I want to start with the physical pendulum,
which is exactly the same one that I discussed during the
first lecture. This is a hoop with mass M and
radius R. And, we were calculating the
period of this hoop as it oscillates.
And we did that using the famous stork equation from 8.01,
the torque relative to point P is the moment of inertia
relative to point, P, times the angular
acceleration, alpha.
Today, I will do this again, but I will use the conservation
of energy to show you that in case there is no damping,
when mechanical energy is conserved, but you can find the
correct to fragile equations through the conservation of
energy. If the thing is swinging,
in general there are two components.
You have kinetic energy and you have potential energy.
And the kinetic energy, K, is one half times the moment
of inertia about point P times omega squared.
You remember that from 8.01. And, this omega is theta dot.
We call that the angular velocity.
This angular velocity changes with time.
When the object goes through equilibrium, to a new or
velocity is at maximum. But when object comes to a
halt, the angular velocity is zero.
Do not confuse this omega with omega zero.
I will give that a zero, now, to distinguish it from
omega, which is the angular frequency.
The angular frequency is a constant of the motion,
and that is two pi divided by T zero if T zero is the period of
oscillation. So, this is the kinetic energy,
and this is the square of the angular velocity.
And then, we have potential energy.
Let this be point A, and when we are here,
the center of mass is at point B.
And so, the potential energy, U, or actually I should say the
potential energy at point B minus the potential energy at
point, A, it's always the difference in potential energy
that matters. That equals MGH,
H being the difference in height between point B and point
A. So, this here is MGH,
Massachusetts General Hospital. That's the way to remember it.
Now, H is very easy. H is the same as R times one
minus the cosine of theta. We went through that many
times, so you can easily confirm that.
R is the radius of this circle. This is the radius.
Now, for small angles, the cosine of theta is theta
squared divided by two. And so, I can rewrite this
differential equation now, that E total.
I see nothing wrong with that. I'm sorry.
Is there anything wrong with it?
The cosine theta alone, you're right,
is one minus theta squared over two.
Thank you very much. Thank you.
So, I'm going to write it now as a total energy is one half IP
times theta dot squared. And then, I get plus MGR times
theta squared over two. And, what you do now,
since this is a constant of the motion, you always do that if
you work with the conservation of energy.
You take the time derivative of this equation,
which must be zero because mechanical energy is conserved.
And, out pops the differential equation that we also found
during my lecture number one when I used the different
method. So, I take the time derivative.
So, this [T?] each of this half, and so we get I if P times
theta dot times theta double dot.
I had used the chain rule. And then, I get plus MGR,
these two, each of these two, and so I get theta times theta
dot, and now this equals zero because DE DT is zero.
And whenever you do that, you will always see that the
theta dot term disappears. Or if you have the equation in
X, then the X dot term disappears.
And, you see that. This term goes.
You can divide it out. And so, your differential
equation takes on, now, a very familiar form.
Let me write these two here. So, I get theta double dot plus
MGR divided by I of P times theta equals zero.
And this differential equation you should recognize.
The solution to that equation is immediately obvious.
Omega zero squared, I use now the omega zero equals
MG times R divided by I of P. And so, the general solution
for this oscillation then becomes that theta is theta
zero. That is the amplitude times the
cosine or the sine, if you prefer that,
omega zero T plus some phase angle, phi.
That is the general solution. And if you knew the initial
conditions, what the situation was at T equals zero,
if you know the velocity, the angular velocity at T
equals zero, and if you know where it is at T equals zero,
you can solve for this theta zero, and you can solve for this
phi. But, omega zero is independent
of the initial conditions. Very well.
Now, I would like to cover a case whereby I am going to
introduce damping. Whenever we deal with damping,
there are two terms that are important in physics.
That is the speed itself. That is a damping term which
opposes the velocity, which is when you're
proportional with the speed. And then there is a damping
term which opposes the velocity, which is proportional with the
square of the speed. We will always leave the square
out with 8.03 because the differential equations become
impossible to solve. Maybe numerically you can do
it, but not analytically. However, if you're interested
in all the physics, which is wonderful,
with a V squared and V, my lecture number 12 on OCW
Open Courseware from 1999, Newtonian mechanics,
I deal with the V squared and with the V term,
and I do many demonstrations to show you that there are certain
domains where the V term is important .
We call that the viscous term, and there are certain domains
in physics where the V squared term is important.
So, I will now simply restrict myself, then,
to the damping force, which is linearly proportional
with the velocity, and we will write it down as F
equals minus B times V. We will use a shorthand
notation that gamma equals B over M.
That is only a shorthand notation.
I will erase this. We don't need this anymore.
We need so many blackboards today that I'll use this
blackboard from a damping problem.
When you deal with damping, we recognize three different
domains. One domain, whereby gamma is
smaller than the omega zero, we call that underdamped.
Then we have a domain whereby gamma is larger than omega zero.
We call that overdamped. And then, you have a very
special case where gamma equals of omega zero.
[SOUND OFF/THEN ON]
And the behavior of these three different kinds is very
different. I will only discuss with you
today the underdamped case. So, I have a pendulum,
and the pendulum has mass M and length L.
And I will assume that the mass in the string is negligibly
small. I have damping,
and this is the case that gamma is smaller than omega zero.
