I just recalling some of the notation we are going to need
for today, and a couple of the facts that we're going to use,
plus trying to clear up a couple of confusions that the
recitations report. This can be thought of two
ways. It's a formal polynomial in D,
in the letter D. It just has the shape of the
polynomial, D squared plus AD plus B.
A and B are constant coefficients.
But, it's also, at the same time,
if you think what it does, it's a linear operator on
functions. It's a linear operator on
functions like y of t. You think of it both ways:
formal polynomial because we want to do things like factoring
it, substituting two for D and things like that.
Those are things you do with polynomials.
You do them algebraically. You can take the formal
derivative of the polynomial because it's just sums of
powers. On the other hand,
as a linear operator, it does something to functions.
It differentiates them, multiplies them by constants or
something like that. So it's, so to speak,
has a dual aspect this way. And, that's one of the things
we are exploiting what we use operator methods to solve
differential equations. Now, let me remind you of the
key thing we were interested in. f of t:
not any old function, we'll get to that next time,
but f of t, exponentials.
So, it should be an exponential or something like an
exponential, or pretty close to it, for example,
something with sine t and cosine t,
or e to the, that could be thought of as
part of the real or imaginary part of a complex exponential.
And, maybe by the end of today, we will have generalized that
even little more. But basically,
I'm interested in exponentials. Let's make it alpha complex.
That will at least take care of the cases, e to the ax times
cosine bx, sin bx,
which are the main cases. Those are the main cases.
Then, remember the little table we made.
I simply gave you the formula for the particular solution.
So, what we're looking for is we already know how to solve the
homogeneous equation. What we want is that particular
solution. And then, the recipe for it I
gave you, these things were proved by the substitution rules
and exponential shift rules. The recipe was that if f of t
was, let's make a little table.
f of t is, well, it's always e to the a t.
So, in other words,
it's e to the a t. The cases are,
so yp, what is the yp? Well, it is the normal case is
yp equals e to that alpha t divided by the
polynomial where you substitute, you take that polynomial,
and wherever you see a D, you substitute the complex
number, alpha. There, I'm thinking of it as a
formal polynomial. I'm not thinking of it as an
operator. Now, this breaks down.
So, that's the formula for the particular solution.
The only trouble is, it breaks down if p of alpha is
zero. So, we have to assume that it's
not. Now, if p of alpha is zero,
that means alpha is a root of the polynomial,
a zero of the polynomial is a better word.
So, in that case, it will be e to the alpha t
divided by p prime of alpha.
Differentiate formally the polynomials, --
-- and you will get 2D plus A.
And now, substitute in the alpha.
And, this will be okay provided p prime of alpha
is not zero. That means that alpha is the
simple root, simple zero of p. And then, there's one more
case, which, since I won't need today, I won't write on the
board. But, you'll need it for
homework. So, make sure you know it.
Another words, if this is zero,
then you've got a double root. And, there is still a different
formula. And, this is wrong because I
forgot the t. Yes?
I could tell on your faces. That was before,
and now we are up to today. What we are interested in
talking about today is what this has to do with the phenomenon of
resonance. Everybody knows at least one
case of resonance, I hope.
A little kid is on his swing, right?
Back and forth, and they are very,
very little, so they want a push.
Okay, well, everybody knows that to make the swing go,
a swing has a certain natural frequency.
It swings back and forth like that.
It's a simple pendulum. It's actually damped,
but let's pretend that it isn't.
Everybody knows you want to push a kid on a swing so that
they go high. You have to push with
essentially the same frequency that the natural frequency of
the spring, of the swing is. It's automatic,
because when you come back here, it gets to there,
and that's where you push. So, automatically,
you time your pushes. But if you want the kid to
stop, you just do the opposite. Push at the wrong time.
So anyway, that's resonance. Of course, there are more
serious applications of it. It's what made the Tacoma
Bridge fall down, and I think movies of that are
now being shown not merely on television, but in elementary
school. Resonance is what made,
okay, more resonance stories later.
So, my aim is, what is this physical
phenomenon, that to get a big amplitude you should have it
match the frequency? What does that have to do with
a differential equation? Well, the differential equation
for that simple pendulum, let's assume it's undamped,
will be of the type y double prime plus,
I'm using t now since t is time.
