oscillate with the
same frequency,
149
00:09:59,080 --> 00:10:06,000
and they're coming to a halt at
the same moment in time.
150
00:10:06,000 --> 00:10:10,071
But, if one is here,
the other is there.
151
00:10:10,071 --> 00:10:14,978
And so, therefore,
this one here must be minus X
152
00:10:14,978 --> 00:10:17,692
zero plus.
That minus sign,
153
00:10:17,692 --> 00:10:23,016
now, gives me the 180°.
So I can still keep that phi
154
00:10:23,016 --> 00:10:25,000
plus there.
155
00:10:25,000 --> 00:10:31,000
156
00:10:31,000 --> 00:10:34,194
It is utterly trivial,
without any work,
157
00:10:34,194 --> 00:10:37,553
to calculate,
or tell you what omega minus
158
00:10:37,553 --> 00:10:41,485
is, that just the frequency of a
single pendulum,
159
00:10:41,485 --> 00:10:45,090
so that's easy.
So, omega minus is simply the
160
00:10:45,090 --> 00:10:48,531
square root of G over L.
There's no spring,
161
00:10:48,531 --> 00:10:51,070
right?
Each one is doing its own
162
00:10:51,070 --> 00:10:55,084
thing, and the spring is never
pushing or pulling.
163
00:10:55,084 --> 00:11:00,000
But we should be able to
calculate omega plus.
164
00:11:00,000 --> 00:11:04,862
So let's do that.
And I will only make a drawing
165
00:11:04,862 --> 00:11:09,931
of one of those pendulums.
I don't need them both.
166
00:11:09,931 --> 00:11:13,344
So, here's one of those
pendulums.
167
00:11:13,344 --> 00:11:18,931
And let this separation
distance be X from equilibrium.
168
00:11:18,931 --> 00:11:24,827
And, I call this angle theta.
And so, here is that spring.
169
00:11:24,827 --> 00:11:30,000
And, the other one is on the
other side.
170
00:11:30,000 --> 00:11:37,482
Let's pull in all the forces
that we can think of.
171
00:11:37,482 --> 00:11:40,995
So, there's gravity,
MG.
172
00:11:40,995 --> 00:11:43,896
There is tension,
T.
173
00:11:43,896 --> 00:11:49,394
But there's more.
What else is there?
174
00:11:49,394 --> 00:11:54,891
Spring.
How much longer is the spring
175
00:11:54,891 --> 00:12:01,000
than wants to be?
2X, exactly.
176
00:12:01,000 --> 00:12:04,649
In other words,
there is a force here.
177
00:12:04,649 --> 00:12:10,369
I call that the spring force,
FS, which is bringing it also
178
00:12:10,369 --> 00:12:14,413
back to equilibrium.
And, that one is 2KX.
179
00:12:14,413 --> 00:12:17,964
Now, the tangent is very close
to MG.
180
00:12:17,964 --> 00:12:21,515
We have discussed that several
times.
181
00:12:21,515 --> 00:12:25,756
I'm going to introduce a
shorthand notation.
182
00:12:25,756 --> 00:12:31,378
Omega zero squared is G divided
by L, and omega S squared,
183
00:12:31,378 --> 00:12:37,000
making reference to the spring
is K over M.
184
00:12:37,000 --> 00:12:42,647
And now I have to set up the
differential equation.
185
00:12:42,647 --> 00:12:46,601
I have to apply Newton's Second
Law.
186
00:12:46,601 --> 00:12:51,684
And to remind you,
the magnitude of the spring
187
00:12:51,684 --> 00:12:55,863
force is 2KX.
So, Newton's Second Law,
188
00:12:55,863 --> 00:13:01,624
MX double dot equals,
now, there are two forces that
189
00:13:01,624 --> 00:13:07,498
I have to take into account:
first, the spring force,
190
00:13:07,498 --> 00:13:13,360
which is minus 2KX.
And then I need the horizontal
191
00:13:13,360 --> 00:13:16,114
component of the
[tension/tangent?].
192
00:13:16,114 --> 00:13:19,891
Remember, that's only one that
is important here,
193
00:13:19,891 --> 00:13:22,645
which is T sine theta.
But, T is MG,
194
00:13:22,645 --> 00:13:27,131
and sine theta is X divided by
L if L is the length of the
195
00:13:27,131 --> 00:13:29,334
pendulum.
So, we get minus T,
196
00:13:29,334 --> 00:13:35,000
which is MG times the sine of
theta, which is X divided by L.
197
00:13:35,000 --> 00:13:39,495
And so, this is the
differential equation.
198
00:13:39,495 --> 00:13:44,758
I'm going to divide M out,
and I'm going to bring
199
00:13:44,758 --> 00:13:50,460
everything to one side.
So, I'm going to get X double
200
00:13:50,460 --> 00:13:56,929
dot plus two omega S squared
times X plus omega zero squared
201
00:13:56,929 --> 00:14:02,824
times X equals zero.
And our task now is to solve
202
00:14:02,824 --> 00:14:07,564
this differential equation.
And that, of course,
203
00:14:07,564 --> 00:14:12,910
you can do in three seconds
because you recognize this
204
00:14:12,910 --> 00:14:17,449
differential equation.
It is X double dot plus
205
00:14:17,449 --> 00:14:21,887
something times X.
And so, the new frequency,
206
00:14:21,887 --> 00:14:27,737
which I will call omega plus,
that new frequency omega plus
207
00:14:27,737 --> 00:14:35,000
is the square root of two omega
S squared plus omega squared.
208
00:14:35,000 --> 00:14:40,444
And, what you see is something
that I already anticipated,
209
00:14:40,444 --> 00:14:44,265
which was consistent with your
intuition.
210
00:14:44,265 --> 00:14:49,710
It is larger than omega zero
because this is the effect of
211
00:14:49,710 --> 00:14:53,626
the spring.
And so, omega zero is the one.
212
00:14:53,626 --> 00:14:58,976
Here, an omega plus now is the
square root of two omega S
213
00:14:58,976 --> 00:15:04,994
squared plus omega zero squared.
And, omega S squared is defined
214
00:15:04,994 --> 00:15:07,000
this way.
215
00:15:07,000 --> 00:15:13,000
216
00:15:13,000 --> 00:15:18,982
So, if I now turn to my general
solution, if for now you accept
217
00:15:18,982 --> 00:15:23,035
the fact that that is the
general solution,
218
00:15:23,035 --> 00:15:28,728
the superposition of the normal
modes, then the key point is
219
00:15:28,728 --> 00:15:35,000
that independent of your initial
conditions is omega minus.
220
00:15:35,000 --> 00:15:37,875
That was the square root of G
over L.
221
00:15:37,875 --> 00:15:40,750
I never put in any initial
condition.
222
00:15:40,750 --> 00:15:44,744
Independent of the initial
condition is omega plus.
223
00:15:44,744 --> 00:15:48,099
That's this one.
I never put in any initial
224
00:15:48,099 --> 00:15:51,054
conditions.
Independent of the initial
225
00:15:51,054 --> 00:15:54,488
conditions is this ratio,
which is plus one.
226
00:15:54,488 --> 00:16:00,000
And, independent of the initial
conditions is this ratio.
227
00:16:00,000 --> 00:16:02,718
That is minus one.
In other words,
228
00:16:02,718 --> 00:16:06,507
if I didn't tell you at the
initial conditions,
229
00:16:06,507 --> 00:16:10,214
which I haven't,
I can predict that this ratio
230
00:16:10,214 --> 00:16:13,427
is plus one.
And I can predict that this
231
00:16:13,427 --> 00:16:16,722
ratio is minus one.
If I make this three,
232
00:16:16,722 --> 00:16:20,429
then this is minus three.
If I make this five,
233
00:16:20,429 --> 00:16:24,300
then this is minus five.
The ratio is minus one.
234
00:16:24,300 --> 00:16:28,501
So, the ratios are independent
of initial condition,
235
00:16:28,501 --> 00:16:32,208
and the frequencies are
independent of initial
236
00:16:32,208 --> 00:16:35,091
condition.
If I tell you the initial
237
00:16:35,091 --> 00:16:39,375
conditions, then of course you
can also calculate the
238
00:16:39,375 --> 00:16:44,566
individual amplitudes.
Suppose, now,
239
00:16:44,566 --> 00:16:51,437
I start this system off in some
way at time T equals zero,
240
00:16:51,437 --> 00:16:56,861
which I can choose.
And it used the following.
241
00:16:56,861 --> 00:17:00,357
At T equals zero,
I make X1 C,
242
00:17:00,357 --> 00:17:05,540
just some number C that we can
choose, 3 cm,
243
00:17:05,540 --> 00:17:11,862
whatever you want to choose.
But, I make V1 zero.
244
00:17:11,862 --> 00:17:14,562
So, as I release it:
no speed.
245
00:17:14,562 --> 00:17:18,846
And, suppose I make X2 zero and
I make V2 zero.
246
00:17:18,846 --> 00:17:24,433
So, if you don't know what that
means, of course I'm going to
247
00:17:24,433 --> 00:17:28,437
demonstrate it.
It means that this one here,
248
00:17:28,437 --> 00:17:33,000
the other one I offset,
that's all I do.
249
00:17:33,000 --> 00:17:36,483
And I let this one go,
pull my hands off,
250
00:17:36,483 --> 00:17:40,488
and then I want to know what's
going to happen,
251
00:17:40,488 --> 00:17:43,710
right?
Because I release this one with
252
00:17:43,710 --> 00:17:47,193
zero speed.
That's what you see in there.
253
00:17:47,193 --> 00:17:51,983
There is one of an infinite
number of initial conditions
254
00:17:51,983 --> 00:17:55,640
that you may choose.
So, this now has to be
255
00:17:55,640 --> 00:17:58,775
substituted into my general
solution.
256
00:17:58,775 --> 00:18:04,000
And so, you have to take the
derivative of X1.
257
00:18:04,000 --> 00:18:07,989
So, you have to do X1 dot and
put that equal to zero.
258
00:18:07,989 --> 00:18:10,751
I leave that in your competent
hands.
259
00:18:10,751 --> 00:18:14,818
Then you have to take X2,
and you have to take X2 dot,
260
00:18:14,818 --> 00:18:17,810
and you have to make that equal
to zero.
261
00:18:17,810 --> 00:18:21,033
And I leave that into your
competent hands.
262
00:18:21,033 --> 00:18:23,565
That's easy.
This and you'll find,
263
00:18:23,565 --> 00:18:27,631
then, that in this particular
case for this particular
264
00:18:27,631 --> 00:18:30,930
example, phi minus equals zero,
and phi two,
265
00:18:30,930 --> 00:18:36,035
and phi plus equal zero.
It will not take you more than
266
00:18:36,035 --> 00:18:39,834
a few minutes to find that.