This is the equilibrium position of the pendulum.
I will therefore call this position X, and I'm going to put
all the forces on this object in this picture,
which is MG. And, that is T.
And, there are no other forces on that mass.
Now, if we deal with small angles, as we have done before
more than once, then the tension,
T, is very close to MG. So, the only force that is
driving it back to equilibrium, the only restoring force,
then, is the horizontal component of T.
And so, there is this horizontal component.
That's the only one that we are concerned about.
So, the differential equation, then, in terms of Newton's
second law becomes MX double dot.
And then, we get minus T times the sine of theta minus B times
X dot. That's the damping,
and this is the restoring force due to the tension in the
string. And, the sine of theta is X
divided by L. So, I get MX double dot plus MG
divided by L times X. That is the sine of theta,
plus B times X dot equals zero. Make sure I have this M double
dot. I have the plus sign here.
I have MG X over L. That's fine.
So now, I divide M out, and I also use the shorthand
notation that omega zero squared is G divided by L.
These are shorthand notations which give you a little bit more
insight when you see the solutions.
And so, that gives me, now, the differential equation
that X double dot plus gamma times X dot plus I divide M out,
right, plus omega zero squared times X equals zero.
And, that is the differential equation that you'll recognize.
And, I would never want you to derive the solution to this
differential equation. That's, of course,
way too time consuming to do that during an exam.
The solutions to this are given on your formula sheet,
which will be part of your exam.
And so, let me write down here what that solution is.
So, X as a function of time is a certain amplitude times E to
the minus gamma over two times T.
And then, we have a cosine or a sine, cosine omega T plus some
phase angle alpha. If you knew the initial
conditions, then you can find what the amplitude is,
and you can find with the angle, alpha,
is. If you don't know the initial
conditions, then you do not know this.
There is another way that you can write this form,
which is sometimes better. And, I cannot tell you when it
is better and when it is not better.
It depends on the initial conditions.
But, I want you to appreciate that you can also write this,
for instance, as E to the minus gamma over
two times T, and then B times cosine omega T plus C times the
sine of omega T. From the physics point of view,
there is no difference. But from the math point of
view, there is a difference. You now have these as the two
adjustable constants depending upon the initial condition.
And sometimes, if you assume this,
it works faster than if you assume that.
And as I said, it really depends on the
initial conditions which goes faster, that they are very
similar, of course. Omega, which is now the
frequency with which this object, angular frequency,
is going to move, that omega is always lower than
omega zero. And, that shouldn't surprise
you because when there is damping, there is something that
is opposing the motion. And so, it shouldn't surprise
you that when you solve for omega, that omega squared is
omega zero squared minus gamma squared divided by four.
That is also something that we would probably give you on the
formula sheet because you will only find that if you substitute
this back into the differential equation.
Often, particularly French, likes to write down Q equals
omega divided by gamma. This is the quality omega zero
divided by gamma. And, if you do that,
then omega squared can also be written as omega zero squared
times one minus one over four Q squared.
And so, you see that if Q is for instance ten,
which is by no means absurdly high, that omega is only one
eighth of a percent lower than omega zero.
So, there that close. And, even when Q is two,
the difference is only 3% between omega and omega zero.
And so, what happens here is that the amplitude is decaying
at a time constant, a one over E time constant,
which is two divided by gamma seconds.
If you put in T, two divided by gamma,
the amplitude goes down by a factor of E.
Since energy is always proportional to amplitude
squared, the decay time of the energy is not two over gamma but
is one divided by gamma. Now, I'm going to make a
change. I'm going to drive this system.
This is no longer the equilibrium position,
but the equilibrium position was really here.
And, I am driving the top of this pendulum with a function,
eta, is eta zero times the cosine of omega T.
So, I am driving it now. So, this eta is in terms of
inches, millimeters. This is the motion of my hand.
And, this omega is no longer negotiable.
This omega has nothing to do with this omega.
This is the frequency with which the system likes to
oscillate. That, now, is the frequency
with which I want the system to oscillate.
They are totally unrelated. This is my will.
This is nonnegotiable. I can make this anything I
want. I can make it zero.
I can make it large. I can make it also that value,
of course. So, now, the equilibrium
position is not here.
And I will call, now, this distance from the
equilibrium position, [eta?].
You always use a U coordinate system, the equilibrium
position, as your zero. And so, the only thing that is
going to change now is that the sine of theta is no longer X
divided by L. But, the sine of theta is X
minus eta divided by L because, notice, if this is eta,
which is the position of my hand, then the sine of this
angle is now X minus eta divided by L.
In this, now, you had to carry through your
differential equations. And, what you'll see,
then, if you do that, that this now is not zero.
But this now becomes eta zero times omega zero squared times
the cosine of omega T. My omega, so,
instead of sine theta, X divided by L,
all you have to do is this. You know that eta is eta zero
cosine omega T. You carry that through,
and you will see that your differential equation now
changes. Notice that X double dot is an
acceleration. Notice that gamma times X dot
is an acceleration. Gamma is one over second,
and X dot is meters per second. This is an acceleration.
And, X dot is meters per second.
This is an acceleration. Notice that this is an
acceleration that has the dimension omega squared,
which is one second squared times meters.
That is an acceleration. Notice that this is an
acceleration. Eta zero is meters,
and omega squared is one divided by second squared.
