To avoid possible misunderstandings,
my lectures start at 9:30 Eastern Standard Time,
which is different from lobby seven time, which you may have
noticed. The clock in lobby seven is
seven minutes slow. Today, we're going to cover
coupled oscillators, which is a big part in 8.03,
though we leave out damping in order to avoid a major
complication. Imagine that I have a pendulum
length L mass M, and I have another pendulum,
also mass M length L. And I connect a spring between
them, as you see there, spring constant K.
Imagine now at time T equals zero that I give this object,
which I call object number one, and this is object number two,
give it a certain position. I give it a certain velocity,
so I have four choices. And, I let the system go.
What you're going to see is something extremely chaotic,
and our task today is to predict what the position of
this one is at any moment in time, and what the position of
that one is at any moment in time.
And, to show you how chaotic that motion is,
I will just show you this. So, I take this one and just
displace it from equilibrium. I displace this one from
equilibrium, and at time T equals zero, I will give the
one, my left hand, just a certain velocity.
And, if you now look at the position of the individual
object, it would seem nearly impossible to come with an
analytic solution, which tells you what these
motions are. You'll see that amplitudes
build up. Of certain ones,
amplitude goes down. This one is hardly moving at
all now. Now it's picking up again.
And so, our task today is to work on that.
What is by no means obvious that I will show that to you,
that any motion no matter how you start it off is going to be
the superposition of two normal [mode?] solutions.
Anyway you start it can always be written as the superposition
of two normal mode solutions. What is a normal mode?
A normal mode is, in this case,
that both objects have exactly the same frequency that is
fundamental to normal mode and that they are either in phase
with each other or out of phase with each other,
nothing in between because there's no damping.
So, it's either in phase or it's out of phase.
That is the normal mode. In other words,
if I call one of those frequencies omega minus,
minus means it is the lowest frequency.
There are two frequencies in the system because there are two
objects. Omega minus,
I call that the lowest frequency.
Then it would mean, if they are in phase with each
other, that they come to a halt at the same moment in time.
That means in phase and in the same direction.
So, they come to a halt at the same moment in time in the same
direction, and at the same frequency.
If, then, I have another frequency which is higher there
are two normal modes because we have two objects.
If we have three objects there are three normal modes.
If we go to the higher normal mode, the higher frequency,
they have the same frequency that there are 180° out of
phase. So, when one comes to a halt
here at the other one comes to a halt there.
That's what it means 180° out of phase.
I can excite, and I will excite,
this system into its normal modes only.
I can excite this normal mode alone and this normal mode alone
if I choose the correct initial conditions.
So, for any randomly chosen initial condition,
the motion of each object can be written as a linear
combination of these two modes. If you take my word for that
for now, but of course I will demonstrate to you and I will
prove that to you, it would mean,
then, that [SOUND OFF/THEN ON] function of time can be written
as having some kind of an amplitude, X zero,
I give it a minus sign. Of course, it's related to that
normal mode frequency times the cosine of omega minus T plus
some [phi?] minus, plus some other amplitude,
which I call X zero plus, which is going to be related to
this frequency times the cosine of omega plus T plus phi plus.
Now, let's look at this. Let's try to see through this,
what this means. It means that if I know my
initial conditions, that I can determine this
amplitude. I can determine this phase.
I can determine this amplitude, and I can determine this phase.
There have to be four adjustable constants because I
have the choice between the positions at T zero,
and the velocity. So, there must be four.
What you see is that omega minus and omega plus,
which are these normal mode frequencies, are independent of
the initial condition. So, now I'm going to write down
the position in time for object number two.
And I know that in this mode, it must have the same
frequency. So there must be here a cosine
minus T plus the same, exactly the same phase because
I told you, normal mode means same frequency in phase,
or same frequency out of phase. And, it is the next one which
is going to be out of phase. But, you will see that shortly.
So, this term must become omega plus T.
For now, I will say plus phi, plus what you're going to see
how the out of phase comes in very shortly.
Notice this omega minus must be that omega minus.
Otherwise it wouldn't be a normal mode.
This omega plus must be this omega plus.
Otherwise it wouldn't be a normal mode.
These phi's are the same to allow them to be in phase.
These phi's are the same, even though they are out of
phase. But you will see very shortly
that I will get a minus sign here, which will take care of
the 180°.
I can, with this system, excite the lowest mode,
so that it only oscillates in that lowest mode.
So, this term is not there, and that this term is not
there. I can do that.
I can always oscillates something in one of the normal
modes. But, if I set it off at random,
then it is going to be a superposition.
Even though it is early for you, can you show me using your
hands and your legs, whatever you want to,
when I set this off at just the right initial conditions,
what will that lowest normal mode look like?
How will these pendulums oscillate?
Very good, very good. So you have a good instinct.
Very good. Let's do that.
So, let's set them off in the same direction and let them go.
Those are ideal initial conditions for this mode.
And you will see that it is a normal mode, namely,
they have the same frequency, and they are in phase with each
other. They come to a halt at the same
moment in time. So, what you see already is
that this value here must, also, be X zero minus.
That's a must. You have just seen it.
