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I want to start with the
physical pendulum,
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which is exactly the same one
that I discussed during the
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00:00:40,310 --> 00:00:44,490
first lecture.
This is a hoop with mass M and
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radius R.
And, we were calculating the
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period of this hoop as it
oscillates.
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And we did that using the
famous stork equation from 8.01,
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the torque relative to point P
is the moment of inertia
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00:01:01,969 --> 00:01:05,579
relative to point,
P, times the angular
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acceleration,
alpha.
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Today, I will do this again,
but I will use the conservation
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of energy to show you that in
case there is no damping,
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when mechanical energy is
conserved, but you can find the
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correct to fragile equations
through the conservation of
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energy.
If the thing is swinging,
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in general there are two
components.
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You have kinetic energy and you
have potential energy.
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And the kinetic energy,
K, is one half times the moment
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of inertia about point P times
omega squared.
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You remember that from 8.01.
And, this omega is theta dot.
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We call that the angular
velocity.
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This angular velocity changes
with time.
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When the object goes through
equilibrium, to a new or
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velocity is at maximum.
But when object comes to a
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halt, the angular velocity is
zero.
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Do not confuse this omega with
omega zero.
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I will give that a zero,
now, to distinguish it from
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omega, which is the angular
frequency.
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The angular frequency is a
constant of the motion,
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and that is two pi divided by T
zero if T zero is the period of
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oscillation.
So, this is the kinetic energy,
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and this is the square of the
angular velocity.
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And then, we have potential
energy.
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Let this be point A,
and when we are here,
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the center of mass is at point
B.
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And so, the potential energy,
U, or actually I should say the
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potential energy at point B
minus the potential energy at
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point, A, it's always the
difference in potential energy
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that matters.
That equals MGH,
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H being the difference in
height between point B and point
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A.
So, this here is MGH,
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Massachusetts General Hospital.
That's the way to remember it.
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Now, H is very easy.
H is the same as R times one
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minus the cosine of theta.
We went through that many
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times, so you can easily confirm
that.
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R is the radius of this circle.
This is the radius.
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Now, for small angles,
the cosine of theta is theta
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squared divided by two.
And so, I can rewrite this
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differential equation now,
that E total.
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I see nothing wrong with that.
I'm sorry.
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Is there anything wrong with
it?
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The cosine theta alone,
you're right,
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is one minus theta squared over
two.
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Thank you very much.
Thank you.
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So, I'm going to write it now
as a total energy is one half IP
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times theta dot squared.
And then, I get plus MGR times
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theta squared over two.
And, what you do now,
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since this is a constant of the
motion, you always do that if
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you work with the conservation
of energy.
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You take the time derivative of
this equation,
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which must be zero because
mechanical energy is conserved.
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And, out pops the differential
equation that we also found
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during my lecture number one
when I used the different
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method.
So, I take the time derivative.
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So, this [T?] each of this
half, and so we get I if P times
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theta dot times theta double
dot.
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I had used the chain rule.
And then, I get plus MGR,
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these two, each of these two,
and so I get theta times theta
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dot, and now this equals zero
because DE DT is zero.
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And whenever you do that,
you will always see that the
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theta dot term disappears.
Or if you have the equation in
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X, then the X dot term
disappears.
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And, you see that.
This term goes.
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You can divide it out.
And so, your differential
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equation takes on,
now, a very familiar form.
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Let me write these two here.
So, I get theta double dot plus
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MGR divided by I of P times
theta equals zero.
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And this differential equation
you should recognize.
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The solution to that equation
is immediately obvious.
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Omega zero squared,
I use now the omega zero equals
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MG times R divided by I of P.
And so, the general solution
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for this oscillation then
becomes that theta is theta
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zero.
That is the amplitude times the
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cosine or the sine,
if you prefer that,
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omega zero T plus some phase
angle, phi.
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That is the general solution.
And if you knew the initial
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conditions, what the situation
was at T equals zero,
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if you know the velocity,
the angular velocity at T
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equals zero, and if you know
where it is at T equals zero,
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you can solve for this theta
zero, and you can solve for this
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phi.
But, omega zero is independent
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of the initial conditions.
Very well.
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Now, I would like to cover a
case whereby I am going to
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introduce damping.
Whenever we deal with damping,
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there are two terms that are
important in physics.
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That is the speed itself.
That is a damping term which
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opposes the velocity,
which is when you're
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proportional with the speed.
And then there is a damping
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term which opposes the velocity,
which is proportional with the
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square of the speed.
We will always leave the square
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out with 8.03 because the
differential equations become
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impossible to solve.
Maybe numerically you can do
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it, but not analytically.
However, if you're interested
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in all the physics,
which is wonderful,
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with a V squared and V,
my lecture number 12 on OCW
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Open Courseware from 1999,
Newtonian mechanics,
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I deal with the V squared and
with the V term,
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and I do many demonstrations to
show you that there are certain
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domains where the V term is
important .
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We call that the viscous term,
and there are certain domains
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in physics where the V squared
term is important.
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So, I will now simply restrict
myself, then,
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to the damping force,
which is linearly proportional
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with the velocity,
and we will write it down as F
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equals minus B times V.
We will use a shorthand
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notation that gamma equals B
over M.
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That is only a shorthand
notation.
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I will erase this.
We don't need this anymore.
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We need so many blackboards
today that I'll use this
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blackboard from a damping
problem.
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When you deal with damping,
we recognize three different
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domains.
One domain, whereby gamma is
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smaller than the omega zero,
we call that underdamped.
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Then we have a domain whereby
gamma is larger than omega zero.
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We call that overdamped.
And then, you have a very
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special case where gamma equals
of omega zero.
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[SOUND OFF/THEN ON]
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And the behavior of these three
different kinds is very
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different.
I will only discuss with you
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today the underdamped case.
So, I have a pendulum,
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and the pendulum has mass M and
length L.
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And I will assume that the mass
in the string is negligibly
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small.
I have damping,
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and this is the case that gamma
is smaller than omega zero.
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This is the equilibrium
position of the pendulum.
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I will therefore call this
position X, and I'm going to put
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all the forces on this object in
this picture,
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which is MG.
And, that is T.
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And, there are no other forces
on that mass.
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Now, if we deal with small
angles, as we have done before
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more than once,
then the tension,
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T, is very close to MG.
So, the only force that is
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driving it back to equilibrium,
the only restoring force,
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then, is the horizontal
component of T.
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And so, there is this
horizontal component.
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That's the only one that we are
concerned about.
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So, the differential equation,
then, in terms of Newton's
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second law becomes MX double
dot.
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And then, we get minus T times
the sine of theta minus B times
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X dot.
That's the damping,
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and this is the restoring force
due to the tension in the
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string.
And, the sine of theta is X
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divided by L.
So, I get MX double dot plus MG
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divided by L times X.
That is the sine of theta,
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plus B times X dot equals zero.
Make sure I have this M double
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dot.
I have the plus sign here.
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I have MG X over L.
That's fine.
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So now, I divide M out,
and I also use the shorthand
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notation that omega zero squared
is G divided by L.
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00:12:20,769 --> 00:12:25,461
These are shorthand notations
which give you a little bit more
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insight when you see the
solutions.
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And so, that gives me,
now, the differential equation
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that X double dot plus gamma
times X dot plus I divide M out,
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right, plus omega zero squared
times X equals zero.
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And, that is the differential
equation that you'll recognize.
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00:12:47,974 --> 00:12:52,846
And, I would never want you to
derive the solution to this
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00:12:52,846 --> 00:12:56,264
differential equation.
That's, of course,
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way too time consuming to do
that during an exam.
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The solutions to this are given
on your formula sheet,
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which will be part of your
exam.
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And so, let me write down here
what that solution is.
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So, X as a function of time is
a certain amplitude times E to
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the minus gamma over two times
T.
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00:13:27,701 --> 00:13:35,623
And then, we have a cosine or a
sine, cosine omega T plus some
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phase angle alpha.
If you knew the initial
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00:13:40,716 --> 00:13:44,522
conditions, then you can find
what the amplitude is,
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and you can find with the
angle, alpha,
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is.
If you don't know the initial
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conditions, then you do not know
this.
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00:13:52,507 --> 00:13:56,164
There is another way that you
can write this form,
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which is sometimes better.
And, I cannot tell you when it
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00:14:00,343 --> 00:14:04,000
is better and when it is not
better.
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00:14:04,000 --> 00:14:06,827
It depends on the initial
conditions.
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00:14:06,827 --> 00:14:11,383
But, I want you to appreciate
that you can also write this,
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00:14:11,383 --> 00:14:14,604
for instance,
as E to the minus gamma over
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two times T, and then B times
cosine omega T plus C times the
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00:14:19,317 --> 00:14:23,009
sine of omega T.
From the physics point of view,
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00:14:23,009 --> 00:14:26,858
there is no difference.
But from the math point of
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00:14:26,858 --> 00:14:32,209
view, there is a difference.
You now have these as the two
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00:14:32,209 --> 00:14:36,552
adjustable constants depending
upon the initial condition.
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00:14:36,552 --> 00:14:39,066
And sometimes,
if you assume this,
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00:14:39,066 --> 00:14:42,038
it works faster than if you
assume that.
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00:14:42,038 --> 00:14:44,933
And as I said,
it really depends on the
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00:14:44,933 --> 00:14:49,123
initial conditions which goes
faster, that they are very
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00:14:49,123 --> 00:14:52,323
similar, of course.
Omega, which is now the
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00:14:52,323 --> 00:14:56,209
frequency with which this
object, angular frequency,
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00:14:56,209 --> 00:14:59,866
is going to move,
that omega is always lower than
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00:14:59,866 --> 00:15:04,056
omega zero.
And, that shouldn't surprise
200
00:15:04,056 --> 00:15:08,244
you because when there is
damping, there is something that
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00:15:08,244 --> 00:15:12,063
is opposing the motion.
And so, it shouldn't surprise
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00:15:12,063 --> 00:15:16,104
you that when you solve for
omega, that omega squared is
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00:15:16,104 --> 00:15:20,071
omega zero squared minus gamma
squared divided by four.
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00:15:20,071 --> 00:15:24,478
That is also something that we
would probably give you on the
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00:15:24,478 --> 00:15:29,033
formula sheet because you will
only find that if you substitute
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00:15:29,033 --> 00:15:33,000
this back into the differential
equation.
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00:15:33,000 --> 00:15:37,389
Often, particularly French,
likes to write down Q equals
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00:15:37,389 --> 00:15:41,618
omega divided by gamma.
This is the quality omega zero
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00:15:41,618 --> 00:15:44,571
divided by gamma.
And, if you do that,
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00:15:44,571 --> 00:15:49,279
then omega squared can also be
written as omega zero squared
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00:15:49,279 --> 00:15:52,391
times one minus one over four Q
squared.
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00:15:52,391 --> 00:15:55,982
And so, you see that if Q is
for instance ten,
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00:15:55,982 --> 00:16:00,531
which is by no means absurdly
high, that omega is only one
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00:16:00,531 --> 00:16:05,000
eighth of a percent lower than
omega zero.
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00:16:05,000 --> 00:16:08,818
So, there that close.
And, even when Q is two,
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00:16:08,818 --> 00:16:13,399
the difference is only 3%
between omega and omega zero.