That will be our new independent variable,
plus omega nought squared is the natural
frequency of the pendulum or of the spring, or whatever it is
that's doing the vibrating. Yeah, any questions?
What we're doing is driving that with the cosine,
with something of a different frequency.
So, this is the input, or the driving term as it's
often called, or it's sometimes called the
forcing term. And, the point is I'm going to
assume that the frequency is different.
The driving frequency is different from the natural
frequency. So, this is the input
frequency. Okay, and now let's simply
solve the equation and see what we get.
So, it's if I write it using the operator,
it's D squared plus omega nought squared applied to y
is equal to cosine.
It's a good idea to do this because the formulas are going
to ask you to substitute into a polynomial.
So, it's good to have the polynomial right in front of you
to avoid the possibility of error.
Well, really what I want is the particular solution.
It's the particular solution that's going to give me a pure
oscillation. And, the thing to do is,
of course, since this cosine, you want to make it complex.
So, we are going to complexify the equation in order to be able
to solve it more easily, and in order to be able to use
those formulas. So, the complex equation is
going to be D squared plus omega nought squared.
Well, it's going to be a
complex, particular solution. So, I'll call it y tilde.
And, on the right-hand side, that's going to be e to the i
omega1 t. Cosine is the real part of
this. So, when we get our answer,
we want to be sure to take the real part of the answer.
I don't want the complex answer, I want its real part.
I want the real answer, in other words,
the really real answer, the real real answer.
So, now without further ado, because of those beautiful,
the problem has been solved once and for all by using the
substitution rule. I did that for you on Monday.
The answer is simply e to the i omega1 t
divided by what? This polynomial with omega one
substituted in for D. So, sorry, i omega one,
the complex coefficient of t.
So, it is substitute i omega for D, I omega one for D,
and you get (i omega one) squared plus omega nought
squared.
Well, let's make that look a little bit better.
This should be e to the (i omega one t)
divided by, now, what's this?
This is simply omega nought squared minus omega
one squared. But, I want the real part of
it. So, as one final,
last step, the real part of that is what we call just the
real particular solution, so, yp without the tilde
anymore. And, the real part of this,
well, this cosine plus i sine. And, the denominator,
luckily, turns out to be real. So, it's simply going to be
cosine omega one t.
That's the top, divided by this thing,
omega nought squared minus omega one squared.
In other words,
that's the response. This is the input,
and that's what came out. Well, in other words,
what one sees is, regardless of what natural
frequency this system wanted to use for itself,
at least for this solution, what it responds to is the
driving frequency, the input frequency.
The only thing is that the amplitude has changed,
and in a rather dramatic way, if omega1, depending on the
relative sizes of omega1 and omega2.
Now, the interesting case is when omega one is very close to
omega, the natural frequency. When you push it with
approximately it's natural frequency, then the solution is
big amplitude. The amplitude is large.
So, the solution looks like the frequency.
The input might have looked like this.
Well, it's cosine, so it ought to start up here.
The input might have looked like this, but the response will
be a curve with the same frequency and still a pure
oscillation. But, it will have much,
much bigger amplitude. And, it's because the
denominator, omega nought squared minus omega
one squared, is always zero.
So, the response will, instead, look like this.
Now, to all intents and purposes, that's resonance.
You are pushing something with approximately the same
frequency, something that wants to oscillate.
And, you are pushing it with approximately the same frequency
that it would like to oscillate by itself.
And, what that does is it builds up the amplitude
Well, what happens if omega one is actually equal
to omega zero? So, that's the case I'd like to
analyze for you now. Suppose the two are equal,
in other words. Well, the problem is,
of course, I can't use that same solution.
It isn't applicable. But that's why I gave you,
derived for you using the exponential shift law last time,
the second version, when it is a root.
So, if omega one equals omega nought,
so now our equation looks like D squared plus omega nought
squared, the natural frequency, y.
But this time, the driving frequency,
the input frequency, is omega nought itself.
Then, the same analysis, a lot of it is,
well, I'd better be careful. I'd better be careful.
Let's go through the analysis again very rapidly.