In this case, what it means is that you might
as well remove the spring. The spring is not doing
anything. The spring never gets any
longer, never gets any shorter. What do you think the pendulums
will do when I excite the other mode, which has a higher
frequency? Very good, can I see some other
hands or heads or legs? Good.
That's this, symmetry.
There is a symmetry in the system.
And that's why you people are so smart and see immediately the
connection. So, let me make here a drawing
of the omega minus, and then the amplitudes are the
same. And here, I make a drawing of
the omega plus. Higher frequency,
and the amplitudes are now the same, but 180° out of phase.
So, it's easy to excite that. There we go.
Notice they oscillate with the same frequency,
and they're coming to a halt at the same moment in time.
But, if one is here, the other is there.
And so, therefore, this one here must be minus X
zero plus. That minus sign,
now, gives me the 180°. So I can still keep that phi
plus there.
It is utterly trivial, without any work,
to calculate, or tell you what omega minus
is, that just the frequency of a single pendulum,
so that's easy. So, omega minus is simply the
square root of G over L. There's no spring,
right? Each one is doing its own
thing, and the spring is never pushing or pulling.
But we should be able to calculate omega plus.
So let's do that. And I will only make a drawing
of one of those pendulums. I don't need them both.
So, here's one of those pendulums.
And let this separation distance be X from equilibrium.
And, I call this angle theta. And so, here is that spring.
And, the other one is on the other side.
Let's pull in all the forces that we can think of.
So, there's gravity, MG.
There is tension, T.
But there's more. What else is there?
Spring. How much longer is the spring
than wants to be? 2X, exactly.
In other words, there is a force here.
I call that the spring force, FS, which is bringing it also
back to equilibrium. And, that one is 2KX.
Now, the tangent is very close to MG.
We have discussed that several times.
I'm going to introduce a shorthand notation.
Omega zero squared is G divided by L, and omega S squared,
making reference to the spring is K over M.
And now I have to set up the differential equation.
I have to apply Newton's Second Law.
And to remind you, the magnitude of the spring
force is 2KX. So, Newton's Second Law,
MX double dot equals, now, there are two forces that
I have to take into account: first, the spring force,
which is minus 2KX. And then I need the horizontal
component of the [tension/tangent?].
Remember, that's only one that is important here,
which is T sine theta. But, T is MG,
and sine theta is X divided by L if L is the length of the
pendulum. So, we get minus T,
which is MG times the sine of theta, which is X divided by L.
And so, this is the differential equation.
I'm going to divide M out, and I'm going to bring
everything to one side. So, I'm going to get X double
dot plus two omega S squared times X plus omega zero squared
times X equals zero. And our task now is to solve
this differential equation. And that, of course,
you can do in three seconds because you recognize this
differential equation. It is X double dot plus
something times X. And so, the new frequency,
which I will call omega plus, that new frequency omega plus
is the square root of two omega S squared plus omega squared.
And, what you see is something that I already anticipated,
which was consistent with your intuition.
It is larger than omega zero because this is the effect of
the spring. And so, omega zero is the one.
Here, an omega plus now is the square root of two omega S
squared plus omega zero squared. And, omega S squared is defined
this way.
So, if I now turn to my general solution, if for now you accept
the fact that that is the general solution,
the superposition of the normal modes, then the key point is
that independent of your initial conditions is omega minus.
That was the square root of G over L.
I never put in any initial condition.
Independent of the initial condition is omega plus.
That's this one. I never put in any initial
conditions. Independent of the initial
conditions is this ratio, which is plus one.
And, independent of the initial conditions is this ratio.
That is minus one. In other words,
if I didn't tell you at the initial conditions,
which I haven't, I can predict that this ratio
is plus one. And I can predict that this
ratio is minus one. If I make this three,
then this is minus three. If I make this five,
then this is minus five. The ratio is minus one.
So, the ratios are independent of initial condition,
and the frequencies are independent of initial
condition. If I tell you the initial
conditions, then of course you can also calculate the
individual amplitudes. Suppose, now,
I start this system off in some way at time T equals zero,
which I can choose. And it used the following.
At T equals zero, I make X1 C,
just some number C that we can choose, 3 cm,
whatever you want to choose. But, I make V1 zero.
So, as I release it: no speed.
And, suppose I make X2 zero and I make V2 zero.
So, if you don't know what that means, of course I'm going to
demonstrate it. It means that this one here,
the other one I offset, that's all I do.
And I let this one go, pull my hands off,
and then I want to know what's going to happen,
right? Because I release this one with
zero speed. That's what you see in there.
There is one of an infinite number of initial conditions
that you may choose. So, this now has to be
substituted into my general solution.
And so, you have to take the derivative of X1.
So, you have to do X1 dot and put that equal to zero.
I leave that in your competent hands.
Then you have to take X2, and you have to take X2 dot,
and you have to make that equal to zero.
And I leave that into your competent hands.
That's easy. This and you'll find,
then, that in this particular case for this particular
example, phi minus equals zero, and phi two,
and phi plus equal zero. It will not take you more than
a few minutes to find that. I didn't want waste your time
on taking a derivative of such a simple function.
So, if we take this right now, then I will substitute these
results in the equation. So, you did the hard work.
You did the velocities. I will do the positions.
