I will start today calculating for you the normal mode
frequencies of a double pendulum.
And then, I will drive that double pendulum.
And then we will see very dramatic things that have
changed. So, let us start here with a
double pendulum. This is the equilibrium
position if it hangs straight, length L, mass is M.
This angle [be?] theta one. And this angle is theta two.
I call this position X1, and I call this position X2.
I'm going to introduce a shorthand notation at omega zero
squared, G divided by L. I want to remind you that the
sine of theta one is X1 divided by L, and that the sine of theta
two is going to be X2 minus X1 divided by L.
For small angle approximation, we do know the tangents.
Here we have a tangent which I will call T2.
That is here, of course, MG.
So, there is also a tangent, T2 here.
Action equals minus reaction, and then there is,
here, the tangent T1. And, here is MG.
But, there are the only forces on these objects.
In small angles, I do know that T1 must be
approximately 2MG because it's carrying boy loads,
so to speak. And we know that T2 is very
close to MG. And we're going to use that in
our approximation. So, let me first start with the
bottom one that has only two forces on it so that maybe the
easiest M2 X2 double dot. And so, the only force that is
driving it back to equilibrium in the small angle approximation
is, then, the horizontal component of the T2.
So, that equals minus T2 times the sine of theta two.
And, for that, I can write minus MG divided by
L times theta two X2 minus X1. We divided out M,
and we use our shorthand notation.
And so, we get that X2 double dot.
And now we have here minus MG over LX2.
So, that comes in, and that becomes a plus.
So, I got plus omega zero squared times X2.
I have here plus X1. When that comes in,
that becomes a minus. So, I get minus omega zero
squared X1, and that equals zero.
So, this is my first differential equation for the
second object. So, now I'm going to my first
object. So, now we get MX1 double dot,
and now there is one [SOUND OFF/THEN ON] that is driving it
back to equilibrium, and that is the horizontal
component of T1. But, the horizontal component
of T2 is driving it away from equilibrium in the drawing that
I have made here. So, you get minus T1 times the
sign of theta one plus T2 times the sine of theta.
And so, I'm going to substitute in there now sines,
and I'm going to substitute in their T1 equals 2MG,
and T2 equals MG. So, this becomes equal to minus
2MG times the sine of theta one, which is X1 divided by L.
And then, I get plus T2, which is MG times the sine X2
minus X1 divided by L. And, I'm going to divide by M,
and I'm going to use my shorthand notation.
And, when I do that, I will come out here,
X1 double dot. It's this one,
here the N goes. Go over L becomes omega zero
squared. But now, look closely.
There is here a two times X1. And here, there is a one times
X1. And, both have a minus sign.
So, when they come out, I get three omega zero squared.
So, I get plus three omega zero squared times X1.
And then, I have to bring X2 out.
That becomes minus omega zero squared times X2.
And, that is a zero. And so, we have to solve now
this differential equation coupled with X1 and X2 together
with this one. My solution has to satisfy
both. I want to go over that one to
me sure that it is correct. [SOUND OFF/THEN ON] pays off
because one sign wrong and you're hung.
The whole thing falls apart. You're dead in the water,
so it pays off to think about it again.
So, we have an X2 double dot. I can live with that,
plus omega zero squared X2. That has the right smell for
me, minus omega zero squared X1. I can live with that
differential equation. I go to this one.
Of course, the three is well-known in a system like this
that you get to three. I have an X1 double dot plus
three omega zero squared X1. I think we are in good shape.
And so, now we're going to put in our trial solutions.
X1 is C1 times cosine omega T. And, X2 is C2 times cosine
omega T. And, we're going to search for
the frequencies. So, this omega,
we have to solve for this omega.
This is not a given. Since we are looking for normal
mode solutions, these two omegas must be the
same. I cannot write down omega one
and omega two. And, I don't have to worry
about any phase angles because since we have no damping,
either they are in phase or they are 180° out of phase.
And, 180° out of phase sibling means a minus sign.
So, that's the great thing about this.
The signs will take care, then, of the possible phase
angles. So, now we're going to
substitute this solution in these two differential
equations. And, I'm going to write it down
in a form that I put the C1's to the left, and the C2's to the
right. And so, I'm first going to this
one. That is my object number one.
I take the second derivative. So, I get C1 times minus omega
squared because the second derivative here gives me a minus
omega squared. I ditch the cosine omega T
terms because I'm going to have a cosine omega T everywhere.
So, I'm not going to write down the cosine omega T.
I have here plus three omega zero squared times C1.
I have here minus omega zero squared times C2.
And, that is zero. Notice I put the C1's here and
the C2 there. I'm going to the other
equation. And, I'm going to put the C1 on
the left. So, we get minus omega zero
squared times C1. That's the term you have here.
And then, we have plus C2 times omega zero squared minus omega
squared. And, this minus omega squared
comes from the second derivative of this one, second derivative
of this one minus omega squared, right?
You get omega squared comes out, and then the cosine goes to
a sine that gives you the minus sign.
So, now we have here two equations with three unknowns.
That is typical for normal mode solutions of a system with two
oscillations. We don't know C1;
we don't know C2, and we do not know omega.
And so, we remember from last time that you can always solve
for the ratio C1 over C2, and you can solve for omega.
You can only find C1 if you also know the initial
conditions, which I have not given.
Instead of solving it the high school way that I did last week,
the simple way, the fast way,
I'm going to do it now using Kramer's rule.
And the reason why I want to do this once, even though in this
case it really is not necessary, when you have three coupled
oscillators or four, there is just no way that the
high school method will do it for you.
You have to have a more general approach.
