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I just recalling some of the
notation we are going to need
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for today, and a couple of the
facts that we're going to use,
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plus trying to clear up a
couple of confusions that the
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recitations report.
This can be thought of two
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ways.
It's a formal polynomial in D,
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in the letter D.
It just has the shape of the
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polynomial, D squared plus AD
plus B.
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A and B are constant
coefficients.
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But, it's also,
at the same time,
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if you think what it does,
it's a linear operator on
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functions.
It's a linear operator on
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functions like y of t.
You think of it both ways:
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formal polynomial because we
want to do things like factoring
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it, substituting two for D and
things like that.
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Those are things you do with
polynomials.
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You do them algebraically.
You can take the formal
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derivative of the polynomial
because it's just sums of
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powers.
On the other hand,
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as a linear operator,
it does something to functions.
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It differentiates them,
multiplies them by constants or
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something like that.
So it's, so to speak,
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has a dual aspect this way.
And, that's one of the things
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we are exploiting what we use
operator methods to solve
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differential equations.
Now, let me remind you of the
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key thing we were interested in.
f of t:
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not any old function,
we'll get to that next time,
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but f of t,
exponentials.
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So, it should be an exponential
or something like an
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exponential, or pretty close to
it, for example,
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something with sine t
and cosine t,
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or e to the,
that could be thought of as
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part of the real or imaginary
part of a complex exponential.
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And, maybe by the end of today,
we will have generalized that
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even little more.
But basically,
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I'm interested in exponentials.
Let's make it alpha complex.
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That will at least take care of
the cases, e to the ax times
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cosine bx, sin bx,
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which are the main cases.
Those are the main cases.
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Then, remember the little table
we made.
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I simply gave you the formula
for the particular solution.
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So, what we're looking for is
we already know how to solve the
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homogeneous equation.
What we want is that particular
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solution.
And then, the recipe for it I
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gave you, these things were
proved by the substitution rules
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and exponential shift rules.
The recipe was that if f of t
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was, let's make a
little table.
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f of t is, well,
it's always e to the a t.
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So, in other words,
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it's e to the a t.
The cases are,
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so yp, what is the yp?
Well, it is the normal case is
51
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yp equals e to that alpha t
divided by the
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polynomial where you substitute,
you take that polynomial,
53
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and wherever you see a D,
you substitute the complex
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number, alpha.
There, I'm thinking of it as a
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formal polynomial.
I'm not thinking of it as an
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operator.
Now, this breaks down.
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So, that's the formula for the
particular solution.
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The only trouble is,
it breaks down if p of alpha is
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zero.
So, we have to assume that it's
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not.
Now, if p of alpha is zero,
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that means alpha is a root of
the polynomial,
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a zero of the polynomial is a
better word.
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So, in that case,
it will be e to the alpha t
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divided by p prime of alpha.
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Differentiate formally the
polynomials, --
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-- and you will get 2D plus A.
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And now, substitute in the
alpha.
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And, this will be okay provided
p prime of alpha
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is not zero.
That means that alpha is the
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simple root, simple zero of p.
And then, there's one more
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case, which, since I won't need
today, I won't write on the
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board.
But, you'll need it for
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homework.
So, make sure you know it.
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Another words,
if this is zero,
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then you've got a double root.
And, there is still a different
76
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formula.
And, this is wrong because I
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forgot the t.
Yes?
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I could tell on your faces.
That was before,
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and now we are up to today.
What we are interested in
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talking about today is what this
has to do with the phenomenon of
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resonance.
Everybody knows at least one
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case of resonance,
I hope.
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A little kid is on his swing,
right?
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Back and forth,
and they are very,
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very little,
so they want a push.
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Okay, well, everybody knows
that to make the swing go,
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a swing has a certain natural
frequency.
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It swings back and forth like
that.
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It's a simple pendulum.
It's actually damped,
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but let's pretend that it
isn't.
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Everybody knows you want to
push a kid on a swing so that
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they go high.
You have to push with
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essentially the same frequency
that the natural frequency of
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the spring, of the swing is.
It's automatic,
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because when you come back
here, it gets to there,
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and that's where you push.
So, automatically,
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you time your pushes.
But if you want the kid to
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stop, you just do the opposite.
Push at the wrong time.
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So anyway, that's resonance.
Of course, there are more
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serious applications of it.
It's what made the Tacoma
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Bridge fall down,
and I think movies of that are
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now being shown not merely on
television, but in elementary
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school.
Resonance is what made,
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okay, more resonance stories
later.
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So, my aim is,
what is this physical
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phenomenon, that to get a big
amplitude you should have it
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match the frequency?
What does that have to do with
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a differential equation?
Well, the differential equation
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for that simple pendulum,
let's assume it's undamped,
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will be of the type y double
prime plus,
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I'm using t now since t is
time.
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That will be our new
independent variable,
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plus omega nought squared
is the natural
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frequency of the pendulum or of
the spring, or whatever it is
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that's doing the vibrating.
Yeah, any questions?
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What we're doing is driving
that with the cosine,
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with something of a different
frequency.
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So, this is the input,
or the driving term as it's
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often called,
or it's sometimes called the
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forcing term.
And, the point is I'm going to
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assume that the frequency is
different.
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The driving frequency is
different from the natural
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frequency.
So, this is the input
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frequency.
Okay, and now let's simply
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solve the equation and see what
we get.
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So, it's if I write it using
the operator,
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it's D squared plus omega
nought squared applied to y
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is equal to cosine.
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It's a good idea to do this
because the formulas are going
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to ask you to substitute into a
polynomial.
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So, it's good to have the
polynomial right in front of you
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to avoid the possibility of
error.
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Well, really what I want is the
particular solution.
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It's the particular solution
that's going to give me a pure
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oscillation.