So, I will substitute in that equation T equals zero.
I know already that the phi's are not there.
And so, then I get that C at time T equals zero is going to
be X zero minus T zero. So, this is one.
T zero, so this is one, plus X zero plus.
That's this initial condition. Then I go through the second
initial condition that is zero, is X zero minus,
minus X zero plus. There is a minus sign here.
And so, what do I find I have solved now?
The general solution, I will find that X zero minus
is one half C, and F zero plus is also one
half C, which of course should not come as a surprise to you.
So, let me write down, now, the general solution that
we have for this specific initial condition.
So, we know everything. We know X zero one.
We know X zero one minus, X zero minus,
not one X zero minus. We know X zero plus.
We know phi's. We know everything.
So, I'm going to write it down here.
So, X1 is going to be one half C.
Remember, we found it at one half C times the cosine of omega
T equals phi zero plus one half C times, this is minus,
by the way, cosine omega minus T cosine omega plus T.
That is X1, and X2 is one half C times the cosine omega minus T
minus one half C times the cosine omega plus T.
Take a deep breath, substitute in there T equals
zero, and you see immediately that X is C and substitute T
equals zero into seconds, and then you see,
indeed, that X2 is zero: no surprise because that's my
initial condition. Now, I remember from my high
school days that the cosine alpha plus the cosine of beta is
twice the cosine of half the sum times the cosine of half the
difference. So, I can write down X1 as
twice the cosine of half the sum times the cosine of half the
difference. So, that two that I get eats up
these one halves. So, I get C times the cosine
omega minus plus omega plus divided by two times T times the
cosine of omega minus, minus omega plus divided by two
times T. I've just rewritten it in a
different form. And, X2 as a function of time,
I now have the cosine of alpha minus the cosine of beta.
That is twice the sign, half the sum times the sign of
half the difference. So, now I get here the sign of
omega minus plus omega plus divided by two times T times the
sign of omega minus, minus omega plus divided by two
times T. Notice when T is zero,
that the sign of this one is zero, consistent with what our
initial conditions, X2 zero, remember?
It's all there. Initial conditions are all
there. Initial conditions are all
there, just written in a different form.
Now, imagine now in your mind that these two frequencies are
not too far apart. Then, these equations smell of
what? Beats.
See, this here is the fast term.
This one and that one, and this one is the slow one.
And so, if these two are close, then what you'll see is
something quite remarkable. At T equals zero,
this one stands still. And, the cosine term is going
to be one because T is one. This one is going to oscillate
happily with this frequency. But, this cosine term is very
gradually going to zero. And, as that cosine term
gradually goes to zero, this one will stop oscillating.
But, this sine term becomes plus one.
And so, the other one will start to oscillate.
And then, a little later in time the cosine term will become
minus one. So, it starts to oscillate.
But when that happens, the sine term is zero again.
So, it stops. So, you see a beautiful beat
phenomenon. The first one will gradually
come to a halt, and the other one will pick up.
And then, the other one will come to a halt,
and then transfers in [a way?] energy.
It is, of course, consistent with the
conservation of mechanical energy.
And, I want to demonstrate that.
So, we have this year. All I have to do is offset X1
over a distance, C, that we can choose,
and then we release this one at zero speed.
And then we'll just watch it. And then, you should see that
strange beat phenomenon. You ready for this?
This will have to hold in place.
Three, two, one, zero.
Look, this one is standing still.
That's beating. And you see that the energy is
transferred from one to the other.
And that is a beat phenomenon that follows immediately from
this. What would happen if I moved
the spring up? Suppose I moved the spring here
higher, say, halfway. Anyone without looking too much
at the blackboard, sort of use your intuition.
What would happen? Would the same phenomenon
happen? Yeah?
The spring what?
Yes, so would the same phenomenon happen?
But what would happen with the beat period?
Now you may look at the blackboard.
Do you think these have a certain separation?
Right, you make that separation smaller.
You make the effect of the spring smaller.
So, the omega minus becomes even closer to the omega plus.
And so, the big period will increase.
It will take longer for one to stop.
Let's try that. So, are we going to move these
up? I'll put it roughly halfway.
OK, are we ready for this? And so we're doing it again.
And then you'll see the same phenomenon, except it will take
longer for the first one to come to a halt.
So, I have decreased the coupling.
It's still swinging happily. Now it's beginning to change
its mind. You see, and now it standing
still. It took longer.
Do we agree? Now, there is something
mind-boggling, something that you really want
to see. Suppose I bring this all the
way to the top. That's interesting because then
omega minus is exactly the same as omega plus because,
look, this term goes away. What now will happen?
It's almost cutting out the spring.
There is no spring anymore. What now will happen if I start
one here? Three, two, one,
zero.
What do you think? Yeah?
Ah, brilliant. Brilliant.
One will swing, and the other won't.
Isn't that shocking? You have two pendulums,
and they're not even connected anymore.
And you start this one swinging, and the other one will
never start to swing. Isn't that amazing?
For $30,000 tuition, you learn something fantastic
except that the big period is infinitely long.
So, with a bit of patience, so let's demonstrate that.
So let's demonstrate this. So, there we go.
Unbelievable. Physics works.
