WEBVTT
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This is a brief,
so, the equation,
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and we got the characteristic
equation from the last time.
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The general topic for today is
going to be oscillations,
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which are extremely important
in the applications and in
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everyday life.
But, the oscillations,
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we know, are associated with a
complex root.
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So, they correspond to complex
roots of the characteristic
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equation.
r squared plus br plus k equals
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zero.
I'd like to begin.
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Most of the lecture will be
about discussing the relations
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between these numbers,
these constants,
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and the various properties that
the solutions,
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oscillatory solutions,
have.
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But, before that,
I'd like to begin by clearing
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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
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the solutions.
That's the relation,
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and that's why I'm talking
about this for the first few
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minutes.
Now, what is the problem?
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The complex roots,
of course, there will be two
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roots, and they occur at the
complex conjugates of each
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other.
So, they will be of the form a
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plus or minus bi.
Last time, I showed you,
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I took the root r equals a plus
bi,
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which leads to the solution.
The corresponding solution is a
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complex solution which is e to
the at, (a plus i b)t.
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And, what we did was the
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problem was to get real
solutions out of that.
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We needed two real solutions,
and the way I got them was by
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separating this into its real
part and its imaginary part.
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And, I proved a little theorem
for you that said both of those
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give solutions.
So, the real part was e to the
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a t times cosine b t,
and the imaginary
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part was e to the at sine b t.
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And, those were the two
solutions.
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So, here was y1.
And, the point was those,
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out of the complex solutions,
we got real solutions.
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We have to have real solutions
because we live in the real
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world.
The equation is real.
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Its coefficients are real.
They represent real quantities.
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That's the way the solutions,
therefore, have to be.
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So, these, the point is,
these are now real solutions,
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these two guys,
y1 and y2.
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Now, the first question almost
everybody has,
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and I was pleased to see at the
end of the lecture,
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a few people came up and asked
me, yeah, well,
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you took a plus bi,
but there was another root,
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a minus bi.
You didn't use that one.
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That would give two more
solutions, right?
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Of course, they didn't say
that.
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They were too smart.
They just said,
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what about that other root?
Well, what about it?
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The reason I don't have to talk
about the other root is because
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although it does give to
solutions, it doesn't give two
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new ones.
Maybe I can indicate that most
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clearly here even though you
won't be able to take notes by
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just using colored chalk.
Suppose, instead of plus bi,
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I used a minus bi.
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What would have changed?
Well, this would now become
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minus here.
Would this change?
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No, because e to the minus ibt
is the cosine of
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minus b, but that's the same as
the cosine of b.
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How about here?
This would become the sine of
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minus bt.
But that's simply the negative
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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.
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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,
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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
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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
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question.
And, I'm visiting recitations
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not this semester,
but in previous semesters.
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In 18.03, so many recitations
do this.
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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,
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and therefore you probably
should learn it also.
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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
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roots.
That gives us two solutions,
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which are different.
Neither one is a constant
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multiple of the other.
So, the other approach is,
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use, as a general solution,
y equals, now,
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I'm going to put a capital C
here.
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You will see why in just a
second, times e to the (a plus b
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i) times t.
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And then, I will use the other
solution: C2 times e to the (a
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minus b i) t.
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These are two independent
solutions.
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And therefore,
can't I get the general
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solution in that form?
Now, in a sense,
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you can.
The whole problem is the
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following, of course,
that I'm only interested in
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real solutions.
This is a complex function.
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This is another complex
function.
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It's got an i in it,
in other words,
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when I write it out as u plus
iv.
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If I expect to be able to get a
real solution out of that,
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that means I have to make,
allow these coefficients to be
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complex numbers,
and not real numbers.
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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,
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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,
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u plus iv.
How do I know when a complex
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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?
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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,
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then it must have been real,
if the expression doesn't
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change when I change I to minus
I.
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Well, sure.
But you will see it works.
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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,
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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,
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and that's going to give me
what conditions,
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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
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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,
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I'll recopy it quickly,
plus C2 e to the (a minus b i).
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Okay, we're going to change i
to negative i.
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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?
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What do we change it to?
Its complex conjugate.
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And, the notation for complex
conjugate is you put a bar over
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it.
So, in other words,
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when I do that,
the C1 changes to C1 bar,
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complex conjugate,
the complex conjugate of C1.
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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
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conjugate of C2 now,
times e to the (a plus b i) t.
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Well, I want these two to be
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the same.
I want the two expressions the
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same.
Why do I want them the same?
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Because, if there's no change,
that will mean that it's real.
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Now, when is that going to
happen?
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That happens if,
well, here is this,
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that.
If C2 should be equal to C1
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bar, that's only one condition.
There's another condition.
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C2 bar should equal C1.
So, I get two conditions,
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but there's really only one
condition there because if this
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is true, that's true too.
I simply put bars over both
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things, and two bars cancel each
other out.
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If you take the complex
conjugate and do it again,
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you get back where you started.
Change i to minus i,
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and then i to minus i again.
Well, never mind.
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Anyway, these are the same.
This equation doesn't say
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anything that the first one
didn't say already.
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So, this one is redundant.
And, our conclusion is that the
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real solutions to the equation
are, in their entirety,
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I now don't need both C2 and
C1.
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One of them will do,
and since I'm going to write it
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out as a complex number,
I will write it out in terms of
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its coefficient.
So, it's C1.
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Let's just simply write it.
C plus i times d,
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that's the coefficient.
That's what I called C1 before.
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And, that's times e to the (a
plus b i) t.
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There's no reason why I put bi
here and id there,
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in case you're wondering,
sheer caprice.