I didn't want waste your time
267
00:18:39,834 --> 00:18:43,090
on taking a derivative of such a
simple function.
268
00:18:43,090 --> 00:18:47,025
So, if we take this right now,
then I will substitute these
269
00:18:47,025 --> 00:18:50,417
results in the equation.
So, you did the hard work.
270
00:18:50,417 --> 00:18:53,605
You did the velocities.
I will do the positions.
271
00:18:53,605 --> 00:18:57,133
So, I will substitute in that
equation T equals zero.
272
00:18:57,133 --> 00:19:01,000
I know already that the phi's
are not there.
273
00:19:01,000 --> 00:19:08,303
And so, then I get that C at
time T equals zero is going to
274
00:19:08,303 --> 00:19:13,214
be X zero minus T zero.
So, this is one.
275
00:19:13,214 --> 00:19:18,251
T zero, so this is one,
plus X zero plus.
276
00:19:18,251 --> 00:19:25,555
That's this initial condition.
Then I go through the second
277
00:19:25,555 --> 00:19:31,474
initial condition that is zero,
is X zero minus,
278
00:19:31,474 --> 00:19:37,508
minus X zero plus.
There is a minus sign here.
279
00:19:37,508 --> 00:19:41,225
And so, what do I find I have
solved now?
280
00:19:41,225 --> 00:19:45,871
The general solution,
I will find that X zero minus
281
00:19:45,871 --> 00:19:49,681
is one half C,
and F zero plus is also one
282
00:19:49,681 --> 00:19:56,000
half C, which of course should
not come as a surprise to you.
283
00:19:56,000 --> 00:20:02,000
284
00:20:02,000 --> 00:20:06,398
So, let me write down,
now, the general solution that
285
00:20:06,398 --> 00:20:10,036
we have for this specific
initial condition.
286
00:20:10,036 --> 00:20:13,589
So, we know everything.
We know X zero one.
287
00:20:13,589 --> 00:20:16,803
We know X zero one minus,
X zero minus,
288
00:20:16,803 --> 00:20:20,271
not one X zero minus.
We know X zero plus.
289
00:20:20,271 --> 00:20:23,063
We know phi's.
We know everything.
290
00:20:23,063 --> 00:20:26,024
So, I'm going to write it down
here.
291
00:20:26,024 --> 00:20:30,000
So, X1 is going to be one half
C.
292
00:20:30,000 --> 00:20:38,480
Remember, we found it at one
half C times the cosine of omega
293
00:20:38,480 --> 00:20:46,113
T equals phi zero plus one half
C times, this is minus,
294
00:20:46,113 --> 00:20:53,462
by the way, cosine omega minus
T cosine omega plus T.
295
00:20:53,462 --> 00:21:02,226
That is X1, and X2 is one half
C times the cosine omega minus T
296
00:21:02,226 --> 00:21:10,000
minus one half C times the
cosine omega plus T.
297
00:21:10,000 --> 00:21:13,503
Take a deep breath,
substitute in there T equals
298
00:21:13,503 --> 00:21:17,753
zero, and you see immediately
that X is C and substitute T
299
00:21:17,753 --> 00:21:20,884
equals zero into seconds,
and then you see,
300
00:21:20,884 --> 00:21:24,835
indeed, that X2 is zero:
no surprise because that's my
301
00:21:24,835 --> 00:21:28,264
initial condition.
Now, I remember from my high
302
00:21:28,264 --> 00:21:32,663
school days that the cosine
alpha plus the cosine of beta is
303
00:21:32,663 --> 00:21:37,136
twice the cosine of half the sum
times the cosine of half the
304
00:21:37,136 --> 00:21:41,993
difference.
So, I can write down X1 as
305
00:21:41,993 --> 00:21:48,901
twice the cosine of half the sum
times the cosine of half the
306
00:21:48,901 --> 00:21:53,736
difference.
So, that two that I get eats up
307
00:21:53,736 --> 00:21:58,917
these one halves.
So, I get C times the cosine
308
00:21:58,917 --> 00:22:05,710
omega minus plus omega plus
divided by two times T times the
309
00:22:05,710 --> 00:22:11,812
cosine of omega minus,
minus omega plus divided by two
310
00:22:11,812 --> 00:22:16,478
times T.
I've just rewritten it in a
311
00:22:16,478 --> 00:22:20,608
different form.
And, X2 as a function of time,
312
00:22:20,608 --> 00:22:25,657
I now have the cosine of alpha
minus the cosine of beta.
313
00:22:25,657 --> 00:22:30,521
That is twice the sign,
half the sum times the sign of
314
00:22:30,521 --> 00:22:36,709
half the difference.
So, now I get here the sign of
315
00:22:36,709 --> 00:22:44,003
omega minus plus omega plus
divided by two times T times the
316
00:22:44,003 --> 00:22:50,309
sign of omega minus,
minus omega plus divided by two
317
00:22:50,309 --> 00:22:54,018
times T.
Notice when T is zero,
318
00:22:54,018 --> 00:23:01,189
that the sign of this one is
zero, consistent with what our
319
00:23:01,189 --> 00:23:07,000
initial conditions,
X2 zero, remember?
320
00:23:07,000 --> 00:23:11,205
It's all there.
Initial conditions are all
321
00:23:11,205 --> 00:23:14,487
there.
Initial conditions are all
322
00:23:14,487 --> 00:23:18,487
there, just written in a
different form.
323
00:23:18,487 --> 00:23:24,538
Now, imagine now in your mind
that these two frequencies are
324
00:23:24,538 --> 00:23:29,461
not too far apart.
Then, these equations smell of
325
00:23:29,461 --> 00:23:31,741
what?
Beats.
326
00:23:31,741 --> 00:23:35,574
See, this here is the fast
term.
327
00:23:35,574 --> 00:23:41,880
This one and that one,
and this one is the slow one.
328
00:23:41,880 --> 00:23:48,556
And so, if these two are close,
then what you'll see is
329
00:23:48,556 --> 00:23:53,996
something quite remarkable.
At T equals zero,
330
00:23:53,996 --> 00:24:00,301
this one stands still.
And, the cosine term is going
331
00:24:00,301 --> 00:24:06,973
to be one because T is one.
This one is going to oscillate
332
00:24:06,973 --> 00:24:10,723
happily with this frequency.
But, this cosine term is very
333
00:24:10,723 --> 00:24:13,881
gradually going to zero.
And, as that cosine term
334
00:24:13,881 --> 00:24:17,434
gradually goes to zero,
this one will stop oscillating.
335
00:24:17,434 --> 00:24:19,802
But, this sine term becomes
plus one.
336
00:24:19,802 --> 00:24:22,763
And so, the other one will
start to oscillate.
337
00:24:22,763 --> 00:24:26,644
And then, a little later in
time the cosine term will become
338
00:24:26,644 --> 00:24:30,000
minus one.
So, it starts to oscillate.
339
00:24:30,000 --> 00:24:33,753
But when that happens,
the sine term is zero again.
340
00:24:33,753 --> 00:24:36,831
So, it stops.
So, you see a beautiful beat
341
00:24:36,831 --> 00:24:39,758
phenomenon.
The first one will gradually
342
00:24:39,758 --> 00:24:43,211
come to a halt,
and the other one will pick up.
343
00:24:43,211 --> 00:24:46,439
And then, the other one will
come to a halt,
344
00:24:46,439 --> 00:24:49,217
and then transfers in [a way?]
energy.
345
00:24:49,217 --> 00:24:51,919
It is, of course,
consistent with the
346
00:24:51,919 --> 00:24:54,396
conservation of mechanical
energy.
347
00:24:54,396 --> 00:24:58,000
And, I want to demonstrate
that.
348
00:24:58,000 --> 00:25:03,000
349
00:25:03,000 --> 00:25:07,464
So, we have this year.
All I have to do is offset X1
350
00:25:07,464 --> 00:25:10,791
over a distance,
C, that we can choose,
351
00:25:10,791 --> 00:25:14,468
and then we release this one at
zero speed.
352
00:25:14,468 --> 00:25:19,545
And then we'll just watch it.
And then, you should see that
353
00:25:19,545 --> 00:25:23,309
strange beat phenomenon.
You ready for this?
354
00:25:23,309 --> 00:25:26,023
This will have to hold in
place.
355
00:25:26,023 --> 00:25:29,000
Three, two, one,
zero.
356
00:25:29,000 --> 00:25:34,000
357
00:25:34,000 --> 00:25:37,343
Look, this one is standing
still.
358
00:25:37,343 --> 00:25:41,000
Look, this one is standing
still.
359
00:25:41,000 --> 00:25:46,000
360
00:25:46,000 --> 00:25:49,534
That's beating.
And you see that the energy is
361
00:25:49,534 --> 00:25:52,125
transferred from one to the
other.
362
00:25:52,125 --> 00:25:56,680
And that is a beat phenomenon
that follows immediately from
363
00:25:56,680 --> 00:25:59,272
this.
What would happen if I moved
364
00:25:59,272 --> 00:26:05,490
the spring up?
Suppose I moved the spring here
365
00:26:05,490 --> 00:26:13,023
higher, say, halfway.
Anyone without looking too much
366
00:26:13,023 --> 00:26:19,542
at the blackboard,
sort of use your intuition.
367
00:26:19,542 --> 00:26:25,771
What would happen?
Would the same phenomenon
368
00:26:25,771 --> 00:26:27,509
happen?
Yeah?
369
00:26:27,509 --> 00:26:32,000
Yeah?
The spring what?
370
00:26:32,000 --> 00:26:34,823
Yes, so would the same
phenomenon happen?
371
00:26:34,823 --> 00:26:38,000
But what would happen with the
beat period?
372
00:26:38,000 --> 00:26:44,000
373
00:26:44,000 --> 00:26:46,206
Now you may look at the
blackboard.
374
00:26:46,206 --> 00:26:49,061
Do you think these have a
certain separation?
375
00:26:49,061 --> 00:26:51,591
Right, you make that separation
smaller.
376
00:26:51,591 --> 00:26:54,251
You make the effect of the
spring smaller.
377
00:26:54,251 --> 00:26:57,950
So, the omega minus becomes
even closer to the omega plus.
378
00:26:57,950 --> 00:27:01,000
And so, the big period will
increase.
379
00:27:01,000 --> 00:27:05,827
It will take longer for one to
stop.
380
00:27:05,827 --> 00:27:12,034
Let's try that.
So, are we going to move these
381
00:27:12,034 --> 00:27:17,000
up?
I'll put it roughly halfway.
382
00:27:17,000 --> 00:27:22,000
383
00:27:22,000 --> 00:27:26,153
OK, are we ready for this?
And so we're doing it again.
384
00:27:26,153 --> 00:27:30,692
And then you'll see the same
phenomenon, except it will take
385
00:27:30,692 --> 00:27:33,923
longer for the first one to come
to a halt.