I've sent this to you by e-mail, all of you,
and I also assume that you have worked on this a little bit
since I requested that you would prepare for that.
So, I'm first going to write down what D is,
which is the determinant of [NOISE OBSCURES] C1's,
and we have the C2's. So, let me write down here what
D is. So, I'm going to get three
omega zeroes squared. [SOUND OFF/THEN ON] that is the
C1. Then I get here minus omega
squared. That's the C2.
And then I get here minus omega zero squared.
And here I get omega zero squared minus omega squared.
And, the determinant of this, that's D.
And, I presume you all know how to get the determinant of this
very simple matrix. We'll do that together,
of course. So, following Kramer's rule,
then, my C1, which is X there,
that's the one I want to solve for, so my C1 I now have to take
this zero. This is also a zero,
by the way. I forgot to mention that this
is a zero. This equation is a zero.
So, this column, now, zero-zero has to come
first. So, I get here zero-zero.
And then, the second column is the same as it was here minus
omega zero squared. And then I get omega zero
squared minus omega squared. And, that's divided by D.
So, we now know that that is C1.
And then, we go to C2. So now, the zero-zero column
shifts towards the right, goes here, and the first column
is unchanged. So, we have here three omega
zero squared minus omega squared minus omega zero squared zero,
zero, divided by D. You've got to admit that the
upstairs here, the determinant of this matrix
is zero because you have two zeroes here.
And, it's also a zero for this one.
But, clearly zero solutions for C1 and C2 are meaningless.
They are not incorrect, because you will get in this
differential equation that zero is zero, which is rather
obvious. So, we don't want the solutions
that C1 and C2 are zero. And, the only way that we can
avoid that is to demand that D become zero because now you get
zero divided by zero. And that's a whole different
story. That's not necessarily zero.
And that, then, is the idea behind getting the
solutions to the searched-for normal mode frequencies.
So, we must demand in this case that this become zero.
Otherwise, you get trivial solutions which are of no
interest. So, when we make D equals zero,
we do get that the determinant of that matrix becomes three
omega zero squared minus omega squared times omega zero squared
minus omega squared. And then, minus the product,
that becomes a minus, not a plus, because minus-minus
is already plus. So, I get minus omega zero to
the fourth equals zero. And this equation is an
equation in omega to the fourth. You can solve that.
You can solve that for omega squared.
I will leave you with that solution.
This is utterly trivial. And, [NOISE OBSCURES] omega
minus squared [NOISE OBSCURES] is two minus the square root of
two times omega zero squared, and omega plus squared is two
plus the square root of two times omega zero squared.
So, this step for you will take you no more than maybe half a
minute. Be careful because you can
easily slip up, of course.
So, we now have a solution that omega minus is approximately,
if I calculate the two minus square root of two,
it's approximately 0.77, 0.76 omega zero.
Not intuitive at all. So, it's lower than the
resonance frequency of a single pendulum.
And then, I can substitute this value for omega E,
or back into my equations. Or, I substitute it back in
this if you want to. You get zero divided by zero,
which is not going to be zero. And, that you're going to find,
then that C2 divided by C1 in that minus mode,
in that lowest possible mode, you will find that that is one
plus the square root of two, which is approximately plus
2.4. And, the omega plus solution
gives you a frequency that is about 0.85 omega zero.
That solution you can put back into your differential equations
here, or not the differential equations, I mean,
into this one, or you put it back into this if
you want to. And you will find,
now, that C2 over C1 for that plus mode; I call this the plus
mode [with?] my shorthand notation is going to be minus
one divided by one plus the square root of two.
So, it's going to be minus one divided by 2.4.
So, it's going to be roughly -0.41.
So, those are, then, the formal solutions for
the normal modes. And if I gave you the initial
conditions, then of course you could calculate C1.
And then you know everything because you know the ratio C2
over C1. But, without the initial
conditions, you cannot do that. Each of these normal mode
solutions satisfies both differential equations.
So, therefore, a linear superposition of the
two normal mode solutions is a general solution.
So, when you start that system off, at time T equals zero,
you specify X1. You specify the velocity of
object one. You specify X2.
And, you specify the velocity of that object.
Then it's going to oscillate in the superposition,
the linear superposition of two normal mode solutions.
One, you see here has omega minus.
And this will be the ratio of amplitudes.
And this is the omega plus. And that will be the ratio of
amplitudes. And that is nonnegotiable.
That's what the system will do. Now, we're going to make a
dramatic change. Now we are going to drive the
system. So, now this is the equilibrium
position of the whole system. I'm going to drive it back and
forth holding in my hand like this.
And I'm going to shake it. I call the position of my hand
[eta?] equals eta zero times the cosine [NOISE OBSCURES] is the
amplitude of my motion of my hand.
That is what I'm going to do. And so, when I look,
now, at this equation, at the figure,
not the equation but at the figure, I call this now X1.
I call this now X2 because you should always call X1 and X2 the
distance from equilibrium. And this is now equilibrium.
If I don't shake at all, I have the pendulum here,
and so it's hanging straight down.
So, this is now equilibrium. And this location here,
at a random moment in time, is now eta.
What has changed? Well, at first sight,
very little has changed. The only thing that has changed
now is that the sine of theta one is now X1 minus eta divided
by L. So, I have here a minus eta.
It looks rather innocent with the consequences will be
unbelievable. So, I can leave everything on
the blackboard the way I have it.
All I have to do is to put instead of the sine theta X1
divided by L, I have to put in X1 minus eta
divided by L. See, nature was very kind to
me. It already left some space
there. You see that?
