1
00:00:26 --> 00:00:32
This is a brief,
so, the equation,

2
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and we got the characteristic
equation from the last time.

3
00:00:35 --> 00:00:41
The general topic for today is
going to be oscillations,

4
00:00:41 --> 00:00:47
which are extremely important
in the applications and in

5
00:00:47 --> 00:00:53
everyday life.
But, the oscillations,

6
00:00:50 --> 00:00:56
we know, are associated with a
complex root.

7
00:00:55 --> 00:01:01
So, they correspond to complex
roots of the characteristic

8
00:01:01 --> 00:01:07
equation.
r squared plus br plus k equals

9
00:01:06 --> 00:01:12
zero.
I'd like to begin.

10
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Most of the lecture will be
about discussing the relations

11
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between these numbers,
these constants,

12
00:01:16 --> 00:01:22
and the various properties that
the solutions,

13
00:01:19 --> 00:01:25
oscillatory solutions,
have.

14
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But, before that,
I'd like to begin by clearing

15
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up a couple of questions almost
everybody has at some point or

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other when they study the case
of complex roots.

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Complex roots are the case
which produce oscillations in

18
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the solutions.
That's the relation,

19
00:01:40 --> 00:01:46
and that's why I'm talking
about this for the first few

20
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minutes.
Now, what is the problem?

21
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The complex roots,
of course, there will be two

22
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roots, and they occur at the
complex conjugates of each

23
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other.
So, they will be of the form a

24
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plus or minus bi.
Last time, I showed you,

25
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I took the root r equals a plus
bi,

26
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which leads to the solution.
The corresponding solution is a

27
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complex solution which is e to
the at, (a plus i b)t.

28
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And, what we did was the

29
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problem was to get real
solutions out of that.

30
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We needed two real solutions,
and the way I got them was by

31
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separating this into its real
part and its imaginary part.

32
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And, I proved a little theorem
for you that said both of those

33
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give solutions.
So, the real part was e to the

34
00:02:49 --> 00:02:55
a t times cosine b t,
and the imaginary

35
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part was e to the at sine b t.

36
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And, those were the two
solutions.

37
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So, here was y1.
And, the point was those,

38
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out of the complex solutions,
we got real solutions.

39
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We have to have real solutions
because we live in the real

40
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world.
The equation is real.

41
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Its coefficients are real.
They represent real quantities.

42
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That's the way the solutions,
therefore, have to be.

43
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So, these, the point is,
these are now real solutions,

44
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these two guys,
y1 and y2.

45
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Now, the first question almost
everybody has,

46
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and I was pleased to see at the
end of the lecture,

47
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a few people came up and asked
me, yeah, well,

48
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you took a plus bi,
but there was another root,

49
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a minus bi.
You didn't use that one.

50
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That would give two more
solutions, right?

51
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Of course, they didn't say
that.

52
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They were too smart.
They just said,

53
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what about that other root?
Well, what about it?

54
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The reason I don't have to talk
about the other root is because

55
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although it does give to
solutions, it doesn't give two

56
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new ones.
Maybe I can indicate that most

57
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clearly here even though you
won't be able to take notes by

58
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just using colored chalk.
Suppose, instead of plus bi,

59
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I used a minus bi.

60
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What would have changed?
Well, this would now become

61
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minus here.
Would this change?

62
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No, because e to the minus ibt
is the cosine of

63
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minus b, but that's the same as
the cosine of b.

64
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How about here?
This would become the sine of

65
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minus bt.
But that's simply the negative

66
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of the sine of bt.
So, the only change would have

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been to put a minus sign there.
Now, I don't care if I get y2

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or negative y2 because what am I
going to do with it?

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When I get it,
I'm going to write y,

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the general solution,
as c1 y1 plus c2 y2.

71
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So, if I get negative y2,
that just changes that

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arbitrary constant from c2 to
minus c2, which is just as

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arbitrary a constant.
So, in other words,

74
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there's no reason to use the
other root because it doesn't

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give anything new.
Now, there the story could

76
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stop.
And, I would like it to stop,

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frankly, but I don't dare
because there's a second

78
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question.
And, I'm visiting recitations

79
00:05:37 --> 00:05:43
not this semester,
but in previous semesters.

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00:05:40 --> 00:05:46
In 18.03, so many recitations
do this.

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00:05:42 --> 00:05:48
I have to partly inoculate you
against it, and partly tell you

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that some of the engineering
courses do do it,

83
00:05:50 --> 00:05:56
and therefore you probably
should learn it also.

84
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So, there is another way of
proceeding, which is what you

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might have thought.
Hey, look, we got two complex

86
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roots.
That gives us two solutions,

87
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which are different.
Neither one is a constant

88
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multiple of the other.
So, the other approach is,

89
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use, as a general solution,
y equals, now,

90
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I'm going to put a capital C
here.

91
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You will see why in just a
second, times e to the (a plus b

92
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i) times t.

93
00:06:27 --> 00:06:33
And then, I will use the other
solution: C2 times e to the (a

94
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minus b i) t.

95
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These are two independent
solutions.

96
00:06:39 --> 00:06:45
And therefore,
can't I get the general

97
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solution in that form?
Now, in a sense,

98
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you can.
The whole problem is the

99
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following, of course,
that I'm only interested in

100
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real solutions.
This is a complex function.

101
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This is another complex
function.

102
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It's got an i in it,
in other words,

103
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when I write it out as u plus
iv.

104
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If I expect to be able to get a
real solution out of that,

105
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that means I have to make,
allow these coefficients to be

106
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complex numbers,
and not real numbers.

107
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So, in other words,
what I'm saying is that an

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expression like this,
where the a plus bi and a minus

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bi are complex roots of that
characteristic equation,

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is formally a very general,
complex solution to the

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equation.
And therefore,

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the problem becomes,
how, from this expression,

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do I get the real solutions?
So, the problem is,

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I accept these as the complex
solutions.

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My problem is,
to find among all these guys

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where C1 and C2 are allowed to
be complex, the problem is,

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which of the green solutions
are real?

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Now, there are many ways of
getting the answer.