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00:16:13,399 --> 00:16:18,321
And so, what happens here is
that the amplitude is decaying
218
00:16:18,321 --> 00:16:22,224
at a time constant,
a one over E time constant,
219
00:16:22,224 --> 00:16:25,363
which is two divided by gamma
seconds.
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00:16:25,363 --> 00:16:28,503
If you put in T,
two divided by gamma,
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00:16:28,503 --> 00:16:33,000
the amplitude goes down by a
factor of E.
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00:16:33,000 --> 00:16:37,605
Since energy is always
proportional to amplitude
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00:16:37,605 --> 00:16:43,680
squared, the decay time of the
energy is not two over gamma but
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00:16:43,680 --> 00:16:48,384
is one divided by gamma.
Now, I'm going to make a
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00:16:48,384 --> 00:16:52,108
change.
I'm going to drive this system.
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00:16:52,108 --> 00:16:56,223
This is no longer the
equilibrium position,
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00:16:56,223 --> 00:17:00,535
but the equilibrium position
was really here.
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00:17:00,535 --> 00:17:06,218
And, I am driving the top of
this pendulum with a function,
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00:17:06,218 --> 00:17:12,000
eta, is eta zero times the
cosine of omega T.
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00:17:12,000 --> 00:17:16,806
So, I am driving it now.
So, this eta is in terms of
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00:17:16,806 --> 00:17:21,518
inches, millimeters.
This is the motion of my hand.
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00:17:21,518 --> 00:17:25,193
And, this omega is no longer
negotiable.
233
00:17:25,193 --> 00:17:30,000
This omega has nothing to do
with this omega.
234
00:17:30,000 --> 00:17:33,493
This is the frequency with
which the system likes to
235
00:17:33,493 --> 00:17:36,027
oscillate.
That, now, is the frequency
236
00:17:36,027 --> 00:17:38,835
with which I want the system to
oscillate.
237
00:17:38,835 --> 00:17:41,780
They are totally unrelated.
This is my will.
238
00:17:41,780 --> 00:17:45,068
This is nonnegotiable.
I can make this anything I
239
00:17:45,068 --> 00:17:46,712
want.
I can make it zero.
240
00:17:46,712 --> 00:17:50,136
I can make it large.
I can make it also that value,
241
00:17:50,136 --> 00:17:52,465
of course.
So, now, the equilibrium
242
00:17:52,465 --> 00:17:55,000
position is not here.
243
00:17:55,000 --> 00:18:01,000
244
00:18:01,000 --> 00:18:05,034
And I will call,
now, this distance from the
245
00:18:05,034 --> 00:18:07,661
equilibrium position,
[eta?].
246
00:18:07,661 --> 00:18:12,540
You always use a U coordinate
system, the equilibrium
247
00:18:12,540 --> 00:18:17,512
position, as your zero.
And so, the only thing that is
248
00:18:17,512 --> 00:18:23,048
going to change now is that the
sine of theta is no longer X
249
00:18:23,048 --> 00:18:26,800
divided by L.
But, the sine of theta is X
250
00:18:26,800 --> 00:18:31,867
minus eta divided by L because,
notice, if this is eta,
251
00:18:31,867 --> 00:18:36,933
which is the position of my
hand, then the sine of this
252
00:18:36,933 --> 00:18:42,000
angle is now X minus eta divided
by L.
253
00:18:42,000 --> 00:18:46,918
In this, now,
you had to carry through your
254
00:18:46,918 --> 00:18:52,072
differential equations.
And, what you'll see,
255
00:18:52,072 --> 00:18:57,576
then, if you do that,
that this now is not zero.
256
00:18:57,576 --> 00:19:04,486
But this now becomes eta zero
times omega zero squared times
257
00:19:04,486 --> 00:19:09,287
the cosine of omega T.
My omega, so,
258
00:19:09,287 --> 00:19:12,952
instead of sine theta,
X divided by L,
259
00:19:12,952 --> 00:19:18,500
all you have to do is this.
You know that eta is eta zero
260
00:19:18,500 --> 00:19:22,264
cosine omega T.
You carry that through,
261
00:19:22,264 --> 00:19:27,316
and you will see that your
differential equation now
262
00:19:27,316 --> 00:19:31,103
changes.
Notice that X double dot is an
263
00:19:31,103 --> 00:19:34,047
acceleration.
Notice that gamma times X dot
264
00:19:34,047 --> 00:19:37,131
is an acceleration.
Gamma is one over second,
265
00:19:37,131 --> 00:19:40,987
and X dot is meters per second.
This is an acceleration.
266
00:19:40,987 --> 00:19:43,160
And, X dot is meters per
second.
267
00:19:43,160 --> 00:19:46,385
This is an acceleration.
Notice that this is an
268
00:19:46,385 --> 00:19:49,820
acceleration that has the
dimension omega squared,
269
00:19:49,820 --> 00:19:52,624
which is one second squared
times meters.
270
00:19:52,624 --> 00:19:55,849
That is an acceleration.
Notice that this is an
271
00:19:55,849 --> 00:19:58,092
acceleration.
Eta zero is meters,
272
00:19:58,092 --> 00:20:03,000
and omega squared is one
divided by second squared.
273
00:20:03,000 --> 00:20:06,814
So, these are apples.
These are apples.
274
00:20:06,814 --> 00:20:10,628
These are apples.
And these are apples.
275
00:20:10,628 --> 00:20:14,643
So, the equation looks very
kosher to me.
276
00:20:14,643 --> 00:20:19,059
Now, the solutions:
that's a different story.
277
00:20:19,059 --> 00:20:24,178
That is something that I
wouldn't want you to derive
278
00:20:24,178 --> 00:20:30,000
either, but the solutions now
become as follows.
279
00:20:30,000 --> 00:20:35,861
I now get X as a function of T
is an amplitude,
280
00:20:35,861 --> 00:20:40,957
which is a very strong function
of omega.
281
00:20:40,957 --> 00:20:47,073
But, I will simply write it
down as an amplitude,
282
00:20:47,073 --> 00:20:53,571
A, times the cosine of my
omega, T, minus some phase
283
00:20:53,571 --> 00:20:58,922
angle, delta.
And, this is what we call the
284
00:20:58,922 --> 00:21:05,293
steady state solution.
And this A becomes,
285
00:21:05,293 --> 00:21:09,517
then, a rather complicated
function.
286
00:21:09,517 --> 00:21:15,189
Upstairs, I get the eta zero
omega zero squared.
287
00:21:15,189 --> 00:21:19,051
I get this part.
And, downstairs,
288
00:21:19,051 --> 00:21:25,568
I get the square root of omega
zero squared minus omega
289
00:21:25,568 --> 00:21:30,637
squared, squared,
plus omega gamma squared.
290
00:21:30,637 --> 00:21:38,000
And, that is the amplitude as a
function of omega.
291
00:21:38,000 --> 00:21:47,226
And the tangent of delta
becomes omega gamma divided by
292
00:21:47,226 --> 00:21:53,718
omega zero squared minus omega
squared.
293
00:21:53,718 --> 00:22:02,432
If we plot that function,
A, as a function of omega,
294
00:22:02,432 --> 00:22:10,291
so here's omega,
and here I'm going to plot the
295
00:22:10,291 --> 00:22:14,932
A.
Then, I can recognize that if
296
00:22:14,932 --> 00:22:19,887
omega goes to zero,
that means if I move this very
297
00:22:19,887 --> 00:22:25,651
slowly, that for sure the
amplitude of this object must be
298
00:22:25,651 --> 00:22:30,000
the same as my hand,
must be eta zero.
299
00:22:30,000 --> 00:22:34,325
If it takes me one week to go
from here to here,
300
00:22:34,325 --> 00:22:39,754
then of course the pendulum is
always hanging below my hand.
301
00:22:39,754 --> 00:22:45,000
So, when omega goes to zero,
I can always check the result
302
00:22:45,000 --> 00:22:49,049
here that A must go to eta zero.
And, indeed,
303
00:22:49,049 --> 00:22:54,202
if you put in omega equals
zero, you will see that is the
304
00:22:54,202 --> 00:22:57,055
case.
If omega goes to infinity,
305
00:22:57,055 --> 00:23:02,194
then A goes to zero.
And then, there is a very
306
00:23:02,194 --> 00:23:05,880
special case.
That is when omega happens to
307
00:23:05,880 --> 00:23:09,655
be omega zero,
which is the frequency of the
308
00:23:09,655 --> 00:23:12,463
system in the absence of
damping.
309
00:23:12,463 --> 00:23:15,974
This is my omega zero.
It's not this one.
310
00:23:15,974 --> 00:23:20,714
But it is the omega zero.
Then, you'll get an amplitude
311
00:23:20,714 --> 00:23:25,630
which is Q times eta zero.
And, again, I just happened to
312
00:23:25,630 --> 00:23:30,553
remember that.
If you substitute this back in
313
00:23:30,553 --> 00:23:33,702
here, this omega equals of omega
zero.
314
00:23:33,702 --> 00:23:37,872
You will see that.
And, that's always very nice to
315
00:23:37,872 --> 00:23:40,255
remember.
At that omega zero,
316
00:23:40,255 --> 00:23:44,000
you always get Q times more
than your driver.
317
00:23:44,000 --> 00:23:47,659
If Q is 100,
you get 100 times the amplitude
318
00:23:47,659 --> 00:23:49,957
of the driver.
And so, here,
319
00:23:49,957 --> 00:23:52,765
you get eta zero if omega is
zero.
320
00:23:52,765 --> 00:23:57,021
And then, when you reach that
omega zero frequency,
321
00:23:57,021 --> 00:24:02,697
you come out high.
And then it goes back to zero.
322
00:24:02,697 --> 00:24:06,204
And this point here is Q times
eta zero.
323
00:24:06,204 --> 00:24:11,059
And, we discussed before,
I'll never make too much of a
324
00:24:11,059 --> 00:24:15,376
deal out of that,
that the actual maximum of this
325
00:24:15,376 --> 00:24:19,692
curve is not at omega zero but
is always a lower.
326
00:24:19,692 --> 00:24:24,727
That is really more of an
algebraic interest than it is a
327
00:24:24,727 --> 00:24:28,414
physical interest because if
gamma is low,
328
00:24:28,414 --> 00:24:33,000
if Q is high,
the two almost coincide.
329
00:24:33,000 --> 00:24:37,492
So, the steady state solution
has no adjustable constants.
330
00:24:37,492 --> 00:24:41,039
In other words,
the system has lost its memory
331
00:24:41,039 --> 00:24:44,428
of what happened when we started
driving it.
332
00:24:44,428 --> 00:24:47,738
So, at T equals zero,
if we know what X is,
333
00:24:47,738 --> 00:24:52,467
and if we know what X dot is,
and if we know what the driving
334
00:24:52,467 --> 00:24:56,724
term is at T equals zero,
then the general solution for
335
00:24:56,724 --> 00:25:01,532
all times larger than zero is
the sum of the trangent solution
336
00:25:01,532 --> 00:25:06,500
and the steady state solution.
And so, here you see the
337
00:25:06,500 --> 00:25:09,687
[trangent?] solution.
You can write it in this form,
338
00:25:09,687 --> 00:25:11,625
or you can read it in this
form.