[SOUND OFF/THEN ON] anticipated that I was going to do that.
So, you have a minus eta there. Well, if now you divide M out,
and you're going to use the shorthand notation,
then leaving eta on the right side, which is nice to do,
you get minus eta. So, this zero now becomes two
omega zero squared times eta zero times the cosine of omega
T. It becomes two,
omega zero squared times eta. [You see it's two?].
So, it's no longer a zero. And so, when we now substitute
in there these trial functions, they now have a completely
different meaning. Omega is no longer negotiable.
Omega [as set?] by me is a driver.
So we're not going to solve for omega.
Omega is a given. That means if omega is a given
that we are going to end up not with two equations with three
unknowns, C1, C2, and omega,
but we're going to end up with two equations with two unknowns,
only C1 and C2 because omega is a given.
And so, in a steady state solution, you get a number for
C1 and you get a number for C2 in terms of eta zero,
of course. You will see how that works.
So, when we put in these functions now,
these omegas are fixed and no longer something that we search
for. They are my omegas.
I can make them zero. I can make them infinity.
I can make them anything I want to.
And that's what we want to study.
So, if we want to go back, now, to these two equations,
what changes here since we have put in this trial function,
the only thing that disappears is this cosine omega T.
So, we end up here with two omega zero squared.
Times eta zero. And, nothing else changes.
But now, we're looking now at our solution for C1.
And we're going to look at our solution for C2 in steady state.
Ah, it's easy, right?
We apply Kramer's rule, and the only thing that changes
is this one, which has to be replaced by this,
2 omega zero squared times eta zero.
Eta zero, remember, is the amplitude of my hand.
Eta is the displacement of my hand at any moment in time.
Eta zero is the amplitude. And so, here,
we get, then, also, two omega zero squared
times eta zero. And so, now,
we can solve for C1 and C2. We get an answer,
not just a ratio only. We get an answer.
We know exactly what C1 is going to be, and we know exactly
what C2 is going to be because omega is nonnegotiable.
Omega is now unknown. When we solve this,
we were searching for an omega, and out came these omegas.
That's not the case anymore. We know omega.
It's called omega, and that's it.
So, I can write down now C1. So, I take the determinant
there of the upstairs. So, that gives me omega zero
squared times eta zero times omega zero squared minus omega
squared. That is this diagonal,
and this one is zero. And, I have to divide that by
D. Now, I could write D in that
forum, and I could do that. There's nothing wrong with
that. But, I can write it in a form
which is a little bit more transparent.
We do know that omega minus and omega plus will make D zero.
So, you can write this, then, in the following way.
You can write here omega squared minus omega minus
squared times omega squared minus omega plus squared.
That must be the same as what I have there because,
you see, this one becomes omega minus.
And this goes to zero. And then, this one becomes
omega plus also. So, it's a different way of
writing. It gives you a little bit more
insight. It reminds you that the
downstairs will go to zero when you hit those resonance
frequencies. And so, I will also write down
C2, then. This one is zero.
I get minus this one. So, I get plus two omega zero
to the fourth times eta zero divided by that same D.
And, you can write this for it. So, let me check that,
see whether I'm happy with that, two omega zero squared,
eta zero, omega zero squared. Well, it's omega squared that
looked good. This looks good.
And, I have here I'm happy.
Our task, now, is not to look at these
equations, but to see through them.
And they are by no means trivial what they're going to do
as a function of omega. It's an extraordinarily
complicated dependence on omega. And, I have plotted for you
these values of C1 and C2 for which you get an answer now.
You also know C1 over C2 of course because you know C1 and
you know C2. And, I'm going to show you this
as a function of omega, and then we will try to digest
that together. And this plot,
like the other plot that I'll show you later today will be on
the 8.03 website. They will be part of lecture
notes. So, you have to click on
lecture notes. And then you'll see these
plots. This is the first one,
which is a double pendulum. What is plotted here
horizontally is omega divided by omega zero.
So, that's the omega zero that we mentioned there.
And so you see the first resonance here is about at 0.76
omega zero. And, the second resonance here
is at 1.85 omega zero. And, what we plot here is the
amplitude, C, divided by eta zero because
obviously if you make eta zero larger, you expect that has an
effect on the C's, of course.
You know, if you drive it to the larger amplitude,
of course object will also respond accordingly.
And so, therefore, we have it as a function of eta
zero. You see the eta zero here?
C1 is linearly proportional with eta zero.
C2 is linearly proportional to eta zero.
That's no surprise. So, we divided by eta zero.
When we plot it up there, it means that the amplitude is
in phase with the driver. That's the way we have written
it. And if it is below the zero
line, it means that the amplitude to the object is out
of phase with the driver. That's all it means.
That's the sine convention. So, let us now look at this and
try to digest it, and use to some degree our
intuition, and see whether that agrees with our intuition.
And, let us start when omega goes to zero.
And don't look now at the solution.
If omega is zero, I have a double pendulum in my
hand. And, I'm going to move it to
the left. And, 25 years from now,
I'm going to go back. And, I move it to the right
again. So, the pendulum is always
straight. And it is clear that C1 and C2
both must be eta zero. And, you'd better believe it
that if you substitute in there omega zero, that's what you will
find. So, C1 must be C2,
and must be eta zero. And, it must even be plus eta
zero. It must be in phase with the
driver. And you see that,
not the ratio, but just C divided by eta zero.
That is a ratio that is plotted that plus one.
And so, both are in phase with the driver.
So that's a trivial result. It means that the pendulum,
which is here, 25 years from now is here.
That's omega zero, almost zero.