What we want to do is first complexify it,
and then solve. So, the complex equation will
be D squared plus omega nought squared times y tilde equals e
to the i omega nought t, this time.
But now, i omega is zero of this polynomial.
That's why I picked it, right?
If I plug in i omega zero, I get i omega zero
quantity squared plus omega nought squared.
That's zero.
So, I'm in the second case. So, i omega nought is a simple
root, simple zero, of D squared plus
omega nought, that polynomial squared.
Therefore, the complex particular solution is now t e
to the i omega nought t divided by p prime,
where you plug in that root, the i omega nought.
Now, what's p prime?
p prime is 2D, right?
If I differentiate this formally, as if D were a
variable, the way you differentiate polynomials,
the derivative, this is a constant,
and the derivative is 2D. So, the denominator should have
two times for D. You are going to plug in i
omega zero. So, it's 2 i omega zero.
And now, I want the real part
of that, which is what? Well, think about it.
The top is cosine plus i sine. The real part is now going to
come from the sine, right, because it's cosine plus
i sine. But this i is going to divide
out the i that goes with this sine.
And, therefore, the real part is going to be t
times the sine, this time, of omega nought t.
And, that's going to be divided
by, well, the i canceled out the i that was in front of the sine
function. And therefore,
what's left is two omega nought down below.
So, that's our particular solution now.
Well, it looks different from that guy.
It doesn't look like that anymore.
What does it look like? Well, it shows the way to plot
such things is basically it's an oscillation of frequency omega
nought. But, its amplitude is changing.
So, the way to do it is, as always, if you have a basic
oscillation which is neither too fast nor too slow,
think of that as the thing, and the other stuff multiplying
it, think of it as changing the amplitude of that oscillation
with time. So, the amplitude is that
function, t divided by two omega zero.
So, just as we did when we talked about damping,
you plot that and it's negative on the picture.
So, this is the function whose graph is t divided by two omega
nought. That's the changing amplitude,
as it were. And then, the function itself
does what oscillation it can, but it has to stay within those
lines. So, the thing that's
oscillating is sine omega nought t,
which would like to be a pure oscillation, but can't because
its amplitude is being changed by that thing.
So, it's doing this, and now the rest I have to
leave to your imagination. In other words,
what happens when omega nought is equal to, when the driving
frequency is actually equal to omega nought,
mathematically this turns into a different looking solution,
one with steadily increasing amplitude.
The amplitude increases linearly like the function t
divided by two omega nought.
Well, many people are upset by this, slightly,
in the sense that there is a funny feeling.
How is it that that solution can turn into this one?
If I simply let omega one go to omega zero, what happens?
Well, the pink curve just gets taller and taller,
and after a while all you see of it is just a bunch of
vertical lines which seem to be spaced at whatever the right
period is for that function. It's sort of like being in a
first story window and watching a giraffe go by.
All you see is that. Okay.
So, my concern is how does that function turn into this one?
I have something in mind to remind you of,
and that's why we'll go through this little exercise.
It's a simple exercise. But the function of it is,
of course that as omega one goes to omega zero cannot
possibly turn into this. It's doing the wrong thing near
zero. It's already zooming up.
But, the point is, this is not the only particular
solution on the block. Any solution whatsoever of the
differential equation, the inhomogeneous equation,
is a particular solution. It's like Fred Rogers:
everybody is special. Okay, so all solutions are
special. We don't have to use that one.
So, I will use, where are all the other
solutions? So, I'm going back to the
equation D squared plus omega zero squared,
applied to y,
is equal to cosine omega one t.
Now, the particular solution we found was that one,
cosine omega one t divided by that omega nought squared minus
omega one squared.
What do the other particular solutions look like?
Well, in general, any particular solution will
look like that one we found, what is it, omega nought
squared minus omega one squared,
plus I'm allowed to add to it any piece of the complementary
solution. Equally particular,
and equally good, as a particular solution is
this plus anything which solved the homogeneous equation.
Now, all I'm going to do is pick out one good function which
solves the homogeneous equation, and here it is.