339
00:25:11,625 --> 00:25:14,250
And here you see the steady
state solution.
340
00:25:14,250 --> 00:25:16,500
This A has nothing to do with
that A.
341
00:25:16,500 --> 00:25:20,375
This A follows from the initial
conditions just like the alpha,
342
00:25:20,375 --> 00:25:23,500
this A does not follow from the
initial conditions.
343
00:25:23,500 --> 00:25:26,937
This A follows from eta zero,
what my amplitude is of my
344
00:25:26,937 --> 00:25:30,000
driver, and it follows from
omega.
345
00:25:30,000 --> 00:25:33,756
So, the two A's don't confuse
the two.
346
00:25:33,756 --> 00:25:39,544
And so, the general solution
is, then, the sum of the two.
347
00:25:39,544 --> 00:25:45,433
The trangent one will die out.
If you weigh it a few times,
348
00:25:45,433 --> 00:25:49,089
two over gamma,
the trangent is gone.
349
00:25:49,089 --> 00:25:54,267
And so, you would end up,
then, only with the steady
350
00:25:54,267 --> 00:25:58,532
state solution.
And, you had some chance in
351
00:25:58,532 --> 00:26:05,002
your homework to work with that.
If we are not driving the
352
00:26:05,002 --> 00:26:09,169
system, and if we have one
object on a spring or on a
353
00:26:09,169 --> 00:26:13,816
pendulum, or a floating object
in liquid, then there is one
354
00:26:13,816 --> 00:26:17,821
frequency, one normal mode
frequency, one resonance
355
00:26:17,821 --> 00:26:21,667
frequency, we call it also the
natural frequency.
356
00:26:21,667 --> 00:26:26,234
The moment that you couple
oscillators, if you couple two,
357
00:26:26,234 --> 00:26:30,000
you get two normal mode
frequencies.
358
00:26:30,000 --> 00:26:32,692
You get to resonance
frequencies.
359
00:26:32,692 --> 00:26:37,319
And if you have three objects,
you get three normal mode
360
00:26:37,319 --> 00:26:40,600
frequencies.
And so, now I would like to
361
00:26:40,600 --> 00:26:44,387
discuss with you the case
whereby I couple two
362
00:26:44,387 --> 00:26:47,920
oscillators.
If I gave you on an exam three
363
00:26:47,920 --> 00:26:51,622
coupled oscillators,
that would be very nasty
364
00:26:51,622 --> 00:26:54,735
because it's extremely time
consuming.
365
00:26:54,735 --> 00:26:59,110
If I gave you four coupled
oscillators, that would be
366
00:26:59,110 --> 00:27:05,000
criminal because you cannot
finish that in 85 minutes.
367
00:27:05,000 --> 00:27:07,870
So, two is certainly within
reason.
368
00:27:07,870 --> 00:27:10,656
Three is marginally within
reason.
369
00:27:10,656 --> 00:27:14,709
Four is out of the question.
When we have coupled
370
00:27:14,709 --> 00:27:18,087
oscillators, we always leave
damping out.
371
00:27:18,087 --> 00:27:21,126
And yet, we will learn a lot
from it.
372
00:27:21,126 --> 00:27:24,588
Even without damping,
we will learn a lot.
373
00:27:24,588 --> 00:27:29,823
And so, what I've chosen to do
with you is a spring system with
374
00:27:29,823 --> 00:27:34,298
two objects, two masses,
and the springs have no mass,
375
00:27:34,298 --> 00:27:39,817
negligible mass.
The spring constants are K,
376
00:27:39,817 --> 00:27:43,469
and the masses of the objects
are M.
377
00:27:43,469 --> 00:27:49,417
So, this is the equilibrium
position of these two objects,
378
00:27:49,417 --> 00:27:53,173
and the ends are fixed of the
string.
379
00:27:53,173 --> 00:27:59,330
I will introduce a shorthand
notation, zero squared equals K
380
00:27:59,330 --> 00:28:02,452
over M.
Now, what I do,
381
00:28:02,452 --> 00:28:07,974
I offset both these objects
from equilibrium just as I did
382
00:28:07,974 --> 00:28:11,558
that here.
I always offset them in the
383
00:28:11,558 --> 00:28:15,433
same direction,
and I call that positive.
384
00:28:15,433 --> 00:28:17,176
Is that a must?
No.
385
00:28:17,176 --> 00:28:19,017
Is that useful?
Yes.
386
00:28:19,017 --> 00:28:22,891
So, I offset them.
And so, this position,
387
00:28:22,891 --> 00:28:28,703
now this is object number one,
and this is object number two.
388
00:28:28,703 --> 00:28:35,000
This is now a distance,
X1, away from equilibrium.
389
00:28:35,000 --> 00:28:38,493
And this one is now at a
position, X2,
390
00:28:38,493 --> 00:28:43,590
away from its equilibrium.
You always want to know what
391
00:28:43,590 --> 00:28:48,405
the displacement is relative to
its own equilibrium.
392
00:28:48,405 --> 00:28:54,069
And, its own equilibrium for
one is here, and equilibrium for
393
00:28:54,069 --> 00:28:57,562
two is there.
I now make the following
394
00:28:57,562 --> 00:29:02,000
assumption that X2 is larger
than X1.
395
00:29:02,000 --> 00:29:03,719
Is that a must?
No.
396
00:29:03,719 --> 00:29:07,828
Is it useful?
Yes, but if you want to assume
397
00:29:07,828 --> 00:29:11,554
that X2 is smaller than X1,
be my guest.
398
00:29:11,554 --> 00:29:15,280
I will assume that X2 is larger
than X1.
399
00:29:15,280 --> 00:29:20,439
Now, follow me closely now.
So, we have an object here,
400
00:29:20,439 --> 00:29:25,598
and we have an object here,
and this is now one spring.
401
00:29:25,598 --> 00:29:32,000
This is the other spring,
and this is the third spring.
402
00:29:32,000 --> 00:29:35,806
It is immediately obvious that
this spring is longer than it
403
00:29:35,806 --> 00:29:38,193
wants to be.
So, there is a force that
404
00:29:38,193 --> 00:29:41,677
drives it back to equilibrium.
If X2 is larger than X1,
405
00:29:41,677 --> 00:29:44,645
this spring is also longer than
it wants to be.
406
00:29:44,645 --> 00:29:48,000
So, it wants to contract,
so there is a force in that
407
00:29:48,000 --> 00:29:50,516
direction.
If this spring is longer than
408
00:29:50,516 --> 00:29:52,838
it wants to be,
it wants to contract.
409
00:29:52,838 --> 00:29:55,677
So, there is a force in X2 in
this direction.
410
00:29:55,677 --> 00:30:00,000
This spring is clearly shorter
than it wants to be.
411
00:30:00,000 --> 00:30:05,632
So, it's pushing.
So there is a force in this
412
00:30:05,632 --> 00:30:10,751
direction.
And so, now I'm going to write
413
00:30:10,751 --> 00:30:18,175
down the differential equation
first for object number one.
414
00:30:18,175 --> 00:30:25,087
So I get M X1 double dot.
So, this one is minus K times
415
00:30:25,087 --> 00:30:29,183
X1.
And, this one is plus K times
416
00:30:29,183 --> 00:30:34,278
X2 minus X1.
Is it a disaster if it turns
417
00:30:34,278 --> 00:30:36,963
out that X2 is not larger than
X1?
418
00:30:36,963 --> 00:30:38,997
Not at all.
This equation,
419
00:30:38,997 --> 00:30:41,763
now, is correct for all
situations.
420
00:30:41,763 --> 00:30:46,482
The fact that I have assumed
that X2 is larger than X1 gave
421
00:30:46,482 --> 00:30:49,981
me a plus sign here,
and so, my differential
422
00:30:49,981 --> 00:30:54,212
equation is safe no matter what
X1 is relative to X2.
423
00:30:54,212 --> 00:30:57,385
So, you can always make that
assumption.
424
00:30:57,385 --> 00:31:03,000
And you don't have to worry
later anymore about signs.
425
00:31:03,000 --> 00:31:07,190
I can divide,
now, M out, and then I get that
426
00:31:07,190 --> 00:31:11,952
X1 double dot plus two omega
zero squared times X1.
427
00:31:11,952 --> 00:31:16,714
But you notice I have one here
and I have one here.
428
00:31:16,714 --> 00:31:21,952
They have both a minus sign.
And then, I have here plus,
429
00:31:21,952 --> 00:31:26,619
so that becomes minus omega
zero squared times X2.
430
00:31:26,619 --> 00:31:29,761
That is zero.
So, that is my first
431
00:31:29,761 --> 00:31:35,965
differential equation.
So, now I go to the second one.
432
00:31:35,965 --> 00:31:41,034
I get MX two double dot.
Now I have two forces both in
433
00:31:41,034 --> 00:31:45,626
the negative direction.
First I have the one that
434
00:31:45,626 --> 00:31:49,165
drives this one away from
equilibrium.
435
00:31:49,165 --> 00:31:52,417
So I get minus K times X2 minus
X1.
436
00:31:52,417 --> 00:31:57,965
And then, I have the force due
to the fact that this one is
437
00:31:57,965 --> 00:32:02,173
shorter than it wants to be by
an amount, K2.
438
00:32:02,173 --> 00:32:08,524
So, I get minus K times X2.
It's shorter by an amount,
439
00:32:08,524 --> 00:32:13,477
X2, than it wants to be.
And so, I can divide M out,
440
00:32:13,477 --> 00:32:19,207
and so I get MX2 double dot.
And then, I get plus omega zero
441
00:32:19,207 --> 00:32:23,480
squared times X2.
And then, I get minus omega
442
00:32:23,480 --> 00:32:26,685
zero squared times X1 equals
zero.
443
00:32:26,685 --> 00:32:30,083
Now, compare this one with this
one.
444
00:32:30,083 --> 00:32:36,007
Notice the incredible symmetry.
I could have found this one by
445
00:32:36,007 --> 00:32:43,000
changing a one to two here,
and by changing the two to one.
446
00:32:43,000 --> 00:32:46,034
Of course, the system is so
symmetric that,
447
00:32:46,034 --> 00:32:49,069
yes, thank you very much for
the two right?
448
00:32:49,069 --> 00:32:52,465
Yeah, thank you.
The system is so symmetric that
449
00:32:52,465 --> 00:32:56,945
clearly nature cannot make any
distinction between one and two.
450
00:32:56,945 --> 00:33:01,063
So, it is in this particular
case because of the beautiful
451
00:33:01,063 --> 00:33:05,398
symmetry, it is obvious that
these two differential equations
452
00:33:05,398 --> 00:33:10,230
look very, very similar.
Now, I'm going to make an
453
00:33:10,230 --> 00:33:14,334
important step.
I'm going to substitute in here
454
00:33:14,334 --> 00:33:18,617
X1 is C times cosine omega T,
and X2, this is C1,
455
00:33:18,617 --> 00:33:23,702
is C2 times cosine omega T.
I want to know what the normal
456
00:33:23,702 --> 00:33:26,914
mode frequencies are for this
system.
457
00:33:26,914 --> 00:33:32,000
I want to solve for omega.
I want to find omega.