So, notice that we now know the ratio, C2 over C1,
which is plus one. The ratio C1 over C2 or C2 over
C1 are now entirely dictated by this.
Nothing to do with normal modes anymore.
Don't be surprised that C1 over C2 is now plus one.
Now, look what happens. I'm going to increase my omega.
And what you see is that the red curve, which is the second
object, it's the lowest one of the two, is going to have a
large amplitude than the top one.
You see, it already begins to grow very rapidly.
And, when you reach resonance, the ratio is going to be plus
2.4, of course. That's obvious.
Now, at resonance, you get an infinite amplitude
for each. That, of course,
is nonsense. There is physically no meaning.
So, you shouldn't really think of it as going to infinity.
For one thing, if C1 became infinity,
that means if this point here ends up there in the hole,
it's hard to argue that it's a small angle, right?
So in any case, the solution wouldn't even hold
apart from the fact, of course, that there's always
some damping. And so, we never go completely
off scale so to speak. However, you'll see that when
you drive the system, the double pendulum,
and you approach resonance, but you will very quickly see
that the racial of the amplitude of the second one over the top
one will grow, will become one and one half,
will become two, and then in the limiting case,
you will get the plus 2.4. So, as omega goes up,
you're going to see that C2 over C1 is going to be larger
than one right there, this one.
And then, ultimately it will reach 2.4.
But that is that extreme case of resonance.
And when you look at the motion of the pendulum,
if you drive it somewhere here. Notice that they're both in
phase with the driver, and C2 is larger than C1.
So, what you'll see is this is C1.
And then, this is C2. So, C2 is larger than C1.
And, they are in phase with the driver.
And so, when they return, you will see a pendulum like
this. That's the [sweet?] thing that
you will see. Now is something truly bizarre.
I go a little higher in frequency.
I go over the resonance, and I see here a point whereby
the frequency happens to be exactly the frequency of a
single pendulum. Omega divided by omega zero is
one. And the top one refuses to
move. But the bottom one does.
The bottom one here has an amplitude which is twice the
amplitude of the driver. Look, you see a two here.
And, it is out of phase with the driver.
That is unimaginable. It is unimaginable what you're
going to see. So, in omega is omega zero.
C1 becomes zero. C2 becomes minus two eta zero.
And so, this pendulum is going to look, then,
as follows. I'll make a drawing here.
I have a little bit more space. So, if my hand is here at eta
zero, then number one will stand still.
It won't do anything. But number two will have an
amplitude which is twice eta zero, but it's on the other
side. So this is two eta zero.
And then, if you look half a period later,
it will look like this. So, this is eta zero,
and then this is two eta zero. And, this one doesn't move.
That's what it tells you. How on Earth can the lower
pendulum oscillate if the upper one stands still?
I want you to think about that. I'm still having sleepless
nights about it. And so, maybe you'll have some
too. And, if you have some clever
ideas, come to see me. But the logical consequence of
what we did is that this one will stand still,
and the other one will still oscillate and be driven by my
hand. I have to keep moving this,
otherwise this will go to pieces.
I must keep doing this all the time at that frequency which is
exactly the resonance frequency of a single pendulum.
It is that omega zero. I must keep doing this,
and this one does nothing. And this one has doubled the
swing of this, it is out of phase.
If, then, you go even higher, then you'll see that the two
objects will go out of phase. The upper one will be in phase
with the driver, and the lower one will be out
of phase with the driver. And then you get that second
resonance when things will get out of hand.
And you get that ratio, -0.42 [back?] of course.
I will want to demonstrate to you the situation here and the
situation to see whether they make sense.
And so, now, I'm going to use a double
pendulum, and drive it with my own frequency,
which I determine. I'm the boss.
I determine omega. I'm not looking for normal mode
solutions. I determine omega.
And I'm going to first drive it with omega zero.
Omega is about zero. In other words,
I'm going to drive it here. What you're going to see is,
of course, fantastic, absolutely fantastic.
It's hanging straight down now. And I'm moving it over a
distance of 1 foot. So, eta zero is one foot.
That C1 is one foot, and C2 is one foot.
And, they are in phase with the driver.
Physics works. If I go a little higher up
here, so I go somewhere here, approaching resonance,
then you'll see that C2 becomes larger than C1.
This is C1 and this is C1. And, you get to see a picture
which is very much like this. And I'll try that.
So, I'll drive it below resonance, but not too far
below. And there you see it.
You see, C1 is smaller than C2. No longer one plus one.
This was C1 over C2 is one. That's no longer the case now.
You really see that C2 is getting ahead,
ahead not in terms of phase ahead, but in terms of
amplitude, very clear. Now I'm going to attempt to do
the impossible. And the impossible is to try to
hit this point. The point whereby the upper one
stands still, and whereby the lower one will
have an amplitude which is twice that of my hands but out of
phase with my hands, how on Earth can I ever drive
this system with that frequency, omega zero, which is the
frequency of a single pendulum? Well, maybe I can't.
But I will try. And the way that I'm going to
try this is the following. I know what the resonance
frequency of a single pendulum is.
That's this. I can feel it in my hands.
I can feel it in my stomach. I can feel it in my brain.
I feel it all over my body. I can burn this frequency into
my chips here. Then I can close my eyes while
you are looking, and generate that frequency
which was burned here and drive the system as a double pendulum.
If I succeed, you'll see that upper one stand
still. And the bottom one will have
twice the amplitude of my hands. So, the success of this depends
exclusively on how accurately I can burn this frequency into my
chips. So, you have to be quiet.
So I'm going to count: one, two, three,
I'm burning now. One, two, three,
four, one, two, three, four.