It's the function minus cosine. In fact, what does solve the
homogeneous equation? Well, it's solved by sine omega
nought t, cosine omega nought t,
and any linear combination of
those. So, out of all those functions,
the one I'm going to pick is cosine omega nought t.
And, I'm going to divide it by this same guy.
So, this is part of the complementary solution.
That's what we call the complementary solution,
the solution to the associated homogeneous equation,
to the reduced equation. Call it what you like.
So, this is one of the guys in there, and it's still a
particular solution to take the one I first found,
and add to it anything which solves the homogeneous equation.
I showed you that when we first set out to solve the
inhomogeneous equation in general.
Now, why do I pick that? Well, I'm going to now
calculate, what's the limit? So, these guys are also good
solutions to that. This is a good solution to that
equation, this equation. All I'm going to do now is
calculate the limit as omega one approaches omega zero of this
function. Well, what is that?
It's cosine omega one t minus cosine omega zero t divided by
omega nought squared minus omega one squared.
Now, you see why I did that. If I let just this guy,
omega one approaches omega zero,
I get infinity. I don't get anything.
But, this is different here because I fixed it up,
now. The denominator becomes zero,
but so does the numerator. In other words,
I've put myself in position to use L'Hopital rule.
So, let's L'Hopital it. It's the limit.
As omega one approaches omega zero, and what do you do?
You differentiate the top and the bottom with respect to what?
Right, with respect to omega one.
Omega one is the variable. That's what's changing.
The t that I'm thinking of is, I'm thinking,
for the temporary fixed. This has a fixed value.
Omega nought is fixed. All that's changing in this
limit operation is omega one. And therefore,
it's with respect to omega one that I differentiate it.
You got that? Well, you are in no position to
say yes or no, so I shouldn't even ask the
question, but okay, rhetorical question.
All right, let's differentiate this expression,
the top and bottom with respect to omega one.
So, the derivative of the top with respect to omega one is
negative sine omega one t.
But, I have to use the chain rule.
That's differentiating with respect to this argument,
this variable. But now, I must take times the
derivative of this thing with respect to omega one.
And that is t is the constant, so times t.
And, how about the bottom? The derivative of the bottom
with respect to omega one is, well, that's a constant.
So, it becomes zero. And, this becomes negative two
omega one. So, it's the limit of this
expression as omega one approaches omega zero.
And now it's not indeterminate
anymore. The answer is,
the negative signs cancel. It's simply t sine omega nought
t divided by two omega nought.
So, that's how we get that
solution. It is a limit as omega one,
but not of the particular solution we found
first, but of this other one. Now, it's still too much
algebra. I mean, what's going on here?
Well, that's something else you should know.
Okay, so my question is, therefore, what does this mean?
What's the geometric meaning of all this?
In other words, what does that function look
like? Well, that's another
trigonometric identity, which in your book is just
buried as half of one line sort of casual as if everybody knows
it, and I know that virtually no one knows it.
But, here's your chance. So, the cosine of B minus the
cosine of A can be expressed as a product of
signs. It's the sine of (A minus B)
over two times the sine of (A plus B) over two,
I believe.
My only uncertainty: is there a two in front of
that? I think there has to be.
Let me check. Sorry.
Is there a two? I wouldn't trust my memory
anyway. I'd look it up.
I did look it up, two, yes.
If you had to prove that, you could use the sine formula
to expand this out. That would be a bad way to do
it. The best way is to use complex
numbers. Express the sign in terms of
complex numbers, exponentials,
you know, the backwards Euler formula.
Then do it here, and then just multiply those
two expressions involving exponentials together,
and cancel, cancel, cancel, cancel,
cancel, and this is what you will end up with.
You see why I did this. It's because this has that
form. So, let's apply that formula to
it. So, what's the left-hand side?
B is omega one t, and A is omega nought t.
So, this is omega one t,
and this is omega nought t
All right, so what we get is
that the cosine of omega one t minus the cosine of omega nought
t, which is exactly the
numerator of this function that I'm trying to get a handle on.
Then we will divide it by its amplitude.
So, that's this constant factor that's real.
It's a small number because I'm thinking of omega one
as being rather close to omega zero,
and getting closer and closer. What does this tell us about
the right-hand side? Well, the right-hand side is
twice the sine of A minus B.
