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00:00:21,380 --> 00:00:23,240
PROFESSOR: Welcome back.

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00:00:23,240 --> 00:00:27,660
Today, we will
consider problems where

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00:00:27,660 --> 00:00:30,940
the solutions are
standing waves.

11
00:00:30,940 --> 00:00:36,550
So the first problem I'm going
to do is shown over here.

12
00:00:36,550 --> 00:00:47,298
What we have is a string
attached at both ends.

13
00:00:47,298 --> 00:00:50,990
The length of the
string is L. And it

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00:00:50,990 --> 00:00:55,800
is distorted in
this strange shape.

15
00:00:55,800 --> 00:00:57,720
The dimensions are
given everywhere.

16
00:00:57,720 --> 00:01:02,210
It's symmetric about
the middle, OK?

17
00:01:02,210 --> 00:01:07,040
Somehow miraculously,
it's held fixed.

18
00:01:07,040 --> 00:01:10,120
You could imagine maybe sort
of nails nailed here and here

19
00:01:10,120 --> 00:01:11,890
and fingers holding there.

20
00:01:11,890 --> 00:01:17,020
And then miraculously,
we let go, all right?

21
00:01:17,020 --> 00:01:20,950
And the question
is after we let go,

22
00:01:20,950 --> 00:01:25,210
what is the wavelength of
the lowest mode that is not

23
00:01:25,210 --> 00:01:32,860
excited, the amplitude of the
lowest mode which is excited?

24
00:01:32,860 --> 00:01:36,110
And finally, the
question is, what

25
00:01:36,110 --> 00:01:41,640
is the shape of this string
at this strange time?

26
00:01:41,640 --> 00:01:44,712
Now, then what do they tell us?

27
00:01:44,712 --> 00:01:49,180
That this is an ideal
lossless string,

28
00:01:49,180 --> 00:01:53,800
which has a mass per
unit length of mu.

29
00:01:53,800 --> 00:01:59,460
It has a uniform tension T.
For all the equations to work

30
00:01:59,460 --> 00:02:01,760
and the derivation
of the wave equation,

31
00:02:01,760 --> 00:02:08,090
you have to assume that the
angle is sine angle, et cetera,

32
00:02:08,090 --> 00:02:11,720
is equal like this, which
is-- you look at this

33
00:02:11,720 --> 00:02:13,100
and say, holy smoke.

34
00:02:13,100 --> 00:02:14,930
How can this be the case?

35
00:02:14,930 --> 00:02:19,870
Well, of course, this is
an idealized situation.

36
00:02:19,870 --> 00:02:23,310
In reality, this would
be very, very small

37
00:02:23,310 --> 00:02:25,240
compared to that, all right?

38
00:02:25,240 --> 00:02:31,042
And one would have to have
some curvature at these points.

39
00:02:31,042 --> 00:02:35,450
The differences will
be, the real situation

40
00:02:35,450 --> 00:02:38,420
from this idealised
one, will only

41
00:02:38,420 --> 00:02:41,150
differ in some very
high frequencies.

42
00:02:41,150 --> 00:02:45,760
So the first approximation,
one can almost

43
00:02:45,760 --> 00:02:49,520
imagine that it's
possible to set this up.

44
00:02:49,520 --> 00:02:52,010
And this assumption, as
I told you, initially

45
00:02:52,010 --> 00:02:58,970
is the system, at t
equals 0, is stationary.

46
00:02:58,970 --> 00:02:59,470
OK.

47
00:03:03,220 --> 00:03:10,430
The above, as I was just saying,
is really highly idealized.

48
00:03:10,430 --> 00:03:14,500
It is almost a
mathematical representation

49
00:03:14,500 --> 00:03:18,620
already of a real situation.

50
00:03:18,620 --> 00:03:20,790
So you could imagine
this almost being

51
00:03:20,790 --> 00:03:24,700
a mathematical description
of this situation.

52
00:03:24,700 --> 00:03:30,470
And what makes it physics
rather than the mathematics

53
00:03:30,470 --> 00:03:35,040
is that string obeys
the wave equation.

54
00:03:35,040 --> 00:03:38,470
It obeys this wave equation.

55
00:03:38,470 --> 00:03:46,010
Once I've told you that
that picture on that diagram

56
00:03:46,010 --> 00:03:50,420
satisfies the wave equation
where V is a constant given

57
00:03:50,420 --> 00:03:53,590
by this quantity, we've
now converted this problem

58
00:03:53,590 --> 00:03:54,750
completely to mathematics.

59
00:03:58,610 --> 00:04:03,930
One thing to notice, this
is the phase velocity

60
00:04:03,930 --> 00:04:08,080
of propagation of a
progressive wave on a string.

61
00:04:08,080 --> 00:04:11,150
So here, that time,
they ask us in part

62
00:04:11,150 --> 00:04:19,290
C-- is the shape of the string
at a time L over V from time

63
00:04:19,290 --> 00:04:21,209
equals 0.

64
00:04:21,209 --> 00:04:25,350
So we're now solving this
problem, mathematics problem.