And, the thing to do is,
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of course, since this cosine,
you want to make it complex.
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So, we are going to complexify
the equation in order to be able
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to solve it more easily,
and in order to be able to use
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those formulas.
So, the complex equation is
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going to be D squared plus omega
nought squared.
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Well, it's going to be a
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complex, particular solution.
So, I'll call it y tilde.
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And, on the right-hand side,
that's going to be e to the i
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omega1 t.
Cosine is the real part of
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this.
So, when we get our answer,
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we want to be sure to take the
real part of the answer.
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I don't want the complex
answer, I want its real part.
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I want the real answer,
in other words,
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the really real answer,
the real real answer.
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So, now without further ado,
because of those beautiful,
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the problem has been solved
once and for all by using the
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substitution rule.
I did that for you on Monday.
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The answer is simply e to the i
omega1 t
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divided by what?
This polynomial with omega one
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substituted in for D.
So, sorry, i omega one,
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the complex
coefficient of t.
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So, it is substitute i omega
for D, I omega one for D,
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and you get (i omega one)
squared plus omega nought
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squared.
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Well, let's make that look a
little bit better.
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This should be e to the (i
omega one t)
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divided by, now,
what's this?
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This is simply omega nought
squared minus omega
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one squared.
But, I want the real part of
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it.
So, as one final,
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last step, the real part of
that is what we call just the
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real particular solution,
so, yp without the tilde
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anymore.
And, the real part of this,
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well, this cosine plus i sine.
And, the denominator,
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luckily, turns out to be real.
So, it's simply going to be
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cosine omega one t.
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That's the top,
divided by this thing,
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omega nought squared minus
omega one squared.
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In other words,
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that's the response.
This is the input,
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and that's what came out.
Well, in other words,
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what one sees is,
regardless of what natural
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frequency this system wanted to
use for itself,
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at least for this solution,
what it responds to is the
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driving frequency,
the input frequency.
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The only thing is that the
amplitude has changed,
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and in a rather dramatic way,
if omega1, depending on the
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relative sizes of omega1 and
omega2.
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Now, the interesting case is
when omega one is very close to
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omega, the natural frequency.
When you push it with
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approximately it's natural
frequency, then the solution is
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big amplitude.
The amplitude is large.
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So, the solution looks like the
frequency.
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The input might have looked
like this.
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Well, it's cosine,
so it ought to start up here.
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The input might have looked
like this, but the response will
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be a curve with the same
frequency and still a pure
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oscillation.
But, it will have much,
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much bigger amplitude.
And, it's because the
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denominator, omega nought
squared minus omega
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one squared,
is always zero.
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So, the response will,
instead, look like this.
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Now, to all intents and
purposes, that's resonance.
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You are pushing something with
approximately the same
200
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frequency, something that wants
to oscillate.
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And, you are pushing it with
approximately the same frequency
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that it would like to oscillate
by itself.
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And, what that does is it
builds up the amplitude
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Well, what happens if omega one
is actually equal
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to omega zero?
So, that's the case I'd like to
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analyze for you now.
Suppose the two are equal,
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in other words.
Well, the problem is,
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of course, I can't use that
same solution.
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00:13:24 --> 00:13:30
It isn't applicable.
But that's why I gave you,
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00:13:28 --> 00:13:34
derived for you using the
exponential shift law last time,
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00:13:33 --> 00:13:39
the second version,
when it is a root.
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00:13:38 --> 00:13:44
So, if omega one equals omega
nought,
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00:13:42 --> 00:13:48
so now our equation looks like
D squared plus omega nought
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00:13:47 --> 00:13:53
squared, the natural frequency,
y.
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00:13:51 --> 00:13:57
But this time,
the driving frequency,
216
00:13:54 --> 00:14:00
the input frequency,
is omega nought itself.
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00:13:57 --> 00:14:03
Then, the same analysis,
a lot of it is,
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00:14:00 --> 00:14:06
well, I'd better be careful.
I'd better be careful.
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00:14:04 --> 00:14:10
Let's go through the analysis
again very rapidly.
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What we want to do is first
complexify it,
221
00:14:13 --> 00:14:19
and then solve.
So, the complex equation will
222
00:14:17 --> 00:14:23
be D squared plus omega nought
squared times y tilde equals e
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00:14:22 --> 00:14:28
to the i omega nought t,
this time.
224
00:14:24 --> 00:14:30
225
00:14:29 --> 00:14:35
But now, i omega is zero of
this polynomial.
226
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That's why I picked it,
right?
227
00:14:35 --> 00:14:41
If I plug in i omega zero,
I get i omega zero
228
00:14:40 --> 00:14:46
quantity squared plus omega
nought squared.
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That's zero.
230
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So, I'm in the second case.
So, i omega nought is a simple
231
00:14:53 --> 00:14:59
root, simple zero,
of D squared plus
232
00:14:58 --> 00:15:04
omega nought,
that polynomial squared.
233
00:15:04 --> 00:15:10
Therefore, the complex
particular solution is now t e
234
00:15:08 --> 00:15:14
to the i omega nought t
divided by p prime,
235
00:15:13 --> 00:15:19
where you plug in that root,
the i omega nought.
236
00:15:17 --> 00:15:23
Now, what's p prime?
237
00:15:19 --> 00:15:25
p prime is 2D, right?
238
00:15:22 --> 00:15:28
If I differentiate this
formally, as if D were a
239
00:15:26 --> 00:15:32
variable, the way you
differentiate polynomials,
240
00:15:29 --> 00:15:35
the derivative,
this is a constant,
241
00:15:32 --> 00:15:38
and the derivative is 2D.
So, the denominator should have
242
00:15:38 --> 00:15:44
two times for D.
You are going to plug in i
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00:15:42 --> 00:15:48
omega zero.