So, these are apples. These are apples.
And these are apples.
So, the equation looks very kosher to me.
Now, the solutions: that's a different story.
That is something that I wouldn't want you to derive
either, but the solutions now become as follows.
I now get X as a function of T is an amplitude,
which is a very strong function of omega.
But, I will simply write it down as an amplitude,
A, times the cosine of my omega, T, minus some phase
angle, delta. And, this is what we call the
steady state solution. And this A becomes,
then, a rather complicated function.
Upstairs, I get the eta zero omega zero squared.
I get this part. And, downstairs,
I get the square root of omega zero squared minus omega
squared, squared, plus omega gamma squared.
And, that is the amplitude as a function of omega.
And the tangent of delta becomes omega gamma divided by
omega zero squared minus omega squared.
If we plot that function, A, as a function of omega,
so here's omega, and here I'm going to plot the
A. Then, I can recognize that if
omega goes to zero, that means if I move this very
slowly, that for sure the amplitude of this object must be
the same as my hand, must be eta zero.
If it takes me one week to go from here to here,
then of course the pendulum is always hanging below my hand.
So, when omega goes to zero, I can always check the result
here that A must go to eta zero. And, indeed,
if you put in omega equals zero, you will see that is the
case. If omega goes to infinity,
then A goes to zero. And then, there is a very
special case. That is when omega happens to
be omega zero, which is the frequency of the
system in the absence of damping.
This is my omega zero. It's not this one.
But it is the omega zero. Then, you'll get an amplitude
which is Q times eta zero. And, again, I just happened to
remember that. If you substitute this back in
here, this omega equals of omega zero.
You will see that. And, that's always very nice to
remember. At that omega zero,
you always get Q times more than your driver.
If Q is 100, you get 100 times the amplitude
of the driver. And so, here,
you get eta zero if omega is zero.
And then, when you reach that omega zero frequency,
you come out high. And then it goes back to zero.
And this point here is Q times eta zero.
And, we discussed before, I'll never make too much of a
deal out of that, that the actual maximum of this
curve is not at omega zero but is always a lower.
That is really more of an algebraic interest than it is a
physical interest because if gamma is low,
if Q is high, the two almost coincide.
So, the steady state solution has no adjustable constants.
In other words, the system has lost its memory
of what happened when we started driving it.
So, at T equals zero, if we know what X is,
and if we know what X dot is, and if we know what the driving
term is at T equals zero, then the general solution for
all times larger than zero is the sum of the trangent solution
and the steady state solution. And so, here you see the
[trangent?] solution. You can write it in this form,
or you can read it in this form.
And here you see the steady state solution.
This A has nothing to do with that A.
This A follows from the initial conditions just like the alpha,
this A does not follow from the initial conditions.
This A follows from eta zero, what my amplitude is of my
driver, and it follows from omega.
So, the two A's don't confuse the two.
And so, the general solution is, then, the sum of the two.
The trangent one will die out. If you weigh it a few times,
two over gamma, the trangent is gone.
And so, you would end up, then, only with the steady
state solution. And, you had some chance in
your homework to work with that. If we are not driving the
system, and if we have one object on a spring or on a
pendulum, or a floating object in liquid, then there is one
frequency, one normal mode frequency, one resonance
frequency, we call it also the natural frequency.
The moment that you couple oscillators, if you couple two,
you get two normal mode frequencies.
You get to resonance frequencies.
And if you have three objects, you get three normal mode
frequencies. And so, now I would like to
discuss with you the case whereby I couple two
oscillators. If I gave you on an exam three
coupled oscillators, that would be very nasty
because it's extremely time consuming.
If I gave you four coupled oscillators, that would be
criminal because you cannot finish that in 85 minutes.
So, two is certainly within reason.
Three is marginally within reason.
Four is out of the question. When we have coupled
oscillators, we always leave damping out.
And yet, we will learn a lot from it.
Even without damping, we will learn a lot.
And so, what I've chosen to do with you is a spring system with
two objects, two masses, and the springs have no mass,
negligible mass. The spring constants are K,
and the masses of the objects are M.
So, this is the equilibrium position of these two objects,
and the ends are fixed of the string.
I will introduce a shorthand notation, zero squared equals K
over M. Now, what I do,
I offset both these objects from equilibrium just as I did
that here. I always offset them in the
same direction, and I call that positive.
Is that a must? No.
Is that useful? Yes.
So, I offset them. And so, this position,
now this is object number one, and this is object number two.
This is now a distance, X1, away from equilibrium.
And this one is now at a position, X2,
away from its equilibrium. You always want to know what
the displacement is relative to its own equilibrium.
And, its own equilibrium for one is here, and equilibrium for
two is there. I now make the following
assumption that X2 is larger than X1.
Is that a must? No.
Is it useful? Yes, but if you want to assume
that X2 is smaller than X1, be my guest.
I will assume that X2 is larger than X1.
Now, follow me closely now. So, we have an object here,
and we have an object here, and this is now one spring.
This is the other spring, and this is the third spring.
It is immediately obvious that this spring is longer than it
wants to be. So, there is a force that
drives it back to equilibrium. If X2 is larger than X1,
this spring is also longer than it wants to be.