Now, that's good because these guys sort of resemble each
other. So, that's (omega nought minus
omega one) times t.
That's A minus B, and I'm supposed to divide that
by two. And then, the other one will be
the same thing with plus: sine omega nought plus omega
one over two times t.
Now, how big is this, approximately?
Remember, think of omega one as close to omega zero.
Then, this is approximately
omega zero. So this part is approximately
sine of omega zero t.
This part, on the other hand, that's a very small thing.
Okay, now what I want to know is what does this function look
like? The interest in knowing what
the function looks like it is because we want to be able to
see that it's limited is that thing.
You can't tell what's what its limit is, geometrically,
unless you know it looks like. So, what does it look like?
Well, again, the way to analyze it is the
thing, that thing. What you think of is,
yeah, of course you cannot divide one side of equality
without dividing the equation by the other side.
So, that's got to be there, too.
Now, what does that look like? Well, the way to think of it
is, here is something with a normal sort of frequency,
omega nought. It's doing its thing.
It's a sine curve. It's doing that.
What's this? Think of all this part as
varying amplitude. It's just another example of
what I gave you before. Here is a basic,
pure oscillation, and now, think of everything
else that's multiplying it as varying its amplitude.
All right, so what does that thing look like?
Well, first what we want to do is plot the amplitude lines.
Now, what will they be? This is sine of an extremely
small number times t. The frequency is small.
How does the sine curve look if its frequency is very low,
very close to zero? Well, that must mean its period
is very large. Here's something with a big
frequency. Here's something with a very,
very low frequency. Now, with a low frequency,
it would hardly get off the ground and get up to one here,
and it would do that. But, it's made to look a little
more presentable because of this coefficient in front,
which is rather large. And so, what this thing looks
like, I won't pause to analyze it more exactly.
It's something which goes up at a reasonable rate for quite a
while, and let's say that's quite awhile.
And then it comes down, and then it goes,
and so on. Of course, in figuring out its
amplitude, we have to be willing to draw its negative,
too. And since I didn't figure
things out right, I can at least make it cross,
right? Okay.
So, this is a picture of this slowly varying amplitude.
And in between, this is the function which is
doing the oscillation, as well as it can.
But, it has to stay within that amplitude.
So, it's doing this. Now, what happens?
As omega one approaches omega zero,
this frequency gets closer and closer to zero,
which means the period of that dotted line gets further and
further out, goes to infinity, and you never do ultimately get
a chance to come down again. All you can see is the initial
part, where it's rising and rising.
And, that's how this curve turns into that one.
Now, of course, this curve is enormously
interesting. You must have had this
somewhere. That's the phenomenon of what
are called beats. Too frequencies--
Your book has half a page explaining this.
That's the half a page where he gives you this identity,
except it gives it in a wrong form, so that it's hard to
figure out. But anyway, the beats are two
frequencies when you combine them, the two frequencies being
two combined pure oscillations where the frequencies are very
close to each other. What you get is a curve which
looks like that. And, of course,
what you hear is the envelope of the curve.
You hear the dotted lines. Well, you hear this.
You hear that, too.
But, what you hear is-- And, that's how good violinists and
cellists, and so on, tune their instruments.
They get one string right, and then the other strings are
tuned by listening. They don't actually listen for
the sound of the note. They listened just for the
beats, wah, wah, wah, wah, and they turn the peg
and it goes wah, wah, wah, wah,
and then finally as soon as the wahs disappear,
they know that the two strings are in tune.
A piano tuner does the same thing.
Of course, I, being a very bad cellist,
use a tuner. That's another solution,
a more modern solution. Okay.
Oh well. Let's give it a try.
The bad news is that problem six in your problem set,
I didn't ask you about the undamped case.
I thought, since you are mature citizens, you could be asked
about the damped case.
I warn you, first of all you have to get the notation.
This is probably the most important thing I'll do with
this. Your book uses this, resonance.
I'm optimistic. [LAUGHTER] Let's say zero or f
of t. It doesn't matter.
In other words, the constants,
the book uses two sets of constants to describe these
equations. If it's a spring,
and not even talking about RLC circuits, the spring mass,
damping, k, spring constant. Then you divide out by m and
you get this. You're familiar with that.