So, it's 2 i omega zero.
244
00:15:47 --> 00:15:53
And now, I want the real part
245
00:15:51 --> 00:15:57
of that, which is what?
Well, think about it.
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00:15:55 --> 00:16:01
The top is cosine plus i sine.
The real part is now going to
247
00:16:00 --> 00:16:06
come from the sine,
right, because it's cosine plus
248
00:16:05 --> 00:16:11
i sine.
But this i is going to divide
249
00:16:09 --> 00:16:15
out the i that goes with this
sine.
250
00:16:11 --> 00:16:17
And, therefore,
the real part is going to be t
251
00:16:15 --> 00:16:21
times the sine,
this time, of omega nought t.
252
00:16:18 --> 00:16:24
And, that's going to be divided
253
00:16:22 --> 00:16:28
by, well, the i canceled out the
i that was in front of the sine
254
00:16:27 --> 00:16:33
function.
And therefore,
255
00:16:28 --> 00:16:34
what's left is two omega nought
down below.
256
00:16:34 --> 00:16:40
So, that's our particular
solution now.
257
00:16:36 --> 00:16:42
Well, it looks different from
that guy.
258
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It doesn't look like that
anymore.
259
00:16:42 --> 00:16:48
What does it look like?
Well, it shows the way to plot
260
00:16:46 --> 00:16:52
such things is basically it's an
oscillation of frequency omega
261
00:16:50 --> 00:16:56
nought.
But, its amplitude is changing.
262
00:16:54 --> 00:17:00
So, the way to do it is,
as always, if you have a basic
263
00:16:58 --> 00:17:04
oscillation which is neither too
fast nor too slow,
264
00:17:02 --> 00:17:08
think of that as the thing,
and the other stuff multiplying
265
00:17:06 --> 00:17:12
it, think of it as changing the
amplitude of that oscillation
266
00:17:10 --> 00:17:16
with time.
So, the amplitude is that
267
00:17:14 --> 00:17:20
function, t divided by two omega
zero.
268
00:17:18 --> 00:17:24
So, just as we did when we
talked about damping,
269
00:17:21 --> 00:17:27
you plot that and it's negative
on the picture.
270
00:17:25 --> 00:17:31
So, this is the function whose
graph is t divided by two omega
271
00:17:29 --> 00:17:35
nought.
That's the changing amplitude,
272
00:17:34 --> 00:17:40
as it were.
And then, the function itself
273
00:17:37 --> 00:17:43
does what oscillation it can,
but it has to stay within those
274
00:17:41 --> 00:17:47
lines.
So, the thing that's
275
00:17:43 --> 00:17:49
oscillating is sine omega nought
t,
276
00:17:47 --> 00:17:53
which would like to be a pure
oscillation, but can't because
277
00:17:52 --> 00:17:58
its amplitude is being changed
by that thing.
278
00:17:55 --> 00:18:01
So, it's doing this,
and now the rest I have to
279
00:17:58 --> 00:18:04
leave to your imagination.
In other words,
280
00:18:03 --> 00:18:09
what happens when omega nought
is equal to, when the driving
281
00:18:07 --> 00:18:13
frequency is actually equal to
omega nought,
282
00:18:11 --> 00:18:17
mathematically this turns into
a different looking solution,
283
00:18:15 --> 00:18:21
one with steadily increasing
amplitude.
284
00:18:18 --> 00:18:24
The amplitude increases
linearly like the function t
285
00:18:22 --> 00:18:28
divided by two omega nought.
286
00:18:25 --> 00:18:31
Well, many people are upset by
this, slightly,
287
00:18:29 --> 00:18:35
in the sense that there is a
funny feeling.
288
00:18:32 --> 00:18:38
How is it that that solution
can turn into this one?
289
00:18:38 --> 00:18:44
If I simply let omega one go to
omega zero, what happens?
290
00:18:43 --> 00:18:49
Well, the pink curve just gets
taller and taller,
291
00:18:47 --> 00:18:53
and after a while all you see
of it is just a bunch of
292
00:18:52 --> 00:18:58
vertical lines which seem to be
spaced at whatever the right
293
00:18:57 --> 00:19:03
period is for that function.
It's sort of like being in a
294
00:19:03 --> 00:19:09
first story window and watching
a giraffe go by.
295
00:19:08 --> 00:19:14
All you see is that.
Okay.
296
00:19:12 --> 00:19:18
297
00:19:20 --> 00:19:26
So, my concern is how does that
function turn into this one?
298
00:19:24 --> 00:19:30
I have something in mind to
remind you of,
299
00:19:27 --> 00:19:33
and that's why we'll go through
this little exercise.
300
00:19:31 --> 00:19:37
It's a simple exercise.
But the function of it is,
301
00:19:35 --> 00:19:41
of course that as omega one
goes to omega zero cannot
302
00:19:39 --> 00:19:45
possibly turn into this.
It's doing the wrong thing near
303
00:19:43 --> 00:19:49
zero.
It's already zooming up.
304
00:19:45 --> 00:19:51
But, the point is,
this is not the only particular
305
00:19:49 --> 00:19:55
solution on the block.
Any solution whatsoever of the
306
00:19:52 --> 00:19:58
differential equation,
the inhomogeneous equation,
307
00:19:56 --> 00:20:02
is a particular solution.
It's like Fred Rogers:
308
00:20:01 --> 00:20:07
everybody is special.
Okay, so all solutions are
309
00:20:05 --> 00:20:11
special.
We don't have to use that one.
310
00:20:08 --> 00:20:14
So, I will use,
where are all the other
311
00:20:11 --> 00:20:17
solutions?