So, it wants to contract, so there is a force in that
direction. If this spring is longer than
it wants to be, it wants to contract.
So, there is a force in X2 in this direction.
This spring is clearly shorter than it wants to be.
So, it's pushing. So there is a force in this
direction. And so, now I'm going to write
down the differential equation first for object number one.
So I get M X1 double dot. So, this one is minus K times
X1. And, this one is plus K times
X2 minus X1. Is it a disaster if it turns
out that X2 is not larger than X1?
Not at all. This equation,
now, is correct for all situations.
The fact that I have assumed that X2 is larger than X1 gave
me a plus sign here, and so, my differential
equation is safe no matter what X1 is relative to X2.
So, you can always make that assumption.
And you don't have to worry later anymore about signs.
I can divide, now, M out, and then I get that
X1 double dot plus two omega zero squared times X1.
But you notice I have one here and I have one here.
They have both a minus sign. And then, I have here plus,
so that becomes minus omega zero squared times X2.
That is zero. So, that is my first
differential equation. So, now I go to the second one.
I get MX two double dot. Now I have two forces both in
the negative direction. First I have the one that
drives this one away from equilibrium.
So I get minus K times X2 minus X1.
And then, I have the force due to the fact that this one is
shorter than it wants to be by an amount, K2.
So, I get minus K times X2. It's shorter by an amount,
X2, than it wants to be. And so, I can divide M out,
and so I get MX2 double dot. And then, I get plus omega zero
squared times X2. And then, I get minus omega
zero squared times X1 equals zero.
Now, compare this one with this one.
Notice the incredible symmetry. I could have found this one by
changing a one to two here, and by changing the two to one.
Of course, the system is so symmetric that,
yes, thank you very much for the two right?
Yeah, thank you. The system is so symmetric that
clearly nature cannot make any distinction between one and two.
So, it is in this particular case because of the beautiful
symmetry, it is obvious that these two differential equations
look very, very similar. Now, I'm going to make an
important step. I'm going to substitute in here
X1 is C times cosine omega T, and X2, this is C1,
is C2 times cosine omega T. I want to know what the normal
mode frequencies are for this system.
I want to solve for omega. I want to find omega.
I'm not driving the system. Since I have no damping,
in which case at normal mode solutions, either of the two
objects are in phase with each other or they are out of phase
with each other. But if there are out of phase,
we can always take care of that with a minus sign.
So, I'm now going to substitute that in here.
And I may want to continue working on the centerboard,
in which case I might as well erase this so that it stays a
little compact.
So, I'm going to substitute this trial function into my
differential equations. And, every term will have a
cosine omega T, so I dump all the cosine omega
T's. So, I go to the equation,
which has a big one there. And so, that one becomes C1.
X1 double dot gives me a minus omega squared,
right? If I take cosine omega T and I
do the second derivative, I get minus omega squared in
front of it. So, I get C1 times two omega
zero squared. That is that two omega zero
squared. And then, I get minus omega
squared. And then I get plus,
not plus, then I get minus omega zero squared times C2.
And that equals zero. And then I go to my second
differential equation which is this one.
But I'm going to rearrange the C1's and the C2's.
So, I'm going to put the C1's here.
So, I'm going to get minus omega zero squared times C1.
And here, I'm going to get plus two omega zero squared minus
omega zero squared times C2. And that equals zero.
So, I can leave this plus out. So, what do I have here now?
I have here two equations with three unknowns.
I want to know what omega is in the normal modes,
but I also would like to know what C1 and C2 is.
Well, that is tough luck. You can't have it both ways.
We do not know at all what C1 and C2 is separately.
But you can always find in the normal modes,
without knowing anything else what the ratios of those
amplitudes, C1 and C2. And, you can always solve for
omega, and you will see how that works.
So, with two equations and three unknowns,
you can only solve for omega and for the ratio,
C1 over C2. So now, you may do that anyway
you want to. This is not so difficult to
solve this. I will, however,
use Kramer's rule, and I will introduce D,
which is the determinant of the following: two omega zero
squared minus omega squared. And then I have here minus
omega zero squared. And then, I have here minus
omega zero squared minus omega zero squared.
And this, now, must be zero if I'm not looking
for solutions, which I am not,
C1 is zero and C2 is zero. It's clear that if C1 and C2
are zero, the two equations are satisfied.
Zero is zero. But that's not an interesting
case. And, the only way that you can
get a solution which is interesting is value's for C1
and C2 which are not zero is by making this determinant zero.
And so, that means that D, which now becomes two omega
zero squared minus omega squared, squared,
minus omega zero to the fourth, you have to make that zero.
And when you do that, you find two values for omega.
You find that omega squared is two omega zero squared plus or
minus omega zero squared. Those are the two solutions.
And, when we evaluate those two solutions, you find a result
which is so embarrassingly simple that you could almost
have said that without any work, almost.
Notice that omega minus, which is the lowest one of the
two, is the same as omega zero because that's the minus sign.
What does that mean if omega minus is omega zero?
Well, it means, of course, that the two objects
are just oscillating like this. The inner spring is never
stretched, is never longer than it wants to be,
is never shorter than it wants to be, so each one is only
driven, so to speak, by the outer spring.
So, that's immediately obvious that that's a normal mode.