And look at this equation. When omega minus is omega plus,
this term is always zero. So, you see X2 remains zero.
Ha, ha, ha, that's what you see.
And, when omega minus is omega plus, this cosine term is always
plus one. Ha, ha, ha, that's what you
see. Isn't it amazing the power of
physics?
So, it is truly remarkable that we can describe for any initial
condition the motion in terms of the linear superposition of the
two normal modes. And so, what originally looked
like an impossibility when I started the first 30 sections of
my lecture with the demonstration,
when I started them off in a random way, it looked so chaotic
that it was almost unimaginable that we would be able to short
that out and be able to predict the motion of each one
individually as a function of time.
I want you to appreciate that we have two coupled oscillators
here. That gives you to normal modes.
If you have three coupled oscillators, we will get back to
that, then there are three normal modes.
And if you have for, then you have four normal
modes. Now, clearly what we need is a
general recipe. In this case,
the problem was a beautifully symmetric we could do all mega
minus just a split second. Omega plus: you could see
exactly the motion. So, in no time [you get?] that
the ratio of the amplitude was plus one and was minus one.
But let me tell you, if I break the symmetry,
oh man, all hell breaks loose. For instance,
if you change the ratio of the masses, you make one longer than
the other, you have three beats connected with springs.
If you break the symmetry, it is very, very hard work.
And, for that, we need a general recipe.
And, I'm going to give you the general recipe.
And then, I'm going to apply that general recipe to this
case. So, everything that you have
seen will then come out after a lot of algebra,
but at least you know that the recipe is working.
So, the first thing that you do that's number one,
you give each object a displacement from equilibrium.
Even though you are free to choose the direction,
I always, matter of discipline, set them all off in the same
direction. That's not a must.
But, that reduces the chance of mistakes.
But, you give them a displacement,
and I always do them in the same direction.
That's number one. Once you have done that,
you write down Newton's Second Law for each,
which means if you have two objects, you have three
unknowns. Well, that is three unknowns.
So, you want to find one of the normal mode frequencies.
You want to find this one, and independently you want to
find that one, which has to come out of that
recipe, what are in principle the three unknowns.
This ratio plus one holds here, but that doesn't hold in other
cases. And so, in principle you have,
as an uncertainty, this amplitude and omega.
So, you always end up with two equations with three unknowns.
So if I see two objects, then you have three unknowns.
And you have two equations. So, that looks like a problem.
Number three: you're going to put in the
condition for normal modes, which means that X1 is C1 times
cosine omega T. But, X2 can be some other
amplitude, which we had to calculate also times the same
omega T. I don't have to put the phase
in there because it's either in phase or out of phase,
and out of phase means that you get minus signs.
And so, you see immediately that if you substitute this in
Newton's Second Law, that you get for two equations,
three unknowns. C1 is unknown,
C2 is unknown, and omega is unknown.
So, you substitute this in here, and then comes the
problem. How are you going to deal with
two equations with three unknowns?
Well, think about what I told you earlier.
If you don't know the initial conditions, you can never find X
zero minus, so you can never find C1.
But the ratio is independent of the initial conditions.
So that means the ratio C1 over C2 cannot be dependent on the
initial conditions. In other words,
you can always solve these equations in terms of C1 divided
by C2 and omega. And, I will show you that that
works. So, you follow this recipe,
and we will do that religiously together.
We will find, then, in this case two
variables for omega We will find an omega minus,
which has an associated value of C over C2.
I'll put a minus there. And, we will find an omega
plus, which has its own ratio associated with it.
In the case of our simple symmetric system,
this would be plus one and this would be minus one.
But that is not the case when you break symmetry.
So, my task is, now, to apply this recipe in
its most general form to the system.
It will be 16 minutes of grinding.
We will go through each step, and out comes something that we
already know. But at least you'll see that
I've made no assumption which I did there about symmetry.
Therefore, this is an ideal moment for you to have your
five-minute break so you can take a deep breath to get ready
for this 16 minute marathon. [SOUND OFF/THEN ON]
All right: general recipe. Someone asked me during the
intermission whether this omega is the same as that one.
Yes, of course, otherwise it wouldn't be in
normal mode. We're going to search for the
normal modes. And, in the normal mode,
the frequencies must be the same of all objects,
whether you have five or six or two or three.
But the ratios are different. That's a different issue.
However, they are in phase or out of phase.
So the ratios can be negative, positive, but you never had any
phase angles other than the 180° or zero because we have no
damping. We have taken the damping out.
So, it is essential that when you substitute that in Newton's
Second Law that these omegas are the same because they will then
give you the normal modes. OK, you ready?
I'm going to offset the first one over a distance,
X1. So, they have mass M.
They have length L to remind you.
And, we are going to have omega zero squared equals G over L,
and omega S squared equals K over M.
And, this one I offset over a distance, X2.
Notice, you don't have to do that.
I always offset them in the same direction.
And then, there is this spring that connects them.
I will just make it very thin, this spring.
Otherwise the picture becomes a little bit too complicated.
So let this angle be theta one. And let this angle be theta
two. And so, we have here MG.
We have here MG. We have here the
[tangent/tension?] is roughly MG for small angles,
and we have here a tangent which is roughly MG.