I'm closing my eyes. One, two, three,
four, one, two, three, four,
one, two, three, four, one, two,
three, four, one.
You're not saying anything. Didn't you see this one
standing still? Did you see it?
Did you also see that the other one had twice the amplitude of
my hands and out of phase with my hands?
You didn't see that, right?
Admit it. You didn't see it because you
were not looking for it. Especially for you,
I'll do it again. So, you really have to see,
number one, that this one will practically stand still,
number two, but this one has doubled the amplitude of my hand
but out of phase. The decay time of burning is
only one minute. So I have to burn it again.
One, two, three, four, five, one,
two, three, four, five, one, oh-oh.
Yeah, these things happen. You have to start all over with
the burning. There we go.
One, two, three, four, five, one,
two, three, four, five, one, two,
three, four, five, one, two,
three, are you seeing it? Yes.
All right. [APPLAUSE]
This is an ideal moment for the break.
I'm going to hand out the mini quiz and we will reconvene.
I'll give you six minutes this time so you can even stretch
your legs, and I would like some help handing these out.
And then you bring it back and I will put boxes out there so
that you can help me handing it out here.
You can start right away, hand these out here.
For those of you who have no seats, come forward and get some
seats. Nicole, we're still friends,
right? So, why don't you hand those
out? And, why don't you hand these
out here? You can also give it to people
here. You can start right away.
[SOUND OFF/THEN ON] I'm now going to do something perhaps
even more ambitious. And I'm going to now couple
three oscillators, not pendulums yet.
But I'm going to couple three oscillators, which I connected
with four springs. I'm going to work on this.
Three masses equal masses. Four springs,
spring constant K, and the spring constants are
the same. And I'm going to drive that
system. One, two, three,
four, and this is the end. In other words,
I have here a spring. And, here is the first mass,
second mass, third mass, and here it is
fixed. And I'm going to drive it here
with a displacement, eta, which is eta zero times
cosine omega T. So, at a random moment in time,
this is where my hand will be. So this is eta.
This is where the first mass will be.
Remember, you always call the displacement X1 from its
equilibrium. That is its dotted line.
Here is the spring. This one is here.
So I call this X2. Here is the spring.
And this one is here. So, this is X3.
So, here's the spring and here is the spring.
You may have noticed more than once now that I have a certain
discipline that I always offset them in the same direction.
Do you have to do that? No.
If you don't do it, your chance of a mistake on the
sign slip is much larger than if you always set them off in the
same direction. You'll see shortly why.
So that's certainly something that is not a must.
But, it's a smart thing to do. I do find this is my positive
direction. But that, of course,
is [free?] choice. Now, at the situation at this
moment in time, let X1 be larger than eta.
Let X2 be larger than X1. And let X3 be larger than X2.
And, this assumption will have no consequences for what
follows, at least not for the differential equations.
If X1 is larger than eta, that first spring is longer
than it wants to be because I have assumed that X1 is larger
than eta. And so, that means there will
be a force in this direction because the spring is longer
than it wants to be. If X2 is larger than X1,
the spring is also longer than it wants to be.
[NOISE OBSCURES] So, there is a force in this
direction. And so, I can write down now a
differential equation for my first object.
So that's going to be N, X1 double dot.
That equals minus K times X1 minus eta because that's the
amount by which it is longer it wants to be.
So, times X1 minus eta, that is this force.
And, this force is now in the plus direction.
It's plus K times, the spring here is longer than
a wants to be by an amount X2 minus X1.
Not omega one, but X1.
That's my differential equation for the first object.
And this one is always correct even if X1 is not larger than
eta because if X1 is not larger than eta, then this force flips
over. Well, this will also flip over.
So, that's why it's always kosher and advisable to make
that ascension to start with because, again,
it reduces the probability of making mistakes.
That's all. There's nothing else to add.
Just reduce the chance of slipping up.
So, let's now go to this object.
If this spring is longer than a wants to be, it wants to
contract. So, this object we'll see
forced to the left. But, if the spring is longer
than a wants to be, because X3 is larger than X2,
it will experience a force to the right.
So, I can write down, now, a differential equation
for object number two and X2 double dot.
Notice that the one that is here to the left is the same one
that is here to the right. Action equals minus reaction.
This pool is the same as this pool.
So, it is going to be this term which is now the minus sign.
And, you always see that in coupled oscillators that was as
a plus here is going to come out here as a minus sign.
You see that comes out nicely because the spring is longer
than a wants to be by an amount X2 minus X1, and the force is in
the minus direction. And this one is now going to be
plus K times X3 minus X2. So, now I go to the next spring
to the next object. So, this object here will
experience a force to the left. This spring is longer than it
wants to be. So, it wants to contract.
But, this one is pushing. So, therefore,
the force due to the spring is now also in this direction
because the [end?] is fixed. And so, we get for the third
object M X3 double dot equals minus K times X3 minus X2,
which is this term. But it switches signs.
And then, in addition, I get minus K times X3.
When you reach this point on an exam, you pause.
You take a deep breath, and you go over every term and
every sign. If you slip up on one sign,
one casual mistake that you just, even though you know it
you casually write here, for instance,
a one instead of a three, it's all over.
You are dead in the water. The problem will fall apart,
and it may not even oscillate in a simple harmonic way.
So, therefore, let's look at it.
MX1 double dot. X1 is larger than eta.
Therefore, the force is in this direction.
I love it. The other one:
in this direction. Perfect, X2 minus X1.
That same force here is going to pull on the second one.
So, if this is correct, this is also correct.
