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PROFESSOR: I want to start with
the physical pendulum,
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which is exactly the same one
that I discussed during my
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first lecture.
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This is a hoop.
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It has mass m and 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 torque
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equation from 8.01.
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The torque relative to point
P is the moment of inertia
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relative to point P times the
angular acceleration alpha.
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Today, I will do this again,
but I will use the
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conservation of energy to show
you that in case there is no
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damping, when mechanical energy
is conserved, that you
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can find the correct
differential equations through
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the conservation of energy.
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If the thing is swinging, in
general, there are two
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components.
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You have kinetic energy, and
you have potential energy.
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And the kinetic energy, K, is
1/2 times the moment of
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inertia about point P
times omega squared.
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You remember that from 8.01.
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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 to
equilibrium, the angular
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velocity is at maximum, and
when the object comes to a
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halt, the angular
velocity is 0.
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Do not confuse this omega
with omega 0.
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I will give that as 0 now to
distinguish it from omega,
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which is the angular
frequency.
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The angular frequency is a
constant of the motion, and
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that is 2 pi divided by T0
if T0 is the period of
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oscillation.
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So this is the kinetic energy,
and this is the square of the
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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, the center of
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mass is at point B. And so
the potential energy u--
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or actually, I should say the
potential energy at point B
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minus the potential energy at
point A-- it's always the
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difference in potential
energy that matters--
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that equals mgh, h being the
difference in height between
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point B and point A. So
this here is h-- mgh--
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Massachusetts General
Hospital.
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That's the way to remember it.
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Now, h is very easy.
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h is the same as R times 1 minus
the cosine of theta.
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We went through that many
times, so you can easily
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confirm that.
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R is the radius of
this circle.
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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 2.
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So I can rewrite this
differential
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equation now that E total--
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AUDIENCE: [INAUDIBLE]?
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PROFESSOR: I see nothing
wrong with that.
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I'm sorry.
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Is there anything
wrong with it?
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AUDIENCE: I thought the cosine
was 1 minus theta
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squared over 2.
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PROFESSOR: The cosine's
theta alone--
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you're right, is 1 minus
theta squared over 2.
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Thank you very much.
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Thank you.
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So I'm going to write it now as
the total energy is 1/2 Ip
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times theta dot squared.
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And then I get plus mgR times
theta squared over 2.
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And what you do now, since
this is a constant of the
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motion, you always do that
if you work with the
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conservation of energy.
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You take the time derivative of
this equation, which must
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be 0, because mechanical energy
is conserved, and out
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pops the differential equation
that we also found during my
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lecture number one, when I used
the different method.
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So I take the time
derivative--
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so this T eats up this 1/2,
and so we get I of p times
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theta dot times theta
double dot--
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I have to use the chain rule--
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and then I get plus mgR.
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This 2 eats up this 2, and so
I get theta times theta dot,
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and now this equals 0,
because the EDT is 0.
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And whenever you do that, you
will always see that the theta
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dot term disappears, or if you
have the equation in x, then
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the x dot term disappears.
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And you see that.
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This term goes.
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You can divide it out.
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And so your differential
equation takes on, now, a very
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familiar form.
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Let me write this 2 here.
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So I get theta double dot plus
mgR divided by I of p times
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theta equals 0.
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And this differential equation,
you should recognize
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the solution to that equation
is immediately
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obvious, omega 0 squared.
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I use now the omega 0 equals mg
times R divided by I of p.
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So the general solution for this
oscillation then becomes
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that theta is theta 0--
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that is the amplitude--
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times the cosine--
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or the sine, if you
prefer that--
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omega 0 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 0, if you know
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the angular velocity at t equals
0, and if you know
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where it is, t equals 0, you
can solve for this theta 0,
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and you can solve
for this phi.
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But omega 0 is independent of
the initial conditions.
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Very well.
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Now I would like to cover a
case whereby I'm going to
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introduce damping.
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Whenever we deal with damping,
there are two terms that are
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important in physics.
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That is the speed itself, there
is a damping term, which
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opposes the velocity,
which is linearly
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proportional with the speed.
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And there is a damping term
which opposes the velocity,
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which is proportional with
the square of the speed.