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And what's the other term?
Now, the other term is
00:14:34.000 --> 00:14:40.000
completely determined.
Its coefficient must be C minus
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i d times e to the
(a minus b i) t.
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In other words,
this thing is perfectly
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general.
Any complex number times that
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first root you use,
exponentiated,
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and the second term can be
described as the complex
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conjugate of the first.
The coefficient is the complex
00:15:03.000 --> 00:15:09.000
conjugate, and this part is the
complex conjugate of that.
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Now, it's in this form,
many engineers write the
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solution this way,
and physicists,
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too, so, scientists and
engineers we will include.
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Write the solution this way.
Write the real solutions this
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way in that complex form.
Well, why do they do something
00:15:35.000 --> 00:15:41.000
so perverse?
You will have to ask them.
00:15:38.000 --> 00:15:44.000
But, in fact,
when we studied Fourier series,
00:15:41.000 --> 00:15:47.000
we will probably have to do
something, have to do that at
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one point.
If you work a lot with complex
00:15:48.000 --> 00:15:54.000
numbers, it turns out to be,
in some ways,
00:15:51.000 --> 00:15:57.000
a more convenient
representation than the one I've
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given you in terms of sines and
cosines.
00:15:59.000 --> 00:16:05.000
Well, from this,
how would I get,
00:16:01.000 --> 00:16:07.000
suppose I insisted,
well, if someone gave it to me
00:16:05.000 --> 00:16:11.000
in that form,
I don't see how I would convert
00:16:08.000 --> 00:16:14.000
it back into sines and cosines.
And, I'd like to show you how
00:16:13.000 --> 00:16:19.000
to do that efficiently,
too, because,
00:16:16.000 --> 00:16:22.000
again, it's one of the
fundamental techniques that I
00:16:20.000 --> 00:16:26.000
think you should know.
And, I didn't get a chance to
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say it when we studied complex
numbers that first lecture.
00:16:28.000 --> 00:16:34.000
It's in the notes,
but it doesn't prove anything
00:16:32.000 --> 00:16:38.000
since I don't think it made you
use it in an example.
00:16:38.000 --> 00:16:44.000
So, the problem is,
now, by way of finishing this
00:16:43.000 --> 00:16:49.000
up, too, to change this to the
old form, I mean the form
00:16:50.000 --> 00:16:56.000
involving sines and cosines.
Now, again, there are two ways
00:16:56.000 --> 00:17:02.000
of doing it.
The hack way is you write it
00:17:01.000 --> 00:17:07.000
all out.
Well, e to the (a plus b i)t
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turns into e to the a t times
00:17:08.000 --> 00:17:14.000
cosine this plus i sine that.
And, the other term does,
00:17:12.000 --> 00:17:18.000
too.
And then you've got stuff out
00:17:14.000 --> 00:17:20.000
front.
And, the thing stretches over
00:17:17.000 --> 00:17:23.000
two boards.
But you group all the terms
00:17:20.000 --> 00:17:26.000
together.
You finally get it.
00:17:22.000 --> 00:17:28.000
By the way, when you do it,
you'll find that the imaginary
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part disappears completely.
It has to because that's the
00:17:31.000 --> 00:17:37.000
way we chose the coefficients.
So, here's the hack method.
00:17:37.000 --> 00:17:43.000
Write it all out:
blah, blah, blah,
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blah, blah, blah,
blah, and nicer.
00:17:42.000 --> 00:17:48.000
Nicer, and teach you something
you're supposed to know.
00:17:47.000 --> 00:17:53.000
Write it this way.
First of all,
00:17:50.000 --> 00:17:56.000
you notice that both terms have
an e to the a t
00:17:55.000 --> 00:18:01.000
factor.
Let's get rid of that right
00:17:58.000 --> 00:18:04.000
away.
I'm pulling it out front
00:18:02.000 --> 00:18:08.000
because that's automatically
real, and therefore,
00:18:06.000 --> 00:18:12.000
isn't going to affect the rest
of the answer at all.
00:18:10.000 --> 00:18:16.000
So, let's pull out that,
and what's left?
00:18:14.000 --> 00:18:20.000
Well, what's left,
you see, involves just the two
00:18:18.000 --> 00:18:24.000
parameters, C and d,
so I'm going to have a C term.
00:18:22.000 --> 00:18:28.000
And, I'm going to have a d
term.
00:18:25.000 --> 00:18:31.000
What multiplies the arbitrary
constant, C?
00:18:30.000 --> 00:18:36.000
Answer: after I remove the e to
the a t,
00:18:34.000 --> 00:18:40.000
what multiplies it is,
e to the b i t plus e to the --
00:18:39.000 --> 00:18:45.000
e to the b i t.
Let's write it i b t.
00:18:43.000 --> 00:18:49.000
And, the other term is plus e
to the negative i b t.
00:18:48.000 --> 00:18:54.000
See how I got that,
00:18:51.000 --> 00:18:57.000
pulled it out?
And, how about the d?
00:18:54.000 --> 00:19:00.000
What goes with d?
d goes with,
00:18:57.000 --> 00:19:03.000
well, first of all,
there's an I in front that i
00:19:01.000 --> 00:19:07.000
better not forget.
And then, the rest of it is i.
00:19:07.000 --> 00:19:13.000
So, it's i d times,
it's e to the b i t,
00:19:11.000 --> 00:19:17.000
e to the i b t minus,
now, e to the minus i b t.
00:19:16.000 --> 00:19:22.000
So, that's the way the solution
00:19:22.000 --> 00:19:28.000
looks.