65
00:04:29,870 --> 00:04:42,110
We know that if you have
a coupled oscillator,

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00:04:42,110 --> 00:04:46,700
there are normal mode
solutions, solutions

67
00:04:46,700 --> 00:04:49,010
where every oscillator
is oscillating

68
00:04:49,010 --> 00:04:53,170
with the same
frequency and phase.

69
00:04:53,170 --> 00:04:57,050
For a system, a
continuous system

70
00:04:57,050 --> 00:05:00,700
like this is, an infinite
number of identical oscillators

71
00:05:00,700 --> 00:05:06,910
in a straight line for a
string like that, the most

72
00:05:06,910 --> 00:05:12,540
general solution
will also be-- one

73
00:05:12,540 --> 00:05:16,610
will be able to represent
it as a superposition

74
00:05:16,610 --> 00:05:19,110
of normal modes.

75
00:05:19,110 --> 00:05:23,330
Of course, if you
look at such a system

76
00:05:23,330 --> 00:05:25,430
and you distort this
from equilibrium,

77
00:05:25,430 --> 00:05:28,950
you can set up an infinite
different varieties

78
00:05:28,950 --> 00:05:30,220
of solutions.

79
00:05:30,220 --> 00:05:33,330
But what I'm saying is that
any one of those solutions

80
00:05:33,330 --> 00:05:36,180
however complicated,
one could always

81
00:05:36,180 --> 00:05:41,520
reduce it to a superposition
of normal modes.

82
00:05:41,520 --> 00:05:45,180
In particular, if you
study the solutions

83
00:05:45,180 --> 00:05:50,620
of a system like this,
one dimensional system

84
00:05:50,620 --> 00:05:56,740
like that, you will find
that the normal modes are,

85
00:05:56,740 --> 00:06:00,830
in fact, standing waves
which are sinusoidal.

86
00:06:07,610 --> 00:06:16,210
So I've written for you here
a completely general solution

87
00:06:16,210 --> 00:06:17,405
to a string.

88
00:06:20,510 --> 00:06:26,390
And you can write it, as
I said, as a superposition

89
00:06:26,390 --> 00:06:31,855
of an infinite number of normal
modes, pretty complicated.

90
00:06:34,550 --> 00:06:38,270
It tells you at every
position on the string

91
00:06:38,270 --> 00:06:41,130
and at every time what
is the displacement.

92
00:06:41,130 --> 00:06:43,480
You can calculate the
transverse velocity

93
00:06:43,480 --> 00:06:45,710
by differentiating, et cetera.

94
00:06:45,710 --> 00:06:48,230
It just tells you everything
about the oscillations

95
00:06:48,230 --> 00:06:50,260
of a string.

96
00:06:50,260 --> 00:06:52,760
This is not any old string.

97
00:06:52,760 --> 00:06:55,520
Because we are told
some things about it.

98
00:06:55,520 --> 00:06:58,810
We are told it's
attached at both ends

99
00:06:58,810 --> 00:07:02,500
so that string has no
displacement in the Y

100
00:07:02,500 --> 00:07:08,770
direction at x equals 0 and
at x equals L. For all times

101
00:07:08,770 --> 00:07:18,280
t this equation will satisfy
that string-- doesn't matter

102
00:07:18,280 --> 00:07:21,760
even what distortion
it is-- provided

103
00:07:21,760 --> 00:07:27,060
we put some constraints
on these various constants

104
00:07:27,060 --> 00:07:28,510
such that this is true.

105
00:07:31,550 --> 00:07:33,850
Well, you can almost
do it in your head.

106
00:07:33,850 --> 00:07:41,370
If you look at this, all
these-- if for all times, at x

107
00:07:41,370 --> 00:07:49,980
equals 0, y is 0, this
term, B of N, has to be 0.

108
00:07:49,980 --> 00:07:51,790
So this, you could forget about.

109
00:07:51,790 --> 00:07:53,995
And you're down to this.

110
00:07:56,550 --> 00:07:58,310
Furthermore, you
know it has to be

111
00:07:58,310 --> 00:08:06,980
0 at x equals L. That puts
a constraint on k of n.

112
00:08:06,980 --> 00:08:14,410
So your solution
then is of this form.

113
00:08:14,410 --> 00:08:16,960
This is any arbitrary constant.

114
00:08:21,210 --> 00:08:24,320
For the wave equation,
you can show for a string

115
00:08:24,320 --> 00:08:29,880
that these two are related
through the phase velocity.

116
00:08:29,880 --> 00:08:32,440
And so once you've
established this,

117
00:08:32,440 --> 00:08:34,820
then we've got that constant.

118
00:08:34,820 --> 00:08:37,450
So you're still left with
quite a number of constants

119
00:08:37,450 --> 00:08:43,260
and an infinite series where
the An can be anything.

120
00:08:43,260 --> 00:08:44,710
This phase can be anything.

121
00:08:48,200 --> 00:08:50,470
It's always useful
to check yourself.

122
00:08:50,470 --> 00:08:51,980
There are some
things you can check.

123
00:08:51,980 --> 00:08:58,870
So for example, let's consider
the n equals 1 harmonic,

124
00:08:58,870 --> 00:09:03,660
that normal mode, the
first normal mode.