And, you can make a prediction that C1 over C2 must be plus
one. They must go in unison like
this. And you can confirm that by
substituting this omega one. If you substitute that in this
equation, you'll find the ratio see one over C2.
If you prefer this equation, be my guest.
You can do that too. And you'll find no matter which
one of the two you take that it is plus one.
Now, omega plus is less obvious, is the square root of
three times omega zero. That's when you take the plus.
Even though it's not obvious that it is the square root of
three, it is clear what is happening.
Here are the two objects, and now they are doing this.
They are exactly 180° out of phase with each other.
So, now you can make comfortably the prediction that
C1 over C2 is minus one. And indeed, if you substitute
the square at a free omega zero in either this or that equation,
this indeed comes out. So, the general solution now
for a given initial condition, so if now I give you the
initial condition, then the general solution will
be for X1 would be X0 minus, the minus makes reference to
that lowest frequency times the cosine omega minus T plus some
phase angle minus, all these minus signs make
reference to this mode. And then, I have plus X0 plus
times the cosine of omega plus T plus some phase angle plus.
Of course, if you prefer instead of cosine-signs,
that's fine, of course.
And, you have four adjustable constants: one,
two, three, four. And, if you know the initial
conditions, if you know what X1 is, and what X1 dot is,
and what X2 is, and what X2 dot is at time T
equals zero, you can solve for all four in principle.
There is no longer any freedom from number two because all four
adjustable constants have now been consumed.
And so, therefore, you can write down,
now, the solution for X2 by simply making this one a two.
And, this whole term here is the same.
The only difference is that this plus sign now becomes a
minus sign. Nothing else is different.
The frequencies are the same. Otherwise they wouldn't be
normal mode frequencies. And, it is the ratio here of
the C1 over C2 that is plus one. And, it is the ratio here that
is the minus one. That's the reason why this
becomes a minus, and why the plus here remains a
plus. So that, now,
is the general solution if you also know the initial
conditions. So now, we are going to drive
the system. So now, we go back to where we
were, and I'm going to drive this end.
And on going to drive this and with my hand.
This is eta, and eta equals eta zero times
cosine omega T. There is only one term in all
of this on the blackboard that is going to change.
And that is this term. This spring here on the left is
no longer shorter by the amount X1, but is shorter by the amount
X1 minus eta. And so, it's only that term.
It's only this minus KX1 that now have to be changed into
minus K times X1 minus eta. Nothing else changes.
But, that has major consequences,
of course. For one thing,
if you're going to substitute, now, these trial functions,
omega is no longer a negotiable.
Omega is my omega now. I set this omega.
You're not going to solve for that omega.
That would be an insult to me. I dictate what omega is.
So, that means I now get two equations with two unknowns,
C1 and C2. Omega is no longer unknown.
So, now I get solutions for C1 and they get solutions for C2.
And of all the equations on the blackboards that you have,
the one that is going to change is the first one.
And when you carry through, that eta, which is eta zero
cosine omega T, this zero here changes into eta
zero times omega zero squared. The cosine omega T is gone
because I've divided cosine omega T out in all those other
terms. So, now I use creamers rule,
and now I can actually come up with a solution for C1,
no longer ratio C1 over C2. No, I can actually come up now
with a solution. So, C1 now using Kramer's rule,
this now as my first column. Eta zero omega zero squared
zero, and this now becomes my second column minus omega zero
squared. And I get here two omega zero
squared minus omega squared divided by D.
If I pick a random value for omega, D is not zero.
D is only zero at those two resonance frequencies.
And then, C2 becomes, the first column is like this.
That is two omega zero squared minus omega squared minus omega
zero squared. And now, the second column
becomes eta zero omega zero squared, and a zero divided by
D. If I work C1 out a little
further, then I get C1 equals eta zero omega zero squared
times two omega zero squared minus omega zero squared divided
by D. And here, I get that C2,
this is zero. So, I get plus omega zero to
the fourth times eta zero divided by D.
And D, now, is this determinant but is not zero.
It's only zero for those two special frequencies.
If you want to see what the amplitudes now are as a function
of omega, it's a very, very interesting behavior,
then what helps is you go omega first to zero.
And then, you see what happens. And, it is not so intuitive
what happens. If you put in omega equals
zero, you will find that C1 is plus two thirds eta zero.
You can confirm that by substituting that into C1.
You will see that that's what happens.
And, you'll find that C2 is plus one third eta zero.
So, that is at omega equals zero.
And, when you go to infinity with frequencies,
then it should not surprise you that C1 is zero,
and that C2 is zero. And then, there is one case,
which is pathetic. And, that is the case that I
make omega squared, two omega zero squared.
At that frequency, C1 becomes zero,
but C2 is not zero. And, I spent almost a whole
lecture with you on that demonstrating that in three
different ways that, indeed, there is this bizarre
solution. So, when omega squared equals
two omega zero squared, then C1 becomes zero,
but C2 is not zero. And so, now you can make a plot
of C's as a function of omega. And during that lecture,
I showed you three of those plots.
And, I used the convention, then, that when the object is
moving in phase with the driver, I put it positive and out of
phase negative. And, I plot here C divided by
eta zero. And, I will use a color code.
I will do C1 in red, and I will do,
C1 is red and C2 I will do in white chalk.