386
00:27:33,923 --> 00:27:37,000
So, I have decreased the
coupling.
387
00:27:37,000 --> 00:27:48,000
388
00:27:48,000 --> 00:27:54,117
It's still swinging happily.
Now it's beginning to change
389
00:27:54,117 --> 00:27:58,159
its mind.
You see, and now it standing
390
00:27:58,159 --> 00:28:01,000
still.
It took longer.
391
00:28:01,000 --> 00:28:04,874
Do we agree?
Now, there is something
392
00:28:04,874 --> 00:28:09,745
mind-boggling,
something that you really want
393
00:28:09,745 --> 00:28:13,619
to see.
Suppose I bring this all the
394
00:28:13,619 --> 00:28:18,712
way to the top.
That's interesting because then
395
00:28:18,712 --> 00:28:24,579
omega minus is exactly the same
as omega plus because,
396
00:28:24,579 --> 00:28:31,000
look, this term goes away.
What now will happen?
397
00:28:31,000 --> 00:28:33,390
It's almost cutting out the
spring.
398
00:28:33,390 --> 00:28:37,468
There is no spring anymore.
What now will happen if I start
399
00:28:37,468 --> 00:28:39,226
one here?
Three, two, one,
400
00:28:39,226 --> 00:28:40,000
zero.
401
00:28:40,000 --> 00:28:45,000
402
00:28:45,000 --> 00:28:47,005
What do you think?
Yeah?
403
00:28:47,005 --> 00:28:49,098
Ah, brilliant.
Brilliant.
404
00:28:49,098 --> 00:28:52,151
One will swing,
and the other won't.
405
00:28:52,151 --> 00:28:55,901
Isn't that shocking?
You have two pendulums,
406
00:28:55,901 --> 00:29:00,000
and they're not even connected
anymore.
407
00:29:00,000 --> 00:29:05,510
And you start this one
swinging, and the other one will
408
00:29:05,510 --> 00:29:09,591
never start to swing.
Isn't that amazing?
409
00:29:09,591 --> 00:29:14,591
For $30,000 tuition,
you learn something fantastic
410
00:29:14,591 --> 00:29:19,183
except that the big period is
infinitely long.
411
00:29:19,183 --> 00:29:24,591
So, with a bit of patience,
so let's demonstrate that.
412
00:29:24,591 --> 00:29:30,000
So let's demonstrate this.
So, there we go.
413
00:29:30,000 --> 00:29:32,691
Unbelievable.
Physics works.
414
00:29:32,691 --> 00:29:38,373
And look at this equation.
When omega minus is omega plus,
415
00:29:38,373 --> 00:29:43,657
this term is always zero.
So, you see X2 remains zero.
416
00:29:43,657 --> 00:29:46,747
Ha, ha, ha, that's what you
see.
417
00:29:46,747 --> 00:29:52,928
And, when omega minus is omega
plus, this cosine term is always
418
00:29:52,928 --> 00:29:56,517
plus one.
Ha, ha, ha, that's what you
419
00:29:56,517 --> 00:29:59,806
see.
Isn't it amazing the power of
420
00:29:59,806 --> 00:30:02,000
physics?
421
00:30:02,000 --> 00:30:06,000
422
00:30:06,000 --> 00:30:11,314
So, it is truly remarkable that
we can describe for any initial
423
00:30:11,314 --> 00:30:16,714
condition the motion in terms of
the linear superposition of the
424
00:30:16,714 --> 00:30:20,742
two normal modes.
And so, what originally looked
425
00:30:20,742 --> 00:30:25,885
like an impossibility when I
started the first 30 sections of
426
00:30:25,885 --> 00:30:28,714
my lecture with the
demonstration,
427
00:30:28,714 --> 00:30:33,857
when I started them off in a
random way, it looked so chaotic
428
00:30:33,857 --> 00:30:39,085
that it was almost unimaginable
that we would be able to short
429
00:30:39,085 --> 00:30:43,628
that out and be able to predict
the motion of each one
430
00:30:43,628 --> 00:30:48,000
individually as a function of
time.
431
00:30:48,000 --> 00:30:52,505
I want you to appreciate that
we have two coupled oscillators
432
00:30:52,505 --> 00:30:55,209
here.
That gives you to normal modes.
433
00:30:55,209 --> 00:30:59,490
If you have three coupled
oscillators, we will get back to
434
00:30:59,490 --> 00:31:02,418
that, then there are three
normal modes.
435
00:31:02,418 --> 00:31:05,798
And if you have for,
then you have four normal
436
00:31:05,798 --> 00:31:09,243
modes.
Now, clearly what we need is a
437
00:31:09,243 --> 00:31:11,337
general recipe.
In this case,
438
00:31:11,337 --> 00:31:15,750
the problem was a beautifully
symmetric we could do all mega
439
00:31:15,750 --> 00:31:19,565
minus just a split second.
Omega plus: you could see
440
00:31:19,565 --> 00:31:23,229
exactly the motion.
So, in no time [you get?] that
441
00:31:23,229 --> 00:31:27,493
the ratio of the amplitude was
plus one and was minus one.
442
00:31:27,493 --> 00:31:30,783
But let me tell you,
if I break the symmetry,
443
00:31:30,783 --> 00:31:34,997
oh man, all hell breaks loose.
For instance,
444
00:31:34,997 --> 00:31:39,752
if you change the ratio of the
masses, you make one longer than
445
00:31:39,752 --> 00:31:43,893
the other, you have three beats
connected with springs.
446
00:31:43,893 --> 00:31:47,958
If you break the symmetry,
it is very, very hard work.
447
00:31:47,958 --> 00:31:50,949
And, for that,
we need a general recipe.
448
00:31:50,949 --> 00:31:54,401
And, I'm going to give you the
general recipe.
449
00:31:54,401 --> 00:31:58,619
And then, I'm going to apply
that general recipe to this
450
00:31:58,619 --> 00:32:03,636
case.
So, everything that you have
451
00:32:03,636 --> 00:32:09,610
seen will then come out after a
lot of algebra,
452
00:32:09,610 --> 00:32:15,844
but at least you know that the
recipe is working.
453
00:32:15,844 --> 00:32:22,207
So, the first thing that you do
that's number one,
454
00:32:22,207 --> 00:32:30,000
you give each object a
displacement from equilibrium.
455
00:32:30,000 --> 00:32:35,126
Even though you are free to
choose the direction,
456
00:32:35,126 --> 00:32:41,427
I always, matter of discipline,
set them all off in the same
457
00:32:41,427 --> 00:32:44,417
direction.
That's not a must.
458
00:32:44,417 --> 00:32:48,689
But, that reduces the chance of
mistakes.
459
00:32:48,689 --> 00:32:52,213
But, you give them a
displacement,
460
00:32:52,213 --> 00:32:56,699
and I always do them in the
same direction.
461
00:32:56,699 --> 00:33:01,184
That's number one.
Once you have done that,
462
00:33:01,184 --> 00:33:05,776
you write down Newton's Second
Law for each,
463
00:33:05,776 --> 00:33:11,116
which means if you have two
objects, you have three
464
00:33:11,116 --> 00:33:16,000
unknowns.
Well, that is three unknowns.
465
00:33:16,000 --> 00:33:23,000
466
00:33:23,000 --> 00:33:28,288
So, you want to find one of the
normal mode frequencies.
467
00:33:28,288 --> 00:33:33,576
You want to find this one,
and independently you want to
468
00:33:33,576 --> 00:33:37,711
find that one,
which has to come out of that
469
00:33:37,711 --> 00:33:42,326
recipe, what are in principle
the three unknowns.
470
00:33:42,326 --> 00:33:48,192
This ratio plus one holds here,
but that doesn't hold in other
471
00:33:48,192 --> 00:33:51,653
cases.
And so, in principle you have,
472
00:33:51,653 --> 00:33:55,788
as an uncertainty,
this amplitude and omega.
473
00:33:55,788 --> 00:34:03,000
So, you always end up with two
equations with three unknowns.
474
00:34:03,000 --> 00:34:08,760
So if I see two objects,
then you have three unknowns.
475
00:34:08,760 --> 00:34:14,956
And you have two equations.
So, that looks like a problem.
476
00:34:14,956 --> 00:34:19,195
Number three:
you're going to put in the
477
00:34:19,195 --> 00:34:25,500
condition for normal modes,
which means that X1 is C1 times
478
00:34:25,500 --> 00:34:29,847
cosine omega T.
But, X2 can be some other
479
00:34:29,847 --> 00:34:35,826
amplitude, which we had to
calculate also times the same
480
00:34:35,826 --> 00:34:40,038
omega T.
I don't have to put the phase
481
00:34:40,038 --> 00:34:43,763
in there because it's either in
phase or out of phase,
482
00:34:43,763 --> 00:34:47,067
and out of phase means that you
get minus signs.
483
00:34:47,067 --> 00:34:51,073
And so, you see immediately
that if you substitute this in
484
00:34:51,073 --> 00:34:54,658
Newton's Second Law,
that you get for two equations,
485
00:34:54,658 --> 00:34:56,696
three unknowns.
C1 is unknown,
486
00:34:56,696 --> 00:35:00,000
C2 is unknown,
and omega is unknown.
487
00:35:00,000 --> 00:35:04,447
So, you substitute this in
here, and then comes the
488
00:35:04,447 --> 00:35:07,828
problem.
How are you going to deal with
489
00:35:07,828 --> 00:35:10,763
two equations with three
unknowns?
490
00:35:10,763 --> 00:35:14,411
Well, think about what I told
you earlier.
491
00:35:14,411 --> 00:35:19,837
If you don't know the initial
conditions, you can never find X
492
00:35:19,837 --> 00:35:23,039
zero minus, so you can never
find C1.
493
00:35:23,039 --> 00:35:29,000
But the ratio is independent of
the initial conditions.
494
00:35:29,000 --> 00:35:34,663
So that means the ratio C1 over
C2 cannot be dependent on the
495
00:35:34,663 --> 00:35:37,873
initial conditions.
In other words,
496
00:35:37,873 --> 00:35:43,348
you can always solve these
equations in terms of C1 divided
497
00:35:43,348 --> 00:35:47,690
by C2 and omega.
And, I will show you that that
498
00:35:47,690 --> 00:35:50,805
works.
So, you follow this recipe,
499
00:35:50,805 --> 00:35:54,581
and we will do that religiously
together.
500
00:35:54,581 --> 00:35:57,884
We will find,
then, in this case two
501
00:35:57,884 --> 00:36:03,633
variables for omega
We will find an omega minus,
502
00:36:03,633 --> 00:36:07,583
which has an associated value
of C over C2.
503
00:36:07,583 --> 00:36:12,191
I'll put a minus there.
And, we will find an omega
504
00:36:12,191 --> 00:36:16,705
plus, which has its own ratio
associated with it.