125
00:09:03,660 --> 00:09:08,390
If you look at this, it's
clear that the wavelength

126
00:09:08,390 --> 00:09:11,790
of that harmonic is 2L.

127
00:09:11,790 --> 00:09:18,790
If you look at this, the period
of that harmonic is 2L over v.

128
00:09:18,790 --> 00:09:23,900
Period is 1 over frequency,
or frequency is 1 over period.

129
00:09:23,900 --> 00:09:25,210
So the frequency is v/2L.

130
00:09:29,470 --> 00:09:36,080
We know that for harmonic
waves or standing waves

131
00:09:36,080 --> 00:09:40,090
that lambda times F
is the phase velocity.

132
00:09:40,090 --> 00:09:43,360
Let's check it here,
multiply lambda by F,

133
00:09:43,360 --> 00:09:45,980
and you get that
this is satisfied.

134
00:09:45,980 --> 00:09:47,730
So at least this
is a quick check

135
00:09:47,730 --> 00:09:51,070
that we haven't made
some stupid mistake.

136
00:09:51,070 --> 00:09:59,770
All right, so this
satisfies any string

137
00:09:59,770 --> 00:10:06,380
of length L which is
tied at both ends.

138
00:10:06,380 --> 00:10:09,830
But our string is
not any string.

139
00:10:09,830 --> 00:10:17,810
We've said that initially, at
t equals 0 for all positions,

140
00:10:17,810 --> 00:10:20,350
the string is stationary.

141
00:10:20,350 --> 00:10:26,020
So our description of it must
take into account the equation

142
00:10:26,020 --> 00:10:29,215
which describes this
shape of the string

143
00:10:29,215 --> 00:10:35,800
and it must have this
boundary condition, OK?

144
00:10:35,800 --> 00:10:40,510
So we take this
equation-- so far,

145
00:10:40,510 --> 00:10:42,880
we've reduced it to
that-- differentiate it

146
00:10:42,880 --> 00:10:45,980
with respect to time, get this.

147
00:10:45,980 --> 00:10:49,650
And that's an easy-- we
just have to differentiate--

148
00:10:49,650 --> 00:10:52,380
the cosine gives you minus sine.

149
00:10:52,380 --> 00:10:57,230
OK, so you end up with this.

150
00:10:57,230 --> 00:11:00,210
And this, we know is 0.

151
00:11:00,210 --> 00:11:05,705
It's 0 at-- I'm sorry,
this is a t equals 0.

152
00:11:05,705 --> 00:11:07,180
I'm sorry.

153
00:11:07,180 --> 00:11:13,490
We know that at t equals
0 at every position

154
00:11:13,490 --> 00:11:15,510
x-- we said that.

155
00:11:15,510 --> 00:11:19,060
We're holding this string
with nothing moving.

156
00:11:19,060 --> 00:11:20,930
At t equals 0, it's stationary.

157
00:11:20,930 --> 00:11:21,980
We then let go.

158
00:11:21,980 --> 00:11:23,780
So this is 0.

159
00:11:23,780 --> 00:11:26,310
And if you look
at this, this will

160
00:11:26,310 --> 00:11:32,710
be 0 at all values
of x at t equals 0,

161
00:11:32,710 --> 00:11:36,100
only if all values
of phi n is 0.

162
00:11:36,100 --> 00:11:40,370
So notice, gradually, by making
use of all the information we

163
00:11:40,370 --> 00:11:44,590
have about the situation--
what string it is, where it's

164
00:11:44,590 --> 00:11:50,540
attached, how long it is,
what is its motion at t

165
00:11:50,540 --> 00:11:55,740
equals 0-- we're gradually
getting rid of the constants

166
00:11:55,740 --> 00:11:59,400
or determining them.

167
00:11:59,400 --> 00:12:03,370
Now that phi n is
0, I go back to this

168
00:12:03,370 --> 00:12:05,050
and rewrite the equation.

169
00:12:05,050 --> 00:12:12,030
And now, we've ended up that
the general equation which

170
00:12:12,030 --> 00:12:18,530
describes a string
which has length L tied

171
00:12:18,530 --> 00:12:25,800
at both ends, stationary
at t equals 0 is this.

172
00:12:25,800 --> 00:12:29,810
An is still to be determined.

173
00:12:29,810 --> 00:12:34,200
We have not made use, so
far, of the information what

174
00:12:34,200 --> 00:12:36,900
is the shape of the
string at t equals 0.

175
00:12:36,900 --> 00:12:38,560
That's what we'll do next.

176
00:12:38,560 --> 00:12:42,230
But before I do this, I
just want to make a comment.

177
00:12:42,230 --> 00:12:45,360
In many books,
problems, et cetera,

178
00:12:45,360 --> 00:12:49,190
you'll find people starting
with this equation.

179
00:12:49,190 --> 00:12:53,500
And you'll be wondering, how
on Earth-- is this the most

180
00:12:53,500 --> 00:12:56,910
general equation
for string problems?

181
00:12:56,910 --> 00:12:58,480
No.