So, I'm going back to the
312
00:20:14 --> 00:20:20
equation D squared plus omega
zero squared,
313
00:20:18 --> 00:20:24
applied to y,
314
00:20:20 --> 00:20:26
is equal to cosine
omega one t.
315
00:20:24 --> 00:20:30
Now, the particular solution we
found was that one,
316
00:20:27 --> 00:20:33
cosine omega one t divided by
that omega nought squared minus
317
00:20:32 --> 00:20:38
omega one squared.
318
00:20:39 --> 00:20:45
What do the other particular
solutions look like?
319
00:20:44 --> 00:20:50
Well, in general,
any particular solution will
320
00:20:49 --> 00:20:55
look like that one we found,
what is it, omega nought
321
00:20:54 --> 00:21:00
squared minus omega
one squared,
322
00:21:01 --> 00:21:07
plus I'm allowed to add to it
any piece of the complementary
323
00:21:07 --> 00:21:13
solution.
Equally particular,
324
00:21:11 --> 00:21:17
and equally good,
as a particular solution is
325
00:21:14 --> 00:21:20
this plus anything which solved
the homogeneous equation.
326
00:21:18 --> 00:21:24
Now, all I'm going to do is
pick out one good function which
327
00:21:23 --> 00:21:29
solves the homogeneous equation,
and here it is.
328
00:21:26 --> 00:21:32
It's the function minus cosine.
In fact, what does solve the
329
00:21:32 --> 00:21:38
homogeneous equation?
Well, it's solved by sine omega
330
00:21:36 --> 00:21:42
nought t,
cosine omega nought t,
331
00:21:40 --> 00:21:46
and any linear combination of
332
00:21:44 --> 00:21:50
those.
So, out of all those functions,
333
00:21:47 --> 00:21:53
the one I'm going to pick is
cosine omega nought t.
334
00:21:51 --> 00:21:57
And, I'm going to divide it by
this same guy.
335
00:21:55 --> 00:22:01
So, this is part of the
complementary solution.
336
00:22:00 --> 00:22:06
That's what we call the
complementary solution,
337
00:22:02 --> 00:22:08
the solution to the associated
homogeneous equation,
338
00:22:06 --> 00:22:12
to the reduced equation.
Call it what you like.
339
00:22:08 --> 00:22:14
So, this is one of the guys in
there, and it's still a
340
00:22:12 --> 00:22:18
particular solution to take the
one I first found,
341
00:22:15 --> 00:22:21
and add to it anything which
solves the homogeneous equation.
342
00:22:19 --> 00:22:25
I showed you that when we first
set out to solve the
343
00:22:22 --> 00:22:28
inhomogeneous equation in
general.
344
00:22:24 --> 00:22:30
Now, why do I pick that?
Well, I'm going to now
345
00:22:27 --> 00:22:33
calculate, what's the limit?
So, these guys are also good
346
00:22:31 --> 00:22:37
solutions to that.
This is a good solution to that
347
00:22:35 --> 00:22:41
equation, this equation.
All I'm going to do now is
348
00:22:38 --> 00:22:44
calculate the limit as omega one
approaches omega zero of this
349
00:22:42 --> 00:22:48
function.
Well, what is that?
350
00:22:46 --> 00:22:52
It's cosine omega one t minus
cosine omega zero t divided by
351
00:22:50 --> 00:22:56
omega nought squared minus omega
one squared.
352
00:22:54 --> 00:23:00
353
00:22:57 --> 00:23:03
Now, you see why I did that.
If I let just this guy,
354
00:23:02 --> 00:23:08
omega one approaches
omega zero,
355
00:23:07 --> 00:23:13
I get infinity.
I don't get anything.
356
00:23:10 --> 00:23:16
But, this is different here
because I fixed it up,
357
00:23:14 --> 00:23:20
now.
The denominator becomes zero,
358
00:23:17 --> 00:23:23
but so does the numerator.
In other words,
359
00:23:21 --> 00:23:27
I've put myself in position to
use L'Hopital rule.
360
00:23:26 --> 00:23:32
So, let's L'Hopital it.
It's the limit.
361
00:23:29 --> 00:23:35
As omega one approaches omega
zero, and what do you do?
362
00:23:34 --> 00:23:40
You differentiate the top and
the bottom with respect to what?
363
00:23:42 --> 00:23:48
Right, with respect to omega
one.
364
00:23:44 --> 00:23:50
Omega one is the variable.
That's what's changing.
365
00:23:47 --> 00:23:53
The t that I'm thinking of is,
I'm thinking,
366
00:23:50 --> 00:23:56
for the temporary fixed.
This has a fixed value.
367
00:23:53 --> 00:23:59
Omega nought is fixed.
All that's changing in this
368
00:23:57 --> 00:24:03
limit operation is omega one.
And therefore,
369
00:24:01 --> 00:24:07
it's with respect to omega one
that I differentiate it.
370
00:24:05 --> 00:24:11
You got that?
Well, you are in no position to
371
00:24:08 --> 00:24:14
say yes or no,
so I shouldn't even ask the
372
00:24:11 --> 00:24:17
question, but okay,
rhetorical question.
373
00:24:14 --> 00:24:20
All right, let's differentiate
this expression,
374
00:24:17 --> 00:24:23
the top and bottom with respect
to omega one.
375
00:24:20 --> 00:24:26
So, the derivative of the top
with respect to omega one is
376
00:24:24 --> 00:24:30
negative sine omega one t.
377
00:24:28 --> 00:24:34
But, I have to use the chain
rule.
378
00:24:32 --> 00:24:38
That's differentiating with
respect to this argument,
379
00:24:35 --> 00:24:41
this variable.
But now, I must take times the
380
00:24:38 --> 00:24:44
derivative of this thing with
respect to omega one.
381
00:24:43 --> 00:24:49
And that is t is the constant,
so times t.