505
00:36:16,705 --> 00:36:20,655
In the case of our simple
symmetric system,
506
00:36:20,655 --> 00:36:25,357
this would be plus one and this
would be minus one.
507
00:36:25,357 --> 00:36:31,000
But that is not the case when
you break symmetry.
508
00:36:31,000 --> 00:36:34,382
So, my task is,
now, to apply this recipe in
509
00:36:34,382 --> 00:36:37,134
its most general form to the
system.
510
00:36:37,134 --> 00:36:39,730
It will be 16 minutes of
grinding.
511
00:36:39,730 --> 00:36:44,449
We will go through each step,
and out comes something that we
512
00:36:44,449 --> 00:36:47,674
already know.
But at least you'll see that
513
00:36:47,674 --> 00:36:52,078
I've made no assumption which I
did there about symmetry.
514
00:36:52,078 --> 00:36:56,325
Therefore, this is an ideal
moment for you to have your
515
00:36:56,325 --> 00:37:00,966
five-minute break so you can
take a deep breath to get ready
516
00:37:00,966 --> 00:37:06,000
for this 16 minute marathon.
[SOUND OFF/THEN ON]
517
00:37:06,000 --> 00:37:09,831
All right: general recipe.
Someone asked me during the
518
00:37:09,831 --> 00:37:13,807
intermission whether this omega
is the same as that one.
519
00:37:13,807 --> 00:37:16,843
Yes, of course,
otherwise it wouldn't be in
520
00:37:16,843 --> 00:37:19,807
normal mode.
We're going to search for the
521
00:37:19,807 --> 00:37:22,481
normal modes.
And, in the normal mode,
522
00:37:22,481 --> 00:37:25,879
the frequencies must be the
same of all objects,
523
00:37:25,879 --> 00:37:30,000
whether you have five or six or
two or three.
524
00:37:30,000 --> 00:37:33,894
But the ratios are different.
That's a different issue.
525
00:37:33,894 --> 00:37:36,923
However, they are in phase or
out of phase.
526
00:37:36,923 --> 00:37:41,322
So the ratios can be negative,
positive, but you never had any
527
00:37:41,322 --> 00:37:45,504
phase angles other than the 180°
or zero because we have no
528
00:37:45,504 --> 00:37:48,245
damping.
We have taken the damping out.
529
00:37:48,245 --> 00:37:52,572
So, it is essential that when
you substitute that in Newton's
530
00:37:52,572 --> 00:37:57,115
Second Law that these omegas are
the same because they will then
531
00:37:57,115 --> 00:38:02,000
give you the normal modes.
OK, you ready?
532
00:38:02,000 --> 00:38:07,000
533
00:38:07,000 --> 00:38:12,861
I'm going to offset the first
one over a distance,
534
00:38:12,861 --> 00:38:15,732
X1.
So, they have mass M.
535
00:38:15,732 --> 00:38:19,559
They have length L to remind
you.
536
00:38:19,559 --> 00:38:26,736
And, we are going to have omega
zero squared equals G over L,
537
00:38:26,736 --> 00:38:32,000
and omega S squared equals K
over M.
538
00:38:32,000 --> 00:38:36,245
And, this one I offset over a
distance, X2.
539
00:38:36,245 --> 00:38:39,581
Notice, you don't have to do
that.
540
00:38:39,581 --> 00:38:43,826
I always offset them in the
same direction.
541
00:38:43,826 --> 00:38:48,779
And then, there is this spring
that connects them.
542
00:38:48,779 --> 00:38:53,025
I will just make it very thin,
this spring.
543
00:38:53,025 --> 00:39:00,000
Otherwise the picture becomes a
little bit too complicated.
544
00:39:00,000 --> 00:39:05,597
So let this angle be theta one.
And let this angle be theta
545
00:39:05,597 --> 00:39:08,300
two.
And so, we have here MG.
546
00:39:08,300 --> 00:39:11,388
We have here MG.
We have here the
547
00:39:11,388 --> 00:39:16,117
[tangent/tension?] is roughly MG
for small angles,
548
00:39:16,117 --> 00:39:20,557
and we have here a tangent
which is roughly MG.
549
00:39:20,557 --> 00:39:24,707
It's a little bit too big the
way I drew it,
550
00:39:24,707 --> 00:39:27,699
but that's not so important
now.
551
00:39:27,699 --> 00:39:32,139
And now, there is another force
that X on both,
552
00:39:32,139 --> 00:39:38,520
and which force is that?
That's the spring force.
553
00:39:38,520 --> 00:39:44,191
Now, do we agree what the
magnitude of the spring force
554
00:39:44,191 --> 00:39:47,552
is?
And then, we will argue about
555
00:39:47,552 --> 00:39:51,123
the direction.
[Nickel?] magnitude,
556
00:39:51,123 --> 00:39:55,009
look very closely.
I offset one by X2,
557
00:39:55,009 --> 00:39:59,000
and I offset the other by X1.
558
00:39:59,000 --> 00:40:04,000
559
00:40:04,000 --> 00:40:10,566
I was asking you a question.
You can't answer with a
560
00:40:10,566 --> 00:40:13,012
question.
Very good.
561
00:40:13,012 --> 00:40:19,450
The magnitude of that force is
K times X2 minus X1,
562
00:40:19,450 --> 00:40:22,025
nonnegotiable,
right?
563
00:40:22,025 --> 00:40:29,236
Now, if X2 is larger than X1,
do we agree that the spring
564
00:40:29,236 --> 00:40:35,290
force is in this direction?
And do we agree,
565
00:40:35,290 --> 00:40:40,048
then, that this spring force
must be in this direction if X2
566
00:40:40,048 --> 00:40:43,758
is larger than X1?
Well, that means I can leave
567
00:40:43,758 --> 00:40:48,516
everything the way it is now.
As long as I give this a minus
568
00:40:48,516 --> 00:40:53,032
sign that is a plus sign,
I'm OK because if X2 is smaller
569
00:40:53,032 --> 00:40:57,709
than X1, it will automatically
flip over, and my algebra is
570
00:40:57,709 --> 00:41:00,320
fine.
So, therefore,
571
00:41:00,320 --> 00:41:05,603
in my head I can just think of
X to being larger than X1,
572
00:41:05,603 --> 00:41:10,698
set up the differential
equations, and I no longer have
573
00:41:10,698 --> 00:41:16,169
to worry about the fact that
maybe X2 at certain moments in
574
00:41:16,169 --> 00:41:20,603
time is not larger than X1.
So, now I set up the
575
00:41:20,603 --> 00:41:24,094
differential equation,
MX1 double dot.
576
00:41:24,094 --> 00:41:28,622
So, that's this one.
Let's first do the pendulum.
577
00:41:28,622 --> 00:41:34,000
That is minus T times the sign
of theta one.
578
00:41:34,000 --> 00:41:38,765
T is MG, so it is minus MG
times X1 divided by L.
579
00:41:38,765 --> 00:41:43,829
Do we agree there is that
horizontal component here?
580
00:41:43,829 --> 00:41:48,794
The horizontal component of T
is driving it back to
581
00:41:48,794 --> 00:41:53,163
equilibrium minus sign.
FS, the spring force,
582
00:41:53,163 --> 00:41:56,340
is driving away from
equilibrium.
583
00:41:56,340 --> 00:42:02,000
So, it's going to get plus K
times X2 minus X1.
584
00:42:02,000 --> 00:42:07,182
Look at this.
Signs are very important now.
585
00:42:07,182 --> 00:42:11,131
Very important.
And the next one,
586
00:42:11,131 --> 00:42:14,587
X2 double dot,
it has, again,
587
00:42:14,587 --> 00:42:21,868
the restoring force due to the
pendulum which is minus MG X2
588
00:42:21,868 --> 00:42:29,395
divided by L because it's the
sign of theta two now that comes
589
00:42:29,395 --> 00:42:33,203
in.
And now it has the spring
590
00:42:33,203 --> 00:42:36,289
force, which now is a negative
sign.
591
00:42:36,289 --> 00:42:40,168
And so, now we get minus K
times X2 minus X1.
592
00:42:40,168 --> 00:42:44,223
And, when you have this down on
your next exam,
593
00:42:44,223 --> 00:42:48,983
you take a deep breath,
and you go over each individual
594
00:42:48,983 --> 00:42:53,303
step to make absolutely sure
that this is correct.
595
00:42:53,303 --> 00:42:58,151
If there's anything wrong with
this, you are dead in the
596
00:42:58,151 --> 00:43:01,148
waters.
The whole problem will fall
597
00:43:01,148 --> 00:43:05,021
apart.
It may not even be a simple
598
00:43:05,021 --> 00:43:08,690
harmonic oscillator.
You get something ridiculous.
599
00:43:08,690 --> 00:43:11,461
Signs are crucial here.
So, therefore,
600
00:43:11,461 --> 00:43:13,857
look at this again.
Take a pause.
601
00:43:13,857 --> 00:43:18,425
This one, if there's a positive
offset is indeed in a negative
602
00:43:18,425 --> 00:43:21,270
direction, a component of T sine
theta.
603
00:43:21,270 --> 00:43:25,838
This one, if X2 is larger than
X1 is indeed in that direction.
604
00:43:25,838 --> 00:43:29,207
And you notice,
if X2 becomes smaller than X1,
605
00:43:29,207 --> 00:43:34,000
well then, this becomes
automatically a negative.
606
00:43:34,000 --> 00:43:36,947
So, are you OK?
So this is fine.
607
00:43:36,947 --> 00:43:41,702
This one has a restoring force
due to the pendulum,
608
00:43:41,702 --> 00:43:47,027
which is T sine theta two
negative sine and the restoring
609
00:43:47,027 --> 00:43:51,211
force of the spring always
opposite this one:
610
00:43:51,211 --> 00:43:53,208
I'm happy.
I am happy.
611
00:43:53,208 --> 00:43:58,438
So, now, we are going to
rearrange this a little bit and
612
00:43:58,438 --> 00:44:05,000
we're going to introduce the
omega zero squared notation.
613
00:44:05,000 --> 00:44:13,947
So, I'm going to divide M out,
and so I'm going to get X2,
614
00:44:13,947 --> 00:44:20,540
X1 double dot.
And then, I'm going to bring
615
00:44:20,540 --> 00:44:28,389
the X1's to the left.
And so, I get plus omega zero
616
00:44:28,389 --> 00:44:34,392
squared times X1.
But you have a minus KX1.
617
00:44:34,392 --> 00:44:38,221
You divide by M.
So, you get plus omega S
618
00:44:38,221 --> 00:44:41,762
squared.
And, that's [both?] times X1.
619
00:44:41,762 --> 00:44:45,686
You notice that?
You both have an X1 term.
620
00:44:45,686 --> 00:44:50,950
And now, the X2 comes out.
The X2 has a plus sign on the
621
00:44:50,950 --> 00:44:54,300
right side, so it gets a minus
sign.