This one is driving it away from equilibrium,
X3 minus X2. It's got to be right.
This term shows up here with a minus sign, can't go wrong
there. And since this spring is always
shorter here, if it's pushed to the right,
I am happy with my differential equations.
So now, we are going to substitute in here X1 is C1
cosine omega T trial functions X2 is C2 cosine omega T.
And, X3 is C3 cosine omega T. I was looking for omegas,
oh no. Oh no, omega is given by me.
I'm telling you what omega is. You're not going to negotiate
that with me. We are only solving for C1,
C2, and C3. And in the steady state,
you won't be able to do that because omega is nonnegotiable.
You're going to get three equations with three unknowns:
C1, C2, C3. You don't have to settle to
only calculate the ratios with the amplitude.
No, you're going to get a real answer for C1,
for C2, and for C3 which, of course, will depend on eta
zero. Sure, you know eta zero.
Do we worry about phase angles here?
No, because there is no damping.
And, if there is no damping, either the objects are in phase
or they are out of phase because it is the damping that gives
these phase angles in between. And 180° out of phase is a
minus sign. So, we have the power to
introduce some [NOISE OBSCURES] phase changes.
And, zero phase, for that we have plus and minus
signs. Now, you are going to do some
grinding. I did all the grinding,
every detail on the double pendulum.
Now, you're going to do the grinding.
However, I want to make sure that if you go through that
effort to make the grinding, that you indeed end up with the
right solution. So, in that sense,
I'm going to help you a little bit by giving you the D,
which is the D that we have there.
But, you have to bring me to the D.
And so, the D is going to be, so you have to divide by M.
You have to also, omega S squared K over M,
shorthand notation, some of you may want to call it
omega zero. That's fine because there's
only one spring. There are no pendulums and
springs, but I still call it an S to remind you that it is the
resonance frequency of a single spring.
Then, my D becomes minus omega squared.
That's always the result of that second derivative,
remember? You always get that minus omega
squared route. And, you get plus two omega S
squared. Then you get in the second
column minus omega S squared, and in the third column you get
a zero. No surprise that you get a zero
in the third column because the first differential equation has
no connection with X3 at all. And so, you never see anything
in the third column that will be zero.
But, if you look at the second differential equation that has
an X1, X2, and an X3 in it. So, now you don't see zeroes.
And so, what you're going to see is minus omega S squared.
That's going to be the C1 term. And then, you get here minus
omega squared. I think it is plus two omega S
squared. And, your last problem is going
to be minus omega S squared. Now, the third differential
equation, there is no X1. Therefore, that is a zero here.
And then, you get minus omega S squared.
And then, here you get minus omega squared plus two omega S
squared. And, you have to take the
determinant of this matrix. That is D.
Let me check it to make sure that I didn't slip up with a
minus sign, so when you get home that you don't wonder,
why didn't you get that result? And, I think that looks good to
me. Remember, all those omega
squares always come from those second derivatives because you
have to take the second derivative of cosine omega T.
That always brings out the minus omega squared.
So, no surprise that this is a minus, that this is a minus,
and that that is a minus. So, now we want to know what C1
is. And, the first column will
reflect this eta because the right side, now,
is not going to be zero, remember, like the double
pendulum. So, you're going to get here in
the first column, you're going to get omega S
squared times eta zero. And then, you're going to get a
zero and a zero. And then, this column comes
here. And this column comes here.
That is the determinant upstairs.
And you divide it by D. That, then, is C1.
And of course I will only go one step further to go to C2,
but that becomes a little boring now.
C2, then, you get, that's this one.
And then, the second column [NOISE OBSCURES] omega S squared
eta zero, zero, zero.
And then the third one is going to be this.
And then you go, and we divide it by D,
of course. And then, you can write down
C3. You're on your own.
I'll help you. So, when you do this,
you could, if you wanted to, first solve for what we call
the resonance frequencies. The resonance frequencies are
the ones which are the normal mode frequencies.
That's a resonance. And so, you may want to put in
D equals zero. So, you made that determinant
equal zero, which gives you then the three resonance frequencies,
which earlier we would have called normal modes in case we
are not driving. And so, for those of you who
worked this out, the lowest frequency,
resonance frequency, which is the normal mode,
is two minus the square root of two times omega S squared.
The one that follows is two omega zero squared.
And, the one that follows then, which I will call it omega
plus-plus is going to be two plus the square root of two
times omega S squared. None of these are,
of course, [NOISE OBSCURES]. None of the resonance
frequencies are coupled double pendulums or intuitive because
there's no way that you can even look at this and say,
oh yeah, of course. Excuse me?
What did I do wrong? Yes, thank you very much.
I have an omega zero square. You deserve extra credit.
I've called that omega S. If you want to change all the
omega S's in omega zero, that's fine.
But, you cannot have one omega zero, and the other one omega S.
Yes, so this is the square of the frequency of a single spring
oscillating mass M. Thank you very much for
pointing that out. Ah, boy, you also deserve extra
credit. I really set you up with this,
didn't I? I wanted to people to get some
extra credit. You deserve one;
come and see me later. And who was the other one?
You did one. Thank you.
Yeah, there is a square here. There are more people than one
who claimed this one. All right, thank you very much.
OK, so now, for any given value of omega, for any given value of
omega, you find three answers. You get C1, you get C2,
and you get C3 in the steady state solution.
Of course, you get infinite amplitude, which is physically
meaningless. So, you've got to stay away in
your solutions from the infinities.
But, you will see if you don't come too close to infinities
that the results that you get are quite accurate for high Q
systems. And so, now,
I will show you the three amplitudes for this system which
will also be put on the Web this afternoon.