458
00:33:32,000 --> 00:33:35,870
I'm not driving the system.
Since I have no damping,
459
00:33:35,870 --> 00:33:40,120
in which case at normal mode
solutions, either of the two
460
00:33:40,120 --> 00:33:44,674
objects are in phase with each
other or they are out of phase
461
00:33:44,674 --> 00:33:48,165
with each other.
But if there are out of phase,
462
00:33:48,165 --> 00:33:51,883
we can always take care of that
with a minus sign.
463
00:33:51,883 --> 00:33:55,223
So, I'm now going to substitute
that in here.
464
00:33:55,223 --> 00:33:59,245
And I may want to continue
working on the centerboard,
465
00:33:59,245 --> 00:34:03,647
in which case I might as well
erase this so that it stays a
466
00:34:03,647 --> 00:34:06,000
little compact.
467
00:34:06,000 --> 00:34:13,000
468
00:34:13,000 --> 00:34:17,330
So, I'm going to substitute
this trial function into my
469
00:34:17,330 --> 00:34:21,339
differential equations.
And, every term will have a
470
00:34:21,339 --> 00:34:24,948
cosine omega T,
so I dump all the cosine omega
471
00:34:24,948 --> 00:34:27,273
T's.
So, I go to the equation,
472
00:34:27,273 --> 00:34:32,695
which has a big one there.
And so, that one becomes C1.
473
00:34:32,695 --> 00:34:36,930
X1 double dot gives me a minus
omega squared,
474
00:34:36,930 --> 00:34:40,395
right?
If I take cosine omega T and I
475
00:34:40,395 --> 00:34:45,497
do the second derivative,
I get minus omega squared in
476
00:34:45,497 --> 00:34:49,347
front of it.
So, I get C1 times two omega
477
00:34:49,347 --> 00:34:53,197
zero squared.
That is that two omega zero
478
00:34:53,197 --> 00:34:56,566
squared.
And then, I get minus omega
479
00:34:56,566 --> 00:34:59,262
squared.
And then I get plus,
480
00:34:59,262 --> 00:35:06,000
not plus, then I get minus
omega zero squared times C2.
481
00:35:06,000 --> 00:35:10,392
And that equals zero.
And then I go to my second
482
00:35:10,392 --> 00:35:14,037
differential equation which is
this one.
483
00:35:14,037 --> 00:35:18,523
But I'm going to rearrange the
C1's and the C2's.
484
00:35:18,523 --> 00:35:21,700
So, I'm going to put the C1's
here.
485
00:35:21,700 --> 00:35:26,747
So, I'm going to get minus
omega zero squared times C1.
486
00:35:26,747 --> 00:35:32,261
And here, I'm going to get plus
two omega zero squared minus
487
00:35:32,261 --> 00:35:37,886
omega zero squared times C2.
And that equals zero.
488
00:35:37,886 --> 00:35:43,095
So, I can leave this plus out.
So, what do I have here now?
489
00:35:43,095 --> 00:35:47,137
I have here two equations with
three unknowns.
490
00:35:47,137 --> 00:35:51,449
I want to know what omega is in
the normal modes,
491
00:35:51,449 --> 00:35:55,670
but I also would like to know
what C1 and C2 is.
492
00:35:55,670 --> 00:36:00,431
Well, that is tough luck.
You can't have it both ways.
493
00:36:00,431 --> 00:36:06,000
We do not know at all what C1
and C2 is separately.
494
00:36:06,000 --> 00:36:09,498
But you can always find in the
normal modes,
495
00:36:09,498 --> 00:36:13,810
without knowing anything else
what the ratios of those
496
00:36:13,810 --> 00:36:17,959
amplitudes, C1 and C2.
And, you can always solve for
497
00:36:17,959 --> 00:36:21,050
omega, and you will see how that
works.
498
00:36:21,050 --> 00:36:24,386
So, with two equations and
three unknowns,
499
00:36:24,386 --> 00:36:28,128
you can only solve for omega
and for the ratio,
500
00:36:28,128 --> 00:36:32,828
C1 over C2.
So now, you may do that anyway
501
00:36:32,828 --> 00:36:36,505
you want to.
This is not so difficult to
502
00:36:36,505 --> 00:36:39,051
solve this.
I will, however,
503
00:36:39,051 --> 00:36:42,917
use Kramer's rule,
and I will introduce D,
504
00:36:42,917 --> 00:36:48,197
which is the determinant of the
following: two omega zero
505
00:36:48,197 --> 00:36:53,288
squared minus omega squared.
And then I have here minus
506
00:36:53,288 --> 00:36:57,625
omega zero squared.
And then, I have here minus
507
00:36:57,625 --> 00:37:03,000
omega zero squared minus omega
zero squared.
508
00:37:03,000 --> 00:37:06,739
And this, now,
must be zero if I'm not looking
509
00:37:06,739 --> 00:37:09,150
for solutions,
which I am not,
510
00:37:09,150 --> 00:37:13,638
C1 is zero and C2 is zero.
It's clear that if C1 and C2
511
00:37:13,638 --> 00:37:17,045
are zero, the two equations are
satisfied.
512
00:37:17,045 --> 00:37:20,536
Zero is zero.
But that's not an interesting
513
00:37:20,536 --> 00:37:23,445
case.
And, the only way that you can
514
00:37:23,445 --> 00:37:27,766
get a solution which is
interesting is value's for C1
515
00:37:27,766 --> 00:37:34,000
and C2 which are not zero is by
making this determinant zero.
516
00:37:34,000 --> 00:37:39,157
And so, that means that D,
which now becomes two omega
517
00:37:39,157 --> 00:37:43,147
zero squared minus omega
squared, squared,
518
00:37:43,147 --> 00:37:48,791
minus omega zero to the fourth,
you have to make that zero.
519
00:37:48,791 --> 00:37:53,755
And when you do that,
you find two values for omega.
520
00:37:53,755 --> 00:37:59,593
You find that omega squared is
two omega zero squared plus or
521
00:37:59,593 --> 00:38:05,486
minus omega zero squared.
Those are the two solutions.
522
00:38:05,486 --> 00:38:10,726
And, when we evaluate those two
solutions, you find a result
523
00:38:10,726 --> 00:38:15,523
which is so embarrassingly
simple that you could almost
524
00:38:15,523 --> 00:38:18,986
have said that without any work,
almost.
525
00:38:18,986 --> 00:38:23,782
Notice that omega minus,
which is the lowest one of the
526
00:38:23,782 --> 00:38:30,000
two, is the same as omega zero
because that's the minus sign.
527
00:38:30,000 --> 00:38:33,380
What does that mean if omega
minus is omega zero?
528
00:38:33,380 --> 00:38:36,619
Well, it means,
of course, that the two objects
529
00:38:36,619 --> 00:38:40,563
are just oscillating like this.
The inner spring is never
530
00:38:40,563 --> 00:38:43,802
stretched, is never longer than
it wants to be,
531
00:38:43,802 --> 00:38:47,746
is never shorter than it wants
to be, so each one is only
532
00:38:47,746 --> 00:38:50,563
driven, so to speak,
by the outer spring.
533
00:38:50,563 --> 00:38:54,507
So, that's immediately obvious
that that's a normal mode.
534
00:38:54,507 --> 00:38:58,591
And, you can make a prediction
that C1 over C2 must be plus
535
00:38:58,591 --> 00:39:02,057
one.
They must go in unison like
536
00:39:02,057 --> 00:39:04,495
this.
And you can confirm that by
537
00:39:04,495 --> 00:39:08,914
substituting this omega one.
If you substitute that in this
538
00:39:08,914 --> 00:39:12,495
equation, you'll find the ratio
see one over C2.
539
00:39:12,495 --> 00:39:15,542
If you prefer this equation,
be my guest.
540
00:39:15,542 --> 00:39:19,428
You can do that too.
And you'll find no matter which
541
00:39:19,428 --> 00:39:22,704
one of the two you take that it
is plus one.
542
00:39:22,704 --> 00:39:26,742
Now, omega plus is less
obvious, is the square root of
543
00:39:26,742 --> 00:39:32,000
three times omega zero.
That's when you take the plus.
544
00:39:32,000 --> 00:39:36,142
Even though it's not obvious
that it is the square root of
545
00:39:36,142 --> 00:39:38,759
three, it is clear what is
happening.
546
00:39:38,759 --> 00:39:42,611
Here are the two objects,
and now they are doing this.
547
00:39:42,611 --> 00:39:46,245
They are exactly 180° out of
phase with each other.
548
00:39:46,245 --> 00:39:49,952
So, now you can make
comfortably the prediction that
549
00:39:49,952 --> 00:39:53,804
C1 over C2 is minus one.
And indeed, if you substitute
550
00:39:53,804 --> 00:39:58,383
the square at a free omega zero
in either this or that equation,
551
00:39:58,383 --> 00:40:04,015
this indeed comes out.
So, the general solution now
552
00:40:04,015 --> 00:40:09,830
for a given initial condition,
so if now I give you the
553
00:40:09,830 --> 00:40:15,000
initial condition,
then the general solution will
554
00:40:15,000 --> 00:40:21,030
be for X1 would be X0 minus,
the minus makes reference to
555
00:40:21,030 --> 00:40:27,599
that lowest frequency times the
cosine omega minus T plus some
556
00:40:27,599 --> 00:40:32,338
phase angle minus,
all these minus signs make
557
00:40:32,338 --> 00:40:38,393
reference to this mode.
And then, I have plus X0 plus
558
00:40:38,393 --> 00:40:43,262
times the cosine of omega plus T
plus some phase angle plus.
559
00:40:43,262 --> 00:40:47,223
Of course, if you prefer
instead of cosine-signs,
560
00:40:47,223 --> 00:40:49,038
that's fine,
of course.
561
00:40:49,038 --> 00:40:52,669
And, you have four adjustable
constants: one,
562
00:40:52,669 --> 00:40:56,383
two, three, four.
And, if you know the initial
563
00:40:56,383 --> 00:41:00,839
conditions, if you know what X1
is, and what X1 dot is,
564
00:41:00,839 --> 00:41:04,388
and what X2 is,
and what X2 dot is at time T
565
00:41:04,388 --> 00:41:10,000
equals zero, you can solve for
all four in principle.
566
00:41:10,000 --> 00:41:14,558
There is no longer any freedom
from number two because all four
567
00:41:14,558 --> 00:41:17,720
adjustable constants have now
been consumed.
568
00:41:17,720 --> 00:41:20,441
And so, therefore,
you can write down,
569
00:41:20,441 --> 00:41:24,558
now, the solution for X2 by
simply making this one a two.
570
00:41:24,558 --> 00:41:27,279
And, this whole term here is
the same.
571
00:41:27,279 --> 00:41:31,323
The only difference is that
this plus sign now becomes a
572
00:41:31,323 --> 00:41:35,000
minus sign.
Nothing else is different.
573
00:41:35,000 --> 00:41:39,206
The frequencies are the same.
Otherwise they wouldn't be
574
00:41:39,206 --> 00:41:43,184
normal mode frequencies.
And, it is the ratio here of
575
00:41:43,184 --> 00:41:47,926
the C1 over C2 that is plus one.
And, it is the ratio here that
576
00:41:47,926 --> 00:41:51,215
is the minus one.