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There is a super hack way.
The super hack way is to say,

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well, this one is C1 plus i d1.

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This is C2 plus i d2.

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And, I'll write all this out in
terms of what it is,

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you know, cosine plus i sine,
and don't forget the e to the

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at.
And, I will write it all out,

125
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and it will take an entire
board.

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And then, I will just see what
the condition is.

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I'll write its real part,
and its imaginary part.

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And then, I will say the
imaginary part has got to be

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zero.
And, then I will see what it's

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like.
That works fine.

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It just takes too much space.
And also, it doesn't teach you

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a few things that I think you
should know.

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So, I'm going to give another.
So, let's say we can answer

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this two ways:
by hack, in other words,

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multiply everything out.
Multiply all out,

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make the imaginary part equal
zero.

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Now, here's a better way,
in my opinion.

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What I'm trying to do is,
this is some complex function,

139
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u plus iv.
How do I know when a complex

140
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function is real?
I want this to be real.

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Well, the hack method
corresponds to,

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say, v must be equal to zero.
It's real if v is zero.

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So, expand it out,
and see why v is zero.

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There's a slightly more subtle
method, which is to change i to

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minus i.
And, what?

146
00:10:00 --> 00:10:06
And, see if it stays the same
because if I change i to minus i

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and it turns out,
the expression doesn't change,

148
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then it must have been real,
if the expression doesn't

149
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change when I change I to minus
I.

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00:10:33 --> 00:10:39
Well, sure.
But you will see it works.

151
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Now, that's what I'm going to
apply to this.

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If I want this to be real,
I phrase the question,

153
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I rephrase the question for the
green solution as change,

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so I'm going to change i to
minus i in the green thing,

155
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and that's going to give me
what conditions,

156
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and that will give conditions
on the C's.

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Well, let's do it.
In fact, it's easier done than

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talked about.
Let's change,

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take the green solution,
and change.

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Well, I better recopy it,
C1.

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So, these are complex numbers.
That's why I wrote them as

162
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capital letters because little
letters you tend to interpret as

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real numbers.
So, C1 e to the (a plus b i)t,

164
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I'll recopy it quickly,
plus C2 e to the (a minus b i).

165
00:11:33 --> 00:11:39

166

167

168
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Okay, we're going to change i
to negative i.

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00:11:41 --> 00:11:47
Now, here's a complex number.
What happens to it when you

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change i to negative i?
You change it into its-- Class?

171
00:11:51 --> 00:11:57
What do we change it to?
Its complex conjugate.

172
00:11:55 --> 00:12:01
And, the notation for complex
conjugate is you put a bar over

173
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it.
So, in other words,

174
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when I do that,
the C1 changes to C1 bar,

175
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complex conjugate,
the complex conjugate of C1.

176
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What happens to this guy?
This guy changes to e to the (a

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minus b i) t.
This changes to the complex

178
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conjugate of C2 now,
times e to the (a plus b i) t.

179
00:12:27 --> 00:12:33
Well, I want these two to be

180
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the same.
I want the two expressions the

181
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same.
Why do I want them the same?

182
00:12:40 --> 00:12:46
Because, if there's no change,
that will mean that it's real.

183
00:12:46 --> 00:12:52
Now, when is that going to
happen?

184
00:12:49 --> 00:12:55
That happens if,
well, here is this,

185
00:12:53 --> 00:12:59
that.
If C2 should be equal to C1

186
00:12:56 --> 00:13:02
bar, that's only one condition.
There's another condition.

187
00:13:02 --> 00:13:08
C2 bar should equal C1.
So, I get two conditions,

188
00:13:06 --> 00:13:12
but there's really only one
condition there because if this

189
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is true, that's true too.
I simply put bars over both

190
00:13:14 --> 00:13:20
things, and two bars cancel each
other out.

191
00:13:18 --> 00:13:24
If you take the complex
conjugate and do it again,

192
00:13:21 --> 00:13:27
you get back where you started.
Change i to minus i,

193
00:13:25 --> 00:13:31
and then i to minus i again.
Well, never mind.

194
00:13:30 --> 00:13:36
Anyway, these are the same.
This equation doesn't say

195
00:13:36 --> 00:13:42
anything that the first one
didn't say already.

196
00:13:41 --> 00:13:47
So, this one is redundant.
And, our conclusion is that the

197
00:13:48 --> 00:13:54
real solutions to the equation
are, in their entirety,

198
00:13:55 --> 00:14:01
I now don't need both C2 and
C1.

199
00:14:00 --> 00:14:06
One of them will do,
and since I'm going to write it

200
00:14:03 --> 00:14:09
out as a complex number,
I will write it out in terms of

201
00:14:07 --> 00:14:13
its coefficient.
So, it's C1.

202
00:14:09 --> 00:14:15
Let's just simply write it.
C plus i times d,

203
00:14:14 --> 00:14:20
that's the coefficient.
That's what I called C1 before.

204
00:14:18 --> 00:14:24
And, that's times e to the (a
plus b i) t.

205
00:14:22 --> 00:14:28
There's no reason why I put bi
here and id there,

206
00:14:25 --> 00:14:31
in case you're wondering,
sheer caprice.

207
00:14:30 --> 00:14:36
And what's the other term?
Now, the other term is

208
00:14:34 --> 00:14:40
completely determined.
Its coefficient must be C minus

209
00:14:38 --> 00:14:44
i d times e to the
(a minus b i) t.

210
00:14:43 --> 00:14:49
In other words,
this thing is perfectly

211
00:14:46 --> 00:14:52
general.
Any complex number times that

212
00:14:49 --> 00:14:55
first root you use,
exponentiated,

213
00:14:52 --> 00:14:58
and the second term can be
described as the complex

214
00:14:56 --> 00:15:02
conjugate of the first.
The coefficient is the complex

215
00:15:03 --> 00:15:09
conjugate, and this part is the
complex conjugate of that.

216
00:15:10 --> 00:15:16
Now, it's in this form,
many engineers write the

217
00:15:15 --> 00:15:21
solution this way,
and physicists,

218
00:15:19 --> 00:15:25
too, so, scientists and
engineers we will include.