382
00:24:46 --> 00:24:52
And, how about the bottom?
The derivative of the bottom
383
00:24:49 --> 00:24:55
with respect to omega one is,
well, that's a constant.
384
00:24:53 --> 00:24:59
So, it becomes zero.
And, this becomes negative two
385
00:24:57 --> 00:25:03
omega one.
So, it's the limit of this
386
00:25:01 --> 00:25:07
expression as omega one
approaches omega zero.
387
00:25:05 --> 00:25:11
And now it's not indeterminate
388
00:25:09 --> 00:25:15
anymore.
The answer is,
389
00:25:10 --> 00:25:16
the negative signs cancel.
It's simply t sine omega nought
390
00:25:15 --> 00:25:21
t divided by two omega nought.
391
00:25:19 --> 00:25:25
So, that's how we get that
392
00:25:21 --> 00:25:27
solution.
It is a limit as omega one,
393
00:25:24 --> 00:25:30
but not of the
particular solution we found
394
00:25:28 --> 00:25:34
first, but of this other one.
Now, it's still too much
395
00:25:34 --> 00:25:40
algebra.
I mean, what's going on here?
396
00:25:37 --> 00:25:43
Well, that's something else you
should know.
397
00:25:41 --> 00:25:47
Okay, so my question is,
therefore, what does this mean?
398
00:25:46 --> 00:25:52
What's the geometric meaning of
all this?
399
00:25:50 --> 00:25:56
In other words,
what does that function look
400
00:25:54 --> 00:26:00
like?
Well, that's another
401
00:25:56 --> 00:26:02
trigonometric identity,
which in your book is just
402
00:26:01 --> 00:26:07
buried as half of one line sort
of casual as if everybody knows
403
00:26:07 --> 00:26:13
it, and I know that virtually no
one knows it.
404
00:26:13 --> 00:26:19
But, here's your chance.
So, the cosine of B minus the
405
00:26:17 --> 00:26:23
cosine of A can be expressed
as a product of
406
00:26:22 --> 00:26:28
signs.
It's the sine of (A minus B)
407
00:26:25 --> 00:26:31
over two times the sine of (A
plus B) over two,
408
00:26:29 --> 00:26:35
I believe.
409
00:26:33 --> 00:26:39
My only uncertainty:
is there a two in front of
410
00:26:37 --> 00:26:43
that?
I think there has to be.
411
00:26:40 --> 00:26:46
Let me check.
Sorry.
412
00:26:42 --> 00:26:48
Is there a two?
I wouldn't trust my memory
413
00:26:46 --> 00:26:52
anyway.
I'd look it up.
414
00:26:49 --> 00:26:55
I did look it up,
two, yes.
415
00:26:51 --> 00:26:57
If you had to prove that,
you could use the sine formula
416
00:26:57 --> 00:27:03
to expand this out.
That would be a bad way to do
417
00:27:03 --> 00:27:09
it.
The best way is to use complex
418
00:27:05 --> 00:27:11
numbers.
Express the sign in terms of
419
00:27:08 --> 00:27:14
complex numbers,
exponentials,
420
00:27:10 --> 00:27:16
you know, the backwards Euler
formula.
421
00:27:13 --> 00:27:19
Then do it here,
and then just multiply those
422
00:27:17 --> 00:27:23
two expressions involving
exponentials together,
423
00:27:20 --> 00:27:26
and cancel, cancel,
cancel, cancel,
424
00:27:23 --> 00:27:29
cancel, and this is what you
will end up with.
425
00:27:26 --> 00:27:32
You see why I did this.
It's because this has that
426
00:27:31 --> 00:27:37
form.
So, let's apply that formula to
427
00:27:33 --> 00:27:39
it.
So, what's the left-hand side?
428
00:27:36 --> 00:27:42
B is omega one t, and A is
omega nought t.
429
00:27:39 --> 00:27:45
So, this is omega one t,
430
00:27:42 --> 00:27:48
and this is omega nought t
431
00:27:44 --> 00:27:50
All right, so what we get is
432
00:27:47 --> 00:27:53
that the cosine of omega one t
minus the cosine of omega nought
433
00:27:51 --> 00:27:57
t, which is exactly the
434
00:27:54 --> 00:28:00
numerator of this function that
I'm trying to get a handle on.
435
00:28:00 --> 00:28:06
Then we will divide it by its
amplitude.
436
00:28:02 --> 00:28:08
So, that's this constant factor
that's real.
437
00:28:05 --> 00:28:11
It's a small number because I'm
thinking of omega one
438
00:28:10 --> 00:28:16
as being rather close to omega
zero,
439
00:28:13 --> 00:28:19
and getting closer and closer.
What does this tell us about
440
00:28:17 --> 00:28:23
the right-hand side?
Well, the right-hand side is
441
00:28:21 --> 00:28:27
twice the sine of A minus B.
442
00:28:24 --> 00:28:30
Now, that's good because these
guys sort of resemble each
443
00:28:28 --> 00:28:34
other.
So, that's (omega nought minus
444
00:28:32 --> 00:28:38
omega one) times t.
445
00:28:35 --> 00:28:41
That's A minus B,
and I'm supposed to divide that
446
00:28:39 --> 00:28:45
by two.
And then, the other one will be
447
00:28:42 --> 00:28:48
the same thing with plus:
sine omega nought plus omega
448
00:28:46 --> 00:28:52
one over two times t.
449
00:28:50 --> 00:28:56
Now, how big is this,
approximately?
450
00:28:52 --> 00:28:58
Remember, think of omega one
as close to omega zero.
451
00:28:56 --> 00:29:02
Then, this is approximately
452
00:28:59 --> 00:29:05
omega zero.
So this part is approximately
453
00:29:03 --> 00:29:09
sine of omega zero t.