That's the reason why this
577
00:41:51,215 --> 00:41:54,810
becomes a minus,
and why the plus here remains a
578
00:41:54,810 --> 00:41:56,186
plus.
So that, now,
579
00:41:56,186 --> 00:42:00,087
is the general solution if you
also know the initial
580
00:42:00,087 --> 00:42:06,720
conditions.
So now, we are going to drive
581
00:42:06,720 --> 00:42:13,395
the system.
So now, we go back to where we
582
00:42:13,395 --> 00:42:19,418
were, and I'm going to drive
this end.
583
00:42:19,418 --> 00:42:26,418
And on going to drive this and
with my hand.
584
00:42:26,418 --> 00:42:33,093
This is eta,
and eta equals eta zero times
585
00:42:33,093 --> 00:42:40,387
cosine omega T.
There is only one term in all
586
00:42:40,387 --> 00:42:46,112
of this on the blackboard that
is going to change.
587
00:42:46,112 --> 00:42:52,303
And that is this term.
This spring here on the left is
588
00:42:52,303 --> 00:42:59,663
no longer shorter by the amount
X1, but is shorter by the amount
589
00:42:59,663 --> 00:43:04,722
X1 minus eta.
And so, it's only that term.
590
00:43:04,722 --> 00:43:10,166
It's only this minus KX1 that
now have to be changed into
591
00:43:10,166 --> 00:43:14,833
minus K times X1 minus eta.
Nothing else changes.
592
00:43:14,833 --> 00:43:17,944
But, that has major
consequences,
593
00:43:17,944 --> 00:43:20,277
of course.
For one thing,
594
00:43:20,277 --> 00:43:25,819
if you're going to substitute,
now, these trial functions,
595
00:43:25,819 --> 00:43:30,000
omega is no longer a
negotiable.
596
00:43:30,000 --> 00:43:33,458
Omega is my omega now.
I set this omega.
597
00:43:33,458 --> 00:43:37,004
You're not going to solve for
that omega.
598
00:43:37,004 --> 00:43:41,793
That would be an insult to me.
I dictate what omega is.
599
00:43:41,793 --> 00:43:46,758
So, that means I now get two
equations with two unknowns,
600
00:43:46,758 --> 00:43:50,039
C1 and C2.
Omega is no longer unknown.
601
00:43:50,039 --> 00:43:55,359
So, now I get solutions for C1
and they get solutions for C2.
602
00:43:55,359 --> 00:44:00,413
And of all the equations on the
blackboards that you have,
603
00:44:00,413 --> 00:44:06,000
the one that is going to change
is the first one.
604
00:44:06,000 --> 00:44:12,422
And when you carry through,
that eta, which is eta zero
605
00:44:12,422 --> 00:44:17,894
cosine omega T,
this zero here changes into eta
606
00:44:17,894 --> 00:44:24,555
zero times omega zero squared.
The cosine omega T is gone
607
00:44:24,555 --> 00:44:31,334
because I've divided cosine
omega T out in all those other
608
00:44:31,334 --> 00:44:35,993
terms.
So, now I use creamers rule,
609
00:44:35,993 --> 00:44:41,660
and now I can actually come up
with a solution for C1,
610
00:44:41,660 --> 00:44:47,754
no longer ratio C1 over C2.
No, I can actually come up now
611
00:44:47,754 --> 00:44:52,779
with a solution.
So, C1 now using Kramer's rule,
612
00:44:52,779 --> 00:44:58,660
this now as my first column.
Eta zero omega zero squared
613
00:44:58,660 --> 00:45:04,968
zero, and this now becomes my
second column minus omega zero
614
00:45:04,968 --> 00:45:09,966
squared.
And I get here two omega zero
615
00:45:09,966 --> 00:45:14,059
squared minus omega squared
divided by D.
616
00:45:14,059 --> 00:45:19,072
If I pick a random value for
omega, D is not zero.
617
00:45:19,072 --> 00:45:24,085
D is only zero at those two
resonance frequencies.
618
00:45:24,085 --> 00:45:29,303
And then, C2 becomes,
the first column is like this.
619
00:45:29,303 --> 00:45:35,544
That is two omega zero squared
minus omega squared minus omega
620
00:45:35,544 --> 00:45:41,683
zero squared.
And now, the second column
621
00:45:41,683 --> 00:45:49,758
becomes eta zero omega zero
squared, and a zero divided by
622
00:45:49,758 --> 00:45:53,583
D.
If I work C1 out a little
623
00:45:53,583 --> 00:46:01,516
further, then I get C1 equals
eta zero omega zero squared
624
00:46:01,516 --> 00:46:10,016
times two omega zero squared
minus omega zero squared divided
625
00:46:10,016 --> 00:46:14,923
by D.
And here, I get that C2,
626
00:46:14,923 --> 00:46:19,916
this is zero.
So, I get plus omega zero to
627
00:46:19,916 --> 00:46:24,544
the fourth times eta zero
divided by D.
628
00:46:24,544 --> 00:46:31,000
And D, now, is this determinant
but is not zero.
629
00:46:31,000 --> 00:46:35,076
It's only zero for those two
special frequencies.
630
00:46:35,076 --> 00:46:40,087
If you want to see what the
amplitudes now are as a function
631
00:46:40,087 --> 00:46:44,164
of omega, it's a very,
very interesting behavior,
632
00:46:44,164 --> 00:46:47,986
then what helps is you go omega
first to zero.
633
00:46:47,986 --> 00:46:52,912
And then, you see what happens.
And, it is not so intuitive
634
00:46:52,912 --> 00:46:56,224
what happens.
If you put in omega equals
635
00:46:56,224 --> 00:47:02,000
zero, you will find that C1 is
plus two thirds eta zero.
636
00:47:02,000 --> 00:47:06,142
You can confirm that by
substituting that into C1.
637
00:47:06,142 --> 00:47:09,270
You will see that that's what
happens.
638
00:47:09,270 --> 00:47:13,582
And, you'll find that C2 is
plus one third eta zero.
639
00:47:13,582 --> 00:47:16,287
So, that is at omega equals
zero.
640
00:47:16,287 --> 00:47:20,091
And, when you go to infinity
with frequencies,
641
00:47:20,091 --> 00:47:24,065
then it should not surprise you
that C1 is zero,
642
00:47:24,065 --> 00:47:28,123
and that C2 is zero.
And then, there is one case,
643
00:47:28,123 --> 00:47:32,012
which is pathetic.
And, that is the case that I
644
00:47:32,012 --> 00:47:37,000
make omega squared,
two omega zero squared.
645
00:47:37,000 --> 00:47:40,017
At that frequency,
C1 becomes zero,
646
00:47:40,017 --> 00:47:44,098
but C2 is not zero.
And, I spent almost a whole
647
00:47:44,098 --> 00:47:48,624
lecture with you on that
demonstrating that in three
648
00:47:48,624 --> 00:47:52,972
different ways that,
indeed, there is this bizarre
649
00:47:52,972 --> 00:47:56,344
solution.
So, when omega squared equals
650
00:47:56,344 --> 00:48:00,249
two omega zero squared,
then C1 becomes zero,
651
00:48:00,249 --> 00:48:05,835
but C2 is not zero.
And so, now you can make a plot
652
00:48:05,835 --> 00:48:10,774
of C's as a function of omega.
And during that lecture,
653
00:48:10,774 --> 00:48:13,792
I showed you three of those
plots.
654
00:48:13,792 --> 00:48:18,914
And, I used the convention,
then, that when the object is
655
00:48:18,914 --> 00:48:24,402
moving in phase with the driver,
I put it positive and out of
656
00:48:24,402 --> 00:48:28,426
phase negative.
And, I plot here C divided by
657
00:48:28,426 --> 00:48:33,000
eta zero.
And, I will use a color code.
658
00:48:33,000 --> 00:48:37,348
I will do C1 in red,
and I will do,
659
00:48:37,348 --> 00:48:42,593
C1 is red and C2 I will do in
white chalk.
660
00:48:42,593 --> 00:48:47,965
So, here is zero omega.
Here is omega zero,
661
00:48:47,965 --> 00:48:54,744
which is my omega minus,
and then my omega plus was at
662
00:48:54,744 --> 00:49:01,523
three, the square root of three.
So, that's about 1.7.
663
00:49:01,523 --> 00:49:08,757
So, here's my omega plus.
And then things go nuts.
664
00:49:08,757 --> 00:49:14,961
That is when D goes to zero.
So, C1 is two thirds here.
665
00:49:14,961 --> 00:49:18,982
And then, it goes to infinity
there.
666
00:49:18,982 --> 00:49:23,693
And then, here,
when it is 1.4 times omega
667
00:49:23,693 --> 00:49:30,357
zero, the square root of two,
then it goes to a zero and it
668
00:49:30,357 --> 00:49:35,916
comes up here.
And then, I get a curve here,
669
00:49:35,916 --> 00:49:41,346
without being too precise.
And for C2, I get one third.
670
00:49:41,346 --> 00:49:44,765
And then, it goes up.
And then, C2,
671
00:49:44,765 --> 00:49:49,793
something like this.
And then, I get something like
672
00:49:49,793 --> 00:49:51,000
that.
673
00:49:51,000 --> 00:49:59,000
674
00:49:59,000 --> 00:50:02,825
Yeah, I can live with that.
Now, we ignore damping,
675
00:50:02,825 --> 00:50:06,957
and because we ignore damping,
we get these on physical
676
00:50:06,957 --> 00:50:11,395
infinities when you did the
frequency omega minus and omega
677
00:50:11,395 --> 00:50:13,920
plus, these resonance
frequencies.
678
00:50:13,920 --> 00:50:17,976
Now, if you include damping,
then the solutions become
679
00:50:17,976 --> 00:50:21,802
extremely complicated.
But, of course you avoid the
680
00:50:21,802 --> 00:50:25,168
infinity values.
But, the resonant amplitudes
681
00:50:25,168 --> 00:50:30,213
can still be very high.
And so, these plots are still
682
00:50:30,213 --> 00:50:34,639
very useful provided that you
don't interpret infinities as
683
00:50:34,639 --> 00:50:37,998
being real, but something that
is very large.
684
00:50:37,998 --> 00:50:41,203
If Q is high,
then of course the amplitudes
685
00:50:41,203 --> 00:50:44,332
are enormously high.
And, this can lead to
686
00:50:44,332 --> 00:50:47,309
distraction.
We've seen the movie of the
687
00:50:47,309 --> 00:50:50,133
Tacoma Bridge,
and then we have seen a
688
00:50:50,133 --> 00:50:53,567
dramatic experiment of the
breaking wineglass.
689
00:50:53,567 --> 00:50:56,696
You remember that wineglass
demonstration.
690
00:50:56,696 --> 00:51:00,970
Because of the catastrophic
success, to use Bush's words,
691
00:51:00,970 --> 00:51:04,863
of that demonstration,
students have asked me for an
692
00:51:04,863 --> 00:51:09,241
encore.
They would like to see it
693
00:51:09,241 --> 00:51:12,289
again.
And that's a very reasonable
694
00:51:12,289 --> 00:51:15,965
thing to do.
It fits very nicely into this
695
00:51:15,965 --> 00:51:21,344
concept of coupled oscillators.