219
00:15:25 --> 00:15:31
Write the solution this way.
Write the real solutions this

220
00:15:32 --> 00:15:38
way in that complex form.
Well, why do they do something

221
00:15:35 --> 00:15:41
so perverse?
You will have to ask them.

222
00:15:38 --> 00:15:44
But, in fact,
when we studied Fourier series,

223
00:15:41 --> 00:15:47
we will probably have to do
something, have to do that at

224
00:15:45 --> 00:15:51
one point.
If you work a lot with complex

225
00:15:48 --> 00:15:54
numbers, it turns out to be,
in some ways,

226
00:15:51 --> 00:15:57
a more convenient
representation than the one I've

227
00:15:55 --> 00:16:01
given you in terms of sines and
cosines.

228
00:15:59 --> 00:16:05
Well, from this,
how would I get,

229
00:16:01 --> 00:16:07
suppose I insisted,
well, if someone gave it to me

230
00:16:05 --> 00:16:11
in that form,
I don't see how I would convert

231
00:16:08 --> 00:16:14
it back into sines and cosines.
And, I'd like to show you how

232
00:16:13 --> 00:16:19
to do that efficiently,
too, because,

233
00:16:16 --> 00:16:22
again, it's one of the
fundamental techniques that I

234
00:16:20 --> 00:16:26
think you should know.
And, I didn't get a chance to

235
00:16:24 --> 00:16:30
say it when we studied complex
numbers that first lecture.

236
00:16:28 --> 00:16:34
It's in the notes,
but it doesn't prove anything

237
00:16:32 --> 00:16:38
since I don't think it made you
use it in an example.

238
00:16:38 --> 00:16:44
So, the problem is,
now, by way of finishing this

239
00:16:43 --> 00:16:49
up, too, to change this to the
old form, I mean the form

240
00:16:50 --> 00:16:56
involving sines and cosines.
Now, again, there are two ways

241
00:16:56 --> 00:17:02
of doing it.
The hack way is you write it

242
00:17:01 --> 00:17:07
all out.
Well, e to the (a plus b i)t

243
00:17:03 --> 00:17:09
turns into e to the a t times

244
00:17:08 --> 00:17:14
cosine this plus i sine that.
And, the other term does,

245
00:17:12 --> 00:17:18
too.
And then you've got stuff out

246
00:17:14 --> 00:17:20
front.
And, the thing stretches over

247
00:17:17 --> 00:17:23
two boards.
But you group all the terms

248
00:17:20 --> 00:17:26
together.
You finally get it.

249
00:17:22 --> 00:17:28
By the way, when you do it,
you'll find that the imaginary

250
00:17:26 --> 00:17:32
part disappears completely.
It has to because that's the

251
00:17:31 --> 00:17:37
way we chose the coefficients.
So, here's the hack method.

252
00:17:37 --> 00:17:43
Write it all out:
blah, blah, blah,

253
00:17:40 --> 00:17:46
blah, blah, blah,
blah, and nicer.

254
00:17:42 --> 00:17:48
Nicer, and teach you something
you're supposed to know.

255
00:17:47 --> 00:17:53
Write it this way.
First of all,

256
00:17:50 --> 00:17:56
you notice that both terms have
an e to the a t

257
00:17:55 --> 00:18:01
factor.
Let's get rid of that right

258
00:17:58 --> 00:18:04
away.
I'm pulling it out front

259
00:18:02 --> 00:18:08
because that's automatically
real, and therefore,

260
00:18:06 --> 00:18:12
isn't going to affect the rest
of the answer at all.

261
00:18:10 --> 00:18:16
So, let's pull out that,
and what's left?

262
00:18:14 --> 00:18:20
Well, what's left,
you see, involves just the two

263
00:18:18 --> 00:18:24
parameters, C and d,
so I'm going to have a C term.

264
00:18:22 --> 00:18:28
And, I'm going to have a d
term.

265
00:18:25 --> 00:18:31
What multiplies the arbitrary
constant, C?

266
00:18:30 --> 00:18:36
Answer: after I remove the e to
the a t,

267
00:18:34 --> 00:18:40
what multiplies it is,
e to the b i t plus e to the --

268
00:18:39 --> 00:18:45
e to the b i t.
Let's write it i b t.

269
00:18:43 --> 00:18:49
And, the other term is plus e
to the negative i b t.

270
00:18:48 --> 00:18:54
See how I got that,

271
00:18:51 --> 00:18:57
pulled it out?
And, how about the d?

272
00:18:54 --> 00:19:00
What goes with d?
d goes with,

273
00:18:57 --> 00:19:03
well, first of all,
there's an I in front that i

274
00:19:01 --> 00:19:07
better not forget.
And then, the rest of it is i.

275
00:19:07 --> 00:19:13
So, it's i d times,
it's e to the b i t,

276
00:19:11 --> 00:19:17
e to the i b t minus,
now, e to the minus i b t.

277
00:19:16 --> 00:19:22
So, that's the way the solution

278
00:19:22 --> 00:19:28
looks.
It doesn't look a lot better,

279
00:19:25 --> 00:19:31
but now you must use the magic
formulas, which,

280
00:19:30 --> 00:19:36
I want you to know as well as
you know Euler's formula,

281
00:19:35 --> 00:19:41
even better than you know
Euler's formula.

282
00:19:41 --> 00:19:47
They're a consequence of
Euler's formula.

283
00:19:43 --> 00:19:49
They're Euler's formula read
backwards.

284
00:19:45 --> 00:19:51
Euler's formula says you've got
a complex exponential here.

285
00:19:49 --> 00:19:55
Here's how to write it in terms
of sines and cosines.

286
00:19:52 --> 00:19:58
The backwards thing says you've
got a sine or a cosine.

287
00:19:55 --> 00:20:01
Here is the way to write it in
terms of complex exponentials.

288
00:20:00 --> 00:20:06
And, remember,
the way to do it is,

289
00:20:04 --> 00:20:10
cosine a is equal to e to the i
a t, i a, plus e to the negative

290
00:20:11 --> 00:20:17
i a divided by two.