182
00:12:58,480 --> 00:13:02,090
The answer why they
started with this

183
00:13:02,090 --> 00:13:06,510
is, often without telling
you, doing all this analysis

184
00:13:06,510 --> 00:13:08,580
in their head.

185
00:13:08,580 --> 00:13:14,420
This is only true
for a situation

186
00:13:14,420 --> 00:13:18,550
where this string
is tied at both ends

187
00:13:18,550 --> 00:13:20,720
and initially stationary.

188
00:13:20,720 --> 00:13:23,260
That forces you to that.

189
00:13:23,260 --> 00:13:28,410
And An now is now
not determined.

190
00:13:28,410 --> 00:13:31,750
And in order to find out the
value of An-- in other words,

191
00:13:31,750 --> 00:13:35,340
the amplitude of the
different normal modes--

192
00:13:35,340 --> 00:13:41,340
we have to make use of the shape
of the string at t equals 0.

193
00:13:41,340 --> 00:13:42,710
And that's what we'll do now.

194
00:13:51,390 --> 00:13:55,080
All right, so now,
I'm going to make

195
00:13:55,080 --> 00:13:57,800
use of the last bit
of information I have.

196
00:13:57,800 --> 00:14:02,000
And since that defines the
physical situation completely,

197
00:14:02,000 --> 00:14:04,890
there'd better be enough
to find all the constants.

198
00:14:04,890 --> 00:14:06,550
There should be nothing left.

199
00:14:06,550 --> 00:14:09,720
They should be then
completely determined.

200
00:14:09,720 --> 00:14:14,270
So first of all, I said we
didn't make use of the shape

201
00:14:14,270 --> 00:14:15,630
at t equals 0.

202
00:14:15,630 --> 00:14:19,790
At t equals 0, cosine of 0 is 1.

203
00:14:19,790 --> 00:14:28,790
So what we know is that the
string shape is given by,

204
00:14:28,790 --> 00:14:32,420
from here, by this equation.

205
00:14:32,420 --> 00:14:39,820
So this tells us that--
this is a description

206
00:14:39,820 --> 00:14:43,320
of the shape of
the string in terms

207
00:14:43,320 --> 00:14:46,380
of an infinite series
of sinusoidal functions.

208
00:14:49,290 --> 00:15:00,360
OK, we notice this tells us the
shape of the string at every X

209
00:15:00,360 --> 00:15:03,280
position at t equals 0.

210
00:15:03,280 --> 00:15:07,540
So I look at that,
the original shape.

211
00:15:07,540 --> 00:15:10,870
And I can describe
it mathematically.

212
00:15:10,870 --> 00:15:17,170
y of x at time equals 0 is
0 at all points from one

213
00:15:17,170 --> 00:15:20,190
end of the string to where
the distortion starts.

214
00:15:20,190 --> 00:15:26,760
And it starts at 3/8 L if
you look at the diagram.

215
00:15:26,760 --> 00:15:35,530
Then from position x
of 3/8 L to x is 5/8 L,

216
00:15:35,530 --> 00:15:40,470
the string is straight again,
and its displaced by a distance

217
00:15:40,470 --> 00:15:46,530
H. And afterwards, past
the 5/8 L, it's back to 0.

218
00:15:46,530 --> 00:15:51,170
For a second, let's go
back and just look at that.

219
00:15:51,170 --> 00:15:54,690
So we're talking about
this shape, right?

220
00:15:54,690 --> 00:16:03,890
It's 0, 0 in Y, and then jumps
to H between 3/8 L and 5/8 L.

221
00:16:03,890 --> 00:16:11,040
This shape, when
you let go, that

222
00:16:11,040 --> 00:16:17,260
will determine the amplitude
of the different normal modes

223
00:16:17,260 --> 00:16:24,370
that are important in the
oscillation of that system.

224
00:16:24,370 --> 00:16:31,150
OK, so we must determine these
infinite number of constants.

225
00:16:31,150 --> 00:16:35,970
At first sight, it looks
like an impossible task

226
00:16:35,970 --> 00:16:39,550
if it wasn't for a
brilliant mathematician

227
00:16:39,550 --> 00:16:42,850
in the 18th century,
Fourier, who

228
00:16:42,850 --> 00:16:46,860
made a relatively simple
but incredibly important

229
00:16:46,860 --> 00:16:48,730
observation.

230
00:16:48,730 --> 00:16:53,960
When applied to this
problem, it is the following.

231
00:16:58,410 --> 00:17:03,850
He realized that one can
do the following trick.

232
00:17:08,230 --> 00:17:12,770
If you take this
function, sine n pi

233
00:17:12,770 --> 00:17:17,290
over L x to 1 which is
describing our shape

234
00:17:17,290 --> 00:17:22,220
over there, if you multiply
it-- and by the way,

235
00:17:22,220 --> 00:17:25,780
n-- I'm just reminding you--
goes from 1 to infinity.

236
00:17:25,780 --> 00:17:29,200
It's 1, 2, 3, 4,
5, 6, et cetera.