622
00:44:54,300 --> 00:44:58,128
So, I get minus omega S squared
times X2.
623
00:44:58,128 --> 00:45:01,000
And that is zero.
624
00:45:01,000 --> 00:45:06,000
625
00:45:06,000 --> 00:45:13,461
And then we do the next one.
We get X2 double dot.
626
00:45:13,461 --> 00:45:21,532
Ah, I get the same terms.
I get an omega zero squared.
627
00:45:21,532 --> 00:45:27,167
And I get an omega S squared
times X2.
628
00:45:27,167 --> 00:45:36,000
And then I get this plus;
minus times minus is plus.
629
00:45:36,000 --> 00:45:40,055
So I get minus omega S squared
times X1 equals zero.
630
00:45:40,055 --> 00:45:44,428
When you have reached this
point, you take a deep breath
631
00:45:44,428 --> 00:45:49,278
and you go over each little term
to make sure that you haven't
632
00:45:49,278 --> 00:45:54,207
accidentally slipped on a minus
sign because you slip up on one
633
00:45:54,207 --> 00:45:56,990
minus sign you're dead in the
water.
634
00:45:56,990 --> 00:46:02,000
It may not even become a simple
harmonic oscillation.
635
00:46:02,000 --> 00:46:04,776
So let me do that.
I agree with that.
636
00:46:04,776 --> 00:46:07,553
I agree with this.
I agree with that.
637
00:46:07,553 --> 00:46:11,256
I agree with that.
Differential equation is fine.
638
00:46:11,256 --> 00:46:15,961
Notice that I already did point
number one, and that already I
639
00:46:15,961 --> 00:46:19,663
did even number two.
I've set up the differential
640
00:46:19,663 --> 00:46:23,983
equation, Newton's Second Law.
So now comes number three.
641
00:46:23,983 --> 00:46:28,225
And number three means and
going to substitute in there,
642
00:46:28,225 --> 00:46:34,395
X1 is C1.
Cosine omega T and X2 is C2
643
00:46:34,395 --> 00:46:37,000
cosine omega T.
644
00:46:37,000 --> 00:46:43,000
645
00:46:43,000 --> 00:46:46,822
We go slowly;
I want you to follow each step
646
00:46:46,822 --> 00:46:51,888
and I'm going to work on here so
that you can see it high.
647
00:46:51,888 --> 00:46:56,866
So, I'm going to substitute
this in here so the X1 double
648
00:46:56,866 --> 00:47:00,866
dot will give me a minus omega
squared, right,
649
00:47:00,866 --> 00:47:07,000
because you get twice omega out
minus omega squared times C1.
650
00:47:07,000 --> 00:47:09,520
And, to hell with cosine omega
T.
651
00:47:09,520 --> 00:47:11,962
Why to hell with cosine omega
T?
652
00:47:11,962 --> 00:47:15,349
Because every term will have
cosine omega T.
653
00:47:15,349 --> 00:47:19,208
So you might as well divide out
right away, right?
654
00:47:19,208 --> 00:47:24,092
Every term that you're going to
have in here will have a cosine
655
00:47:24,092 --> 00:47:27,636
omega T in it.
So, I'll leave the cosine omega
656
00:47:27,636 --> 00:47:34,610
T already out now.
So, I get minus omega squared
657
00:47:34,610 --> 00:47:39,220
C1.
Then I get plus omega zero
658
00:47:39,220 --> 00:47:45,102
squared plus omega S squared
times C1.
659
00:47:45,102 --> 00:47:52,733
That is this term,
minus omega S squared times C2
660
00:47:52,733 --> 00:48:01,000
equals zero because,
remember, X2 has the C2.
661
00:48:01,000 --> 00:48:04,099
Am I going to fast?
Beautiful.
662
00:48:04,099 --> 00:48:09,872
So, I go to the second one.
The second one is X2 double
663
00:48:09,872 --> 00:48:13,293
dot.
So, I also get a minus omega
664
00:48:13,293 --> 00:48:17,889
squared times C2.
And then, here I get a C2.
665
00:48:17,889 --> 00:48:23,555
So, I get plus omega zero
squared plus omega S squared
666
00:48:23,555 --> 00:48:27,617
times C2.
And then, I get minus omega S
667
00:48:27,617 --> 00:48:34,400
squared times C1 equals zero.
And, when you have reached this
668
00:48:34,400 --> 00:48:37,653
point in your exam,
you take a deep breath.
669
00:48:37,653 --> 00:48:41,526
And, you make sure that all
your signs are correct.
670
00:48:41,526 --> 00:48:45,940
If not, you're dead in the
water and the problem will fall
671
00:48:45,940 --> 00:48:47,799
apart.
So, let's do that.
672
00:48:47,799 --> 00:48:51,207
Minus omega squared C1,
I can live with that,
673
00:48:51,207 --> 00:48:53,298
plus that term,
which is C1,
674
00:48:53,298 --> 00:48:57,945
yes, minus omega S squared C2.
Even if you make a little slip
675
00:48:57,945 --> 00:49:02,747
of the pen, and you change this
into a one and this into a two,
676
00:49:02,747 --> 00:49:06,000
it's all over,
of course.
677
00:49:06,000 --> 00:49:09,745
And then, the next equation,
minus omega squared C2.
678
00:49:09,745 --> 00:49:13,711
Then I have this times C2,
and then at minus that times
679
00:49:13,711 --> 00:49:15,327
C1.
We're almost there,
680
00:49:15,327 --> 00:49:18,632
even though it doesn't look
that way, does it?
681
00:49:18,632 --> 00:49:22,378
Remember what I said.
You cannot solve two equations
682
00:49:22,378 --> 00:49:26,858
with three unknowns there is no
way that you can solve for C1,
683
00:49:26,858 --> 00:49:30,531
C2, and for this omega,
which is what you're really
684
00:49:30,531 --> 00:49:34,252
after.
This is the omega you are
685
00:49:34,252 --> 00:49:37,405
after.
But you cancel for C1 divided
686
00:49:37,405 --> 00:49:42,810
by C2 and omega knowing that,
already, I'm going to eliminate
687
00:49:42,810 --> 00:49:46,504
C1 over C2.
And, you'll see how I do that.
688
00:49:46,504 --> 00:49:50,468
C1 divided by C2:
I go to the first equation.
689
00:49:50,468 --> 00:49:53,531
This is not so hard what I'm
doing.
690
00:49:53,531 --> 00:49:57,045
I put in my mind this on the
right side.
691
00:49:57,045 --> 00:50:02,000
So, I get a plus omega S
squared times C2.
692
00:50:02,000 --> 00:50:06,906
And then, I divide C1 by C2.
And so, when I get then is the
693
00:50:06,906 --> 00:50:10,374
following.
I get omega S squared up stairs
694
00:50:10,374 --> 00:50:14,435
which is this one.
And all this comes downstairs,
695
00:50:14,435 --> 00:50:19,087
minus omega squared plus omega
zero squared plus omega S
696
00:50:19,087 --> 00:50:20,779
squared.
Do we agree?
697
00:50:20,779 --> 00:50:25,347
That's the first equation.
I've simply written it as C1
698
00:50:25,347 --> 00:50:28,308
divided by C2.
I can always do that,
699
00:50:28,308 --> 00:50:32,531
right?
Bring the C1's to one side,
700
00:50:32,531 --> 00:50:36,000
the C2's to one side,
and divide them.
701
00:50:36,000 --> 00:50:39,843
That's this.
And, I'm going to do the same
702
00:50:39,843 --> 00:50:44,625
with the next equation.
I bring this up to the right
703
00:50:44,625 --> 00:50:48,562
side, that I have C1's there and
C2's here.
704
00:50:48,562 --> 00:50:54,187
And now, this becomes upstairs.
So, I get minus omega squared
705
00:50:54,187 --> 00:50:58,312
plus omega zero squared plus
omega S squared,
706
00:50:58,312 --> 00:51:03,000
and I get downstairs omega S
squared.
707
00:51:03,000 --> 00:51:07,614
And now, I have eliminated C1
and C2 because if I solve this
708
00:51:07,614 --> 00:51:10,586
equation, I get my solutions for
omega.
709
00:51:10,586 --> 00:51:13,871
This is one equation.
There is one unknown.
710
00:51:13,871 --> 00:51:16,452
That's omega.
You've got to admit,
711
00:51:16,452 --> 00:51:19,033
and there better be two
solutions.
712
00:51:19,033 --> 00:51:23,648
There better be an omega minus,
and there better be an omega
713
00:51:23,648 --> 00:51:27,871
plus coming out of this
untouched by young human hands.
714
00:51:27,871 --> 00:51:34,369
I must find two solutions.
Let's first make sure that this
715
00:51:34,369 --> 00:51:37,956
is correct, and the answer is
yes.
716
00:51:37,956 --> 00:51:43,173
So, what do I do now?
Well, I'm going to simplify
717
00:51:43,173 --> 00:51:48,065
this one step further,
which is now very easy.
718
00:51:48,065 --> 00:51:52,739
I multiply this by this,
and this with this.
719
00:51:52,739 --> 00:51:59,043
So, we get omega S through the
power of four is minus omega
720
00:51:59,043 --> 00:52:06,000
squared plus omega zero squared
plus omega S squared.
721
00:52:06,000 --> 00:52:13,033
One equation with one unknown;
omega is the only unknown.
722
00:52:13,033 --> 00:52:17,429
Take the square root left and
right.
723
00:52:17,429 --> 00:52:24,589
So, I get minus omega squared
plus omega zero squared plus
724
00:52:24,589 --> 00:52:32,000
omega S squared equals plus or
minus omega S squared.
725
00:52:32,000 --> 00:52:38,279
Do not forget the plus or minus
because the square root of omega
726
00:52:38,279 --> 00:52:43,064
S to the fourth is plus or minus
omega S squared.
727
00:52:43,064 --> 00:52:47,948
And now, you're going to see
the light of the day.
728
00:52:47,948 --> 00:52:53,032
You won't believe this.
I bring omega squared to the
729
00:52:53,032 --> 00:52:59,012
other side, and I get omega zero
squared plus omega S squared
730
00:52:59,012 --> 00:53:05,106
minus or plus omega S squared.
And, the simplicity is
731
00:53:05,106 --> 00:53:09,127
overpowering.
I feel it over my whole body.
732
00:53:09,127 --> 00:53:13,914
It is unbelievable.
When I have a minus sign here I
733
00:53:13,914 --> 00:53:18,893
find that omega is omega zero.
That's my omega minus.
734
00:53:18,893 --> 00:53:22,436
That's the one.
That's my omega minus.
735
00:53:22,436 --> 00:53:28,276
When there is a plus sign here,
I find exactly what we predict
736
00:53:28,276 --> 00:53:32,916
before.