And so that is the three-car system which we have set up
there. The only difference with the
previous one is we don't plot here omega divided by omega
zero, but the squares. This plot was provided to me by
Professor [Witslaws?], who lectured 8.03 a year ago.
I was a recitation instructor at the time.
And very kind of me that he gave me this plot.
He even added this at my request, which was very nice.
[Car?] one is the first car, green, and then the car two is
the red one, is the second car. And this is supposed to be
blue. You don't see it,
but in any case, if you think it's black that's
fine. That is, then,
the black line. So, horizontally,
it is the ratio of the frequency squared.
So, you see that the second resonance is,
indeed, at a two here. [SOUND OFF/THEN ON] have there
on the blackboard. If we plot it above the zero,
it means that it is in phase with the driver.
If we plotted below the zero, it means that it's out of phase
with the driver. If now you look at this,
then already in omega zero you see something that is by no
means intuitive. Notice that C1,
C2, and C3 are all substantially lower than eta
zero because this is in units of eta zero.
And they are not even the same. They are all three different.
Would I have anticipated that? No.
I wouldn't. Maybe two would be the same for
me, but not all three different. And, that's the case.
They're all three different. When we approached resonance,
things got out of hand. All three are in phase.
And when you just cross over the first resonance,
they are all three again in phase but out of phase with the
driver. That's not so surprising all by
itself. But now, look at this
ridiculous point. If I drive that system with the
resonance frequency of the individual spring with one mass
on it because, remember, if this squared is
one, then omega divided by omega zero is also one.
So that's exactly the square root of K over M.
Then, this one will stand still.
And, these two have roughly the same amplitude.
You could sort of eyeball it here.
It looks like it almost crosses over there.
And it's about minus eta zero because it's about minus one.
So, let me write that on the blackboards.
That's a quite bizarre situation.
So, at that one frequency, so omega equals the square root
of K over M. I get C1 equals zero.
And, my C2 is about C3, maybe even exactly C3.
I never checked that, and that is roughly minus eta
one, eta zero. You can just see that there.
So, it means that this car will stand still.
It's closest to the driver. These two are in phase with
each other, have the same amplitude.
But they are out of phase with the driver.
Crazy, got to be wrong, right?
I mean, how on Earth can this one stand still?
And these happily go hand-in-hand and go,
it can't be right. But I will demonstrate you that
it is right at lease to a high degree of accuracy.
I'm going to concentrate on two points in that graph.
The first point that I want to concentrate on is this one.
It's the hardest one of the two.
I'm going to drive this system at a frequency that we think is
almost there. I'm going to turn it on right
now because it will take three to four minutes for the
[trangents?] to die out. And then in the mean time I'll
explain what you're going to see.
So, I'm driving it now with that frequency.
So don't look at it now. It looks chaotic.
It's causing a ridiculous trangent mode.
This one is oscillating. And it's not supposed to
oscillate. Look there.
It's oscillating, and it looks chaotic.
The amplitudes are changing. Just let it cook.
Now, what are we going to see? If the trangents die out,
which can really take three minutes, you will see that this
is my eta zero. This is two eta zero.
You see this arrow? That is two eta zero.
That's the driver. So, you can calibrate the
amplitude eta zero. You will see,
then, that these two cars have the same displacement,
but out of phase. So when this one goes to the
right for me, they go to the left for me,
and vice versa. They have an amplitude that's
very closely the same. And this one is going to stand
still. What you're going to see is not
something truly spectacular, but it's extremely subtle.
You have never seen it before, and I don't think this has ever
been demonstrated in any lecture hall.
It's extremely subtle. First of all,
to have the patience to wait four minutes for this to happen,
number one, and the number two to make such a daring prediction
that this one can almost come to a halt, and that the other will
have an amplitude which is the same as the driver but 180° out
of phase. So, be patient and it will pay
off. Don't fall asleep now.
So let's look at this middle one to see whether its amplitude
is becoming constant. As long as they amplitudes are
not constant, they are still trangents.
Well, these two are already going hand-in-hand.
You see that? And I would say very much with
the same amplitudes. And look at this.
Hey, they are really out of phase with each other.
You see that? Now this one.
Oh, boy, that amplitude is nowhere nearly as large as this
one. And yet, when we started it,
it was even large isn't this one because of the trangent
phenomenon. Look at this one.
It's almost standing still already.
Can we find that precise frequency?
No. We have to set the dial
somehow. We do the best we can.
I think it's fantastic. It's blowing my mind.
I can't believe it. Physics is working.
This one is [UNINTELLIGIBLE] standing still.
If I had to estimate the amplitude, I would say maybe
1/10 of eta zero. And these, you see the marks
here. This one is really eta zero.
So, those of you who see these marks, that is the same as this.
It's really eta zero. And, boy, they go hand in hand.
Aren't you thrilled? [APPLAUSE] Now I'm going to try
to make you see this point. So, I'm going to start it,
and that I'll explain to you what is so special about that
point because, again, we have to be patient
and we have to wait for the trangents to die out.
I'm going to drive it very close to resonance right below
the resonance. You get a huge amplitude of C1
and C3. But, look what C2 is going to
do. C2 is going to have an
amplitude which you and I may have thought is zero.
Namely, the outer ones do this in the middle ones stand still.
No, no. No, no.
Look at one and three, by the way.
They go nuts already. You see, this is one and this
is three. Now you see the trangent
phenomenon. And now it's picking up again
so we have to be patient. But there is something very
special about this C2. Why is this C2 not zero,
which is what you would expect? You would expect that the outer
two go like this and the middle one would just stand still.