It's a little bit too big the way I drew it,
but that's not so important now.
And now, there is another force that X on both,
and which force is that? That's the spring force.
Now, do we agree what the magnitude of the spring force
is? And then, we will argue about
the direction. [Nickel?] magnitude,
look very closely. I offset one by X2,
and I offset the other by X1.
I was asking you a question. You can't answer with a
question. Very good.
The magnitude of that force is K times X2 minus X1,
nonnegotiable, right?
Now, if X2 is larger than X1, do we agree that the spring
force is in this direction? And do we agree,
then, that this spring force must be in this direction if X2
is larger than X1? Well, that means I can leave
everything the way it is now. As long as I give this a minus
sign that is a plus sign, I'm OK because if X2 is smaller
than X1, it will automatically flip over, and my algebra is
fine. So, therefore,
in my head I can just think of X to being larger than X1,
set up the differential equations, and I no longer have
to worry about the fact that maybe X2 at certain moments in
time is not larger than X1. So, now I set up the
differential equation, MX1 double dot.
So, that's this one. Let's first do the pendulum.
That is minus T times the sign of theta one.
T is MG, so it is minus MG times X1 divided by L.
Do we agree there is that horizontal component here?
The horizontal component of T is driving it back to
equilibrium minus sign. FS, the spring force,
is driving away from equilibrium.
So, it's going to get plus K times X2 minus X1.
Look at this. Signs are very important now.
Very important. And the next one,
X2 double dot, it has, again,
the restoring force due to the pendulum which is minus MG X2
divided by L because it's the sign of theta two now that comes
in. And now it has the spring
force, which now is a negative sign.
And so, now we get minus K times X2 minus X1.
And, when you have this down on your next exam,
you take a deep breath, and you go over each individual
step to make absolutely sure that this is correct.
If there's anything wrong with this, you are dead in the
waters. The whole problem will fall
apart. It may not even be a simple
harmonic oscillator. You get something ridiculous.
Signs are crucial here. So, therefore,
look at this again. Take a pause.
This one, if there's a positive offset is indeed in a negative
direction, a component of T sine theta.
This one, if X2 is larger than X1 is indeed in that direction.
And you notice, if X2 becomes smaller than X1,
well then, this becomes automatically a negative.
So, are you OK? So this is fine.
This one has a restoring force due to the pendulum,
which is T sine theta two negative sine and the restoring
force of the spring always opposite this one:
I'm happy. I am happy.
So, now, we are going to rearrange this a little bit and
we're going to introduce the omega zero squared notation.
So, I'm going to divide M out, and so I'm going to get X2,
X1 double dot. And then, I'm going to bring
the X1's to the left. And so, I get plus omega zero
squared times X1. But you have a minus KX1.
You divide by M. So, you get plus omega S
squared. And, that's [both?] times X1.
You notice that? You both have an X1 term.
And now, the X2 comes out. The X2 has a plus sign on the
right side, so it gets a minus sign.
So, I get minus omega S squared times X2.
And that is zero.
And then we do the next one. We get X2 double dot.
Ah, I get the same terms. I get an omega zero squared.
And I get an omega S squared times X2.
And then I get this plus; minus times minus is plus.
So I get minus omega S squared times X1 equals zero.
When you have reached this point, you take a deep breath
and you go over each little term to make sure that you haven't
accidentally slipped on a minus sign because you slip up on one
minus sign you're dead in the water.
It may not even become a simple harmonic oscillation.
So let me do that. I agree with that.
I agree with this. I agree with that.
Differential equation is fine.
Notice that I already did point number one, and that already I
did even number two. I've set up the differential
equation, Newton's Second Law. So now comes number three.
And number three means and going to substitute in there,
X1 is C1. Cosine omega T and X2 is C2
cosine omega T.
We go slowly; I want you to follow each step
and I'm going to work on here so that you can see it high.
So, I'm going to substitute this in here so the X1 double
dot will give me a minus omega squared, right,
because you get twice omega out minus omega squared times C1.
And, to hell with cosine omega T.
Why to hell with cosine omega T?
Because every term will have cosine omega T.
So you might as well divide out right away, right?
Every term that you're going to have in here will have a cosine
omega T in it. So, I'll leave the cosine omega
T already out now. So, I get minus omega squared
C1. Then I get plus omega zero
squared plus omega S squared times C1.
That is this term, minus omega S squared times C2
equals zero because, remember, X2 has the C2.
Am I going to fast? Beautiful.
So, I go to the second one. The second one is X2 double
dot. So, I also get a minus omega
squared times C2. And then, here I get a C2.
So, I get plus omega zero squared plus omega S squared
times C2. And then, I get minus omega S
squared times C1 equals zero. And, when you have reached this
point in your exam, you take a deep breath.
And, you make sure that all your signs are correct.
If not, you're dead in the water and the problem will fall
apart. So, let's do that.
Minus omega squared C1, I can live with that,
plus that term, which is C1,
yes, minus omega S squared C2. Even if you make a little slip
of the pen, and you change this into a one and this into a two,
it's all over, of course.
And then, the next equation, minus omega squared C2.
Then I have this times C2, and then at minus that times
C1. We're almost there,
even though it doesn't look that way, does it?