I find that omega plus is the
737
00:53:32,916 --> 00:53:38,648
square root of omega zero
squared plus two omega S squared
738
00:53:38,648 --> 00:53:42,167
because the plus sign make this
two.
739
00:53:42,167 --> 00:53:45,988
Isn't that elegant?
Isn't it beautiful?
740
00:53:45,988 --> 00:53:51,117
So, they just pop out.
And I predict that there will
741
00:53:51,117 --> 00:53:54,737
be two solutions.
You have them both.
742
00:53:54,737 --> 00:54:00,368
So, by substituting this in
there, automatically come out
743
00:54:00,368 --> 00:54:06,000
two solutions.
But, what now is C1 over C2?
744
00:54:06,000 --> 00:54:12,620
We haven't solved for that yet.
But I promise you can always
745
00:54:12,620 --> 00:54:18,567
solve it in terms of omega and
in terms of C1 over C2.
746
00:54:18,567 --> 00:54:21,933
Any volunteers?
Any volunteers?
747
00:54:21,933 --> 00:54:27,881
Look at the blackboard.
It's somewhere hidden in those
748
00:54:27,881 --> 00:54:31,471
equations.
I said we are going to
749
00:54:31,471 --> 00:54:37,082
eliminate C1 over C2.
You think C1 over C2 likes to
750
00:54:37,082 --> 00:54:41,884
be eliminated?
Let's call it back.
751
00:54:41,884 --> 00:54:47,242
And look at this equation.
Substitute in here for omega
752
00:54:47,242 --> 00:54:50,615
squared.
Substitute in omega minus.
753
00:54:50,615 --> 00:54:54,285
Then you get an answer for C1
over C2.
754
00:54:54,285 --> 00:54:57,460
What do you think that answer
is?
755
00:54:57,460 --> 00:55:00,634
What do you think that answer
is?
756
00:55:00,634 --> 00:55:02,817
Plus one.
But even one.
757
00:55:02,817 --> 00:55:07,300
Plus one.
So, you get C1 over C2 is plus
758
00:55:07,300 --> 00:55:09,984
one.
If you take this omega zero and
759
00:55:09,984 --> 00:55:13,819
you put it in here,
you get omega zero squared with
760
00:55:13,819 --> 00:55:16,119
a minus sign,
with a plus sign.
761
00:55:16,119 --> 00:55:20,030
You have omega S squared
divided by omega S squared,
762
00:55:20,030 --> 00:55:23,558
and it's plus one.
And now, you take the second
763
00:55:23,558 --> 00:55:26,012
solution, and you put it in
here.
764
00:55:26,012 --> 00:55:30,000
What do you think you're going
to find?
765
00:55:30,000 --> 00:55:33,518
Minus one.
C1 over C2 is now minus one.
766
00:55:33,518 --> 00:55:39,259
And so, what you have seen now
is that the general recipe comes
767
00:55:39,259 --> 00:55:43,981
up with frequencies,
comes up with the ratios of the
768
00:55:43,981 --> 00:55:48,888
amplitudes, not with the
individual amplitudes because
769
00:55:48,888 --> 00:55:52,222
you don't know the initial
condition.
770
00:55:52,222 --> 00:55:56,296
It comes up with the ratio of
the amplitudes,
771
00:55:56,296 --> 00:56:01,574
and now you can write any
solution, provided that you know
772
00:56:01,574 --> 00:56:06,842
the initial conditions.
If you know the initial
773
00:56:06,842 --> 00:56:09,807
conditions, then you can also
find C1.
774
00:56:09,807 --> 00:56:12,932
And therefore,
since you know the ratio,
775
00:56:12,932 --> 00:56:17,099
you automatically have C2.
And, you can find this C1.
776
00:56:17,099 --> 00:56:21,666
So, I'll give this one a minus
sign, and this a plus sign.
777
00:56:21,666 --> 00:56:24,951
And then, you know the ratio in
that mode.
778
00:56:24,951 --> 00:56:30,000
So, now you may think
erroneously that life is easy.
779
00:56:30,000 --> 00:56:35,935
That's far from the truth.
Suppose you have something as
780
00:56:35,935 --> 00:56:38,741
simple as this:
L, M, L, M.
781
00:56:38,741 --> 00:56:44,244
It's called a double pendulum.
There is no symmetry.
782
00:56:44,244 --> 00:56:50,179
I want to test your intuition.
There are going to be two
783
00:56:50,179 --> 00:56:55,791
modes, two normal modes:
a lower one and a higher one
784
00:56:55,791 --> 00:57:02,440
because they are two objects.
In the lowest frequency,
785
00:57:02,440 --> 00:57:08,135
use your hands in your legs.
What will this pendulum look
786
00:57:08,135 --> 00:57:11,593
like?
You are all doing sort of the
787
00:57:11,593 --> 00:57:15,254
right thing.
Would it look like this,
788
00:57:15,254 --> 00:57:19,525
or would it look like this?
In other words,
789
00:57:19,525 --> 00:57:25,627
let me make this one a little
bit, let me make the difference
790
00:57:25,627 --> 00:57:29,694
a little larger.
Would it look like this,
791
00:57:29,694 --> 00:57:36,000
or would it look like this?
Slightly exaggerated.
792
00:57:36,000 --> 00:57:41,636
Who is for this one?
OK, that means that the ratio,
793
00:57:41,636 --> 00:57:46,370
C2 over C1, is going to be plus
two, right?
794
00:57:46,370 --> 00:57:50,992
Who is for this one?
That's the way it is,
795
00:57:50,992 --> 00:57:56,290
believe it or not.
C2 over C1 is going to be one
796
00:57:56,290 --> 00:58:02,152
plus the square root of two.
And, it will take you 30
797
00:58:02,152 --> 00:58:07,000
minutes of grinding to find
that.
798
00:58:07,000 --> 00:58:12,235
And I'm not exaggerating when I
say 30 minutes.
799
00:58:12,235 --> 00:58:18,951
You have to go to the whole
procedure and then you will find
800
00:58:18,951 --> 00:58:24,528
that this ratio is 2.4.
And I will demonstrate it.
801
00:58:24,528 --> 00:58:29,536
The highest mode must be
something like this,
802
00:58:29,536 --> 00:58:33,292
right?
Because they must be out of
803
00:58:33,292 --> 00:58:37,947
pace.
Any idea of the ratio,
804
00:58:37,947 --> 00:58:41,832
C1 over C2?
I'll call this C1,
805
00:58:41,832 --> 00:58:45,583
call this C2?
Any one of you,
806
00:58:45,583 --> 00:58:49,066
any intuition?
Minus, yeah?
807
00:58:49,066 --> 00:58:52,952
Boy, you're good.
You're good.
808
00:58:52,952 --> 00:58:58,980
It is minus 2.4.
This one will be much further
809
00:58:58,980 --> 00:59:05,148
away than this one.
Do any one of you want to make
810
00:59:05,148 --> 00:59:09,086
a guess what omega minus is?
Anyone want to make a guess
811
00:59:09,086 --> 00:59:12,237
what omega plus is?
No way on Earth have you,
812
00:59:12,237 --> 00:59:15,459
or I, or anyone else can look
at this and say,
813
00:59:15,459 --> 00:59:18,180
oh yes, of course,
omega minus is this.
814
00:59:18,180 --> 00:59:21,259
It's a long road.
30 minutes of calculating,
815
00:59:21,259 --> 00:59:24,481
and then out of that popped
these frequencies.
816
00:59:24,481 --> 00:59:27,202
And out of that,
then, pops the ratios.
817
00:59:27,202 --> 00:59:32,000
And now I'm going to
demonstrate you just this case.
818
00:59:32,000 --> 00:59:35,780
And the way I'm going to do
that, even though we have not
819
00:59:35,780 --> 00:59:39,493
discussed driven coupled
oscillators, in order to set it
820
00:59:39,493 --> 00:59:42,869
off in these normal modes,
a normal modes is also a
821
00:59:42,869 --> 00:59:45,097
resonance frequency.
We call that,
822
00:59:45,097 --> 00:59:48,742
often, natural frequency.
It's what the system likes to
823
00:59:48,742 --> 00:59:51,983
do if you leave it alone.
And, it is easy for me,
824
00:59:51,983 --> 00:59:55,628
when I just tease it a little
bit, to excite it in that
825
00:59:55,628 --> 00:59:57,586
resonance mode.
So, therefore,
826
00:59:57,586 --> 1:00:00,624
I am going to drive it,
but only very briefly.
827
1:00:00,624 --> 1:00:04,000
And then, I will leave it
alone.
828
1:00:04,000 --> 1:00:08,302
And I will get it into the
state that you will see this
829
1:00:08,302 --> 1:00:11,170
that is [really?] in the normal
mode.
830
1:00:11,170 --> 1:00:13,799
There's one at only one
frequency.
831
1:00:13,799 --> 1:00:17,464
They come to a halt at the same
moment in time.
832
1:00:17,464 --> 1:00:21,208
They are in phase,
but you will clearly see that
833
1:00:21,208 --> 1:00:24,554
this distance is more than
double this one.
834
1:00:24,554 --> 1:00:27,024
And then, I will go to this
one.
835
1:00:27,024 --> 1:00:30,450
I will try that.
All right, double pendulum.
836
1:00:30,450 --> 1:00:41,000
Oh, my goodness.
This is a double pendulum. Yeah.
837
1:00:41,000 --> 1:00:48,000
838
1:00:48,000 --> 1:00:51,076
So, I have to drive it a
little.
839
1:00:51,076 --> 1:00:55,541
But it's really only very
little at resonance.
840
1:00:55,541 --> 1:00:58,717
And then, I will stop writing
it.
841
1:00:58,717 --> 1:01:03,412
This is the mode.
Do you see that the bottom one
842
1:01:03,412 --> 1:01:06,469
is further away than twice the
top one?
843
1:01:06,469 --> 1:01:10,410
This is a normal mode.
OK, you see that it's not a
844
1:01:10,410 --> 1:01:13,868
straight line?
Now, it's hard to say that it
845
1:01:13,868 --> 1:01:17,487
is one plus the square root of
two, of course,
846
1:01:17,487 --> 1:01:20,544
from your seats.
But that's what it is.
847
1:01:20,544 --> 1:01:24,163
And now, I will excite the
second normal mode,
848
1:01:24,163 --> 1:01:26,656
which is a resonance.
That's it.
849
1:01:26,656 --> 1:01:31,000
And now, I can almost stop
swinging it.
850
1:01:31,000 --> 1:01:35,570
You notice that your upper one
has a much larger amplitude than
851
1:01:35,570 --> 1:01:38,739
the lower one,
and the ratio is one plus the
852
1:01:38,739 --> 1:01:42,351
square root of two.
Did you notice they are out of
853
1:01:42,351 --> 1:01:44,267
phase?
It's the minus sign.
854
1:01:44,267 --> 1:01:47,511
They are out of phase.