A wineglass would have a huge
696
00:51:21,344 --> 00:51:25,468
number of oscillators coupled,
and a wineglass,
697
00:51:25,468 --> 00:51:29,682
I have one here,
can be made to oscillate easily
698
00:51:29,682 --> 00:51:35,506
in its lowest normal mode.
I have to wash my hands to make
699
00:51:35,506 --> 00:51:39,307
you listen to it.
So, because my hands are now a
700
00:51:39,307 --> 00:51:43,755
little greasy from the chalk,
and this is the frequency.
701
00:51:43,755 --> 00:51:47,637
That is the lowest mode.
It is a circle like this
702
00:51:47,637 --> 00:51:50,952
looking from above,
it becomes an ellipse,
703
00:51:50,952 --> 00:51:55,804
and then it becomes an ellipse
like this, and an ellipse like
704
00:51:55,804 --> 00:52:00,575
this, and an ellipse like that.
And if now you dry that with
705
00:52:00,575 --> 00:52:05,347
sound at exactly that frequency
and you put in enough power,
706
00:52:05,347 --> 00:52:09,875
another way of saying it is as
if you make eta zero large
707
00:52:09,875 --> 00:52:14,000
enough, then the system can
break.
708
00:52:14,000 --> 00:52:16,526
It's making a lot of noise.
I'll warn you.
709
00:52:16,526 --> 00:52:19,546
Those were sitting here,
I really think you should
710
00:52:19,546 --> 00:52:22,319
protect your ears.
The sound can be deafening.
711
00:52:22,319 --> 00:52:25,092
So, be careful,
and those who are a little bit
712
00:52:25,092 --> 00:52:27,680
further away,
make sure that you close your
713
00:52:27,680 --> 00:52:29,714
ears.
I have the luxury that I can
714
00:52:29,714 --> 00:52:32,980
turn my hearing aids off,
but without my hearing aids,
715
00:52:32,980 --> 00:52:36,000
believe me, I can still hear a
lot.
716
00:52:36,000 --> 00:52:39,513
So, I also will have to protect
myself.
717
00:52:39,513 --> 00:52:43,582
So, one hearing aid is off.
The other is off,
718
00:52:43,582 --> 00:52:46,818
and I'm going to put this on
anyway.
719
00:52:46,818 --> 00:52:52,366
I'm going to stroke this glass
with a frequency which is only
720
00:52:52,366 --> 00:52:57,174
slightly different from this
frequency of the sounds.
721
00:52:57,174 --> 00:53:03,000
That is about 427 Hz,
which you can very easily hear.
722
00:53:03,000 --> 00:53:09,804
And then you'll see the glass
in slow motion because of the
723
00:53:09,804 --> 00:53:16,374
fact that the stroboscopic light
has a slightly different
724
00:53:16,374 --> 00:53:20,949
frequency.
That should be coming up now.
725
00:53:20,949 --> 00:53:24,000
And you say it will.
726
00:53:24,000 --> 00:53:35,000
727
00:53:35,000 --> 00:53:38,977
They do very much.
I'm going to make it very dark
728
00:53:38,977 --> 00:53:44,033
now because this is important
that you can see this very well.
729
00:53:44,033 --> 00:53:47,762
So, we'll make it completely
dark in the room.
730
00:53:47,762 --> 00:53:52,569
I'm going to protect my ears.
I hope you can still hear me.
731
00:53:52,569 --> 00:53:56,795
I can hardly hear myself.
And I'm going to drive the
732
00:53:56,795 --> 00:54:01,436
system, now, at very low
amplitudes, at a frequency close
733
00:54:01,436 --> 00:54:05,000
to its resonance.
[SOUND PLAYS]
734
00:54:05,000 --> 00:54:09,529
You can already see that the
glass is moving.
735
00:54:09,529 --> 00:54:13,132
I'm going to increase the
amplitude.
736
00:54:13,132 --> 00:54:18,073
You see it moving?
I can go a little bit over the
737
00:54:18,073 --> 00:54:21,470
resonance and under the
resonance.
738
00:54:21,470 --> 00:54:25,382
You will see that it's not
moving then.
739
00:54:25,382 --> 00:54:30,529
So, I do that purposely now.
Now I'm off resonance.
740
00:54:30,529 --> 00:54:36,324
I'm over it.
And now I'm off resonance.
741
00:54:36,324 --> 00:54:40,287
I'm under resonance.
I am below.
742
00:54:40,287 --> 00:54:46,168
I'm now at 423 Hz when the
resonance is at 427.
743
00:54:46,168 --> 00:54:50,643
So, I'm going to put it back at
427.
744
00:54:50,643 --> 00:54:55,630
And now I'm going to increase
the sound.
745
00:54:55,630 --> 00:55:01,000
So I warn you.
And there it goes.
746
00:55:01,000 --> 00:55:04,149
And that's it.
It broke.
747
00:55:04,149 --> 00:55:11,406
And it broke very fast.
[APPLAUSE] So this is an ideal
748
00:55:11,406 --> 00:55:18,116
moment for a break.
Since you had such a good time
749
00:55:18,116 --> 00:55:25,510
with the breaking glass,
let's settle for four minutes.
750
00:55:25,510 --> 00:55:30,029
We will reconvene in four
minutes.
751
00:55:30,029 --> 00:55:37,375
[SOUND OFF/THEN ON]
OK, I now want to discuss with
752
00:55:37,375 --> 00:55:42,102
you continuous media.
If you have N coupled
753
00:55:42,102 --> 00:55:46,603
oscillators, then you get N
normal modes.
754
00:55:46,603 --> 00:55:52,792
When you make N infinitely
high, then you get continuous
755
00:55:52,792 --> 00:55:56,956
media.
And, if we have one dimensional
756
00:55:56,956 --> 00:56:02,583
continuous media like a string,
or a pipe with air,
757
00:56:02,583 --> 00:56:09,447
sound, then we have studied the
transverse motion of a string,
758
00:56:09,447 --> 00:56:15,299
and then we derived that now
you have to use the wave
759
00:56:15,299 --> 00:56:20,758
equation.
We derive the wave equation.
760
00:56:20,758 --> 00:56:26,275
And then you get the D2Y DX
squared is one over V squared
761
00:56:26,275 --> 00:56:31,103
times D2Y DT squared,
Y being now the displacement
762
00:56:31,103 --> 00:56:36,325
away from equilibrium in this
position if X is in this
763
00:56:36,325 --> 00:56:39,792
direction.
And, that is fine.
764
00:56:39,792 --> 00:56:44,471
And so, excuse me?
It's amazing how you can look
765
00:56:44,471 --> 00:56:48,653
at it and think that's fine,
and it is not.
766
00:56:48,653 --> 00:56:54,528
All right, so a V in the case
of a string is the square root
767
00:56:54,528 --> 00:56:58,610
of T over mu.
That was just a special case
768
00:56:58,610 --> 00:57:03,753
for the string.
We also examined longitudinal
769
00:57:03,753 --> 00:57:05,367
motion, again,
1D.
770
00:57:05,367 --> 00:57:10,873
Sound, so longitudinal wave,
and if you deal with pressure,
771
00:57:10,873 --> 00:57:14,955
you think of sound as being a
pressure wave.
772
00:57:14,955 --> 00:57:20,462
Then you get D2P DX squared
equals one over V squared times
773
00:57:20,462 --> 00:57:24,639
D2P DT squared.
P, then, being positive would
774
00:57:24,639 --> 00:57:31,000
be overpressure over and above
the ambient 1 atmosphere.
775
00:57:31,000 --> 00:57:33,877
And if it's negative,
then it is below.
776
00:57:33,877 --> 00:57:38,269
So it's not the total pressure,
but it is the overpressure.
777
00:57:38,269 --> 00:57:40,844
And V, then,
is the speed of sound,
778
00:57:40,844 --> 00:57:45,008
which in air room temperature
is about 340 m per second.
779
00:57:45,008 --> 00:57:49,325
If you prefer not to work and
pressure but in terms of the
780
00:57:49,325 --> 00:57:53,489
actual position of the air
molecules in analogy with the
781
00:57:53,489 --> 00:57:57,276
position of the string,
then you can write down the
782
00:57:57,276 --> 00:58:01,819
same equation in terms of psi or
D2 psi DX squared one over V
783
00:58:01,819 --> 00:58:07,154
squared D2 psi DT squared.
That, then, gives you the
784
00:58:07,154 --> 00:58:11,289
position, the actual motion of
the air molecules.
785
00:58:11,289 --> 00:58:15,942
I often prefer the pressure,
and I will follow the also
786
00:58:15,942 --> 00:58:18,613
today.
So, there are an infinite
787
00:58:18,613 --> 00:58:22,404
number of normal modes.
And the ratios of the
788
00:58:22,404 --> 00:58:27,143
amplitudes of two adjacent
oscillators, they are coupled
789
00:58:27,143 --> 00:58:31,193
now, reflects itself,
of course, in terms of the
790
00:58:31,193 --> 00:58:35,070
overall shape,
which you can best see when you
791
00:58:35,070 --> 00:58:38,000
deal with a string.
792
00:58:38,000 --> 00:58:45,000
793
00:58:45,000 --> 00:58:49,846
As I said, it's easiest to see
with transfers,
794
00:58:49,846 --> 00:58:54,153
oscillations,
what the displacements look
795
00:58:54,153 --> 00:58:57,707
like.
It's harder to see that with
796
00:58:57,707 --> 00:59:01,787
sound.
I also discussed with you a
797
00:59:01,787 --> 00:59:06,227
special situation that I
connected to media,
798
00:59:06,227 --> 00:59:11,492
medium one and medium two,
and I gave these to media
799
00:59:11,492 --> 00:59:14,796
different masses per unit
length.
800
00:59:14,796 --> 00:59:19,339
When you set up a traveling
wave on a string,
801
00:59:19,339 --> 00:59:23,469
or you set it up,
you can set up a pulse,
802
00:59:23,469 --> 00:59:29,044
it reflects at the end.
And, how it reflects depends on
803
00:59:29,044 --> 00:59:34,000
the boundary conditions at the
end.
804
00:59:34,000 --> 00:59:38,730
And when you connect them to
media, then you get not only
805
00:59:38,730 --> 00:59:42,446
your reflection,
but you also get some of the
806
00:59:42,446 --> 00:59:45,149
pulse.
Some of the wave goes into
807
00:59:45,149 --> 00:59:48,359
medium to.
And so, let's assume that we
808
00:59:48,359 --> 00:59:51,822
have an incident wave coming in
like this.
809
00:59:51,822 --> 00:59:56,215
And we have here mu one,
and we have here V1 given by
810
00:59:56,215 --> 00:59:59,340
this equation.
They both have the same
811
00:59:59,340 --> 1:00:05,000
tangent, T, and here you have mu
two and you have V2.
812
1:00:05,000 --> 1:00:09,023
We discussed that.
We used the wave equation to
813
1:00:09,023 --> 1:00:14,534
solve what happens when we have
an incident harmonic wave coming
814
1:00:14,534 --> 1:00:17,333
in.
We derived even this speed of
815
1:00:17,333 --> 1:00:20,394
propagation using the wave
equation.
816
1:00:20,394 --> 1:00:23,281
None of this came out of the
blue.