291
00:20:17 --> 00:20:23
And, sine of a is almost the
same thing, except you use a

292
00:20:24 --> 00:20:30
minus sign.
And, what everybody forgets,

293
00:20:28 --> 00:20:34
you have to divide by i.
So, this is a backwards version

294
00:20:35 --> 00:20:41
of Euler's formula.
The two of them taken together

295
00:20:38 --> 00:20:44
are equivalent to Euler's
formula.

296
00:20:40 --> 00:20:46
If I took cosine a,
multiply this through by i,

297
00:20:44 --> 00:20:50
and added them up,
on the right-hand side I'd get

298
00:20:47 --> 00:20:53
exactly e to the ia.
I'd get Euler's formula,

299
00:20:51 --> 00:20:57
in other words.
All right, so,

300
00:20:53 --> 00:20:59
what does this come out to be,
finally?

301
00:20:55 --> 00:21:01
This particular sum of
exponentials,

302
00:20:58 --> 00:21:04
you should always recognize as
real.

303
00:21:02 --> 00:21:08
You know it's real because when
you change i to minus i,

304
00:21:07 --> 00:21:13
the two terms switch.
And therefore,

305
00:21:10 --> 00:21:16
the expression doesn't change.
What is it?

306
00:21:14 --> 00:21:20
This part is twice the cosine
of bt.

307
00:21:18 --> 00:21:24
What's this part?
This part is 2 i times the sine

308
00:21:22 --> 00:21:28
of bt.
And so, what does the whole

309
00:21:27 --> 00:21:33
thing come to be?
It is e to the a t times 2C

310
00:21:33 --> 00:21:39
cosine bt plus i times,
did I lose possibly a,

311
00:21:38 --> 00:21:44
no it's okay,
minus i times i is minus,

312
00:21:43 --> 00:21:49
so, minus 2d times the sine of
bt.

313
00:21:50 --> 00:21:56
Shall I write that out?

314
00:21:55 --> 00:22:01
So, in other words,
it's e to the a t times 2C

315
00:21:58 --> 00:22:04
cosine b t minus 2d times the
sine of b t,

316
00:22:03 --> 00:22:09
which is, since 2C and negative

317
00:22:07 --> 00:22:13
2d are just arbitrary constants,
just as arbitrary as the

318
00:22:12 --> 00:22:18
constants of C and d themselves
are.

319
00:22:15 --> 00:22:21
This is our old form of writing
the real solution.

320
00:22:19 --> 00:22:25
Here's the way using science
and cosines, and there's the way

321
00:22:24 --> 00:22:30
that uses complex numbers and
complex functions throughout.

322
00:22:30 --> 00:22:36
Notice they both have two
arbitrary constants in them,

323
00:22:33 --> 00:22:39
C and d, two arbitrary
constants.

324
00:22:36 --> 00:22:42
That, you expect.
But that has two arbitrary

325
00:22:39 --> 00:22:45
constants in it,
too, just the real and

326
00:22:42 --> 00:22:48
imaginary parts of that complex
coefficient, C plus i d.

327
00:22:46 --> 00:22:52
Well, that took half the

328
00:22:48 --> 00:22:54
period, and it was a long,
I don't consider it a

329
00:22:52 --> 00:22:58
digression because learning
those ways of dealing with

330
00:22:56 --> 00:23:02
complex numbers of complex
functions is a fairly important

331
00:23:00 --> 00:23:06
goal in this course,
actually.

332
00:23:04 --> 00:23:10
But let's get back now to
studying what the oscillations

333
00:23:07 --> 00:23:13
actually look like.

334
00:23:09 --> 00:23:15

335

336

337
00:23:27 --> 00:23:33
Okay, well, I'd like to save a
little time, but very quickly,

338
00:23:34 --> 00:23:40
you don't have to reproduce
this sketch.

339
00:23:39 --> 00:23:45
I remember very well from
Friday to Monday,

340
00:23:45 --> 00:23:51
but I can't expect you to for a
variety of reasons.

341
00:23:51 --> 00:23:57
I mean, I have to think about
this stuff all weekend.

342
00:23:58 --> 00:24:04
And you, God forbid.
So, here's the picture,

343
00:24:03 --> 00:24:09
and I won't explain anymore
what's in it,

344
00:24:06 --> 00:24:12
except there's the mass.
Here is the spring constant,

345
00:24:09 --> 00:24:15
the spring with its constant
here.

346
00:24:11 --> 00:24:17
Here's the dashpot with its
constant.

347
00:24:13 --> 00:24:19
The equation is from Newton's
law: m x double,

348
00:24:16 --> 00:24:22
so this will be x,
and here's, let's say,

349
00:24:19 --> 00:24:25
the equilibrium point is over
here.

350
00:24:21 --> 00:24:27
It looks like m x double prime;
we derived this last time,

351
00:24:25 --> 00:24:31
plus c x prime plus k x equals
zero.

352
00:24:30 --> 00:24:36
And now, if I put that in
standard form,

353
00:24:33 --> 00:24:39
it's going to look like x
double prime plus c over m x

354
00:24:39 --> 00:24:45
prime plus k over m times x
equals zero.

355
00:24:45 --> 00:24:51
And, finally,

356
00:24:47 --> 00:24:53
the standard form in which your
book writes it,

357
00:24:52 --> 00:24:58
which is good,
it's a standard form in general

358
00:24:56 --> 00:25:02
that is used in the science and
engineering courses.

359
00:25:02 --> 00:25:08
One writes this as,
just to be perverse,

360
00:25:05 --> 00:25:11
I'm going to change x back to
y, okay, mostly just to be

361
00:25:11 --> 00:25:17
eclectic, to get you used to
every conceivable notation.

362
00:25:19 --> 00:25:25
So, I'm going to write this to
change x to y.

363
00:25:22 --> 00:25:28
So, that's going to become y
double prime.

364
00:25:26 --> 00:25:32
And now, this is given a new
name, p, except to get rid of

365
00:25:30 --> 00:25:36
lots of twos,
which would really screw up the

366
00:25:33 --> 00:25:39
formulas, make it 2p.
You will see why in a minute.