454
00:29:06 --> 00:29:12
This part, on the other hand,
that's a very small thing.
455
00:29:09 --> 00:29:15
Okay, now what I want to know
is what does this function look
456
00:29:13 --> 00:29:19
like?
The interest in knowing what
457
00:29:15 --> 00:29:21
the function looks like it is
because we want to be able to
458
00:29:19 --> 00:29:25
see that it's limited is that
thing.
459
00:29:21 --> 00:29:27
You can't tell what's what its
limit is, geometrically,
460
00:29:25 --> 00:29:31
unless you know it looks like.
So, what does it look like?
461
00:29:30 --> 00:29:36
Well, again,
the way to analyze it is the
462
00:29:35 --> 00:29:41
thing, that thing.
What you think of is,
463
00:29:41 --> 00:29:47
yeah, of course you cannot
divide one side of equality
464
00:29:49 --> 00:29:55
without dividing the equation by
the other side.
465
00:29:56 --> 00:30:02
So, that's got to be there,
too.
466
00:30:02 --> 00:30:08
Now, what does that look like?
Well, the way to think of it
467
00:30:06 --> 00:30:12
is, here is something with a
normal sort of frequency,
468
00:30:10 --> 00:30:16
omega nought.
It's doing its thing.
469
00:30:14 --> 00:30:20
It's a sine curve.
It's doing that.
470
00:30:16 --> 00:30:22
What's this?
Think of all this part as
471
00:30:19 --> 00:30:25
varying amplitude.
It's just another example of
472
00:30:23 --> 00:30:29
what I gave you before.
Here is a basic,
473
00:30:26 --> 00:30:32
pure oscillation,
and now, think of everything
474
00:30:29 --> 00:30:35
else that's multiplying it as
varying its amplitude.
475
00:30:35 --> 00:30:41
All right, so what does that
thing look like?
476
00:30:38 --> 00:30:44
Well, first what we want to do
is plot the amplitude lines.
477
00:30:44 --> 00:30:50
Now, what will they be?
This is sine of an extremely
478
00:30:48 --> 00:30:54
small number times t.
The frequency is small.
479
00:30:52 --> 00:30:58
How does the sine curve look if
its frequency is very low,
480
00:30:57 --> 00:31:03
very close to zero?
Well, that must mean its period
481
00:31:02 --> 00:31:08
is very large.
Here's something with a big
482
00:31:05 --> 00:31:11
frequency.
Here's something with a very,
483
00:31:08 --> 00:31:14
very low frequency.
Now, with a low frequency,
484
00:31:11 --> 00:31:17
it would hardly get off the
ground and get up to one here,
485
00:31:15 --> 00:31:21
and it would do that.
But, it's made to look a little
486
00:31:19 --> 00:31:25
more presentable because of this
coefficient in front,
487
00:31:23 --> 00:31:29
which is rather large.
And so, what this thing looks
488
00:31:27 --> 00:31:33
like, I won't pause to analyze
it more exactly.
489
00:31:32 --> 00:31:38
It's something which goes up at
a reasonable rate for quite a
490
00:31:36 --> 00:31:42
while, and let's say that's
quite awhile.
491
00:31:39 --> 00:31:45
And then it comes down,
and then it goes,
492
00:31:42 --> 00:31:48
and so on.
Of course, in figuring out its
493
00:31:45 --> 00:31:51
amplitude, we have to be willing
to draw its negative,
494
00:31:49 --> 00:31:55
too.
And since I didn't figure
495
00:31:51 --> 00:31:57
things out right,
I can at least make it cross,
496
00:31:54 --> 00:32:00
right?
Okay.
497
00:31:55 --> 00:32:01
So, this is a picture of this
slowly varying amplitude.
498
00:32:01 --> 00:32:07
And in between,
this is the function which is
499
00:32:05 --> 00:32:11
doing the oscillation,
as well as it can.
500
00:32:08 --> 00:32:14
But, it has to stay within that
amplitude.
501
00:32:12 --> 00:32:18
So, it's doing this.
Now, what happens?
502
00:32:15 --> 00:32:21
As omega one approaches omega
zero,
503
00:32:21 --> 00:32:27
this frequency gets closer and
closer to zero,
504
00:32:25 --> 00:32:31
which means the period of that
dotted line gets further and
505
00:32:30 --> 00:32:36
further out, goes to infinity,
and you never do ultimately get
506
00:32:35 --> 00:32:41
a chance to come down again.
All you can see is the initial
507
00:32:42 --> 00:32:48
part, where it's rising and
rising.
508
00:32:45 --> 00:32:51
And, that's how this curve
turns into that one.
509
00:32:49 --> 00:32:55
Now, of course,
this curve is enormously
510
00:32:52 --> 00:32:58
interesting.
You must have had this
511
00:32:55 --> 00:33:01
somewhere.
That's the phenomenon of what
512
00:32:59 --> 00:33:05
are called beats.
Too frequencies--
513
00:33:03 --> 00:33:09
Your book has half a page
explaining this.
514
00:33:05 --> 00:33:11
That's the half a page where he
gives you this identity,
515
00:33:09 --> 00:33:15
except it gives it in a wrong
form, so that it's hard to
516
00:33:13 --> 00:33:19
figure out.
But anyway, the beats are two
517
00:33:16 --> 00:33:22
frequencies when you combine
them, the two frequencies being
518
00:33:20 --> 00:33:26
two combined pure oscillations
where the frequencies are very
519
00:33:24 --> 00:33:30
close to each other.
What you get is a curve which
520
00:33:27 --> 00:33:33
looks like that.
And, of course,
521
00:33:31 --> 00:33:37
what you hear is the envelope
of the curve.
522
00:33:34 --> 00:33:40
You hear the dotted lines.
Well, you hear this.