817
1:00:23,281 --> 1:00:26,955
We always derived that.
And then, we found,
818
1:00:26,955 --> 1:00:31,153
by using the boundary
conditions at the junction,
819
1:00:31,153 --> 1:00:36,314
namely that the string is not
breaking, and that DYDX on the
820
1:00:36,314 --> 1:00:42,000
left side is the same as DY DX
on the right side.
821
1:00:42,000 --> 1:00:47,111
[UNINTELLIGIBLE] boundary
conditions, we found that the
822
1:00:47,111 --> 1:00:52,884
amplitude of the reflected wave,
and the same would hold for a
823
1:00:52,884 --> 1:00:58,374
pulse, divided by the amplitude
of the incident wave was V2
824
1:00:58,374 --> 1:01:03,012
minus V1 divided by V1 plus V2.
And, I call that R
825
1:01:03,012 --> 1:01:07,494
reflectivity.
And, we found that the
826
1:01:07,494 --> 1:01:13,845
amplitude of the transmitted
wave or pulse divided by the
827
1:01:13,845 --> 1:01:20,536
amplitude of the incident one
was 2V2 divided by V1 plus V2.
828
1:01:20,536 --> 1:01:27,000
And, I called that shorthand
notation transmitivity.
829
1:01:27,000 --> 1:01:32,782
And when we had done that,
I put in some simple test cases
830
1:01:32,782 --> 1:01:36,028
where our intuition is very
good.
831
1:01:36,028 --> 1:01:41,202
The first thing I did was I
said, suppose mu two was
832
1:01:41,202 --> 1:01:45,869
infinitely high.
So, mu two is infinitely high.
833
1:01:45,869 --> 1:01:50,028
In other words,
that medium two is a wall.
834
1:01:50,028 --> 1:01:55,202
That means, number one,
the string number one cannot
835
1:01:55,202 --> 1:01:57,942
move.
It's fixed at the end.
836
1:01:57,942 --> 1:02:04,296
So, that means V2 is zero.
And, we go to this equation and
837
1:02:04,296 --> 1:02:07,481
we find that R equals minus one.
V2 is zero.
838
1:02:07,481 --> 1:02:11,555
You get minus V1 over V1.
And, we like that because what
839
1:02:11,555 --> 1:02:16,000
it means is that when a mountain
rolls in, it comes back as a
840
1:02:16,000 --> 1:02:18,592
valley.
And then, a valley rolls in.
841
1:02:18,592 --> 1:02:22,740
It comes back as a mountain,
and I demonstrated that with
842
1:02:22,740 --> 1:02:25,555
strings.
It was very pleasing that T of
843
1:02:25,555 --> 1:02:28,444
R is then zero.
Well, it better be zero,
844
1:02:28,444 --> 1:02:32,795
right?
If everything comes back at you
845
1:02:32,795 --> 1:02:38,114
upside down, but nevertheless,
everything comes back at you.
846
1:02:38,114 --> 1:02:41,901
You expect that nothing goes
into the wall.
847
1:02:41,901 --> 1:02:45,868
And, you see when V2 is zero
that TR is zero.
848
1:02:45,868 --> 1:02:50,467
And, we were all very happy.
And we could all sleep.
849
1:02:50,467 --> 1:02:54,163
But then, and you guessed it,
then I said,
850
1:02:54,163 --> 1:02:57,139
let's suppose mu two becomes
zero.
851
1:02:57,139 --> 1:03:03,000
So, I attach that string to
nothing, to empty space.
852
1:03:03,000 --> 1:03:04,579
Is that practical?
Yes.
853
1:03:04,579 --> 1:03:07,595
I can do it.
I can take a nickel wire and I
854
1:03:07,595 --> 1:03:11,976
have here a magnetic field very
strong, and I can pull on that
855
1:03:11,976 --> 1:03:15,063
nickel wire.
And, the end of the nickel wire
856
1:03:15,063 --> 1:03:17,505
ends up in nothing,
in empty space.
857
1:03:17,505 --> 1:03:21,885
But, the I keep the tension on.
So, it's completely practical.
858
1:03:21,885 --> 1:03:24,614
It can be done.
It's not just nonsense.
859
1:03:24,614 --> 1:03:28,204
So, mu two goes to zero.
That means that V2 goes to
860
1:03:28,204 --> 1:03:32,272
infinity.
And then, we looked at R,
861
1:03:32,272 --> 1:03:35,909
and we say, well,
if V2 goes to infinity,
862
1:03:35,909 --> 1:03:40,545
then R equals plus one.
And, we were all very happy.
863
1:03:40,545 --> 1:03:44,727
A mountain comes back as a
mountain, and I even
864
1:03:44,727 --> 1:03:49,090
demonstrated that.
We were still able to sleep at
865
1:03:49,090 --> 1:03:52,727
that point.
But then, then came the awful
866
1:03:52,727 --> 1:03:57,545
thing that if we substitute V2
equals infinity in this
867
1:03:57,545 --> 1:04:00,636
equation, that TR becomes plus
two.
868
1:04:00,636 --> 1:04:07,000
And now, we can no longer sleep
because this is absurd.
869
1:04:07,000 --> 1:04:10,372
All the energy that rolls in
comes back.
870
1:04:10,372 --> 1:04:14,954
But there is something in
addition that goes into that
871
1:04:14,954 --> 1:04:17,375
second medium.
Now, admit it.
872
1:04:17,375 --> 1:04:22,131
Who could not sleep that night?
You should all fail this
873
1:04:22,131 --> 1:04:26,109
course, by the way.
[LAUGHTER] but in any case,
874
1:04:26,109 --> 1:04:30,000
I don't have to feel guilty,
right?
875
1:04:30,000 --> 1:04:36,129
Who thought about this and
said, there is something weird?
876
1:04:36,129 --> 1:04:39,892
I must find an explanation for
that.
877
1:04:39,892 --> 1:04:46,129
In my case, I had to find an
explanation because I couldn't
878
1:04:46,129 --> 1:04:49,462
sleep.
Who found an explanation?
879
1:04:49,462 --> 1:04:54,086
Who could say,
oh yes, don't worry about it?
880
1:04:54,086 --> 1:04:57,634
What was your solution?
Excuse me?
881
1:04:57,634 --> 1:05:01,947
Very good.
That's a very nice way of
882
1:05:01,947 --> 1:05:05,062
looking at it.
So that's probably why you
883
1:05:05,062 --> 1:05:08,178
could sleep.
Well, actually I'll tell you
884
1:05:08,178 --> 1:05:11,761
why I couldn't sleep.
But your solution is even
885
1:05:11,761 --> 1:05:14,565
shorter.
But the reason why I want to
886
1:05:14,565 --> 1:05:19,005
show you what I'm going to show
you is that I want to also
887
1:05:19,005 --> 1:05:22,977
expose you to the idea which you
already alluded to,
888
1:05:22,977 --> 1:05:27,105
namely that there is energy
involved when we deal with
889
1:05:27,105 --> 1:05:31,000
pulses and when we deal with
waves.
890
1:05:31,000 --> 1:05:36,688
You remember when we have a
traveling wave that the total
891
1:05:36,688 --> 1:05:42,479
energy per wavelength lambda,
I only did it per wavelength
892
1:05:42,479 --> 1:05:47,355
that the total energy,
perhaps you remember that,
893
1:05:47,355 --> 1:05:52,638
equals 2A squared times pi
squared times the tension,
894
1:05:52,638 --> 1:05:56,803
T, divided by lambda.
A was the amplitude.
895
1:05:56,803 --> 1:06:03,000
Energy is always proportional
to amplitude squared.
896
1:06:03,000 --> 1:06:06,500
This was the tension,
and this was the wavelength.
897
1:06:06,500 --> 1:06:10,428
And, if V2 goes to infinity,
then the wavelength goes to
898
1:06:10,428 --> 1:06:12,142
infinity.
That's obvious,
899
1:06:12,142 --> 1:06:14,500
right?
If something moves with the
900
1:06:14,500 --> 1:06:18,428
speed of light even faster,
infinity is even faster than
901
1:06:18,428 --> 1:06:21,857
the speed of light,
then lambda goes to infinity.
902
1:06:21,857 --> 1:06:25,357
So, this goes to zero.
And so, we came to the same
903
1:06:25,357 --> 1:06:29,214
conclusion, some decadent
solution, TR equals plus two,
904
1:06:29,214 --> 1:06:34,000
has no meaning because there is
no energy in there.
905
1:06:34,000 --> 1:06:41,307
And so, I was able to sleep.
Let's now turn to normal modes
906
1:06:41,307 --> 1:06:46,850
of continuous media.
And I suggest that we go
907
1:06:46,850 --> 1:06:53,149
longitudinal because we did so
much transfer stuff.
908
1:06:53,149 --> 1:06:59,700
Let's go longitudinal.
I have here a pi has length L,
909
1:06:59,700 --> 1:07:06,000
and it is open here,
and it is open there.
910
1:07:06,000 --> 1:07:09,733
That means the overpressure,
under pressure here,
911
1:07:09,733 --> 1:07:13,622
can never build up.
It's connected to the universe.
912
1:07:13,622 --> 1:07:17,744
So, at these two boundary
conditions, the speed that I
913
1:07:17,744 --> 1:07:21,711
have there must be zero.
If I write down the general
914
1:07:21,711 --> 1:07:26,222
equation for a standing wave,
because normal mode solutions
915
1:07:26,222 --> 1:07:30,033
are standing waves,
I can write down P equals some
916
1:07:30,033 --> 1:07:34,000
amplitude times the sine or the
cosine.
917
1:07:34,000 --> 1:07:38,454
Let me take a sine,
two pi divided by lambda times
918
1:07:38,454 --> 1:07:40,545
X.
I'll be very general.
919
1:07:40,545 --> 1:07:44,181
I will introduce some phase
angle, alpha,
920
1:07:44,181 --> 1:07:48,272
for which I will find a
solution very shortly.
921
1:07:48,272 --> 1:07:53,181
And then, cosine omega T,
or sine omega T if you prefer
922
1:07:53,181 --> 1:07:56,454
that.
This is a standing wave in very
923
1:07:56,454 --> 1:08:00,363
general terms.
Everything here is the space.
924
1:08:00,363 --> 1:08:04,181
X, this is X,
and here, all the information
925
1:08:04,181 --> 1:08:08,909
here deals with time,
which is typical for a standing
926
1:08:08,909 --> 1:08:13,724
wave.
[NOISE OBSCURES] always to
927
1:08:13,724 --> 1:08:17,393
write down for two pi over
lambda K.
928
1:08:17,393 --> 1:08:23,052
And I must observe the
situation that P must be zero at
929
1:08:23,052 --> 1:08:26,930
X equals zero,
and also at X equals L.
930
1:08:26,930 --> 1:08:33,218
If P is zero at X equals zero,
immediately you see that alpha
931
1:08:33,218 --> 1:08:37,765
is zero.
So I'm going to rewrite it now.
932
1:08:37,765 --> 1:08:42,582
I'm going to write down P,
P zero times the sine of KX.
933
1:08:42,582 --> 1:08:45,525
So I'm going to replace this by
K.
934
1:08:45,525 --> 1:08:50,253
I know that alpha is zero times
the cosine of omega T.