367
00:25:38 --> 00:25:44
So, there's 2p times y prime,
and this thing we

368
00:25:42 --> 00:25:48
are going to call omega nought
squared.

369
00:25:46 --> 00:25:52
Now, that's okay.
It's a positive number.

370
00:25:49 --> 00:25:55
Any positive number is the
square of some other positive

371
00:25:53 --> 00:25:59
number.
Take a square root.

372
00:25:55 --> 00:26:01
You will see why,
it makes the formulas much

373
00:25:59 --> 00:26:05
pretty to call it that.
And, it makes also a lot of

374
00:26:04 --> 00:26:10
things much easier to remember.
So, all I'm doing is changing

375
00:26:08 --> 00:26:14
the names of the constants that
way in order to get better

376
00:26:13 --> 00:26:19
formulas, easier to remember
formulas at the end.

377
00:26:16 --> 00:26:22
Now, we are interested in the
case where there is

378
00:26:20 --> 00:26:26
oscillations.
In other words,

379
00:26:22 --> 00:26:28
I only care about the case in
which this has complex roots,

380
00:26:27 --> 00:26:33
because if it has just real
roots, that's the over-damped

381
00:26:31 --> 00:26:37
case.
I don't get any oscillations.

382
00:26:35 --> 00:26:41
By far, oscillations are by far
the more important of the cases,

383
00:26:40 --> 00:26:46
I mean, just because,
I don't know,

384
00:26:43 --> 00:26:49
I could go on for five minutes
listing things that oscillate,

385
00:26:48 --> 00:26:54
oscillations,
you know, like this.

386
00:26:51 --> 00:26:57
So they can oscillate by going
to sleep, and waking up,

387
00:26:56 --> 00:27:02
and going to sleep,
and waking up.

388
00:26:59 --> 00:27:05
They could oscillate.
So, that means we're going to

389
00:27:03 --> 00:27:09
get complex roots.
The characteristic equation is

390
00:27:07 --> 00:27:13
going to be r squared plus 2p.
So, p is a constant,

391
00:27:10 --> 00:27:16
now, right?
Often, p I use in this position

392
00:27:13 --> 00:27:19
to indicate a function of t.
But here, p is a constant.

393
00:27:16 --> 00:27:22
So, r squared plus 2p times r
plus omega nought squared is

394
00:27:20 --> 00:27:26
equal to zero.

395
00:27:23 --> 00:27:29
Now, what are its roots?
Well, you see right away the

396
00:27:27 --> 00:27:33
first advantage in putting in
the two there.

397
00:27:31 --> 00:27:37
When I use the quadratic
formula, it's negative 2p over

398
00:27:34 --> 00:27:40
two.
Remember that two in the

399
00:27:37 --> 00:27:43
denominator.
So, that's simply negative p.

400
00:27:40 --> 00:27:46
And, how about the rest?
Plus or minus the square root

401
00:27:44 --> 00:27:50
of, now do it in your head.
4p squared minus 4 omega nought

402
00:27:48 --> 00:27:54
squared.
So, there's a four in both of

403
00:27:53 --> 00:27:59
those terms.
When I pull it outside becomes

404
00:27:56 --> 00:28:02
two.
And, the two in the denominator

405
00:27:59 --> 00:28:05
is lurking, waiting to
annihilate it.

406
00:28:03 --> 00:28:09
So, that two disappears
entirely, and it will we are

407
00:28:06 --> 00:28:12
left with is,
simply, p squared minus omega

408
00:28:09 --> 00:28:15
nought squared.

409
00:28:11 --> 00:28:17
Now, whenever people write
quadratic equations,

410
00:28:14 --> 00:28:20
and arbitrarily put a two in
there, it's because they were

411
00:28:18 --> 00:28:24
going to want to solve the
quadratic equation using the

412
00:28:21 --> 00:28:27
quadratic formula,
and they don't want all those

413
00:28:24 --> 00:28:30
twos and fours to be cluttering
up the formula.

414
00:28:29 --> 00:28:35
That's what we are doing here.
Okay, now, the first case is

415
00:28:33 --> 00:28:39
where p is equal to zero.
This is going to explain

416
00:28:37 --> 00:28:43
immediately why I wrote that
omega nought squared,

417
00:28:41 --> 00:28:47
as you probably already know
from physics.

418
00:28:44 --> 00:28:50
If p is equal to zero,
the mass isn't zero.

419
00:28:48 --> 00:28:54
Otherwise, nothing good would
be happening here.

420
00:28:52 --> 00:28:58
It must be that the damping is
zero.

421
00:28:55 --> 00:29:01
So, p is equal to zero
corresponds to undamped.

422
00:29:00 --> 00:29:06
There is no dashpot.
The oscillations are undamped.

423
00:29:03 --> 00:29:09
And, the equation,
then, becomes the solutions,

424
00:29:06 --> 00:29:12
then, are, well,
the equation becomes the

425
00:29:09 --> 00:29:15
equation of simple harmonic
motion, which,

426
00:29:12 --> 00:29:18
I think you already are used to
writing in this form.

427
00:29:15 --> 00:29:21
And, the reason you're writing
in this form because you know

428
00:29:19 --> 00:29:25
when you do that,
this becomes the circular

429
00:29:22 --> 00:29:28
frequency of the oscillations.
The solutions are pure

430
00:29:26 --> 00:29:32
oscillations,
and omega nought is

431
00:29:29 --> 00:29:35
the circular frequency.
So, right away from the

432
00:29:33 --> 00:29:39
equation itself,
if you write it in this form,

433
00:29:37 --> 00:29:43
you can read off what the
frequency of the solutions is

434
00:29:41 --> 00:29:47
going to be, the circular
frequency of the solutions.

435
00:29:45 --> 00:29:51
Now, the solutions themselves,
of course, look like,

436
00:29:49 --> 00:29:55
the general solutions look like
y equal, in this particular

437
00:29:54 --> 00:30:00
case, the p part is zero.
This is zero.