Remember what I said. You cannot solve two equations
with three unknowns there is no way that you can solve for C1,
C2, and for this omega, which is what you're really
after. This is the omega you are
after. But you cancel for C1 divided
by C2 and omega knowing that, already, I'm going to eliminate
C1 over C2. And, you'll see how I do that.
C1 divided by C2: I go to the first equation.
This is not so hard what I'm doing.
I put in my mind this on the right side.
So, I get a plus omega S squared times C2.
And then, I divide C1 by C2. And so, when I get then is the
following. I get omega S squared up stairs
which is this one. And all this comes downstairs,
minus omega squared plus omega zero squared plus omega S
squared. Do we agree?
That's the first equation. I've simply written it as C1
divided by C2. I can always do that,
right? Bring the C1's to one side,
the C2's to one side, and divide them.
That's this. And, I'm going to do the same
with the next equation. I bring this up to the right
side, that I have C1's there and C2's here.
And now, this becomes upstairs. So, I get minus omega squared
plus omega zero squared plus omega S squared,
and I get downstairs omega S squared.
And now, I have eliminated C1 and C2 because if I solve this
equation, I get my solutions for omega.
This is one equation. There is one unknown.
That's omega. You've got to admit,
and there better be two solutions.
There better be an omega minus, and there better be an omega
plus coming out of this untouched by young human hands.
I must find two solutions. Let's first make sure that this
is correct, and the answer is yes.
So, what do I do now? Well, I'm going to simplify
this one step further, which is now very easy.
I multiply this by this, and this with this.
So, we get omega S through the power of four is minus omega
squared plus omega zero squared plus omega S squared.
One equation with one unknown; omega is the only unknown.
Take the square root left and right.
So, I get minus omega squared plus omega zero squared plus
omega S squared equals plus or minus omega S squared.
Do not forget the plus or minus because the square root of omega
S to the fourth is plus or minus omega S squared.
And now, you're going to see the light of the day.
You won't believe this. I bring omega squared to the
other side, and I get omega zero squared plus omega S squared
minus or plus omega S squared. And, the simplicity is
overpowering. I feel it over my whole body.
It is unbelievable. When I have a minus sign here I
find that omega is omega zero. That's my omega minus.
That's the one. That's my omega minus.
When there is a plus sign here, I find exactly what we predict
before. I find that omega plus is the
square root of omega zero squared plus two omega S squared
because the plus sign make this two.
Isn't that elegant? Isn't it beautiful?
So, they just pop out. And I predict that there will
be two solutions. You have them both.
So, by substituting this in there, automatically come out
two solutions. But, what now is C1 over C2?
We haven't solved for that yet. But I promise you can always
solve it in terms of omega and in terms of C1 over C2.
Any volunteers? Any volunteers?
Look at the blackboard. It's somewhere hidden in those
equations. I said we are going to
eliminate C1 over C2. You think C1 over C2 likes to
be eliminated? Let's call it back.
And look at this equation. Substitute in here for omega
squared. Substitute in omega minus.
Then you get an answer for C1 over C2.
What do you think that answer is?
Plus one. But even one.
Plus one. So, you get C1 over C2 is plus
one. If you take this omega zero and
you put it in here, you get omega zero squared with
a minus sign, with a plus sign.
You have omega S squared divided by omega S squared,
and it's plus one. And now, you take the second
solution, and you put it in here.
What do you think you're going to find?
Minus one. C1 over C2 is now minus one.
And so, what you have seen now is that the general recipe comes
up with frequencies, comes up with the ratios of the
amplitudes, not with the individual amplitudes because
you don't know the initial condition.
It comes up with the ratio of the amplitudes,
and now you can write any solution, provided that you know
the initial conditions. If you know the initial
conditions, then you can also find C1.
And therefore, since you know the ratio,
you automatically have C2. And, you can find this C1.
So, I'll give this one a minus sign, and this a plus sign.
And then, you know the ratio in that mode.
So, now you may think erroneously that life is easy.
That's far from the truth. Suppose you have something as
simple as this: L, M, L, M.
It's called a double pendulum. There is no symmetry.
I want to test your intuition. There are going to be two
modes, two normal modes: a lower one and a higher one
because they are two objects. In the lowest frequency,
use your hands in your legs. What will this pendulum look
like? You are all doing sort of the
right thing. Would it look like this,
or would it look like this? In other words,
let me make this one a little bit, let me make the difference
a little larger. Would it look like this,
or would it look like this? Slightly exaggerated.
Who is for this one? OK, that means that the ratio,
C2 over C1, is going to be plus two, right?
Who is for this one? That's the way it is,
believe it or not. C2 over C1 is going to be one
plus the square root of two. And, it will take you 30
minutes of grinding to find that.
And I'm not exaggerating when I say 30 minutes.
You have to go to the whole procedure and then you will find
that this ratio is 2.4. And I will demonstrate it.
The highest mode must be something like this,
right? Because they must be out of
pace. Any idea of the ratio,
C1 over C2? I'll call this C1,
call this C2? Any one of you,
any intuition? Minus, yeah?
Boy, you're good. You're good.