237
00:17:29,200 --> 00:17:33,350
If you take this function
and you multiply it

238
00:17:33,350 --> 00:17:39,330
by a similar function, but
with different value of m,

239
00:17:39,330 --> 00:17:46,120
then this integral from one
end of the string to the other,

240
00:17:46,120 --> 00:17:53,390
is equal to 0 if this,
the m you've chosen here,

241
00:17:53,390 --> 00:17:55,010
is different than n.

242
00:17:58,300 --> 00:17:58,970
Try it.

243
00:17:58,970 --> 00:18:02,600
I don't want-- a good
exercise for yourself.

244
00:18:02,600 --> 00:18:05,630
Take this function.

245
00:18:05,630 --> 00:18:10,970
And do the integration, and
you'll find it comes out to 0.

246
00:18:10,970 --> 00:18:13,320
It's not hard to do.

247
00:18:13,320 --> 00:18:17,740
So that is 0 if n
is not equal to m.

248
00:18:17,740 --> 00:18:23,740
On the other hand,
if m equals n--

249
00:18:23,740 --> 00:18:29,200
so this is the integral of
sine squared n pi over L--

250
00:18:29,200 --> 00:18:30,885
it comes out to be L/2.

251
00:18:30,885 --> 00:18:32,010
You can do it for yourself.

252
00:18:35,350 --> 00:18:43,400
This trick allows us to
solve for all values of An.

253
00:18:43,400 --> 00:18:45,880
Now, let me tell
you, I did not do

254
00:18:45,880 --> 00:18:53,320
this in-- the Fourier
trick-- in general.

255
00:18:53,320 --> 00:18:59,330
If you go into books,
you'll find the formulation

256
00:18:59,330 --> 00:19:04,250
of this trick known
as Fourier's theorem

257
00:19:04,250 --> 00:19:06,090
for any periodic function.

258
00:19:09,800 --> 00:19:13,850
The advantage of learning it,
how to do it for any function,

259
00:19:13,850 --> 00:19:17,710
you don't have to do
this integral every time.

260
00:19:17,710 --> 00:19:19,840
Because suppose our
function turned out

261
00:19:19,840 --> 00:19:23,710
there to be a cosine, we
would have to then figure out,

262
00:19:23,710 --> 00:19:28,225
what do I multiply it by to
make this come out so simply?

263
00:19:35,505 --> 00:19:37,240
As I say, the
advantage is then you

264
00:19:37,240 --> 00:19:41,670
can use formulate
in books, et cetera.

265
00:19:41,670 --> 00:19:43,310
There's nothing wrong with that.

266
00:19:43,310 --> 00:19:44,590
It is good.

267
00:19:44,590 --> 00:19:49,940
And the way one changes a
physical situation into one

268
00:19:49,940 --> 00:19:53,840
where you can apply Fourier's
theorem, which only applies

269
00:19:53,840 --> 00:20:00,650
to periodic functions, is
to take your original shape,

270
00:20:00,650 --> 00:20:05,260
and out of it, create
a periodic function.

271
00:20:05,260 --> 00:20:09,450
Once you've done that, you
can use Fourier's theorem.

272
00:20:09,450 --> 00:20:15,140
I've, in this particular
case because I find it easier

273
00:20:15,140 --> 00:20:19,230
to explain it, simply
essentially derived

274
00:20:19,230 --> 00:20:24,380
the Fourier theorem for
this particular problem.

275
00:20:24,380 --> 00:20:29,960
For this particular
problem, the function

276
00:20:29,960 --> 00:20:34,940
by which you multiply
this is sine m pi.

277
00:20:34,940 --> 00:20:39,860
OK, once you've realized
that this is true,

278
00:20:39,860 --> 00:20:50,420
I can go back and use it
to determine every An.

279
00:20:57,890 --> 00:20:59,160
I go to this function.

280
00:21:03,510 --> 00:21:12,390
I multiply both sides in
turn by sine m pi over L

281
00:21:12,390 --> 00:21:21,670
for every n where I want to find
the value of An, for each one.

282
00:21:21,670 --> 00:21:24,070
So I have to do
an infinite number

283
00:21:24,070 --> 00:21:27,010
of integrals in principle.

284
00:21:27,010 --> 00:21:31,100
But often you don't need
to know every value of n.

285
00:21:31,100 --> 00:21:34,920
You may want to
know each one of An

286
00:21:34,920 --> 00:21:39,030
is the amplitude
of the harmonic.

287
00:21:39,030 --> 00:21:41,830
You may be interested
in what is the amplitude

288
00:21:41,830 --> 00:21:45,320
of the first harmonic solution
or the second or third.

289
00:21:45,320 --> 00:21:49,600
However many you want, so
many integrals you have to do.

290
00:21:49,600 --> 00:21:51,730
No magic.

291
00:21:51,730 --> 00:21:54,545
You're basically solving
infinite number of equations.

292
00:21:58,260 --> 00:22:02,890
And if I apply this
to our problem,

293
00:22:02,890 --> 00:22:11,410
we find that An is equal
to 2/L. It's from here.

294
00:22:11,410 --> 00:22:19,400
The integral of 0 of
2L of y sine at dx.