It doesn't look a lot better,
00:19:25.000 --> 00:19:31.000
but now you must use the magic
formulas, which,
00:19:30.000 --> 00:19:36.000
I want you to know as well as
you know Euler's formula,
00:19:35.000 --> 00:19:41.000
even better than you know
Euler's formula.
00:19:41.000 --> 00:19:47.000
They're a consequence of
Euler's formula.
00:19:43.000 --> 00:19:49.000
They're Euler's formula read
backwards.
00:19:45.000 --> 00:19:51.000
Euler's formula says you've got
a complex exponential here.
00:19:49.000 --> 00:19:55.000
Here's how to write it in terms
of sines and cosines.
00:19:52.000 --> 00:19:58.000
The backwards thing says you've
got a sine or a cosine.
00:19:55.000 --> 00:20:01.000
Here is the way to write it in
terms of complex exponentials.
00:20:00.000 --> 00:20:06.000
And, remember,
the way to do it is,
00:20:04.000 --> 00:20:10.000
cosine a is equal to e to the i
a t, i a, plus e to the negative
00:20:11.000 --> 00:20:17.000
i a divided by two.
00:20:17.000 --> 00:20:23.000
And, sine of a is almost the
same thing, except you use a
00:20:24.000 --> 00:20:30.000
minus sign.
And, what everybody forgets,
00:20:28.000 --> 00:20:34.000
you have to divide by i.
So, this is a backwards version
00:20:35.000 --> 00:20:41.000
of Euler's formula.
The two of them taken together
00:20:38.000 --> 00:20:44.000
are equivalent to Euler's
formula.
00:20:40.000 --> 00:20:46.000
If I took cosine a,
multiply this through by i,
00:20:44.000 --> 00:20:50.000
and added them up,
on the right-hand side I'd get
00:20:47.000 --> 00:20:53.000
exactly e to the ia.
I'd get Euler's formula,
00:20:51.000 --> 00:20:57.000
in other words.
All right, so,
00:20:53.000 --> 00:20:59.000
what does this come out to be,
finally?
00:20:55.000 --> 00:21:01.000
This particular sum of
exponentials,
00:20:58.000 --> 00:21:04.000
you should always recognize as
real.
00:21:02.000 --> 00:21:08.000
You know it's real because when
you change i to minus i,
00:21:07.000 --> 00:21:13.000
the two terms switch.
And therefore,
00:21:10.000 --> 00:21:16.000
the expression doesn't change.
What is it?
00:21:14.000 --> 00:21:20.000
This part is twice the cosine
of bt.
00:21:18.000 --> 00:21:24.000
What's this part?
This part is 2 i times the sine
00:21:22.000 --> 00:21:28.000
of bt.
And so, what does the whole
00:21:27.000 --> 00:21:33.000
thing come to be?
It is e to the a t times 2C
00:21:33.000 --> 00:21:39.000
cosine bt plus i times,
did I lose possibly a,
00:21:38.000 --> 00:21:44.000
no it's okay,
minus i times i is minus,
00:21:43.000 --> 00:21:49.000
so, minus 2d times the sine of
bt.
00:21:50.000 --> 00:21:56.000
Shall I write that out?
00:21:55.000 --> 00:22:01.000
So, in other words,
it's e to the a t times 2C
00:21:58.000 --> 00:22:04.000
cosine b t minus 2d times the
sine of b t,
00:22:03.000 --> 00:22:09.000
which is, since 2C and negative
00:22:07.000 --> 00:22:13.000
2d are just arbitrary constants,
just as arbitrary as the
00:22:12.000 --> 00:22:18.000
constants of C and d themselves
are.
00:22:15.000 --> 00:22:21.000
This is our old form of writing
the real solution.
00:22:19.000 --> 00:22:25.000
Here's the way using science
and cosines, and there's the way
00:22:24.000 --> 00:22:30.000
that uses complex numbers and
complex functions throughout.
00:22:30.000 --> 00:22:36.000
Notice they both have two
arbitrary constants in them,
00:22:33.000 --> 00:22:39.000
C and d, two arbitrary
constants.
00:22:36.000 --> 00:22:42.000
That, you expect.
But that has two arbitrary
00:22:39.000 --> 00:22:45.000
constants in it,
too, just the real and
00:22:42.000 --> 00:22:48.000
imaginary parts of that complex
coefficient, C plus i d.
00:22:46.000 --> 00:22:52.000
Well, that took half the
00:22:48.000 --> 00:22:54.000
period, and it was a long,
I don't consider it a
00:22:52.000 --> 00:22:58.000
digression because learning
those ways of dealing with
00:22:56.000 --> 00:23:02.000
complex numbers of complex
functions is a fairly important
00:23:00.000 --> 00:23:06.000
goal in this course,
actually.
00:23:04.000 --> 00:23:10.000
But let's get back now to
studying what the oscillations
00:23:07.000 --> 00:23:13.000
actually look like.
00:23:27.000 --> 00:23:33.000
Okay, well, I'd like to save a
little time, but very quickly,
00:23:34.000 --> 00:23:40.000
you don't have to reproduce
this sketch.
00:23:39.000 --> 00:23:45.000
I remember very well from
Friday to Monday,
00:23:45.000 --> 00:23:51.000
but I can't expect you to for a
variety of reasons.
00:23:51.000 --> 00:23:57.000
I mean, I have to think about
this stuff all weekend.
00:23:58.000 --> 00:24:04.000
And you, God forbid.
So, here's the picture,
00:24:03.000 --> 00:24:09.000
and I won't explain anymore
what's in it,
00:24:06.000 --> 00:24:12.000
except there's the mass.