935
1:08:50,253 --> 1:08:55,070
And now, I must meet the
boundary condition that when X
936
1:08:55,070 --> 1:08:57,924
equals L, that P is,
again, zero.
937
1:08:57,924 --> 1:09:01,938
And that, now,
breaks open a whole spectrum of
938
1:09:01,938 --> 1:09:06,399
possibilities in which I
introduce this normal mode
939
1:09:06,399 --> 1:09:10,502
number N as in Nancy whereby N
can be one, two,
940
1:09:10,502 --> 1:09:15,895
three, etc.
And then I get solutions when K
941
1:09:15,895 --> 1:09:20,939
of N equals N pi divided by L.
You see that immediately
942
1:09:20,939 --> 1:09:26,170
because if I make X now L,
then I get the sine of M pi no
943
1:09:26,170 --> 1:09:30,000
matter what N is.
I always get zero.
944
1:09:30,000 --> 1:09:36,342
And so, my lambda of N,
which is two pi divided by K is
945
1:09:36,342 --> 1:09:41,862
then 2L divided by N.
So, I can rewrite now this
946
1:09:41,862 --> 1:09:48,439
equation as P zero times the
sine of N pi X divided by L.
947
1:09:48,439 --> 1:09:55,251
And now, I have cosine omega NT
because omega N are now the
948
1:09:55,251 --> 1:10:01,241
frequencies which are associated
with the N'th mode,
949
1:10:01,241 --> 1:10:07,437
N being N as in Nancy.
What now is the connection
950
1:10:07,437 --> 1:10:13,250
between this omega N and this K?
Well, that connection you will
951
1:10:13,250 --> 1:10:19,062
find through the wave equation.
You now have to substitute this
952
1:10:19,062 --> 1:10:24,406
result back into the wave
equation, which is this one's to
953
1:10:24,406 --> 1:10:28,343
solve or omega.
And, I want to do that with
954
1:10:28,343 --> 1:10:32,000
you.
It is not that much work.
955
1:10:32,000 --> 1:10:35,428
I go to 2PDX squared.
So, here it is:
956
1:10:35,428 --> 1:10:39,809
D2P DX squared.
So, all I get is I get this out
957
1:10:39,809 --> 1:10:43,238
twice.
So, I get N squared pi squared
958
1:10:43,238 --> 1:10:47,619
over L squared.
I get a minus sign because if I
959
1:10:47,619 --> 1:10:52,761
take twice the derivative,
I always end up with a minus
960
1:10:52,761 --> 1:10:55,904
sign.
But the sine comes back as a
961
1:10:55,904 --> 1:10:58,952
sine.
And, not only does the sine
962
1:10:58,952 --> 1:11:04,000
come back, but all the rest
comes back.
963
1:11:04,000 --> 1:11:06,495
And so, I will just write P
here.
964
1:11:06,495 --> 1:11:11,175
So, this P is exactly that P.
So, this is the only thing that
965
1:11:11,175 --> 1:11:15,153
was added by taking the second
derivative against X.
966
1:11:15,153 --> 1:11:19,130
Now, what is D2PDT squared?
Now I have to go to this
967
1:11:19,130 --> 1:11:22,172
function.
[I have?] partial derivatives.
968
1:11:22,172 --> 1:11:25,526
X is constant here,
but T is the one that is
969
1:11:25,526 --> 1:11:29,426
changing, whereas here,
we had that X was changing.
970
1:11:29,426 --> 1:11:33,964
The T was constant.
So now, we are going to get
971
1:11:33,964 --> 1:11:36,657
minus omega N squared,
again, times P.
972
1:11:36,657 --> 1:11:40,805
The whole function comes back.
And now, we are in business
973
1:11:40,805 --> 1:11:44,880
because now the wave equation
will tell us the connection
974
1:11:44,880 --> 1:11:47,936
between the two.
It tells us that N squared
975
1:11:47,936 --> 1:11:51,866
times pi squared divided by L
squared is now one over V
976
1:11:51,866 --> 1:11:54,631
squared.
That is the one over V squared
977
1:11:54,631 --> 1:11:58,124
that I had there.
And, there's a minus sign here.
978
1:11:58,124 --> 1:12:03,000
And there's a minus sign here
times omega N squared.
979
1:12:03,000 --> 1:12:09,437
So, this is not connected.
And so, you see the solution
980
1:12:09,437 --> 1:12:15,158
for omega N just is being
presented to you [NOISE
981
1:12:15,158 --> 1:12:20,165
OBSCURES] is now N pi times V
divided by L.
982
1:12:20,165 --> 1:12:24,337
That follows from the wave
equation.
983
1:12:24,337 --> 1:12:29,582
And so, if you prefer the
frequency in hertz,
984
1:12:29,582 --> 1:12:34,112
then you have to divide this by
two pi.
985
1:12:34,112 --> 1:12:41,373
So, you get NV divided by 2L.
So, if I want to plot now,
986
1:12:41,373 --> 1:12:45,482
so this is X equals zero,
and this X equals L,
987
1:12:45,482 --> 1:12:49,865
I can plot now here the
pressure in terms of this
988
1:12:49,865 --> 1:12:52,878
overpressure or under pressure,
P.
989
1:12:52,878 --> 1:12:57,352
And, you get a curve which
looks very similar to a
990
1:12:57,352 --> 1:13:01,118
transverse solution.
But, of course,
991
1:13:01,118 --> 1:13:04,550
it is not transverse.
It's really longitudinal.
992
1:13:04,550 --> 1:13:07,534
But you get,
then, that for N equals one,
993
1:13:07,534 --> 1:13:11,487
you would get this mode.
There must be a pressure node
994
1:13:11,487 --> 1:13:15,814
here and there because we can
never build a pressure that's
995
1:13:15,814 --> 1:13:19,693
connected with the universe.
You can never build that
996
1:13:19,693 --> 1:13:22,752
overpressure.
And so, the pressure buildup
997
1:13:22,752 --> 1:13:26,332
here is positive.
And then, later in time it will
998
1:13:26,332 --> 1:13:29,987
be negative, and then positive,
and then negative.
999
1:13:29,987 --> 1:13:34,314
And, when you go to N equals
two, then you get another node
1000
1:13:34,314 --> 1:13:38,903
in pressure here.
And so, now you get this,
1001
1:13:38,903 --> 1:13:43,245
always a pressure node here,
always a pressure node there.
1002
1:13:43,245 --> 1:13:47,129
But now you end up with another
pressure node there.
1003
1:13:47,129 --> 1:13:51,089
And, if you have any
difficulties to see what the air
1004
1:13:51,089 --> 1:13:54,821
molecules are doing,
I would recommend you go back
1005
1:13:54,821 --> 1:14:00,000
to psi space which is the actual
position of the molecules.
1006
1:14:00,000 --> 1:14:03,786
And when you do that,
you will always find that where
1007
1:14:03,786 --> 1:14:06,990
the pressure has an anti-node,
which is here,
1008
1:14:06,990 --> 1:14:10,703
psi always has a node,
and where the pressure has as
1009
1:14:10,703 --> 1:14:13,106
node, psi always has an
anti-node.
1010
1:14:13,106 --> 1:14:16,820
Of course, the molecules can
freely move in and out.
1011
1:14:16,820 --> 1:14:20,315
There is no problem.
So, the molecules can freely
1012
1:14:20,315 --> 1:14:23,956
move in and out here.
So, where psi has its largest
1013
1:14:23,956 --> 1:14:27,669
amplitude, its anti-node,
that is where the pressure
1014
1:14:27,669 --> 1:14:33,687
cannot build up.
But going to psi space actually
1015
1:14:33,687 --> 1:14:41,182
often helps me to see precisely
what is going on with the motion
1016
1:14:41,182 --> 1:14:46,535
of the molecules.
I have here a linear system,
1017
1:14:46,535 --> 1:14:52,126
which is a sound cavity.
It's made of magnesium.
1018
1:14:52,126 --> 1:14:56,646
[There's?] no air,
but it is magnesium.
1019
1:14:56,646 --> 1:15:02,000
And, it is open,
open on both sides.
1020
1:15:02,000 --> 1:15:09,563
You can't have it any better.
I'll make a drawing for you.
1021
1:15:09,563 --> 1:15:14,739
So, here is my magnesium.
This is a rod.
1022
1:15:14,739 --> 1:15:21,241
It's one-dimensional,
and the length is 122 cm and
1023
1:15:21,241 --> 1:15:30,000
the speed of sound in magnesium
is about 5,000 m per second.
1024
1:15:30,000 --> 1:15:36,137
When I hit that magnesium rot
on the side, it wants to go into
1025
1:15:36,137 --> 1:15:40,362
standing waves.
It prefers the lowest mode.
1026
1:15:40,362 --> 1:15:45,493
It almost always does.
But it may also create second
1027
1:15:45,493 --> 1:15:49,920
and third harmonics.
And so, the lowest mode,
1028
1:15:49,920 --> 1:15:55,253
F1, is then V divided by 2L.
So, the lowest frequency,
1029
1:15:55,253 --> 1:16:00,182
F1, is V divided by 2L,
which when I calculate it,
1030
1:16:00,182 --> 1:16:05,591
is about 250 Hz.
And the actual value we measure
1031
1:16:05,591 --> 1:16:08,767
is about 2,044.
And, there may also be,
1032
1:16:08,767 --> 1:16:13,614
when I hit it with a hammer,
there may also be some that is
1033
1:16:13,614 --> 1:16:16,791
twice as high.
So that may be 4,100 Hz.
1034
1:16:16,791 --> 1:16:19,632
That would be double the
frequency.
1035
1:16:19,632 --> 1:16:24,564
That would be the second mode.
And this is quite remarkable.
1036
1:16:24,564 --> 1:16:29,077
So this is not filled with air,
but this is filled with
1037
1:16:29,077 --> 1:16:32,947
magnesium.
But, by exciting it here,
1038
1:16:32,947 --> 1:16:35,751
it's like blowing on the flute
there.
1039
1:16:35,751 --> 1:16:40,346
It goes into these normal mode
solutions, and it is this one
1040
1:16:40,346 --> 1:16:42,917
that you will hear loud and
clear.
1041
1:16:42,917 --> 1:16:46,811
It's a beautiful tone.
And this one you may hear in
1042
1:16:46,811 --> 1:16:50,238
the beginning,
but the higher harmonics often
1043
1:16:50,238 --> 1:16:53,276
die out faster than the lower
harmonics.
1044
1:16:53,276 --> 1:16:56,158
So, are you ready for this?
Open-open.
1045
1:16:56,158 --> 1:17:00,442
It's open on both sides,
and I just bang the hell out of
1046
1:17:00,442 --> 1:17:02,898
it.
2,044.
1047
1:17:02,898 --> 1:17:10,085
Isn't that beautiful?
It's oscillating like this.
1048
1:17:10,085 --> 1:17:18,770
Exactly oscillating the way
that I derived for you for air.
1049
1:17:18,770 --> 1:17:24,459
And here, you see that it
[holds for?].
1050
1:17:24,459 --> 1:17:33,314
Only the fundamental is there.
OK, I wish you luck on
1051
1:17:33,314 --> 1:17:34,670
Thursday.
I'll see you then.