It is minus 2.4. This one will be much further
away than this one. Do any one of you want to make
a guess what omega minus is? Anyone want to make a guess
what omega plus is? No way on Earth have you,
or I, or anyone else can look at this and say,
oh yes, of course, omega minus is this.
It's a long road. 30 minutes of calculating,
and then out of that popped these frequencies.
And out of that, then, pops the ratios.
And now I'm going to demonstrate you just this case.
And the way I'm going to do that, even though we have not
discussed driven coupled oscillators, in order to set it
off in these normal modes, a normal modes is also a
resonance frequency. We call that,
often, natural frequency. It's what the system likes to
do if you leave it alone. And, it is easy for me,
when I just tease it a little bit, to excite it in that
resonance mode. So, therefore,
I am going to drive it, but only very briefly.
And then, I will leave it alone.
And I will get it into the state that you will see this
that is [really?] in the normal mode.
There's one at only one frequency.
They come to a halt at the same moment in time.
They are in phase, but you will clearly see that
this distance is more than double this one.
And then, I will go to this one.
I will try that. All right, double pendulum.
Oh, my goodness. This is a double pendulum. Yeah.
So, I have to drive it a little.
But it's really only very little at resonance.
And then, I will stop writing it.
This is the mode. Do you see that the bottom one
is further away than twice the top one?
This is a normal mode. OK, you see that it's not a
straight line? Now, it's hard to say that it
is one plus the square root of two, of course,
from your seats. But that's what it is.
And now, I will excite the second normal mode,
which is a resonance. That's it.
And now, I can almost stop swinging it.
You notice that your upper one has a much larger amplitude than
the lower one, and the ratio is one plus the
square root of two. Did you notice they are out of
phase? It's the minus sign.
They are out of phase. And in the other mode,
they were in phase. And so, all of that will follow
from the recipe: if I gave this problem on an
exam, you would have reasons to kill me because it would take
you too long. I would certainly wear a
bulletproof vest on campus. That is easy because you have
symmetry. This doesn't have that
symmetry.
Yeah, C2 over C1. This is C2.
I haven't specified what I call one and two.
Is that the problem? I call this one and I call this
two. So, C2 over C1 is plus 2.4.
It's good that you asked that. And here, I changed the
sequence because I knew that this one was larger than this
one. So here, you have C1 over C2 is
minus 2.4. The minus sign means out of
phase. Yeah: good that you asked that.
Very good. Well, I'm slowly,
slowly going to turn you into experts, and now we are going to
try something else to see how good your intuition is.
I want you to understand, though, that your intuition is
no better, no worse than my own. In other words,
there is no way that I could have done any better than you
did. When I was in high school,
or whatever it was, maybe in college when I first
saw this, my first impression was, what the hell?
A straight line? But if you give it a little
thought, you will come to the conclusion it cannot be.
Then, it ultimately comes out this solution.
So, don't feel bad if your intuition lets you down.
I'm as bad as you are when it comes to that.
So, now we're going to evaluate the system, which would take you
an hour and a half to work out with a general recipe.
And that is four springs and three cars.
One car, two cars, three cars, and all springs
have spring constant, K, nicely symmetric,
you would think. And all objects have mass M.
And, I want to know, I want to see whether we have
any intuition without being too quantitative for the three
normal mode solutions. There must be an omega minus
where all three are in phase. Then, there must be an omega
plus, which is more complicated. And then, there must be an
omega plus, plus, which is the highest of all
three. In the highest possible
frequency, every object, two adjacent objects are always
out of phase. So, if you have 20 objects in
the highest mode, number one is out of phase with
number two. Number two is out of phase with
number three. Number three is out of phase
with number four, and so on.
Let's first look at omega minus.
I will even dare asking you whether you have any idea what
omega minus is. You don't have that;
I don't have that. Where would your object be if,
for instance, I displaced the first one over
a distance which I just normalized to be plus one?
I call that plus one. Any idea where this one will
be? Any idea where that one will
be? What to think the second one
will be? Plus one.
Will be the third one? Plus one.
Wrong. Why does it have to be wrong?
Suppose this one is a plus one. And, suppose this one is also a
plus one. Then, this spring is no longer
than it wants to be, and this spring is no longer
than it wants to be. So, there is no force on this
object. So, that's not possible.
The object cannot go anywhere. Any suggestions which of those
is wrong? Is this correct,
and it's plus one? How about the middle one?
Where do you want it, a little further away or a
little less? All have to be in phase.
Don't use the word minus sign. If you use the word minus sign
then, how can you do that? They have to be in phase.
They must be on this side. Do we make the distance larger
or smaller? It's larger.
Should I tell you what it is? Should I tell you what it is?
You want to know? Tell me.
You really want to know, right?
I knew it. Is that obvious?
No. Now, the next one.
The next one is interesting. I will, again,
put this one in plus one because that is my starting
point. I always put that at plus one.
Now what? This one you can guess.
You really can. Think now about the symmetry of
the system. Yeah?
Very good. Very good.
The middle one stays still, and this one comes to minus
one. Now the third one.
Remember, the highest mode, neighbors are always out of
phase with each other no matter how many you have.
So, it's already certain that this one must be on the side,
and it's already certain that this one must be on that side.