Here is the spring constant,
00:24:09.000 --> 00:24:15.000
the spring with its constant
here.
00:24:11.000 --> 00:24:17.000
Here's the dashpot with its
constant.
00:24:13.000 --> 00:24:19.000
The equation is from Newton's
law: m x double,
00:24:16.000 --> 00:24:22.000
so this will be x,
and here's, let's say,
00:24:19.000 --> 00:24:25.000
the equilibrium point is over
here.
00:24:21.000 --> 00:24:27.000
It looks like m x double prime;
we derived this last time,
00:24:25.000 --> 00:24:31.000
plus c x prime plus k x equals
zero.
00:24:30.000 --> 00:24:36.000
And now, if I put that in
standard form,
00:24:33.000 --> 00:24:39.000
it's going to look like x
double prime plus c over m x
00:24:39.000 --> 00:24:45.000
prime plus k over m times x
equals zero.
00:24:45.000 --> 00:24:51.000
And, finally,
00:24:47.000 --> 00:24:53.000
the standard form in which your
book writes it,
00:24:52.000 --> 00:24:58.000
which is good,
it's a standard form in general
00:24:56.000 --> 00:25:02.000
that is used in the science and
engineering courses.
00:25:02.000 --> 00:25:08.000
One writes this as,
just to be perverse,
00:25:05.000 --> 00:25:11.000
I'm going to change x back to
y, okay, mostly just to be
00:25:11.000 --> 00:25:17.000
eclectic, to get you used to
every conceivable notation.
00:25:19.000 --> 00:25:25.000
So, I'm going to write this to
change x to y.
00:25:22.000 --> 00:25:28.000
So, that's going to become y
double prime.
00:25:26.000 --> 00:25:32.000
And now, this is given a new
name, p, except to get rid of
00:25:30.000 --> 00:25:36.000
lots of twos,
which would really screw up the
00:25:33.000 --> 00:25:39.000
formulas, make it 2p.
You will see why in a minute.
00:25:38.000 --> 00:25:44.000
So, there's 2p times y prime,
and this thing we
00:25:42.000 --> 00:25:48.000
are going to call omega nought
squared.
00:25:46.000 --> 00:25:52.000
Now, that's okay.
It's a positive number.
00:25:49.000 --> 00:25:55.000
Any positive number is the
square of some other positive
00:25:53.000 --> 00:25:59.000
number.
Take a square root.
00:25:55.000 --> 00:26:01.000
You will see why,
it makes the formulas much
00:25:59.000 --> 00:26:05.000
pretty to call it that.
And, it makes also a lot of
00:26:04.000 --> 00:26:10.000
things much easier to remember.
So, all I'm doing is changing
00:26:08.000 --> 00:26:14.000
the names of the constants that
way in order to get better
00:26:13.000 --> 00:26:19.000
formulas, easier to remember
formulas at the end.
00:26:16.000 --> 00:26:22.000
Now, we are interested in the
case where there is
00:26:20.000 --> 00:26:26.000
oscillations.
In other words,
00:26:22.000 --> 00:26:28.000
I only care about the case in
which this has complex roots,
00:26:27.000 --> 00:26:33.000
because if it has just real
roots, that's the over-damped
00:26:31.000 --> 00:26:37.000
case.
I don't get any oscillations.
00:26:35.000 --> 00:26:41.000
By far, oscillations are by far
the more important of the cases,
00:26:40.000 --> 00:26:46.000
I mean, just because,
I don't know,
00:26:43.000 --> 00:26:49.000
I could go on for five minutes
listing things that oscillate,
00:26:48.000 --> 00:26:54.000
oscillations,
you know, like this.
00:26:51.000 --> 00:26:57.000
So they can oscillate by going
to sleep, and waking up,
00:26:56.000 --> 00:27:02.000
and going to sleep,
and waking up.
00:26:59.000 --> 00:27:05.000
They could oscillate.
So, that means we're going to
00:27:03.000 --> 00:27:09.000
get complex roots.
The characteristic equation is
00:27:07.000 --> 00:27:13.000
going to be r squared plus 2p.
So, p is a constant,
00:27:10.000 --> 00:27:16.000
now, right?
Often, p I use in this position
00:27:13.000 --> 00:27:19.000
to indicate a function of t.
But here, p is a constant.
00:27:16.000 --> 00:27:22.000
So, r squared plus 2p times r
plus omega nought squared is
00:27:20.000 --> 00:27:26.000
equal to zero.
00:27:23.000 --> 00:27:29.000
Now, what are its roots?
Well, you see right away the
00:27:27.000 --> 00:27:33.000
first advantage in putting in
the two there.
00:27:31.000 --> 00:27:37.000
When I use the quadratic
formula, it's negative 2p over
00:27:34.000 --> 00:27:40.000
two.
Remember that two in the
00:27:37.000 --> 00:27:43.000
denominator.
So, that's simply negative p.
00:27:40.000 --> 00:27:46.000
And, how about the rest?
Plus or minus the square root
00:27:44.000 --> 00:27:50.000
of, now do it in your head.
4p squared minus 4 omega nought
00:27:48.000 --> 00:27:54.000
squared.
So, there's a four in both of
00:27:53.000 --> 00:27:59.000
those terms.
When I pull it outside becomes
00:27:56.000 --> 00:28:02.000
two.
And, the two in the denominator
00:27:59.000 --> 00:28:05.000
is lurking, waiting to
annihilate it.