And in the other mode,
855
1:01:47,511 --> 1:01:51,196
they were in phase.
And so, all of that will follow
856
1:01:51,196 --> 1:01:54,439
from the recipe:
if I gave this problem on an
857
1:01:54,439 --> 1:01:58,862
exam, you would have reasons to
kill me because it would take
858
1:01:58,862 --> 1:02:02,536
you too long.
I would certainly wear a
859
1:02:02,536 --> 1:02:06,119
bulletproof vest on campus.
That is easy because you have
860
1:02:06,119 --> 1:02:08,103
symmetry.
This doesn't have that
861
1:02:08,103 --> 1:02:09,000
symmetry.
862
1:02:09,000 --> 1:02:14,000
863
1:02:14,000 --> 1:02:16,345
Yeah, C2 over C1.
This is C2.
864
1:02:16,345 --> 1:02:19,947
I haven't specified what I call
one and two.
865
1:02:19,947 --> 1:02:24,219
Is that the problem?
I call this one and I call this
866
1:02:24,219 --> 1:02:26,816
two.
So, C2 over C1 is plus 2.4.
867
1:02:26,816 --> 1:02:32,011
It's good that you asked that.
And here, I changed the
868
1:02:32,011 --> 1:02:36,997
sequence because I knew that
this one was larger than this
869
1:02:36,997 --> 1:02:40,058
one.
So here, you have C1 over C2 is
870
1:02:40,058 --> 1:02:43,294
minus 2.4.
The minus sign means out of
871
1:02:43,294 --> 1:02:46,530
phase.
Yeah: good that you asked that.
872
1:02:46,530 --> 1:02:48,892
Very good.
Well, I'm slowly,
873
1:02:48,892 --> 1:02:54,227
slowly going to turn you into
experts, and now we are going to
874
1:02:54,227 --> 1:03:00,000
try something else to see how
good your intuition is.
875
1:03:00,000 --> 1:03:03,458
I want you to understand,
though, that your intuition is
876
1:03:03,458 --> 1:03:06,413
no better, no worse than my own.
In other words,
877
1:03:06,413 --> 1:03:09,997
there is no way that I could
have done any better than you
878
1:03:09,997 --> 1:03:11,884
did.
When I was in high school,
879
1:03:11,884 --> 1:03:14,902
or whatever it was,
maybe in college when I first
880
1:03:14,902 --> 1:03:17,920
saw this, my first impression
was, what the hell?
881
1:03:17,920 --> 1:03:20,624
A straight line?
But if you give it a little
882
1:03:20,624 --> 1:03:23,957
thought, you will come to the
conclusion it cannot be.
883
1:03:23,957 --> 1:03:26,661
Then, it ultimately comes out
this solution.
884
1:03:26,661 --> 1:03:31,000
So, don't feel bad if your
intuition lets you down.
885
1:03:31,000 --> 1:03:35,562
I'm as bad as you are when it
comes to that.
886
1:03:35,562 --> 1:03:42,247
So, now we're going to evaluate
the system, which would take you
887
1:03:42,247 --> 1:03:47,765
an hour and a half to work out
with a general recipe.
888
1:03:47,765 --> 1:03:51,903
And that is four springs and
three cars.
889
1:03:51,903 --> 1:03:56,678
One car, two cars,
three cars, and all springs
890
1:03:56,678 --> 1:04:01,028
have spring constant,
K, nicely symmetric,
891
1:04:01,028 --> 1:04:07,436
you would think.
And all objects have mass M.
892
1:04:07,436 --> 1:04:13,449
And, I want to know,
I want to see whether we have
893
1:04:13,449 --> 1:04:20,445
any intuition without being too
quantitative for the three
894
1:04:20,445 --> 1:04:26,581
normal mode solutions.
There must be an omega minus
895
1:04:26,581 --> 1:04:33,074
where all three are in phase.
Then, there must be an omega
896
1:04:33,074 --> 1:04:37,370
plus, which is more complicated.
And then, there must be an
897
1:04:37,370 --> 1:04:40,629
omega plus, plus,
which is the highest of all
898
1:04:40,629 --> 1:04:42,777
three.
In the highest possible
899
1:04:42,777 --> 1:04:46,851
frequency, every object,
two adjacent objects are always
900
1:04:46,851 --> 1:04:49,962
out of phase.
So, if you have 20 objects in
901
1:04:49,962 --> 1:04:53,518
the highest mode,
number one is out of phase with
902
1:04:53,518 --> 1:04:56,629
number two.
Number two is out of phase with
903
1:04:56,629 --> 1:04:59,666
number three.
Number three is out of phase
904
1:04:59,666 --> 1:05:03,000
with number four,
and so on.
905
1:05:03,000 --> 1:05:05,569
Let's first look at omega
minus.
906
1:05:05,569 --> 1:05:10,292
I will even dare asking you
whether you have any idea what
907
1:05:10,292 --> 1:05:13,193
omega minus is.
You don't have that;
908
1:05:13,193 --> 1:05:17,171
I don't have that.
Where would your object be if,
909
1:05:17,171 --> 1:05:20,734
for instance,
I displaced the first one over
910
1:05:20,734 --> 1:05:24,795
a distance which I just
normalized to be plus one?
911
1:05:24,795 --> 1:05:28,856
I call that plus one.
Any idea where this one will
912
1:05:28,856 --> 1:05:31,425
be?
Any idea where that one will
913
1:05:31,425 --> 1:05:35,453
be?
What to think the second one
914
1:05:35,453 --> 1:05:36,942
will be?
Plus one.
915
1:05:36,942 --> 1:05:39,658
Will be the third one?
Plus one.
916
1:05:39,658 --> 1:05:42,725
Wrong.
Why does it have to be wrong?
917
1:05:42,725 --> 1:05:48,157
Suppose this one is a plus one.
And, suppose this one is also a
918
1:05:48,157 --> 1:05:51,574
plus one.
Then, this spring is no longer
919
1:05:51,574 --> 1:05:55,779
than it wants to be,
and this spring is no longer
920
1:05:55,779 --> 1:06:00,072
than it wants to be.
So, there is no force on this
921
1:06:00,072 --> 1:06:03,866
object.
So, that's not possible.
922
1:06:03,866 --> 1:06:08,533
The object cannot go anywhere.
Any suggestions which of those
923
1:06:08,533 --> 1:06:10,477
is wrong?
Is this correct,
924
1:06:10,477 --> 1:06:13,822
and it's plus one?
How about the middle one?
925
1:06:13,822 --> 1:06:17,477
Where do you want it,
a little further away or a
926
1:06:17,477 --> 1:06:20,277
little less?
All have to be in phase.
927
1:06:20,277 --> 1:06:24,944
Don't use the word minus sign.
If you use the word minus sign
928
1:06:24,944 --> 1:06:30,000
then, how can you do that?
They have to be in phase.
929
1:06:30,000 --> 1:06:37,690
They must be on this side.
Do we make the distance larger
930
1:06:37,690 --> 1:06:40,849
or smaller?
It's larger.
931
1:06:40,849 --> 1:06:48,815
Should I tell you what it is?
Should I tell you what it is?
932
1:06:48,815 --> 1:06:52,248
You want to know?
Tell me.
933
1:06:52,248 --> 1:06:56,369
You really want to know,
right?
934
1:06:56,369 --> 1:06:59,939
I knew it.
Is that obvious?
935
1:06:59,939 --> 1:07:04,276
No.
Now, the next one.
936
1:07:04,276 --> 1:07:09,588
The next one is interesting.
I will, again,
937
1:07:09,588 --> 1:07:16,039
put this one in plus one
because that is my starting
938
1:07:16,039 --> 1:07:20,592
point.
I always put that at plus one.
939
1:07:20,592 --> 1:07:24,640
Now what?
This one you can guess.
940
1:07:24,640 --> 1:07:30,458
You really can.
Think now about the symmetry of
941
1:07:30,458 --> 1:07:34,000
the system.
Yeah?
942
1:07:34,000 --> 1:07:36,006
Very good.
Very good.
943
1:07:36,006 --> 1:07:41,422
The middle one stays still,
and this one comes to minus
944
1:07:41,422 --> 1:07:43,629
one.
Now the third one.
945
1:07:43,629 --> 1:07:49,045
Remember, the highest mode,
neighbors are always out of
946
1:07:49,045 --> 1:07:53,960
phase with each other no matter
how many you have.
947
1:07:53,960 --> 1:07:59,778
So, it's already certain that
this one must be on the side,
948
1:07:59,778 --> 1:08:07,000
and it's already certain that
this one must be on that side.
949
1:08:07,000 --> 1:08:11,847
Any volunteers?
Would you like to try that
950
1:08:11,847 --> 1:08:16,221
again?
Would you do plus one and minus
951
1:08:16,221 --> 1:08:20,004
one?
Help me put this one at plus
952
1:08:20,004 --> 1:08:25,561
one, sorry, minus one,
and this one at plus one,
953
1:08:25,561 --> 1:08:31,000
or shall we change the minus
one here?
954
1:08:31,000 --> 1:08:35,380
Don't feel bad.
I wouldn't have been able to do
955
1:08:35,380 --> 1:08:39,000
either.
Shall I tell you what it is?
956
1:08:39,000 --> 1:08:45,000
957
1:08:45,000 --> 1:08:49,299
Now, I can demonstrate this.
And the way I'm going to
958
1:08:49,299 --> 1:08:53,846
demonstrate this is with the air
track that I have here.
959
1:08:53,846 --> 1:08:58,807
See, the beauty with this air
track is that I can place these
960
1:08:58,807 --> 1:09:02,940
cars wherever I want,
and they're not going to move
961
1:09:02,940 --> 1:09:08,071
because I don't have any air.
Then, I turn on the air and
962
1:09:08,071 --> 1:09:10,757
they are and immediately free to
go.
963
1:09:10,757 --> 1:09:13,826
In other words,
if I offset this one in a
964
1:09:13,826 --> 1:09:16,818
positive direction over a
distance, one,
965
1:09:16,818 --> 1:09:19,196
we've just called this
positive.
966
1:09:19,196 --> 1:09:23,109
I apologize for that because I
called this positive.
967
1:09:23,109 --> 1:09:27,099
And the other one has to be
offset by one in the same
968
1:09:27,099 --> 1:09:31,473
direction, which I just did.
But this one has to be offset
969
1:09:31,473 --> 1:09:36,000
by the square root of two,
which I just did.
970
1:09:36,000 --> 1:09:40,095
And now, turn on the air,
and I just have exactly the
971
1:09:40,095 --> 1:09:43,089
right initial conditions for
this mode.
972
1:09:43,089 --> 1:09:47,578
So, it is going to oscillate in
the superposition of three
973
1:09:47,578 --> 1:09:50,335
normal modes.
But there is only one,
974
1:09:50,335 --> 1:09:54,825
and that's the one you're going
to see because I chose the
975
1:09:54,825 --> 1:09:59,000
initial conditions.