Any volunteers? Would you like to try that
again? Would you do plus one and minus
one? Help me put this one at plus
one, sorry, minus one, and this one at plus one,
or shall we change the minus one here?
Don't feel bad. I wouldn't have been able to do
either. Shall I tell you what it is?
Now, I can demonstrate this. And the way I'm going to
demonstrate this is with the air track that I have here.
See, the beauty with this air track is that I can place these
cars wherever I want, and they're not going to move
because I don't have any air. Then, I turn on the air and
they are and immediately free to go.
In other words, if I offset this one in a
positive direction over a distance, one,
we've just called this positive.
I apologize for that because I called this positive.
And the other one has to be offset by one in the same
direction, which I just did. But this one has to be offset
by the square root of two, which I just did.
And now, turn on the air, and I just have exactly the
right initial conditions for this mode.
So, it is going to oscillate in the superposition of three
normal modes. But there is only one,
and that's the one you're going to see because I chose the
initial conditions. Are you ready for this?
Now, admit it. This is fantastic.
They are all in phase. That was a criterion for normal
mode. They are in phase.
They come to a halt at the same moment in time.
They all have the same frequency, and amplitude here is
the square of two times larger. That is not so obvious,
but you see this is the criterion for normal modes.
Same frequency, and in phase.
This one, OK, we are going to put this one at
zero. I'm going to put this one at
plus one. And I'm going to put this one
at minus one. Now, this one is offset by one
in this direction. This one is offset by one in
this direction, and this one is not offset.
I turn on the air; what do you think will happen?
That's right, it's going to oscillate in a
superposition of three normal modes.
But there is only one. That's the omega plus.
Impressive, isn't it? Next one, this one.
So, we are first going to put this at zero,
this at zero. And we're going to put this at
zero. And now, what we have to do,
this has to be put over plus one plus one.
The other one has to be put minus the square root of two.
And we have measured that here. We've come so close that this
spring is not going to like that.
And now, this one is going to be, again, plus one in the same
direction as this one. That's the highest frequency.
And there we go. Are you ready?
Three, two, one, zero.
One, and only one frequency. And, every adjacent one is out
of phase with the other. These are out of phase.
But these are in phase because these are out of phase.
Now, I had one student calculate this for me.
And I'll be frank with you, I even paid him for that
because I was too lazy to do it myself.
I put the job on the market, and he came within a few days.
There were a few people that said, oh, and an A plus for 8.03
and he was going to do that in no time at all.
20 hours later, he had it right.
It cost me a fortune. Now, I'm going to test you once
more to show you that these things are not so
straightforward, not to make you feel bad at
all. On the contrary,
my intuition is no better. I'm going to make a triple
pendulum. This is a triple pendulum.
Omega minus, what do you think it would look
like? My favorite student is here.
Do you think it will be a straight line?
I don't think so either. I think what you're going to
see without making any predictions about amplitudes,
exaggerated, you're going to see this.
That's what I think. No idea what it is.
How about omega plus?
This is not going to work. It's a whole different system.
Two of them are probably going to be in phase,
and one out of phase. Which two will be in phase?
Will you see this? Now, these two are in phase,
and this is out of phase. Or really, you see this.
Now, these two are in phase, and this one is out of phase.
Who is in favor of this? Who is in favor of that?
I want to see the hands long. Who is in favor of this?
Who is in favor of that? Yeah, well, we'll see.
[LAUGHTER] How about the highest mode?
That's the easiest one because now they must go like this.
Now, of course, the ratio, C1 over C2,
and C1 over C3, that's a different story.
That is your 20 hours of grinding if you get paid.
If you don't get paid, you can probably do it in two
hours. So, I'm going to demonstrate
this to you. I'm going to find these three
modes. Again, I do that by a little
bit of driving because of their resonances.
And then, I want you to eyeball these ratios,
C1 over C2, and C1 over C3. Oh, no problem.
No problem. There we go.
All right, so now, this is not where the length is
the same. So, to get this one into the
lowest mode is easy for me. Just give it a swing,
and let it go. And just do nothing.
Then, I will stop even driving it.
That's it.
Now, it's nicely in that mode. You see, the bottom one [has
it?]. Notice that the lines are never
in the same direction. You see the offset of the
angles? Can you all see that?
So, our first picture on the blackboard is quite accurate.
Now, the second one, that's harder for me to find.
I mean, we first make sure that they are not moving.
OK, so I'm going to try to search for it.
I got it. This is it.
So, the first solution is the correct one.
The upper two are in phase, and the bottom one,
look at the large amplitude of the bottom one.
That's not so intuitive. The bottom one is out of phase
with the upper two. Can you see it?
Normal mode, why is it a normal mode?
Because they all have the same frequency.
And, they are in phase or they are out of phase.
Now, the highest frequency, let me see whether I can get
that one.
Got it. That's it.
Now, look how small the amplitude of the bottom one is.
And, that can all be calculated.
They are beginning to rotate a little bit.
There we go. That's the third mode.
So, now you know everything there is to be known about
coupled oscillators. The most important thing that
you probably have learned is that if you really want to get
the general solution for a system, it is hard work.
See you next Tuesday. Have a good weekend.