00:28:03.000 --> 00:28:09.000
So, that two disappears
entirely, and it will we are
00:28:06.000 --> 00:28:12.000
left with is,
simply, p squared minus omega
00:28:09.000 --> 00:28:15.000
nought squared.
00:28:11.000 --> 00:28:17.000
Now, whenever people write
quadratic equations,
00:28:14.000 --> 00:28:20.000
and arbitrarily put a two in
there, it's because they were
00:28:18.000 --> 00:28:24.000
going to want to solve the
quadratic equation using the
00:28:21.000 --> 00:28:27.000
quadratic formula,
and they don't want all those
00:28:24.000 --> 00:28:30.000
twos and fours to be cluttering
up the formula.
00:28:29.000 --> 00:28:35.000
That's what we are doing here.
Okay, now, the first case is
00:28:33.000 --> 00:28:39.000
where p is equal to zero.
This is going to explain
00:28:37.000 --> 00:28:43.000
immediately why I wrote that
omega nought squared,
00:28:41.000 --> 00:28:47.000
as you probably already know
from physics.
00:28:44.000 --> 00:28:50.000
If p is equal to zero,
the mass isn't zero.
00:28:48.000 --> 00:28:54.000
Otherwise, nothing good would
be happening here.
00:28:52.000 --> 00:28:58.000
It must be that the damping is
zero.
00:28:55.000 --> 00:29:01.000
So, p is equal to zero
corresponds to undamped.
00:29:00.000 --> 00:29:06.000
There is no dashpot.
The oscillations are undamped.
00:29:03.000 --> 00:29:09.000
And, the equation,
then, becomes the solutions,
00:29:06.000 --> 00:29:12.000
then, are, well,
the equation becomes the
00:29:09.000 --> 00:29:15.000
equation of simple harmonic
motion, which,
00:29:12.000 --> 00:29:18.000
I think you already are used to
writing in this form.
00:29:15.000 --> 00:29:21.000
And, the reason you're writing
in this form because you know
00:29:19.000 --> 00:29:25.000
when you do that,
this becomes the circular
00:29:22.000 --> 00:29:28.000
frequency of the oscillations.
The solutions are pure
00:29:26.000 --> 00:29:32.000
oscillations,
and omega nought is
00:29:29.000 --> 00:29:35.000
the circular frequency.
So, right away from the
00:29:33.000 --> 00:29:39.000
equation itself,
if you write it in this form,
00:29:37.000 --> 00:29:43.000
you can read off what the
frequency of the solutions is
00:29:41.000 --> 00:29:47.000
going to be, the circular
frequency of the solutions.
00:29:45.000 --> 00:29:51.000
Now, the solutions themselves,
of course, look like,
00:29:49.000 --> 00:29:55.000
the general solutions look like
y equal, in this particular
00:29:54.000 --> 00:30:00.000
case, the p part is zero.
This is zero.
00:29:57.000 --> 00:30:03.000
It's simply,
so, in this case,
00:29:59.000 --> 00:30:05.000
r is equal to omega nought i
times omega naught plus or
00:30:03.000 --> 00:30:09.000
minus, but as before we don't
bother with the minus sign since
00:30:08.000 --> 00:30:14.000
one of those roots is good
enough.
00:30:13.000 --> 00:30:19.000
And then, the solutions are
simply c1 cosine omega nought t
00:30:16.000 --> 00:30:22.000
plus c2 sine omega nought t.
00:30:20.000 --> 00:30:26.000
That's if you write it out in
00:30:23.000 --> 00:30:29.000
the sign, and if you write it
using the trigonometric
00:30:26.000 --> 00:30:32.000
identity, then the other way of
writing it is a times the cosine
00:30:30.000 --> 00:30:36.000
of omega nought t.
00:30:34.000 --> 00:30:40.000
But now you will have to put it
a phase lag.
00:30:37.000 --> 00:30:43.000
So, you have those two forms of
writing it.
00:30:41.000 --> 00:30:47.000
And, I assume you remember
writing the little triangle,
00:30:45.000 --> 00:30:51.000
which converts one into the
other.
00:30:48.000 --> 00:30:54.000
Okay, so this justifies calling
this omega nought squared
00:30:53.000 --> 00:30:59.000
rather than k over m.
00:30:56.000 --> 00:31:02.000
And now, the question is what
does the damp case look like?
00:31:01.000 --> 00:31:07.000
It requires a somewhat closer
analysis, and it requires a
00:31:06.000 --> 00:31:12.000
certain amount of thinking.
So, let's begin with an epsilon
00:31:13.000 --> 00:31:19.000
bit of thinking.
So, here's my question.
00:31:18.000 --> 00:31:24.000
So, in the damped case,
I want to be sure that I'm
00:31:24.000 --> 00:31:30.000
getting oscillations.
When do I get oscillations if,
00:31:30.000 --> 00:31:36.000
well, we get oscillations if
those roots are really complex,
00:31:37.000 --> 00:31:43.000
and not masquerading.
Now, when are the roots going
00:31:43.000 --> 00:31:49.000
to be really complex?
This has to be,
00:31:46.000 --> 00:31:52.000
the inside has to be negative.
p squared minus omega squared
00:31:52.000 --> 00:31:58.000
must be negative.
00:31:56.000 --> 00:32:02.000
p squared minus omega nought
squared must be less than zero
00:32:01.000 --> 00:32:07.000
so that we are taking a square
root of negative number,
00:32:06.000 --> 00:32:12.000
and we are getting a real
complex roots,
00:32:09.000 --> 00:32:15.000
really complex roots.
In other words,
00:32:14.000 --> 00:32:20.000
now, this says,
remember these numbers are all
00:32:17.000 --> 00:32:23.000
positive, p and omega nought are
positive.