438
00:29:57 --> 00:30:03
It's simply,
so, in this case,

439
00:29:59 --> 00:30:05
r is equal to omega nought i
times omega naught plus or

440
00:30:03 --> 00:30:09
minus, but as before we don't
bother with the minus sign since

441
00:30:08 --> 00:30:14
one of those roots is good
enough.

442
00:30:13 --> 00:30:19
And then, the solutions are
simply c1 cosine omega nought t

443
00:30:16 --> 00:30:22
plus c2 sine omega nought t.

444
00:30:20 --> 00:30:26
That's if you write it out in

445
00:30:23 --> 00:30:29
the sign, and if you write it
using the trigonometric

446
00:30:26 --> 00:30:32
identity, then the other way of
writing it is a times the cosine

447
00:30:30 --> 00:30:36
of omega nought t.

448
00:30:34 --> 00:30:40
But now you will have to put it
a phase lag.

449
00:30:37 --> 00:30:43
So, you have those two forms of
writing it.

450
00:30:41 --> 00:30:47
And, I assume you remember
writing the little triangle,

451
00:30:45 --> 00:30:51
which converts one into the
other.

452
00:30:48 --> 00:30:54
Okay, so this justifies calling
this omega nought squared

453
00:30:53 --> 00:30:59
rather than k over m.

454
00:30:56 --> 00:31:02
And now, the question is what
does the damp case look like?

455
00:31:01 --> 00:31:07
It requires a somewhat closer
analysis, and it requires a

456
00:31:06 --> 00:31:12
certain amount of thinking.
So, let's begin with an epsilon

457
00:31:13 --> 00:31:19
bit of thinking.
So, here's my question.

458
00:31:18 --> 00:31:24
So, in the damped case,
I want to be sure that I'm

459
00:31:24 --> 00:31:30
getting oscillations.
When do I get oscillations if,

460
00:31:30 --> 00:31:36
well, we get oscillations if
those roots are really complex,

461
00:31:37 --> 00:31:43
and not masquerading.
Now, when are the roots going

462
00:31:43 --> 00:31:49
to be really complex?
This has to be,

463
00:31:46 --> 00:31:52
the inside has to be negative.
p squared minus omega squared

464
00:31:52 --> 00:31:58
must be negative.

465
00:31:56 --> 00:32:02
p squared minus omega nought
squared must be less than zero

466
00:32:01 --> 00:32:07
so that we are taking a square
root of negative number,

467
00:32:06 --> 00:32:12
and we are getting a real
complex roots,

468
00:32:09 --> 00:32:15
really complex roots.
In other words,

469
00:32:14 --> 00:32:20
now, this says,
remember these numbers are all

470
00:32:17 --> 00:32:23
positive, p and omega nought are
positive.

471
00:32:21 --> 00:32:27
So, the condition is that p
should be

472
00:32:25 --> 00:32:31
less than omega nought.
In other words,

473
00:32:28 --> 00:32:34
the damping should be less than
the circular frequency,

474
00:32:32 --> 00:32:38
except p is not the damping.
It's half the damping,

475
00:32:38 --> 00:32:44
and it's not really the damping
either because it involved the

476
00:32:43 --> 00:32:49
m, too.
You'd better just call it p.

477
00:32:47 --> 00:32:53
Naturally, I could write the
condition out in terms of c,

478
00:32:52 --> 00:32:58
m, and k.
So, your book does that,

479
00:32:55 --> 00:33:01
but I'm not going to.
It gives it in terms of c,

480
00:32:59 --> 00:33:05
m, and k, which somebody might
want to know.

481
00:33:03 --> 00:33:09
But, you know,
we don't have to do everything

482
00:33:08 --> 00:33:14
here.
Okay, so let's assume that this

483
00:33:12 --> 00:33:18
is true.
What is the solution look like?

484
00:33:15 --> 00:33:21
Well, we already experimented
with that last time.

485
00:33:19 --> 00:33:25
Remember, there was some
guiding thing which was an

486
00:33:23 --> 00:33:29
exponential.
And then, down here,

487
00:33:26 --> 00:33:32
we wrote the negative.
So, this was an exponential.

488
00:33:31 --> 00:33:37
In fact, it was the
exponential, e to the negative

489
00:33:35 --> 00:33:41
pt.
And, in between that,

490
00:33:38 --> 00:33:44
the curve tried to do its
thing.

491
00:33:41 --> 00:33:47
So, the solution looks sort of
like this.

492
00:33:45 --> 00:33:51
It oscillated,
but it had to use that

493
00:33:48 --> 00:33:54
exponential function as its
guidelines, as its amplitude,

494
00:33:53 --> 00:33:59
in other words.
Now, this is a truly terrible

495
00:33:57 --> 00:34:03
picture.
It's so terrible,

496
00:34:01 --> 00:34:07
it's unusable.
Okay, this picture never

497
00:34:05 --> 00:34:11
happened.
Unfortunately,

498
00:34:07 --> 00:34:13
this is not my forte along with
a lot of other things.

499
00:34:12 --> 00:34:18
All right, let's try it better.
Here's our better picture.

500
00:34:18 --> 00:34:24
Okay, there's the exponential.
At this point,

501
00:34:22 --> 00:34:28
I'm supposed to have a lecture
demonstration.

502
00:34:26 --> 00:34:32
It's supposed to go up on the
thing, so you can all see it.

503
00:34:34 --> 00:34:40
But then, you wouldn't be able
to copy it.

504
00:34:37 --> 00:34:43
So, at least we are on even
terms now.

505
00:34:40 --> 00:34:46
Okay, how does the actual curve
look?

506
00:34:43 --> 00:34:49
Well, I'm just trying to be
fair.

507
00:34:45 --> 00:34:51
That's all.
Okay, after a while,

508
00:34:48 --> 00:34:54
the point is,
just so we have something to

509
00:34:51 --> 00:34:57
aim at, let's say,
okay, here we are going to go,

510
00:34:55 --> 00:35:01
we're going to get down through
there.

511
00:35:00 --> 00:35:06
Okay then, this is our better
curve.

512
00:35:03 --> 00:35:09
Okay, so I am a solution,
a particular solution

513
00:35:08 --> 00:35:14
satisfying this initial
condition.