So, here is zero omega. Here is omega zero,
which is my omega minus, and then my omega plus was at
three, the square root of three. So, that's about 1.7.
So, here's my omega plus. And then things go nuts.
That is when D goes to zero. So, C1 is two thirds here.
And then, it goes to infinity there.
And then, here, when it is 1.4 times omega
zero, the square root of two, then it goes to a zero and it
comes up here. And then, I get a curve here,
without being too precise. And for C2, I get one third.
And then, it goes up. And then, C2,
something like this. And then, I get something like
that.
Yeah, I can live with that. Now, we ignore damping,
and because we ignore damping, we get these on physical
infinities when you did the frequency omega minus and omega
plus, these resonance frequencies.
Now, if you include damping, then the solutions become
extremely complicated. But, of course you avoid the
infinity values. But, the resonant amplitudes
can still be very high. And so, these plots are still
very useful provided that you don't interpret infinities as
being real, but something that is very large.
If Q is high, then of course the amplitudes
are enormously high. And, this can lead to
distraction. We've seen the movie of the
Tacoma Bridge, and then we have seen a
dramatic experiment of the breaking wineglass.
You remember that wineglass demonstration.
Because of the catastrophic success, to use Bush's words,
of that demonstration, students have asked me for an
encore. They would like to see it
again. And that's a very reasonable
thing to do. It fits very nicely into this
concept of coupled oscillators. A wineglass would have a huge
number of oscillators coupled, and a wineglass,
I have one here, can be made to oscillate easily
in its lowest normal mode. I have to wash my hands to make
you listen to it. So, because my hands are now a
little greasy from the chalk, and this is the frequency.
That is the lowest mode. It is a circle like this
looking from above, it becomes an ellipse,
and then it becomes an ellipse like this, and an ellipse like
this, and an ellipse like that. And if now you dry that with
sound at exactly that frequency and you put in enough power,
another way of saying it is as if you make eta zero large
enough, then the system can break.
It's making a lot of noise. I'll warn you.
Those were sitting here, I really think you should
protect your ears. The sound can be deafening.
So, be careful, and those who are a little bit
further away, make sure that you close your
ears. I have the luxury that I can
turn my hearing aids off, but without my hearing aids,
believe me, I can still hear a lot.
So, I also will have to protect myself.
So, one hearing aid is off. The other is off,
and I'm going to put this on anyway.
I'm going to stroke this glass with a frequency which is only
slightly different from this frequency of the sounds.
That is about 427 Hz, which you can very easily hear.
And then you'll see the glass in slow motion because of the
fact that the stroboscopic light has a slightly different
frequency. That should be coming up now.
And you say it will.
They do very much. I'm going to make it very dark
now because this is important that you can see this very well.
So, we'll make it completely dark in the room.
I'm going to protect my ears. I hope you can still hear me.
I can hardly hear myself. And I'm going to drive the
system, now, at very low amplitudes, at a frequency close
to its resonance. [SOUND PLAYS]
You can already see that the glass is moving.
I'm going to increase the amplitude.
You see it moving? I can go a little bit over the
resonance and under the resonance.
You will see that it's not moving then.
So, I do that purposely now. Now I'm off resonance.
I'm over it. And now I'm off resonance.
I'm under resonance. I am below.
I'm now at 423 Hz when the resonance is at 427.
So, I'm going to put it back at 427.
And now I'm going to increase the sound.
So I warn you. And there it goes.
And that's it. It broke.
And it broke very fast. [APPLAUSE] So this is an ideal
moment for a break. Since you had such a good time
with the breaking glass, let's settle for four minutes.
We will reconvene in four minutes.
[SOUND OFF/THEN ON] OK, I now want to discuss with
you continuous media. If you have N coupled
oscillators, then you get N normal modes.
When you make N infinitely high, then you get continuous
media. And, if we have one dimensional
continuous media like a string, or a pipe with air,
sound, then we have studied the transverse motion of a string,
and then we derived that now you have to use the wave
equation. We derive the wave equation.
And then you get the D2Y DX squared is one over V squared
times D2Y DT squared, Y being now the displacement
away from equilibrium in this position if X is in this
direction. And, that is fine.
And so, excuse me? It's amazing how you can look
at it and think that's fine, and it is not.
All right, so a V in the case of a string is the square root
of T over mu. That was just a special case
for the string. We also examined longitudinal
motion, again, 1D.
Sound, so longitudinal wave, and if you deal with pressure,
you think of sound as being a pressure wave.
Then you get D2P DX squared equals one over V squared times
D2P DT squared. P, then, being positive would
be overpressure over and above the ambient 1 atmosphere.
And if it's negative, then it is below.
So it's not the total pressure, but it is the overpressure.
And V, then, is the speed of sound,
which in air room temperature is about 340 m per second.
If you prefer not to work and pressure but in terms of the
actual position of the air molecules in analogy with the
position of the string, then you can write down the
same equation in terms of psi or D2 psi DX squared one over V
squared D2 psi DT squared. That, then, gives you the
position, the actual motion of the air molecules.
I often prefer the pressure, and I will follow the also
today. So, there are an infinite
number of normal modes. And the ratios of the
amplitudes of two adjacent oscillators, they are coupled
now, reflects itself, of course, in terms of the
overall shape, which you can best see when you
deal with a string.