00:32:21.000 --> 00:32:27.000
So, the condition is that p
should be
00:32:25.000 --> 00:32:31.000
less than omega nought.
In other words,
00:32:28.000 --> 00:32:34.000
the damping should be less than
the circular frequency,
00:32:32.000 --> 00:32:38.000
except p is not the damping.
It's half the damping,
00:32:38.000 --> 00:32:44.000
and it's not really the damping
either because it involved the
00:32:43.000 --> 00:32:49.000
m, too.
You'd better just call it p.
00:32:47.000 --> 00:32:53.000
Naturally, I could write the
condition out in terms of c,
00:32:52.000 --> 00:32:58.000
m, and k.
So, your book does that,
00:32:55.000 --> 00:33:01.000
but I'm not going to.
It gives it in terms of c,
00:32:59.000 --> 00:33:05.000
m, and k, which somebody might
want to know.
00:33:03.000 --> 00:33:09.000
But, you know,
we don't have to do everything
00:33:08.000 --> 00:33:14.000
here.
Okay, so let's assume that this
00:33:12.000 --> 00:33:18.000
is true.
What is the solution look like?
00:33:15.000 --> 00:33:21.000
Well, we already experimented
with that last time.
00:33:19.000 --> 00:33:25.000
Remember, there was some
guiding thing which was an
00:33:23.000 --> 00:33:29.000
exponential.
And then, down here,
00:33:26.000 --> 00:33:32.000
we wrote the negative.
So, this was an exponential.
00:33:31.000 --> 00:33:37.000
In fact, it was the
exponential, e to the negative
00:33:35.000 --> 00:33:41.000
pt.
And, in between that,
00:33:38.000 --> 00:33:44.000
the curve tried to do its
thing.
00:33:41.000 --> 00:33:47.000
So, the solution looks sort of
like this.
00:33:45.000 --> 00:33:51.000
It oscillated,
but it had to use that
00:33:48.000 --> 00:33:54.000
exponential function as its
guidelines, as its amplitude,
00:33:53.000 --> 00:33:59.000
in other words.
Now, this is a truly terrible
00:33:57.000 --> 00:34:03.000
picture.
It's so terrible,
00:34:01.000 --> 00:34:07.000
it's unusable.
Okay, this picture never
00:34:05.000 --> 00:34:11.000
happened.
Unfortunately,
00:34:07.000 --> 00:34:13.000
this is not my forte along with
a lot of other things.
00:34:12.000 --> 00:34:18.000
All right, let's try it better.
Here's our better picture.
00:34:18.000 --> 00:34:24.000
Okay, there's the exponential.
At this point,
00:34:22.000 --> 00:34:28.000
I'm supposed to have a lecture
demonstration.
00:34:26.000 --> 00:34:32.000
It's supposed to go up on the
thing, so you can all see it.
00:34:34.000 --> 00:34:40.000
But then, you wouldn't be able
to copy it.
00:34:37.000 --> 00:34:43.000
So, at least we are on even
terms now.
00:34:40.000 --> 00:34:46.000
Okay, how does the actual curve
look?
00:34:43.000 --> 00:34:49.000
Well, I'm just trying to be
fair.
00:34:45.000 --> 00:34:51.000
That's all.
Okay, after a while,
00:34:48.000 --> 00:34:54.000
the point is,
just so we have something to
00:34:51.000 --> 00:34:57.000
aim at, let's say,
okay, here we are going to go,
00:34:55.000 --> 00:35:01.000
we're going to get down through
there.
00:35:00.000 --> 00:35:06.000
Okay then, this is our better
curve.
00:35:03.000 --> 00:35:09.000
Okay, so I am a solution,
a particular solution
00:35:08.000 --> 00:35:14.000
satisfying this initial
condition.
00:35:12.000 --> 00:35:18.000
I started here,
and that was my initial
00:35:16.000 --> 00:35:22.000
velocity.
The slope of that thing gave me
00:35:20.000 --> 00:35:26.000
the initial velocity.
Now, the interesting question
00:35:26.000 --> 00:35:32.000
is, the first,
in some ways,
00:35:28.000 --> 00:35:34.000
the most interesting question,
though there will be others,
00:35:35.000 --> 00:35:41.000
too, is what is this spacing?
Well, that's a period.
00:35:42.000 --> 00:35:48.000
And now, it's half a period.
I clearly ought to think of
00:35:47.000 --> 00:35:53.000
this as the whole period.
So, let's call that,
00:35:51.000 --> 00:35:57.000
I'm going to call this pi over,
so this spacing here,
00:35:56.000 --> 00:36:02.000
from there to there,
I will call that pi divided by
00:36:01.000 --> 00:36:07.000
omega one because this,
from here to here,
00:36:05.000 --> 00:36:11.000
should be, I hope,
twice that, two pi over omega
00:36:10.000 --> 00:36:16.000
one.
Now, my question is,
00:36:14.000 --> 00:36:20.000
so this, for a solution,
it's, in fact,
00:36:18.000 --> 00:36:24.000
is going to cross the axis
regularly in that way.
00:36:24.000 --> 00:36:30.000
My question is,
how does this period,
00:36:28.000 --> 00:36:34.000
so this is going to be its half
period.
00:36:34.000 --> 00:36:40.000
I will put period in quotation
marks because this isn't really
00:36:39.000 --> 00:36:45.000
a periodic function because it's
decreasing all the time in
00:36:43.000 --> 00:36:49.000
amplitude.
But, it's trying to be
00:36:46.000 --> 00:36:52.000
periodic.