295
00:22:19,400 --> 00:22:19,910
Right?

296
00:22:19,910 --> 00:22:25,200
All I'm taking, I'm multiplying
both sides of this equation

297
00:22:25,200 --> 00:22:32,540
by sine of m pi L over
x and integrating them

298
00:22:32,540 --> 00:22:37,680
both from 0 to L.
On one side, you

299
00:22:37,680 --> 00:22:42,870
end up with An and on the
other, 2/L times this integral.

300
00:22:42,870 --> 00:22:47,890
Then once you've done this
integral, you know the answer.

301
00:22:47,890 --> 00:22:55,680
In our case, y of x of
0 is particularly easy.

302
00:22:55,680 --> 00:22:59,500
It's 0 in this range,
0 in this range,

303
00:22:59,500 --> 00:23:02,960
and a constant for the rest.

304
00:23:02,960 --> 00:23:07,760
Well, over the part of
the range where it's 0,

305
00:23:07,760 --> 00:23:10,490
if I multiply
something by 0, it's 0.

306
00:23:10,490 --> 00:23:15,240
And so I only have to do
this integral over the range

307
00:23:15,240 --> 00:23:24,780
where y of x for 0 time is H.
So I have to do this integral.

308
00:23:24,780 --> 00:23:31,350
And now I have reduced
the situation that every n

309
00:23:31,350 --> 00:23:33,840
that we are
interested in, we can

310
00:23:33,840 --> 00:23:36,460
calculate by doing
this integral.

311
00:23:36,460 --> 00:23:42,476
So in principle, we've solved
it for all values of n.

312
00:23:46,650 --> 00:23:50,410
So finally, I can
answer the question.

313
00:23:50,410 --> 00:23:54,990
So the first question was
what lowest value of n

314
00:23:54,990 --> 00:23:59,460
for which An is 0.

315
00:23:59,460 --> 00:24:00,220
Well, let's start.

316
00:24:00,220 --> 00:24:03,390
Take number 1, 2, 3.

317
00:24:03,390 --> 00:24:13,870
And for n equals 1, this
integral is not equal to 0,

318
00:24:13,870 --> 00:24:17,360
because what you have
to do is you basically

319
00:24:17,360 --> 00:24:21,260
are integrating
this-- for n equals

320
00:24:21,260 --> 00:24:26,390
1, that's what the sine
pi over L x looks like.

321
00:24:26,390 --> 00:24:30,930
You're integrating that
and multiply it by 0.

322
00:24:30,930 --> 00:24:33,610
And here, you're multiplying
by a constant and 0.

323
00:24:33,610 --> 00:24:37,210
Clearly, this is not
going to give you a 0.

324
00:24:37,210 --> 00:24:40,840
Let's take the second
one, the second harmonic

325
00:24:40,840 --> 00:24:42,090
in this calculation.

326
00:24:42,090 --> 00:24:48,260
You're multiplying
this y by this function

327
00:24:48,260 --> 00:24:50,910
and integrating it, OK?

328
00:24:50,910 --> 00:24:55,830
Notice the sine function
is symmetric about 0

329
00:24:55,830 --> 00:24:57,940
and so is this symmetric.

330
00:24:57,940 --> 00:25:01,710
Except here, this is an
anti-symmetric function.

331
00:25:01,710 --> 00:25:06,370
So the sine is negative
to the right of the middle

332
00:25:06,370 --> 00:25:08,110
and positive to the left.

333
00:25:08,110 --> 00:25:12,320
If I multiply this positive by
that, I get a positive number.

334
00:25:12,320 --> 00:25:15,370
If I multiply this by that,
I get a negative number.

335
00:25:15,370 --> 00:25:17,030
The two are equal.

336
00:25:17,030 --> 00:25:20,900
And so the integral
across there is 0.

337
00:25:20,900 --> 00:25:28,200
So n equals 2 is the lowest
harmonic for which An is 0.

338
00:25:28,200 --> 00:25:37,940
When I let that string
go, the second harmonic

339
00:25:37,940 --> 00:25:40,150
will not be excited.

340
00:25:40,150 --> 00:25:44,750
Which is the lowest n for which
it is excited-- in other words

341
00:25:44,750 --> 00:25:46,010
that An is not equal 0.

342
00:25:46,010 --> 00:25:52,370
Well, I told you when n
equals 1, clearly An is not 0.

343
00:25:52,370 --> 00:25:54,300
So that will be excited.

344
00:25:54,300 --> 00:25:58,920
So the first harmonic
will be excited.

345
00:25:58,920 --> 00:26:00,250
What is the amplitude?

346
00:26:00,250 --> 00:26:02,630
Well, we have to calculate it.

347
00:26:02,630 --> 00:26:05,130
I told you, every An--
I told you what it is.

348
00:26:05,130 --> 00:26:06,880
I have to do some work.

349
00:26:06,880 --> 00:26:12,210
So you take A of 1 is 2/L. The
integral from that to the H

350
00:26:12,210 --> 00:26:17,560
times this function
where n is 1, all right?

351
00:26:17,560 --> 00:26:21,090
And that, fortunately, is
an easy integral to do.