Are you ready for this?
976
1:09:59,000 --> 1:10:01,974
Now, admit it.
This is fantastic.
977
1:10:01,974 --> 1:10:06,901
They are all in phase.
That was a criterion for normal
978
1:10:06,901 --> 1:10:09,039
mode.
They are in phase.
979
1:10:09,039 --> 1:10:13,315
They come to a halt at the same
moment in time.
980
1:10:13,315 --> 1:10:18,335
They all have the same
frequency, and amplitude here is
981
1:10:18,335 --> 1:10:23,354
the square of two times larger.
That is not so obvious,
982
1:10:23,354 --> 1:10:28,002
but you see this is the
criterion for normal modes.
983
1:10:28,002 --> 1:10:32,000
Same frequency,
and in phase.
984
1:10:32,000 --> 1:10:35,297
This one, OK,
we are going to put this one at
985
1:10:35,297 --> 1:10:37,770
zero.
I'm going to put this one at
986
1:10:37,770 --> 1:10:40,617
plus one.
And I'm going to put this one
987
1:10:40,617 --> 1:10:43,839
at minus one.
Now, this one is offset by one
988
1:10:43,839 --> 1:10:47,286
in this direction.
This one is offset by one in
989
1:10:47,286 --> 1:10:50,434
this direction,
and this one is not offset.
990
1:10:50,434 --> 1:10:54,031
I turn on the air;
what do you think will happen?
991
1:10:54,031 --> 1:10:57,103
That's right,
it's going to oscillate in a
992
1:10:57,103 --> 1:11:01,000
superposition of three normal
modes.
993
1:11:01,000 --> 1:11:06,000
But there is only one.
That's the omega plus.
994
1:11:06,000 --> 1:11:11,000
995
1:11:11,000 --> 1:11:16,588
Impressive, isn't it?
Next one, this one.
996
1:11:16,588 --> 1:11:22,455
So, we are first going to put
this at zero,
997
1:11:22,455 --> 1:11:28,463
this at zero.
And we're going to put this at
998
1:11:28,463 --> 1:11:32,182
zero.
And now, what we have to do,
999
1:11:32,182 --> 1:11:35,379
this has to be put over plus
one plus one.
1000
1:11:35,379 --> 1:11:39,744
The other one has to be put
minus the square root of two.
1001
1:11:39,744 --> 1:11:44,422
And we have measured that here.
We've come so close that this
1002
1:11:44,422 --> 1:11:46,916
spring is not going to like
that.
1003
1:11:46,916 --> 1:11:51,594
And now, this one is going to
be, again, plus one in the same
1004
1:11:51,594 --> 1:11:55,569
direction as this one.
That's the highest frequency.
1005
1:11:55,569 --> 1:11:59,000
And there we go.
Are you ready?
1006
1:11:59,000 --> 1:12:06,000
1007
1:12:06,000 --> 1:12:07,919
Three, two, one,
zero.
1008
1:12:07,919 --> 1:12:13,220
One, and only one frequency.
And, every adjacent one is out
1009
1:12:13,220 --> 1:12:17,516
of phase with the other.
These are out of phase.
1010
1:12:17,516 --> 1:12:23,000
But these are in phase because
these are out of phase.
1011
1:12:23,000 --> 1:12:34,000
1012
1:12:34,000 --> 1:12:37,623
Now, I had one student
calculate this for me.
1013
1:12:37,623 --> 1:12:41,823
And I'll be frank with you,
I even paid him for that
1014
1:12:41,823 --> 1:12:44,952
because I was too lazy to do it
myself.
1015
1:12:44,952 --> 1:12:49,729
I put the job on the market,
and he came within a few days.
1016
1:12:49,729 --> 1:12:54,670
There were a few people that
said, oh, and an A plus for 8.03
1017
1:12:54,670 --> 1:12:58,376
and he was going to do that in
no time at all.
1018
1:12:58,376 --> 1:13:02,000
20 hours later,
he had it right.
1019
1:13:02,000 --> 1:13:07,665
It cost me a fortune.
Now, I'm going to test you once
1020
1:13:07,665 --> 1:13:12,459
more to show you that these
things are not so
1021
1:13:12,459 --> 1:13:17,143
straightforward,
not to make you feel bad at
1022
1:13:17,143 --> 1:13:19,322
all.
On the contrary,
1023
1:13:19,322 --> 1:13:24,988
my intuition is no better.
I'm going to make a triple
1024
1:13:24,988 --> 1:13:30,000
pendulum.
This is a triple pendulum.
1025
1:13:30,000 --> 1:13:37,000
1026
1:13:37,000 --> 1:13:41,203
Omega minus,
what do you think it would look
1027
1:13:41,203 --> 1:13:44,428
like?
My favorite student is here.
1028
1:13:44,428 --> 1:13:48,240
Do you think it will be a
straight line?
1029
1:13:48,240 --> 1:13:53,323
I don't think so either.
I think what you're going to
1030
1:13:53,323 --> 1:13:58,308
see without making any
predictions about amplitudes,
1031
1:13:58,308 --> 1:14:03,000
exaggerated,
you're going to see this.
1032
1:14:03,000 --> 1:14:13,064
That's what I think.
No idea what it is.
1033
1:14:13,064 --> 1:14:19,000
How about omega plus?
1034
1:14:19,000 --> 1:14:26,000
1035
1:14:26,000 --> 1:14:31,816
This is not going to work.
It's a whole different system.
1036
1:14:31,816 --> 1:14:36,489
Two of them are probably going
to be in phase,
1037
1:14:36,489 --> 1:14:41,474
and one out of phase.
Which two will be in phase?
1038
1:14:41,474 --> 1:14:46,252
Will you see this?
Now, these two are in phase,
1039
1:14:46,252 --> 1:14:51,341
and this is out of phase.
Or really, you see this.
1040
1:14:51,341 --> 1:14:57,261
Now, these two are in phase,
and this one is out of phase.
1041
1:14:57,261 --> 1:15:03,588
Who is in favor of this?
Who is in favor of that?
1042
1:15:03,588 --> 1:15:09,303
I want to see the hands long.
Who is in favor of this?
1043
1:15:09,303 --> 1:15:14,264
Who is in favor of that?
Yeah, well, we'll see.
1044
1:15:14,264 --> 1:15:18,254
[LAUGHTER] How about the
highest mode?
1045
1:15:18,254 --> 1:15:24,401
That's the easiest one because
now they must go like this.
1046
1:15:24,401 --> 1:15:28,392
Now, of course,
the ratio, C1 over C2,
1047
1:15:28,392 --> 1:15:34,000
and C1 over C3,
that's a different story.
1048
1:15:34,000 --> 1:15:38,812
That is your 20 hours of
grinding if you get paid.
1049
1:15:38,812 --> 1:15:43,821
If you don't get paid,
you can probably do it in two
1050
1:15:43,821 --> 1:15:47,160
hours.
So, I'm going to demonstrate
1051
1:15:47,160 --> 1:15:51,187
this to you.
I'm going to find these three
1052
1:15:51,187 --> 1:15:54,526
modes.
Again, I do that by a little
1053
1:15:54,526 --> 1:15:58,651
bit of driving because of their
resonances.
1054
1:15:58,651 --> 1:16:02,973
And then, I want you to eyeball
these ratios,
1055
1:16:02,973 --> 1:16:08,431
C1 over C2, and C1 over C3.
Oh, no problem.
1056
1:16:08,431 --> 1:16:10,625
No problem.
There we go.
1057
1:16:10,625 --> 1:16:15,301
All right, so now,
this is not where the length is
1058
1:16:15,301 --> 1:16:18,832
the same.
So, to get this one into the
1059
1:16:18,832 --> 1:16:23,412
lowest mode is easy for me.
Just give it a swing,
1060
1:16:23,412 --> 1:16:26,656
and let it go.
And just do nothing.
1061
1:16:26,656 --> 1:16:29,805
Then, I will stop even driving
it.
1062
1:16:29,805 --> 1:16:32,000
That's it.
1063
1:16:32,000 --> 1:16:38,000
1064
1:16:38,000 --> 1:16:42,461
Now, it's nicely in that mode.
You see, the bottom one [has
1065
1:16:42,461 --> 1:16:45,230
it?].
Notice that the lines are never
1066
1:16:45,230 --> 1:16:48,846
in the same direction.
You see the offset of the
1067
1:16:48,846 --> 1:16:51,000
angles?
Can you all see that?
1068
1:16:51,000 --> 1:16:55,384
So, our first picture on the
blackboard is quite accurate.
1069
1:16:55,384 --> 1:17:00,000
Now, the second one,
that's harder for me to find.
1070
1:17:00,000 --> 1:17:04,070
I mean, we first make sure that
they are not moving.
1071
1:17:04,070 --> 1:17:07,262
OK, so I'm going to try to
search for it.
1072
1:17:07,262 --> 1:17:08,859
I got it.
This is it.
1073
1:17:08,859 --> 1:17:12,131
So, the first solution is the
correct one.
1074
1:17:12,131 --> 1:17:15,802
The upper two are in phase,
and the bottom one,
1075
1:17:15,802 --> 1:17:19,394
look at the large amplitude of
the bottom one.
1076
1:17:19,394 --> 1:17:23,704
That's not so intuitive.
The bottom one is out of phase
1077
1:17:23,704 --> 1:17:26,417
with the upper two.
Can you see it?
1078
1:17:26,417 --> 1:17:29,291
Normal mode,
why is it a normal mode?
1079
1:17:29,291 --> 1:17:34,000
Because they all have the same
frequency.
1080
1:17:34,000 --> 1:17:38,904
And, they are in phase or they
are out of phase.
1081
1:17:38,904 --> 1:17:44,643
Now, the highest frequency,
let me see whether I can get
1082
1:17:44,643 --> 1:17:46,000
that one.
1083
1:17:46,000 --> 1:17:52,000
1084
1:17:52,000 --> 1:18:05,063
Got it.
That's it.
1085
1:18:05,063 --> 1:18:46,557
Now, look how small the
amplitude of the bottom one is.
1086
1:18:46,557 --> 1:19:10,378
And, that can all be
calculated.
1087
1:19:10,378 --> 1:19:41,883
They are beginning to rotate a
little bit.
1088
1:19:41,883 --> 1:20:08,009
There we go.
That's the third mode.
1089
1:20:08,009 --> 1:20:48,735
So, now you know everything
there is to be known about
1090
1:20:48,735 --> 1:21:26,387
coupled oscillators.
The most important thing that
1091
1:21:26,387 --> 1:22:10,955
you probably have learned is
that if you really want to get
1092
1:22:10,955 --> 1:22:49,376
the general solution for a
system, it is hard work.
1093
1:22:49,376 --> 1:23:05,513
See you next Tuesday.
Have a good weekend.