514
00:35:12 --> 00:35:18
I started here,
and that was my initial

515
00:35:16 --> 00:35:22
velocity.
The slope of that thing gave me

516
00:35:20 --> 00:35:26
the initial velocity.
Now, the interesting question

517
00:35:26 --> 00:35:32
is, the first,
in some ways,

518
00:35:28 --> 00:35:34
the most interesting question,
though there will be others,

519
00:35:35 --> 00:35:41
too, is what is this spacing?
Well, that's a period.

520
00:35:42 --> 00:35:48
And now, it's half a period.
I clearly ought to think of

521
00:35:47 --> 00:35:53
this as the whole period.
So, let's call that,

522
00:35:51 --> 00:35:57
I'm going to call this pi over,
so this spacing here,

523
00:35:56 --> 00:36:02
from there to there,
I will call that pi divided by

524
00:36:01 --> 00:36:07
omega one because this,
from here to here,

525
00:36:05 --> 00:36:11
should be, I hope,
twice that, two pi over omega

526
00:36:10 --> 00:36:16
one.
Now, my question is,

527
00:36:14 --> 00:36:20
so this, for a solution,
it's, in fact,

528
00:36:18 --> 00:36:24
is going to cross the axis
regularly in that way.

529
00:36:24 --> 00:36:30
My question is,
how does this period,

530
00:36:28 --> 00:36:34
so this is going to be its half
period.

531
00:36:34 --> 00:36:40
I will put period in quotation
marks because this isn't really

532
00:36:39 --> 00:36:45
a periodic function because it's
decreasing all the time in

533
00:36:43 --> 00:36:49
amplitude.
But, it's trying to be

534
00:36:46 --> 00:36:52
periodic.
At lease it's doing something

535
00:36:49 --> 00:36:55
periodically.
It's crossing the axis

536
00:36:52 --> 00:36:58
periodically.
So, this is the half period.

537
00:36:55 --> 00:37:01
Two pi over omega one
would be its full

538
00:37:00 --> 00:37:06
period.
What I want to know is,

539
00:37:02 --> 00:37:08
how does that half period,
or how does-- omega one is

540
00:37:07 --> 00:37:13
called its pseudo-frequency.
This should really be called

541
00:37:13 --> 00:37:19
its pseudo-period.
Everything is pseudo.

542
00:37:16 --> 00:37:22
Everything is fake here.
Like, the amoeba has its fake

543
00:37:21 --> 00:37:27
foot and stuff like that.
Okay, so this is its

544
00:37:24 --> 00:37:30
pseudo-period,
pseudo-frequency,

545
00:37:27 --> 00:37:33
pseudo-circular frequency,
but that's hopeless.

546
00:37:31 --> 00:37:37
I guess it should be circular
pseudo-frequency,

547
00:37:35 --> 00:37:41
or I don't know how you say
that.

548
00:37:39 --> 00:37:45
I don't think pseudo is a word
all by itself,

549
00:37:46 --> 00:37:52
not even in 18.03,
circular.

550
00:37:50 --> 00:37:56
Okay, here's my question.
If the damping goes up,

551
00:37:58 --> 00:38:04
this is the damping term.
If the damping goes up,

552
00:38:06 --> 00:38:12
what happens to the
pseudo-frequency?

553
00:38:11 --> 00:38:17
The frequency is how often the
curve, this is high-frequency,

554
00:38:19 --> 00:38:25
and this is low-frequency,
okay?

555
00:38:23 --> 00:38:29
So, my question is,
which way does the frequency

556
00:38:29 --> 00:38:35
go?
If the damping goes up,

557
00:38:33 --> 00:38:39
does the frequency go up or
down?

558
00:38:38 --> 00:38:44
Down.
I mean, I'm just asking you to

559
00:38:42 --> 00:38:48
answer intuitively on the basis
of your intuition about how this

560
00:38:50 --> 00:38:56
thing explains,
how this thing goes,

561
00:38:55 --> 00:39:01
and now let's get the formula.
What, in fact,

562
00:39:01 --> 00:39:07
is omega one?
What is omega one?

563
00:39:05 --> 00:39:11
The answer is when I solve the
equation, so,

564
00:39:09 --> 00:39:15
r is now, so in other words,
if omega one is,

565
00:39:13 --> 00:39:19
sorry, if I have p,
if p is no longer zero as it

566
00:39:18 --> 00:39:24
was in the undamped case,
what is the root,

567
00:39:22 --> 00:39:28
now?
Okay, well, the root is minus p

568
00:39:25 --> 00:39:31
plus or minus the square root of
p squared, --

569
00:39:31 --> 00:39:37
-- now I'm going to write it
this way, minus,

570
00:39:34 --> 00:39:40
to indicate that it's really a
negative number,

571
00:39:38 --> 00:39:44
omega squared minus p squared.

572
00:39:42 --> 00:39:48
Now, I'm going to call this,
because you see when I change

573
00:39:47 --> 00:39:53
this to sines and cosines,
the square root of this number

574
00:39:52 --> 00:39:58
is what's going to become that
new frequency.

575
00:39:55 --> 00:40:01
I'm going to call that minus p
plus or minus the square root of

576
00:40:00 --> 00:40:06
minus omega one squared.
That's going to be the new

577
00:40:06 --> 00:40:12
frequency.
And therefore,

578
00:40:08 --> 00:40:14
the root is going to change so
that the corresponding solution

579
00:40:13 --> 00:40:19
is going to look,
how?

580
00:40:15 --> 00:40:21
Well, it's going to be e to the
negative pt times,

581
00:40:20 --> 00:40:26
let's write it out first in
terms of sines and cosines,

582
00:40:25 --> 00:40:31
times the cosine of,
well, the square root of omega

583
00:40:29 --> 00:40:35
one squared is omega one.