That's one answer. Yes, of course.
But isn't it so that this must be zero?
You don't believe that this upstairs is going to be zero?
Just say you don't believe it. This zero.
Why now, even though the upstairs is zero,
why do we get an amplitude which is half eta zero but with
a minus sign? Because we get zero divided by
zero because right at that frequency here,
there is a resonance frequency. So, D is zero.
So, you get an amazing coincidence that the upstairs is
zero, the downstairs is zero, but that's what we have 18.01
for. The ratio is not zero.
The ratio is minus one half eta zero.
And, that's what you see. It's minus one half eta zero.
Now, look, one in three are happy.
They oscillate happily out of phase with each other.
Number one has to be in phase with the driver.
It's doing that. Number three,
out of phase with the driver. It's doing that.
And look at the middle one. The middle one for those of you
who can see, the arrows are given eta zero to eta zero.
Its right half of it. Just walk up to it,
and you can see. It's amazing,
this. We hit that right on the nose.
The reason why this is easy is, look.
My omega doesn't have to be precisely here.
Even if it's a little bit to the left, I'm still OK.
That amplitude of number two is not changing very much.
You see that? This is so nicely horizontal.
So, this for me was a piece of cake.
This was hard. Piece of cake.
So, you see there?
Now, I'm going to move to the triple pendulum.
In other words, we have now seen this structure
of my talk. We first did the double
pendulum. The double pendulum was,
I worked out all the way for you.
Then we did the three cars with the four springs.
I set it up for you so that you can't go wrong,
go wrong, go wrong anymore. I gave you the final solution,
and I demonstrated that indeed it is working,
the way we calculate it. Now, I'm going to simply show
you the result of a triple pendulum.
And, the plots that I'm going to show you will be on the Web,
but no calculations at all. So, that triple pendulum or
that, yeah, let me put these down so that we have not so much
[shadow?] on this board, triple pendulum.
This is a triple pendulum. And we are going to drive it
here: eta equals eta zero times the cosine of omega T.
The top one is going to be green.
The middle one is going to be red.
And, the bottom one is going to be blue even though it may look
black there. I'll make it blue.
That is the color code that we have on our plot,
which will be put on the Web this afternoon.
So, here it comes. There it is.
Horizontally, we plot again omega divided by
omega zero where omega zero is the square root of G over L,
the resonance frequency of a single pendulum.
And vertically we do the same thing.
We want C divided by eta zero. Everything has the same
meaning, namely plus one means in phase with the driver and the
same amplitude as eta zero, and minus is out of phase with
the driver. Now, if you take this pendulum,
and you move it with zero frequency, you don't have to
know any differential equations. That will be here.
And 20 years from now it will be there.
And this separation is, then, eta zero.
And so you expect that C1, C2, and C3 will all three be
eta zero. And, they will be plus eta
zero, not minus, but plus, in phase with the
driver. And look.
That's what you see there. So that is by no means a
surprise. You go closer to resonance,
and you see that the bottom one, which is the blue one,
it looks black here, is picking up an amplitude
which is larger than the middle one, than the red one,
and the red one larger than the top one.
So, you are going, then, into the domain where you
are going to see something like this.
So, this one: C1, C2, C3, and then finally
when you hit resonance, I do not know what the ratios
are. And everything gets out of hand
anyhow. Now there comes a point which
boggles the mind right here. The top one is not moving.
It stands still. And there is nothing really so
special about that frequency. That frequency,
for which the top one stands still, so one goes to zero,
that omega, if I try to eyeball it, I would say it's about
0.75-0.77 times omega zero. And somehow at that frequency
the top one will not move. The other two will move.
They have an amplitude which is not stunning.
It's nothing to write your mother about.
But it is about eta zero. It is a minus sign.
So, it's out of phase with the driver.
And this one is a little more. It's bizarre.
Could I demonstrate this? No way on Earth can I generate
0.77 times omega zero. I can burn into my chip omega
zero as I did. But I cannot do 0.77.
There's no way. So, forget it.
I have to disappoint you. But now comes the good news.
Look at this point here. That is a point where the
middle one will stand still. And that is truly amazing.
The middle one stands still. The top one has an amplitude of
about 0.7 eta zero. I just eyeball that.
And the bottom one has an amplitude of about minus 1.5 eta
zero. Let me try to make a drawing of
that. So, this now is an omega equals
omega zero. So my hand is here,
which is eta zero. The top one has an amplitude
which is about 0.7 eta zero, this one.
So, it is roughly here in phase.
This is connected. So, this is roughly 0.7 eta
zero. The next one stands still,
believe it or not here. And then the next one is out of
phase with me and has roughly an amplitude of one and a half
times this. So, that's one and a half.
So, I call that -1.5 times eta zero.
And then half a period later, my hand is here.
And this object is here. And this one stands still.
And this one is here. Truly, these are almost not
believable. And then, there is another
frequency whereby the top one stands still.
I will attempt, I will make a daring attempt to
aim for this solution. And the reason why I can try
that is because the frequency is omega zero, which I can burn
into my chips. That's the last attempt I will
make today. This is the triple pendulum.
So, I have to go through the exercise of learning again,
the period, and then as I close my eyes and drive it with this
frequency, the idea then is that you will see the top one move in
phase with my hand. The one below that wouldn't
stand still. And then the one below that
will have an even larger amplitude.
So, one, two, three, four,
five, six, seven. One, two, three,
four, five. One, two, three,
four, five. One, you see it?
Three, you see it? Did you see that one stand
still? Isn't it amazing that physics
works? OK, see you Thursday.