At lease it's doing something
00:36:49.000 --> 00:36:55.000
periodically.
It's crossing the axis
00:36:52.000 --> 00:36:58.000
periodically.
So, this is the half period.
00:36:55.000 --> 00:37:01.000
Two pi over omega one
would be its full
00:37:00.000 --> 00:37:06.000
period.
What I want to know is,
00:37:02.000 --> 00:37:08.000
how does that half period,
or how does-- omega one is
00:37:07.000 --> 00:37:13.000
called its pseudo-frequency.
This should really be called
00:37:13.000 --> 00:37:19.000
its pseudo-period.
Everything is pseudo.
00:37:16.000 --> 00:37:22.000
Everything is fake here.
Like, the amoeba has its fake
00:37:21.000 --> 00:37:27.000
foot and stuff like that.
Okay, so this is its
00:37:24.000 --> 00:37:30.000
pseudo-period,
pseudo-frequency,
00:37:27.000 --> 00:37:33.000
pseudo-circular frequency,
but that's hopeless.
00:37:31.000 --> 00:37:37.000
I guess it should be circular
pseudo-frequency,
00:37:35.000 --> 00:37:41.000
or I don't know how you say
that.
00:37:39.000 --> 00:37:45.000
I don't think pseudo is a word
all by itself,
00:37:46.000 --> 00:37:52.000
not even in 18.03,
circular.
00:37:50.000 --> 00:37:56.000
Okay, here's my question.
If the damping goes up,
00:37:58.000 --> 00:38:04.000
this is the damping term.
If the damping goes up,
00:38:06.000 --> 00:38:12.000
what happens to the
pseudo-frequency?
00:38:11.000 --> 00:38:17.000
The frequency is how often the
curve, this is high-frequency,
00:38:19.000 --> 00:38:25.000
and this is low-frequency,
okay?
00:38:23.000 --> 00:38:29.000
So, my question is,
which way does the frequency
00:38:29.000 --> 00:38:35.000
go?
If the damping goes up,
00:38:33.000 --> 00:38:39.000
does the frequency go up or
down?
00:38:38.000 --> 00:38:44.000
Down.
I mean, I'm just asking you to
00:38:42.000 --> 00:38:48.000
answer intuitively on the basis
of your intuition about how this
00:38:50.000 --> 00:38:56.000
thing explains,
how this thing goes,
00:38:55.000 --> 00:39:01.000
and now let's get the formula.
What, in fact,
00:39:01.000 --> 00:39:07.000
is omega one?
What is omega one?
00:39:05.000 --> 00:39:11.000
The answer is when I solve the
equation, so,
00:39:09.000 --> 00:39:15.000
r is now, so in other words,
if omega one is,
00:39:13.000 --> 00:39:19.000
sorry, if I have p,
if p is no longer zero as it
00:39:18.000 --> 00:39:24.000
was in the undamped case,
what is the root,
00:39:22.000 --> 00:39:28.000
now?
Okay, well, the root is minus p
00:39:25.000 --> 00:39:31.000
plus or minus the square root of
p squared, --
00:39:31.000 --> 00:39:37.000
-- now I'm going to write it
this way, minus,
00:39:34.000 --> 00:39:40.000
to indicate that it's really a
negative number,
00:39:38.000 --> 00:39:44.000
omega squared minus p squared.
00:39:42.000 --> 00:39:48.000
Now, I'm going to call this,
because you see when I change
00:39:47.000 --> 00:39:53.000
this to sines and cosines,
the square root of this number
00:39:52.000 --> 00:39:58.000
is what's going to become that
new frequency.
00:39:55.000 --> 00:40:01.000
I'm going to call that minus p
plus or minus the square root of
00:40:00.000 --> 00:40:06.000
minus omega one squared.
That's going to be the new
00:40:06.000 --> 00:40:12.000
frequency.
And therefore,
00:40:08.000 --> 00:40:14.000
the root is going to change so
that the corresponding solution
00:40:13.000 --> 00:40:19.000
is going to look,
how?
00:40:15.000 --> 00:40:21.000
Well, it's going to be e to the
negative pt times,
00:40:20.000 --> 00:40:26.000
let's write it out first in
terms of sines and cosines,
00:40:25.000 --> 00:40:31.000
times the cosine of,
well, the square root of omega
00:40:29.000 --> 00:40:35.000
one squared is omega one.
00:40:35.000 --> 00:40:41.000
But, there's an i out front
because of the negative sign in
00:40:39.000 --> 00:40:45.000
front of that.
So, it's going to be the cosine
00:40:42.000 --> 00:40:48.000
of omega one t
plus c2 times the sine of omega
00:40:47.000 --> 00:40:53.000
one t.
Or, if you prefer to write it
00:40:51.000 --> 00:40:57.000
out in the other form,
it's e to the minus p t times
00:40:54.000 --> 00:41:00.000
some amplitude,
which depends on c1 and c2,
00:40:57.000 --> 00:41:03.000
times the cosine of omega one t
minus the phase lag.
00:41:02.000 --> 00:41:08.000
Now, when I do that,
00:41:07.000 --> 00:41:13.000
you see omega one is
this pseudo-frequency.
00:41:13.000 --> 00:41:19.000
In other words,
this number omega one is the
00:41:17.000 --> 00:41:23.000
same one that I identified here.
And, why is that?
00:41:23.000 --> 00:41:29.000
Well, because,
what are two successive times?
00:41:27.000 --> 00:41:33.000
Suppose it crosses,
suppose the solution crosses
00:41:33.000 --> 00:41:39.000
the x-axis, sorry,
y-- the t-axis.