As I said, it's easiest to see with transfers,
oscillations, what the displacements look
like. It's harder to see that with
sound. I also discussed with you a
special situation that I connected to media,
medium one and medium two, and I gave these to media
different masses per unit length.
When you set up a traveling wave on a string,
or you set it up, you can set up a pulse,
it reflects at the end. And, how it reflects depends on
the boundary conditions at the end.
And when you connect them to media, then you get not only
your reflection, but you also get some of the
pulse. Some of the wave goes into
medium to. And so, let's assume that we
have an incident wave coming in like this.
And we have here mu one, and we have here V1 given by
this equation. They both have the same
tangent, T, and here you have mu two and you have V2.
We discussed that. We used the wave equation to
solve what happens when we have an incident harmonic wave coming
in. We derived even this speed of
propagation using the wave equation.
None of this came out of the blue.
We always derived that. And then, we found,
by using the boundary conditions at the junction,
namely that the string is not breaking, and that DYDX on the
left side is the same as DY DX on the right side.
[UNINTELLIGIBLE] boundary conditions, we found that the
amplitude of the reflected wave, and the same would hold for a
pulse, divided by the amplitude of the incident wave was V2
minus V1 divided by V1 plus V2. And, I call that R
reflectivity. And, we found that the
amplitude of the transmitted wave or pulse divided by the
amplitude of the incident one was 2V2 divided by V1 plus V2.
And, I called that shorthand notation transmitivity.
And when we had done that, I put in some simple test cases
where our intuition is very good.
The first thing I did was I said, suppose mu two was
infinitely high. So, mu two is infinitely high.
In other words, that medium two is a wall.
That means, number one, the string number one cannot
move. It's fixed at the end.
So, that means V2 is zero. And, we go to this equation and
we find that R equals minus one. V2 is zero.
You get minus V1 over V1. And, we like that because what
it means is that when a mountain rolls in, it comes back as a
valley. And then, a valley rolls in.
It comes back as a mountain, and I demonstrated that with
strings. It was very pleasing that T of
R is then zero. Well, it better be zero,
right? If everything comes back at you
upside down, but nevertheless, everything comes back at you.
You expect that nothing goes into the wall.
And, you see when V2 is zero that TR is zero.
And, we were all very happy. And we could all sleep.
But then, and you guessed it, then I said,
let's suppose mu two becomes zero.
So, I attach that string to nothing, to empty space.
Is that practical? Yes.
I can do it. I can take a nickel wire and I
have here a magnetic field very strong, and I can pull on that
nickel wire. And, the end of the nickel wire
ends up in nothing, in empty space.
But, the I keep the tension on. So, it's completely practical.
It can be done. It's not just nonsense.
So, mu two goes to zero. That means that V2 goes to
infinity. And then, we looked at R,
and we say, well, if V2 goes to infinity,
then R equals plus one. And, we were all very happy.
A mountain comes back as a mountain, and I even
demonstrated that. We were still able to sleep at
that point. But then, then came the awful
thing that if we substitute V2 equals infinity in this
equation, that TR becomes plus two.
And now, we can no longer sleep because this is absurd.
All the energy that rolls in comes back.
But there is something in addition that goes into that
second medium. Now, admit it.
Who could not sleep that night? You should all fail this
course, by the way. [LAUGHTER] but in any case,
I don't have to feel guilty, right?
Who thought about this and said, there is something weird?
I must find an explanation for that.
In my case, I had to find an explanation because I couldn't
sleep. Who found an explanation?
Who could say, oh yes, don't worry about it?
What was your solution? Excuse me?
Very good. That's a very nice way of
looking at it. So that's probably why you
could sleep. Well, actually I'll tell you
why I couldn't sleep. But your solution is even
shorter. But the reason why I want to
show you what I'm going to show you is that I want to also
expose you to the idea which you already alluded to,
namely that there is energy involved when we deal with
pulses and when we deal with waves.
You remember when we have a traveling wave that the total
energy per wavelength lambda, I only did it per wavelength
that the total energy, perhaps you remember that,
equals 2A squared times pi squared times the tension,
T, divided by lambda. A was the amplitude.
Energy is always proportional to amplitude squared.
This was the tension, and this was the wavelength.
And, if V2 goes to infinity, then the wavelength goes to
infinity. That's obvious,
right? If something moves with the
speed of light even faster, infinity is even faster than
the speed of light, then lambda goes to infinity.
So, this goes to zero. And so, we came to the same
conclusion, some decadent solution, TR equals plus two,
has no meaning because there is no energy in there.
And so, I was able to sleep. Let's now turn to normal modes
of continuous media. And I suggest that we go
longitudinal because we did so much transfer stuff.
Let's go longitudinal. I have here a pi has length L,
and it is open here, and it is open there.
That means the overpressure, under pressure here,
can never build up. It's connected to the universe.
So, at these two boundary conditions, the speed that I
have there must be zero. If I write down the general
equation for a standing wave, because normal mode solutions
are standing waves, I can write down P equals some
amplitude times the sine or the cosine.
Let me take a sine, two pi divided by lambda times
X. I'll be very general.
I will introduce some phase angle, alpha,
for which I will find a solution very shortly.
And then, cosine omega T, or sine omega T if you prefer
that. This is a standing wave in very
general terms. Everything here is the space.