And, it's only after you divided out by the m that you're
allowed to call this the square of the natural frequency.
So, omega naught is the natural frequency, the natural undamped
frequency. If this term were not there,
that omega nought would give the frequency with
which the system, the little spring would like to
vibrate by itself. Now, further complication is
that the visual uses neither of these.
The visual uses x double dot plus b times x prime,
I think we will have to fix this in the future,
but for now, just live with it,
plus kx, and that's some function,
again, a function. So, in other words,
the problem is that b is okay, can't be confused with c.
On the other hand, this is not the same k as that.
What I'm trying to say is, don't automatically go to a
formula one place, and assume it's the same
formula in another place. You have to use these
equivalences. You have to look and see how
the basic equation was written, and then figure out what the
constant should be. Now, there was something
called, when we analyzed this before, and this has happened in
recitation, there was the natural, damped frequency.
I'll call it the natural, damped frequency.
The book calls it the pseudo-frequency.
It's called pseudo-frequency because the function,
if you have zero on the right hand side, but have damping,
the function isn't periodic. It decays.
It does this. Nonetheless,
it still crosses the t-axis at regular intervals,
and therefore, almost everybody just casually
refers to it as the frequency, and understands it's the
natural damped frequency. Now, the relation between them
is given by the little picture I drew you once.
But, I didn't emphasize it enough.
Here is omega nought.
Here is the right angle. The side is omega one,
and this side is the damping.
So, in other words, this is fixed because it's
fixed by the spring. That's the natural frequency of
the spring, by itself. If you are damping near the
motion, then the more you damped it, the bigger this side gets,
and therefore the smaller omega one is, the bigger the damping,
then the smaller the frequency with which the damped thing
vibrates. That sort of intuitive,
and vice versa. If you decrease the damping to
almost zero, well, then you'll make omega one
almost the same size as omega zero.
This must be a right angle, and therefore,
if there's very little damping, the natural damped frequency
will be almost the same as the original frequency,
the natural frequency. So, the relation between them
is that omega one squared is equal to omega nought squared
minus p squared,
and this comes from the characteristic roots from the
characteristic roots of the damped equation.
So, we did that before. I'm just reminding you of it.
Now, the third frequency which now enters, and that I'm asking
you about on the problem set is if you've got a damped spring,
okay, what happens when you impose a motion on it with yet a
third frequency? In other words,
drive the damped spring. I don't care.
I switched to y, since I'm in y mode.
So, our equation looks like this, just as it did before,
except now going to drive that with an undetermined frequency,
cosine omega t.
And, my question, now, is, see,
it's not going to be able to resonate in the correct-- you
really only get true resonance when you don't have damping.
That's the only time where the amplitude can build up
indefinitely. But nonetheless,
for all practical purposes, and there's always some damping
unless you are a perfect vacuum or something,
there's almost always some damping.
So, p isn't zero, can't be exactly zero.
So, the problem is, which omega gives,
which frequency in the input, which input frequency gives the
maximal amplitude for the response?
We solved that problem when it was undamped,
and the answer was easy. Omega should equal omega zero.
But, when it's damped, the answer is different.
And, I'm not asking you to do it in general.
I'm giving you some numbers. But nonetheless,
it still must be the case. So, I'm giving you,
I give you specific values of p and omega zero.
That's on the problem set. Of course, one of them is tied
to your recitation. But, the answer is,
I'm going to give you the general formula for the answer
to make sure that you don't get wildly astray.
Let's call that omega r,
the resonant omega. This isn't true resonance.
Your book calls it practical resonance.
Again, most people just call it resonance.
So, you know what I mean, type of thing.
It is omega r is very much like that.
Maybe I should have written this one down in the same form.
Omega one is the square root of omega nought squared minus p
squared.
What would you expect? Well, what I would expect is
that omega r should be omega one.
The damped system has a natural frequency.
The resonant frequency should be the same as that natural
frequency with which the damped system wants to do its thing.
And the answer is, that's not right.
It is the square root. It's a little lower.
It is omega nought squared
minus two p squared.