00:41:38.000 --> 00:41:44.000
For the first time,
at the point t1,
00:41:42.000 --> 00:41:48.000
what's the next time it crosses
t2?
00:41:46.000 --> 00:41:52.000
Let's jump to the two times
across it.
00:41:51.000 --> 00:41:57.000
So, I want this to be a whole
period, not a half period.
00:41:57.000 --> 00:42:03.000
What's t2?
Well, I say that t2 is nothing
00:42:02.000 --> 00:42:08.000
but 2 pi divided by omega one.
00:42:07.000 --> 00:42:13.000
And, you can see that because
when I plug in,
00:42:10.000 --> 00:42:16.000
if it's zero,
if I have a point where it's
00:42:14.000 --> 00:42:20.000
zero, so, omega one t minus phi,
00:42:18.000 --> 00:42:24.000
when will it be zero for the
first time?
00:42:22.000 --> 00:42:28.000
Well, that will be when the
cosine has to be zero.
00:42:26.000 --> 00:42:32.000
So, it will be some multiple
of, it will be,
00:42:30.000 --> 00:42:36.000
say, pi over two.
Then, the next time this
00:42:35.000 --> 00:42:41.000
happens will be,
if that happens at t1,
00:42:39.000 --> 00:42:45.000
then the next time it happens
will be at t1 plus 2 pi divided
00:42:45.000 --> 00:42:51.000
by omega one.
00:42:49.000 --> 00:42:55.000
That will also be pi over two
plus how much?
00:42:54.000 --> 00:43:00.000
Plus 2 pi, which is the next
time the cosine gets around and
00:43:00.000 --> 00:43:06.000
is doing its thing,
becoming zero as it goes down,
00:43:05.000 --> 00:43:11.000
not as it's coming up again.
In other words,
00:43:11.000 --> 00:43:17.000
this is what you should add to
the first time to get this
00:43:17.000 --> 00:43:23.000
second time that the cosine
becomes zero coming in the
00:43:23.000 --> 00:43:29.000
direction from top to the
bottom.
00:43:26.000 --> 00:43:32.000
So, this is,
in fact, the frequency with
00:43:30.000 --> 00:43:36.000
which it's crossing the axis.
Now, notice,
00:43:36.000 --> 00:43:42.000
I'm running out of boards.
What a disaster!
00:43:41.000 --> 00:43:47.000
In that expression,
take a look at it.
00:43:46.000 --> 00:43:52.000
I want to know what depends on
what.
00:43:50.000 --> 00:43:56.000
So, p, in that,
we got constants.
00:43:54.000 --> 00:44:00.000
We got p.
We got phi.
00:43:57.000 --> 00:44:03.000
We got A.
What else we got?
00:44:00.000 --> 00:44:06.000
Omega one.
What do these things depend
00:44:07.000 --> 00:44:13.000
upon?
You've got to keep it firmly in
00:44:11.000 --> 00:44:17.000
mind.
This depends only on the ODE.
00:44:14.000 --> 00:44:20.000
It's basically the damping.
It depends on c and m.
00:44:19.000 --> 00:44:25.000
Essentially, it's c over 2m
00:44:23.000 --> 00:44:29.000
actually.
How about phi?
00:44:25.000 --> 00:44:31.000
Well, phi, what else depends
only on the ODE?
00:44:30.000 --> 00:44:36.000
Omega one depends
only on the ODE.
00:44:36.000 --> 00:44:42.000
What's the formula for omega
one?
00:44:38.000 --> 00:44:44.000
Omega one squared.
00:44:40.000 --> 00:44:46.000
Where do we have it?
Omega one squared,
00:44:43.000 --> 00:44:49.000
I never wrote the formula for
you.
00:44:46.000 --> 00:44:52.000
So, we have omega nought
squared minus p squared equals
00:44:50.000 --> 00:44:56.000
omega one squared.
00:44:53.000 --> 00:44:59.000
What's the relation between
them?
00:44:56.000 --> 00:45:02.000
That's the Pythagorean theorem.
If this is omega nought,
00:45:00.000 --> 00:45:06.000
then this omega one,
this is p.
00:45:04.000 --> 00:45:10.000
They make a little,
right triangle in other words.
00:45:09.000 --> 00:45:15.000
The omega one depends on the
spring.
00:45:13.000 --> 00:45:19.000
So, it's equal to,
well, it's equal to that thing.
00:45:19.000 --> 00:45:25.000
So, it depends on the damping.
It depends upon the damping,
00:45:25.000 --> 00:45:31.000
and it depends on the spring
constant.
00:45:30.000 --> 00:45:36.000
How about the phi and the A?
What do they depend on?
00:45:36.000 --> 00:45:42.000
They depend upon the initial
conditions.
00:45:42.000 --> 00:45:48.000
So, the mass of constants,
they have different functions.
00:45:47.000 --> 00:45:53.000
What's making this complicated
is that our answer needs four
00:45:53.000 --> 00:45:59.000
parameters to describe it.
This tells you how fast it's
00:45:59.000 --> 00:46:05.000
coming down.
This tells you the phase lag.
00:46:03.000 --> 00:46:09.000
This amplitude modifies,
it tells you whether the
00:46:08.000 --> 00:46:14.000
exponential curve starts going,
is like that or goes like this.
00:46:15.000 --> 00:46:21.000
And, finally,
the omega one is this
00:46:18.000 --> 00:46:24.000
pseudo-frequency,
which tells you how it's
00:46:22.000 --> 00:46:28.000
bobbing up and down.