X, this is X, and here, all the information
here deals with time, which is typical for a standing
wave. [NOISE OBSCURES] always to
write down for two pi over lambda K.
And I must observe the situation that P must be zero at
X equals zero, and also at X equals L.
If P is zero at X equals zero, immediately you see that alpha
is zero. So I'm going to rewrite it now.
I'm going to write down P, P zero times the sine of KX.
So I'm going to replace this by K.
I know that alpha is zero times the cosine of omega T.
And now, I must meet the boundary condition that when X
equals L, that P is, again, zero.
And that, now, breaks open a whole spectrum of
possibilities in which I introduce this normal mode
number N as in Nancy whereby N can be one, two,
three, etc. And then I get solutions when K
of N equals N pi divided by L. You see that immediately
because if I make X now L, then I get the sine of M pi no
matter what N is. I always get zero.
And so, my lambda of N, which is two pi divided by K is
then 2L divided by N. So, I can rewrite now this
equation as P zero times the sine of N pi X divided by L.
And now, I have cosine omega NT because omega N are now the
frequencies which are associated with the N'th mode,
N being N as in Nancy. What now is the connection
between this omega N and this K? Well, that connection you will
find through the wave equation. You now have to substitute this
result back into the wave equation, which is this one's to
solve or omega. And, I want to do that with
you. It is not that much work.
I go to 2PDX squared. So, here it is:
D2P DX squared. So, all I get is I get this out
twice. So, I get N squared pi squared
over L squared. I get a minus sign because if I
take twice the derivative, I always end up with a minus
sign. But the sine comes back as a
sine. And, not only does the sine
come back, but all the rest comes back.
And so, I will just write P here.
So, this P is exactly that P. So, this is the only thing that
was added by taking the second derivative against X.
Now, what is D2PDT squared? Now I have to go to this
function. [I have?] partial derivatives.
X is constant here, but T is the one that is
changing, whereas here, we had that X was changing.
The T was constant. So now, we are going to get
minus omega N squared, again, times P.
The whole function comes back. And now, we are in business
because now the wave equation will tell us the connection
between the two. It tells us that N squared
times pi squared divided by L squared is now one over V
squared. That is the one over V squared
that I had there. And, there's a minus sign here.
And there's a minus sign here times omega N squared.
So, this is not connected. And so, you see the solution
for omega N just is being presented to you [NOISE
OBSCURES] is now N pi times V divided by L.
That follows from the wave equation.
And so, if you prefer the frequency in hertz,
then you have to divide this by two pi.
So, you get NV divided by 2L. So, if I want to plot now,
so this is X equals zero, and this X equals L,
I can plot now here the pressure in terms of this
overpressure or under pressure, P.
And, you get a curve which looks very similar to a
transverse solution. But, of course,
it is not transverse. It's really longitudinal.
But you get, then, that for N equals one,
you would get this mode. There must be a pressure node
here and there because we can never build a pressure that's
connected with the universe. You can never build that
overpressure. And so, the pressure buildup
here is positive. And then, later in time it will
be negative, and then positive, and then negative.
And, when you go to N equals two, then you get another node
in pressure here. And so, now you get this,
always a pressure node here, always a pressure node there.
But now you end up with another pressure node there.
And, if you have any difficulties to see what the air
molecules are doing, I would recommend you go back
to psi space which is the actual position of the molecules.
And when you do that, you will always find that where
the pressure has an anti-node, which is here,
psi always has a node, and where the pressure has as
node, psi always has an anti-node.
Of course, the molecules can freely move in and out.
There is no problem. So, the molecules can freely
move in and out here. So, where psi has its largest
amplitude, its anti-node, that is where the pressure
cannot build up. But going to psi space actually
often helps me to see precisely what is going on with the motion
of the molecules. I have here a linear system,
which is a sound cavity. It's made of magnesium.
[There's?] no air, but it is magnesium.
And, it is open, open on both sides.
You can't have it any better. I'll make a drawing for you.
So, here is my magnesium. This is a rod.
It's one-dimensional, and the length is 122 cm and
the speed of sound in magnesium is about 5,000 m per second.
When I hit that magnesium rot on the side, it wants to go into
standing waves. It prefers the lowest mode.
It almost always does. But it may also create second
and third harmonics. And so, the lowest mode,
F1, is then V divided by 2L. So, the lowest frequency,
F1, is V divided by 2L, which when I calculate it,
is about 250 Hz. And the actual value we measure
is about 2,044. And, there may also be,
when I hit it with a hammer, there may also be some that is
twice as high. So that may be 4,100 Hz.
That would be double the frequency.
That would be the second mode. And this is quite remarkable.
So this is not filled with air, but this is filled with
magnesium. But, by exciting it here,
it's like blowing on the flute there.
It goes into these normal mode solutions, and it is this one
that you will hear loud and clear.
It's a beautiful tone. And this one you may hear in
the beginning, but the higher harmonics often
die out faster than the lower harmonics.
So, are you ready for this? Open-open.
It's open on both sides, and I just bang the hell out of
it. 2,044.
Isn't that beautiful? It's oscillating like this.
Exactly oscillating the way that I derived for you for air.
And here, you see that it [holds for?].
Only the fundamental is there. OK, I wish you luck on
Thursday. I'll see you then.