352
00:26:21,090 --> 00:26:23,140
Integrate the sine,
you get the cosine.

353
00:26:23,140 --> 00:26:26,480
And putting in the limits,
you get this answer.

354
00:26:26,480 --> 00:26:31,490
So that is our answer to
that part of the problem.

355
00:26:31,490 --> 00:26:34,860
The final part of
the problem was

356
00:26:34,860 --> 00:26:39,350
they said, OK, I
have this shape.

357
00:26:39,350 --> 00:26:40,800
I let go.

358
00:26:40,800 --> 00:26:46,160
What will that shape
look like at a time

359
00:26:46,160 --> 00:26:52,360
was L-- in the problem,
the time was L times

360
00:26:52,360 --> 00:26:57,550
the square root of mu over T.
T's the tension in the string.

361
00:26:57,550 --> 00:27:00,570
But I know that square
root of mu over t

362
00:27:00,570 --> 00:27:03,490
is 1/d, the phase velocity.

363
00:27:03,490 --> 00:27:08,900
So they ask us to calculate
the shape of that string

364
00:27:08,900 --> 00:27:12,560
at the time which is the
length of the string divided

365
00:27:12,560 --> 00:27:15,470
by the phase velocity
of progressive waves

366
00:27:15,470 --> 00:27:17,410
on the string.

367
00:27:17,410 --> 00:27:27,720
OK, now we know completely what
is the shape of the string.

368
00:27:27,720 --> 00:27:28,980
It is this formula.

369
00:27:28,980 --> 00:27:33,330
This is the formula
that tells us

370
00:27:33,330 --> 00:27:37,545
what the string looks like
at all places, all times.

371
00:27:41,130 --> 00:27:46,650
From our knowledge
of this integral,

372
00:27:46,650 --> 00:27:52,520
depending where it's
symmetric or anti-symmetric

373
00:27:52,520 --> 00:27:56,290
sinusoidal function
about the center,

374
00:27:56,290 --> 00:28:00,050
we know that every
second one will be 0.

375
00:28:00,050 --> 00:28:05,330
So this sum now,
I've only summed

376
00:28:05,330 --> 00:28:09,830
for n equals 1, 3,
5, 7, et cetera.

377
00:28:09,830 --> 00:28:17,880
The terms n equals 2, 4, 6, et
cetera, will have 0 amplitude.

378
00:28:17,880 --> 00:28:25,540
But this now describes our
situation in its entirety--

379
00:28:25,540 --> 00:28:29,950
because I've now not only know
the spatial shape, but also

380
00:28:29,950 --> 00:28:32,380
as a function of time.

381
00:28:32,380 --> 00:28:38,240
So this describes
what it was at time t.

382
00:28:38,240 --> 00:28:45,130
What will it be if I increase
the time-- let's start from 0,

383
00:28:45,130 --> 00:28:48,690
and I go to a time L/v?

384
00:28:48,690 --> 00:28:56,280
All right, well, if this
changes from 0 to L/v,

385
00:28:56,280 --> 00:29:01,150
this changes to cosine n pi.

386
00:29:01,150 --> 00:29:05,550
But you know that for n
equals 1, 3, 5, et cetera,

387
00:29:05,550 --> 00:29:09,540
cosine of n pi-- in other
words, at pi, at 3 pi,

388
00:29:09,540 --> 00:29:12,710
et cetera-- is minus 1.

389
00:29:12,710 --> 00:29:21,690
So at this later time,
every term in this expansion

390
00:29:21,690 --> 00:29:23,500
has changed signs.

391
00:29:23,500 --> 00:29:30,250
But it's exactly the same
series, except it's minus 1

392
00:29:30,250 --> 00:29:31,220
here.

393
00:29:31,220 --> 00:29:33,190
So what has happened
to the string?

394
00:29:33,190 --> 00:29:38,210
It is exactly the same
shape with an opposite sign.

395
00:29:38,210 --> 00:29:42,660
So with the string, instead
of being up here like that

396
00:29:42,660 --> 00:29:47,800
like it was over here, it's
flipped over like that.

397
00:29:47,800 --> 00:29:50,900
Without doing any more work,
I can conclude that simply

398
00:29:50,900 --> 00:29:54,460
because I have the same series
as before with a negative sign.

399
00:29:54,460 --> 00:29:56,360
So that is the answer.

400
00:29:56,360 --> 00:29:57,550
So I've solved the problem.

401
00:29:57,550 --> 00:30:00,960
But at this instance, I just
want to digress for a second

402
00:30:00,960 --> 00:30:04,900
and tell you this
actually is a problem

403
00:30:04,900 --> 00:30:10,000
that we could have almost
done all by thinking alone.

404
00:30:10,000 --> 00:30:15,160
It wasn't necessary to do big
fraction of the work we did.

405
00:30:15,160 --> 00:30:18,680
The only part which
there was no choice

406
00:30:18,680 --> 00:30:21,330
is this amplitude [INAUDIBLE].

407
00:30:21,330 --> 00:30:28,140
Because I could have gone
back, scratched my memory

408
00:30:28,140 --> 00:30:30,310
and knowledge, and
said, look, hold on.