523
00:33:37 --> 00:33:43
You hear that,
too.
524
00:33:39 --> 00:33:45
But, what you hear is-- And,
that's how good violinists and
525
00:33:43 --> 00:33:49
cellists, and so on,
tune their instruments.
526
00:33:46 --> 00:33:52
They get one string right,
and then the other strings are
527
00:33:51 --> 00:33:57
tuned by listening.
They don't actually listen for
528
00:33:54 --> 00:34:00
the sound of the note.
They listened just for the
529
00:33:58 --> 00:34:04
beats, wah, wah,
wah, wah, and they turn the peg
530
00:34:02 --> 00:34:08
and it goes wah,
wah, wah, wah,
531
00:34:04 --> 00:34:10
and then finally as soon as the
wahs disappear,
532
00:34:07 --> 00:34:13
they know that the two strings
are in tune.
533
00:34:13 --> 00:34:19
A piano tuner does the same
thing.
534
00:34:16 --> 00:34:22
Of course, I,
being a very bad cellist,
535
00:34:19 --> 00:34:25
use a tuner.
That's another solution,
536
00:34:22 --> 00:34:28
a more modern solution.
Okay.
537
00:34:26 --> 00:34:32
538
00:34:49 --> 00:34:55
Oh well.
Let's give it a try.
539
00:34:51 --> 00:34:57
The bad news is that problem
six in your problem set,
540
00:34:55 --> 00:35:01
I didn't ask you about the
undamped case.
541
00:34:58 --> 00:35:04
I thought, since you are mature
citizens, you could be asked
542
00:35:03 --> 00:35:09
about the damped case.
543
00:35:06 --> 00:35:12
544
00:35:21 --> 00:35:27
I warn you, first of all you
have to get the notation.
545
00:35:26 --> 00:35:32
This is probably the most
important thing I'll do with
546
00:35:31 --> 00:35:37
this.
Your book uses this, resonance.
547
00:35:42 --> 00:35:48
548
00:36:02 --> 00:36:08
I'm optimistic.
[LAUGHTER] Let's say zero or f
549
00:36:06 --> 00:36:12
of t.
It doesn't matter.
550
00:36:09 --> 00:36:15
In other words,
the constants,
551
00:36:12 --> 00:36:18
the book uses two sets of
constants to describe these
552
00:36:17 --> 00:36:23
equations.
If it's a spring,
553
00:36:20 --> 00:36:26
and not even talking about RLC
circuits, the spring mass,
554
00:36:26 --> 00:36:32
damping, k, spring constant.
Then you divide out by m and
555
00:36:33 --> 00:36:39
you get this.
You're familiar with that.
556
00:36:36 --> 00:36:42
And, it's only after you
divided out by the m that you're
557
00:36:41 --> 00:36:47
allowed to call this the square
of the natural frequency.
558
00:36:46 --> 00:36:52
So, omega naught is the natural
frequency, the natural undamped
559
00:36:51 --> 00:36:57
frequency.
If this term were not there,
560
00:36:54 --> 00:37:00
that omega nought
would give the frequency with
561
00:36:59 --> 00:37:05
which the system,
the little spring would like to
562
00:37:03 --> 00:37:09
vibrate by itself.
Now, further complication is
563
00:37:08 --> 00:37:14
that the visual uses neither of
these.
564
00:37:11 --> 00:37:17
The visual uses x double dot
plus b times x prime,
565
00:37:16 --> 00:37:22
I think we will have to fix
this in the future,
566
00:37:20 --> 00:37:26
but for now,
just live with it,
567
00:37:22 --> 00:37:28
plus kx,
and that's some function,
568
00:37:27 --> 00:37:33
again, a function.
So, in other words,
569
00:37:30 --> 00:37:36
the problem is that b is okay,
can't be confused with c.
570
00:37:37 --> 00:37:43
On the other hand,
this is not the same k as that.
571
00:37:42 --> 00:37:48
What I'm trying to say is,
don't automatically go to a
572
00:37:48 --> 00:37:54
formula one place,
and assume it's the same
573
00:37:53 --> 00:37:59
formula in another place.
You have to use these
574
00:37:58 --> 00:38:04
equivalences.
You have to look and see how
575
00:38:03 --> 00:38:09
the basic equation was written,
and then figure out what the
576
00:38:08 --> 00:38:14
constant should be.
Now, there was something
577
00:38:12 --> 00:38:18
called, when we analyzed this
before, and this has happened in
578
00:38:17 --> 00:38:23
recitation, there was the
natural, damped frequency.
579
00:38:22 --> 00:38:28
I'll call it the natural,
damped frequency.
580
00:38:25 --> 00:38:31
The book calls it the
pseudo-frequency.
581
00:38:30 --> 00:38:36
It's called pseudo-frequency
because the function,
582
00:38:33 --> 00:38:39
if you have zero on the right
hand side, but have damping,
583
00:38:38 --> 00:38:44
the function isn't periodic.
It decays.
584
00:38:41 --> 00:38:47
It does this.
Nonetheless,
585
00:38:43 --> 00:38:49
it still crosses the t-axis at
regular intervals,
586
00:38:47 --> 00:38:53
and therefore,
almost everybody just casually
587
00:38:50 --> 00:38:56
refers to it as the frequency,
and understands it's the
588
00:38:54 --> 00:39:00
natural damped frequency.
Now, the relation between them
589
00:38:59 --> 00:39:05
is given by the little picture I
drew you once.
590
00:39:02 --> 00:39:08
But, I didn't emphasize it
enough.
591
00:39:07 --> 00:39:13
Here is omega nought.
592
00:39:09 --> 00:39:15
Here is the right angle.
The side is omega one,
593
00:39:12 --> 00:39:18
and this side is the
damping.