584
00:40:35 --> 00:40:41
But, there's an i out front
because of the negative sign in

585
00:40:39 --> 00:40:45
front of that.
So, it's going to be the cosine

586
00:40:42 --> 00:40:48
of omega one t
plus c2 times the sine of omega

587
00:40:47 --> 00:40:53
one t.
Or, if you prefer to write it

588
00:40:51 --> 00:40:57
out in the other form,
it's e to the minus p t times

589
00:40:54 --> 00:41:00
some amplitude,
which depends on c1 and c2,

590
00:40:57 --> 00:41:03
times the cosine of omega one t
minus the phase lag.

591
00:41:02 --> 00:41:08
Now, when I do that,

592
00:41:07 --> 00:41:13
you see omega one is
this pseudo-frequency.

593
00:41:13 --> 00:41:19
In other words,
this number omega one is the

594
00:41:17 --> 00:41:23
same one that I identified here.
And, why is that?

595
00:41:23 --> 00:41:29
Well, because,
what are two successive times?

596
00:41:27 --> 00:41:33
Suppose it crosses,
suppose the solution crosses

597
00:41:33 --> 00:41:39
the x-axis, sorry,
y-- the t-axis.

598
00:41:38 --> 00:41:44
For the first time,
at the point t1,

599
00:41:42 --> 00:41:48
what's the next time it crosses
t2?

600
00:41:46 --> 00:41:52
Let's jump to the two times
across it.

601
00:41:51 --> 00:41:57
So, I want this to be a whole
period, not a half period.

602
00:41:57 --> 00:42:03
What's t2?
Well, I say that t2 is nothing

603
00:42:02 --> 00:42:08
but 2 pi divided by omega one.

604
00:42:07 --> 00:42:13
And, you can see that because
when I plug in,

605
00:42:10 --> 00:42:16
if it's zero,
if I have a point where it's

606
00:42:14 --> 00:42:20
zero, so, omega one t minus phi,

607
00:42:18 --> 00:42:24
when will it be zero for the
first time?

608
00:42:22 --> 00:42:28
Well, that will be when the
cosine has to be zero.

609
00:42:26 --> 00:42:32
So, it will be some multiple
of, it will be,

610
00:42:30 --> 00:42:36
say, pi over two.
Then, the next time this

611
00:42:35 --> 00:42:41
happens will be,
if that happens at t1,

612
00:42:39 --> 00:42:45
then the next time it happens
will be at t1 plus 2 pi divided

613
00:42:45 --> 00:42:51
by omega one.

614
00:42:49 --> 00:42:55
That will also be pi over two
plus how much?

615
00:42:54 --> 00:43:00
Plus 2 pi, which is the next
time the cosine gets around and

616
00:43:00 --> 00:43:06
is doing its thing,
becoming zero as it goes down,

617
00:43:05 --> 00:43:11
not as it's coming up again.
In other words,

618
00:43:11 --> 00:43:17
this is what you should add to
the first time to get this

619
00:43:17 --> 00:43:23
second time that the cosine
becomes zero coming in the

620
00:43:23 --> 00:43:29
direction from top to the
bottom.

621
00:43:26 --> 00:43:32
So, this is,
in fact, the frequency with

622
00:43:30 --> 00:43:36
which it's crossing the axis.
Now, notice,

623
00:43:36 --> 00:43:42
I'm running out of boards.
What a disaster!

624
00:43:41 --> 00:43:47
In that expression,
take a look at it.

625
00:43:46 --> 00:43:52
I want to know what depends on
what.

626
00:43:50 --> 00:43:56
So, p, in that,
we got constants.

627
00:43:54 --> 00:44:00
We got p.
We got phi.

628
00:43:57 --> 00:44:03
We got A.
What else we got?

629
00:44:00 --> 00:44:06
Omega one.
What do these things depend

630
00:44:07 --> 00:44:13
upon?
You've got to keep it firmly in

631
00:44:11 --> 00:44:17
mind.
This depends only on the ODE.

632
00:44:14 --> 00:44:20
It's basically the damping.
It depends on c and m.

633
00:44:19 --> 00:44:25
Essentially, it's c over 2m

634
00:44:23 --> 00:44:29
actually.
How about phi?

635
00:44:25 --> 00:44:31
Well, phi, what else depends
only on the ODE?

636
00:44:30 --> 00:44:36
Omega one depends
only on the ODE.

637
00:44:36 --> 00:44:42
What's the formula for omega
one?

638
00:44:38 --> 00:44:44
Omega one squared.

639
00:44:40 --> 00:44:46
Where do we have it?
Omega one squared,

640
00:44:43 --> 00:44:49
I never wrote the formula for
you.

641
00:44:46 --> 00:44:52
So, we have omega nought
squared minus p squared equals

642
00:44:50 --> 00:44:56
omega one squared.

643
00:44:53 --> 00:44:59
What's the relation between
them?

644
00:44:56 --> 00:45:02
That's the Pythagorean theorem.
If this is omega nought,

645
00:45:00 --> 00:45:06
then this omega one,
this is p.

646
00:45:04 --> 00:45:10
They make a little,
right triangle in other words.

647
00:45:09 --> 00:45:15
The omega one depends on the
spring.

648
00:45:13 --> 00:45:19
So, it's equal to,
well, it's equal to that thing.

649
00:45:19 --> 00:45:25
So, it depends on the damping.
It depends upon the damping,

650
00:45:25 --> 00:45:31
and it depends on the spring
constant.

651
00:45:30 --> 00:45:36
How about the phi and the A?
What do they depend on?

652
00:45:36 --> 00:45:42
They depend upon the initial
conditions.

653
00:45:42 --> 00:45:48
So, the mass of constants,
they have different functions.

654
00:45:47 --> 00:45:53
What's making this complicated
is that our answer needs four

655
00:45:53 --> 00:45:59
parameters to describe it.
This tells you how fast it's

656
00:45:59 --> 00:46:05
coming down.
This tells you the phase lag.

657
00:46:03 --> 00:46:09
This amplitude modifies,
it tells you whether the

658
00:46:08 --> 00:46:14
exponential curve starts going,
is like that or goes like this.

659
00:46:15 --> 00:46:21
And, finally,
the omega one is this

660
00:46:18 --> 00:46:24
pseudo-frequency,
which tells you how it's

661
00:46:22 --> 00:46:28
bobbing up and down.