409
00:30:30,310 --> 00:30:33,440
When I have been a
string, I could always

410
00:30:33,440 --> 00:30:37,560
analyze it in terms
of normal modes.

411
00:30:37,560 --> 00:30:45,380
But also, we know that we
can describe the solutions

412
00:30:45,380 --> 00:30:48,370
in terms of progressive waves.

413
00:30:48,370 --> 00:30:50,890
The two are equivalent
to each other.

414
00:30:50,890 --> 00:30:52,550
It's not extra solutions.

415
00:30:52,550 --> 00:30:56,400
But by adding
appropriately normal modes,

416
00:30:56,400 --> 00:31:02,220
I can get a superposition
of progressive waves.

417
00:31:02,220 --> 00:31:06,080
I can use the uniqueness
theorem to see

418
00:31:06,080 --> 00:31:13,110
if I've got a solution
which represents

419
00:31:13,110 --> 00:31:16,760
the situation at hand.

420
00:31:16,760 --> 00:31:24,580
If you look at the
original problem,

421
00:31:24,580 --> 00:31:31,920
if you look at this problem,
this is stationary, all right?

422
00:31:31,920 --> 00:31:39,530
So I could imagine this
solution being the superposition

423
00:31:39,530 --> 00:31:45,020
of two progressive waves
that look like this but half

424
00:31:45,020 --> 00:31:49,080
the amplitude, one moving
to the left and one

425
00:31:49,080 --> 00:31:51,090
moving to the right.

426
00:31:51,090 --> 00:31:57,190
As they overlap--
in other words,

427
00:31:57,190 --> 00:32:02,750
I have two waves, one like
this moving to the right,

428
00:32:02,750 --> 00:32:06,250
and one like this
moving to the left.

429
00:32:06,250 --> 00:32:11,610
If I add them, I'll
get this distortion,

430
00:32:11,610 --> 00:32:14,610
which is stationary.

431
00:32:14,610 --> 00:32:18,370
Therefore, if I
solved the problem

432
00:32:18,370 --> 00:32:25,860
for each one of these
progressive waves and add them,

433
00:32:25,860 --> 00:32:28,340
I'll get the answer to this.

434
00:32:28,340 --> 00:32:31,300
It'll satisfy all the
boundary conditions.

435
00:32:31,300 --> 00:32:32,840
By the uniqueness
theorem, it will

436
00:32:32,840 --> 00:32:38,720
be the correct description
of what happened.

437
00:32:38,720 --> 00:32:44,320
And then with
that, I can predict

438
00:32:44,320 --> 00:32:49,000
what will be the
shape at any time.

439
00:32:49,000 --> 00:32:54,220
So for example, in
the time they ask L/v,

440
00:32:54,220 --> 00:33:00,220
a progressive wave
will move a distance L.

441
00:33:00,220 --> 00:33:05,010
So the wave over
here, which was going

442
00:33:05,010 --> 00:33:10,310
to do the right would hit this
boundary-- that pulse-- would

443
00:33:10,310 --> 00:33:12,180
flip over.

444
00:33:12,180 --> 00:33:16,940
And you know that this
string is fixed there.

445
00:33:16,940 --> 00:33:21,250
What happens if a pulse
comes to a fixed end?

446
00:33:21,250 --> 00:33:26,440
Imagine you are
causing the pulse.

447
00:33:26,440 --> 00:33:29,340
You are the forcewave
moving along

448
00:33:29,340 --> 00:33:31,300
which causes the distortion.

449
00:33:31,300 --> 00:33:34,810
When you come to the
rigid end, you're pulling.

450
00:33:34,810 --> 00:33:36,300
It won't give.

451
00:33:36,300 --> 00:33:39,550
So the only way to happen,
the string will pull you down.

452
00:33:39,550 --> 00:33:41,450
So you'll flip over.

453
00:33:41,450 --> 00:33:45,040
So this pulse, when
it hits this end,

454
00:33:45,040 --> 00:33:47,340
will flip over
upside down, and then

455
00:33:47,340 --> 00:33:49,600
progress this way upside down.

456
00:33:49,600 --> 00:33:54,710
This one will go to this end,
flip over, and progress back.

457
00:33:54,710 --> 00:34:02,370
By the time t has changed
by L over the velocity,

458
00:34:02,370 --> 00:34:05,670
those flipped pulses would have
come back here and overlapped,

459
00:34:05,670 --> 00:34:07,130
but upside down.

460
00:34:07,130 --> 00:34:08,909
And so this would have flipped.

461
00:34:08,909 --> 00:34:12,730
By this kind of analysis, if
you use your wits about it,

462
00:34:12,730 --> 00:34:20,060
sometimes you can solve
more difficult problems

463
00:34:20,060 --> 00:34:25,159
quickly by using all the
knowledge you have about waves,

464
00:34:25,159 --> 00:34:26,912
the propagation, et cetera.

465
00:34:26,912 --> 00:34:28,120
Probably a good time to stop.

466
00:34:28,120 --> 00:34:29,970
Thank you.