594
00:39:15 --> 00:39:21
So, in other words,
this is fixed because it's
595
00:39:19 --> 00:39:25
fixed by the spring.
That's the natural frequency of
596
00:39:23 --> 00:39:29
the spring, by itself.
If you are damping near the
597
00:39:27 --> 00:39:33
motion, then the more you damped
it, the bigger this side gets,
598
00:39:31 --> 00:39:37
and therefore the smaller omega
one is, the bigger the damping,
599
00:39:36 --> 00:39:42
then the smaller the frequency
with which the damped thing
600
00:39:40 --> 00:39:46
vibrates.
That sort of intuitive,
601
00:39:44 --> 00:39:50
and vice versa.
If you decrease the damping to
602
00:39:47 --> 00:39:53
almost zero, well,
then you'll make omega one
603
00:39:49 --> 00:39:55
almost the same size as omega
zero.
604
00:39:51 --> 00:39:57
This must be a right angle,
and therefore,
605
00:39:54 --> 00:40:00
if there's very little damping,
the natural damped frequency
606
00:39:57 --> 00:40:03
will be almost the same as the
original frequency,
607
00:40:00 --> 00:40:06
the natural frequency.
So, the relation between them
608
00:40:05 --> 00:40:11
is that omega one squared is
equal to omega nought squared
609
00:40:11 --> 00:40:17
minus p squared,
610
00:40:15 --> 00:40:21
and this comes from the
characteristic roots from the
611
00:40:20 --> 00:40:26
characteristic roots of the
damped equation.
612
00:40:26 --> 00:40:32
So, we did that before.
I'm just reminding you of it.
613
00:40:31 --> 00:40:37
Now, the third frequency which
now enters, and that I'm asking
614
00:40:37 --> 00:40:43
you about on the problem set is
if you've got a damped spring,
615
00:40:43 --> 00:40:49
okay, what happens when you
impose a motion on it with yet a
616
00:40:49 --> 00:40:55
third frequency?
In other words,
617
00:40:53 --> 00:40:59
drive the damped spring.
I don't care.
618
00:40:56 --> 00:41:02
I switched to y,
since I'm in y mode.
619
00:41:02 --> 00:41:08
So, our equation looks like
this, just as it did before,
620
00:41:06 --> 00:41:12
except now going to drive that
with an undetermined frequency,
621
00:41:10 --> 00:41:16
cosine omega t.
622
00:41:13 --> 00:41:19
And, my question,
now, is, see,
623
00:41:15 --> 00:41:21
it's not going to be able to
resonate in the correct-- you
624
00:41:19 --> 00:41:25
really only get true resonance
when you don't have damping.
625
00:41:24 --> 00:41:30
That's the only time where the
amplitude can build up
626
00:41:27 --> 00:41:33
indefinitely.
But nonetheless,
627
00:41:31 --> 00:41:37
for all practical purposes,
and there's always some damping
628
00:41:37 --> 00:41:43
unless you are a perfect vacuum
or something,
629
00:41:42 --> 00:41:48
there's almost always some
damping.
630
00:41:46 --> 00:41:52
So, p isn't zero,
can't be exactly zero.
631
00:41:50 --> 00:41:56
So, the problem is,
which omega gives,
632
00:41:54 --> 00:42:00
which frequency in the input,
which input frequency gives the
633
00:42:00 --> 00:42:06
maximal amplitude for the
response?
634
00:42:06 --> 00:42:12
635
00:42:20 --> 00:42:26
We solved that problem when it
was undamped,
636
00:42:22 --> 00:42:28
and the answer was easy.
Omega should equal omega zero.
637
00:42:26 --> 00:42:32
But, when it's damped,
the answer is different.
638
00:42:30 --> 00:42:36
And, I'm not asking you to do
it in general.
639
00:42:33 --> 00:42:39
I'm giving you some numbers.
But nonetheless,
640
00:42:37 --> 00:42:43
it still must be the case.
So, I'm giving you,
641
00:42:40 --> 00:42:46
I give you specific values of p
and omega zero.
642
00:42:45 --> 00:42:51
That's on the problem set.
Of course, one of them is tied
643
00:42:50 --> 00:42:56
to your recitation.
But, the answer is,
644
00:42:53 --> 00:42:59
I'm going to give you the
general formula for the answer
645
00:42:58 --> 00:43:04
to make sure that you don't get
wildly astray.
646
00:43:03 --> 00:43:09
Let's call that omega r,
647
00:43:05 --> 00:43:11
the resonant omega.
This isn't true resonance.
648
00:43:09 --> 00:43:15
Your book calls it practical
resonance.
649
00:43:11 --> 00:43:17
Again, most people just call it
resonance.
650
00:43:15 --> 00:43:21
So, you know what I mean,
type of thing.
651
00:43:18 --> 00:43:24
It is omega r is very much like
that.
652
00:43:20 --> 00:43:26
Maybe I should have written
this one down in the same form.
653
00:43:25 --> 00:43:31
Omega one is the square root of
omega nought squared minus p
654
00:43:29 --> 00:43:35
squared.
655
00:43:34 --> 00:43:40
What would you expect?
Well, what I would expect is
656
00:43:37 --> 00:43:43
that omega r should be omega
one.
657
00:43:40 --> 00:43:46
The damped system has a natural
frequency.
658
00:43:43 --> 00:43:49
The resonant frequency should
be the same as that natural
659
00:43:47 --> 00:43:53
frequency with which the damped
system wants to do its thing.
660
00:43:52 --> 00:43:58
And the answer is,
that's not right.
661
00:43:54 --> 00:44:00
It is the square root.
It's a little lower.
662
00:43:57 --> 00:44:03
It's a little lower.
It is omega nought squared
663
00:44:01 --> 00:44:07
minus two p squared.