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We will always leave the square
out with 8.03, because
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the differential equations
become impossible to solve.
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Maybe numerically you can do
it, but not analytically.
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However, if you're interested
in all the physics, which is
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wonderful, with the v square and
the v, my lecture number
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12 on OCW--
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OpenCourseWare--
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from 1999, Newtonian Mechanics,
I deal with the v
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square and with the v term, and
I do many demonstrations
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to show you that there are
certain domains where the v
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term is important.
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Recall 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, to the damping
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force, which is linearly
proportional with the
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velocity, and we will write it
down as F equals minus b times
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v. We will use a shorthand
notation that gamma
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equals b over m.
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That is only a shorthand
notation.
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I will erase this.
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We don't need this anymore.
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We need so many blackboards
today that I will use this
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blackboard for my
damping problem.
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When you deal with damping,
we recognize
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three different domains.
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One domain, whereby gamma is
smaller than omega 0, we call
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that under-damped.
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Then we have a domain whereby
gamma is larger than omega 0.
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We call that over-damped.
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And then you have a very special
case where gamma
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equals omega 0.
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And the behavior of these
three different
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kinds is very different.
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I will only discuss with you
today the under-damped case.
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So I have a pendulum, and this
pendulum has mass m and length
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l, and I will assume that the
mass in the string is
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negligibly small.
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I have damping, and this is the
case that gamma is smaller
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than omega 0.
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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
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put all the forces on this
object in this picture, which
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is mg, and that is T. And
there are no other
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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
T is very close to mg.
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So the only force that
is driving it back to
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equilibrium, the only restoring
force, then, is the
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horizontal component
of T. So that is
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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
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times x dot.
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That's the damping, and this is
the restoring force due to
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the tangent in the string.
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And the sine of theta
is x divided by l.
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So I get mx double dot plus mg
divided by l times x-- that is
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the sine of theta--
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plus b times x dot equals 0.
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Make sure I have this,
m doubled dot.
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I have the plus sign here.
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I have mg x over l.
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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 0 squared
is g divided by l.
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These are shorthand notations,
which give you a little bit
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more insight when you
see the solutions.
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So that gives me, now, the
differential equation that x
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double dot plus gamma
times x dot plus--
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I divide m out--
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plus omega 0 squared
times x equals 0.
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And that is the differential
equation
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that you will recognize.
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And I would never want you to
derive the solution to this
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00:12:54,700 --> 00:12:56,030
differential equation.
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00:12:56,030 --> 00:12:59,840
That's, of course, way too
time consuming to do that
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during an exam.
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And 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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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
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to the minus gamma
over 2 times t.
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00:13:26,330 --> 00:13:33,830
And then we have a cosine or a
sine, cosine omega t plus some
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phase angle alpha.
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218
00:13:39,960 --> 00:13:43,960
If you knew the initial
conditions, then you can find
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what the amplitude is,
and you can find what
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the angle alpha is.
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If you don't know the initial
conditions, then
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you do not know this.
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00:13:54,090 --> 00:13:57,540
There is another way that you
can write this form, which is
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sometimes better.
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00:13:59,810 --> 00:14:02,550
And I cannot tell you when
it is better and
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00:14:02,550 --> 00:14:03,350
when it is not better.
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00:14:03,350 --> 00:14:05,690
It depends on the initial
conditions.
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00:14:05,690 --> 00:14:08,920
But I want you to appreciate
that you can also write this,
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00:14:08,920 --> 00:14:14,320
for instance, as e to the minus
gamma over 2 times t and
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00:14:14,320 --> 00:14:20,800
then B times cosine omega
t plus C times the
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00:14:20,800 --> 00:14:22,050
sine of omega t.
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00:14:22,050 --> 00:14:25,200
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From the physics point of view,
there is no difference.
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00:14:27,340 --> 00:14:29,860
But from the math point of view,
there is a difference.
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You now have these as the two
adjustable constants depending
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upon the initial conditions.
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And sometimes, if you assume
this, it works faster than if
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you assume that.
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And as I said, it really
depends on the initial
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conditions which goes faster.
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But they are very similar,
of course.
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Omega which is now the
frequency which this
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00:14:55,040 --> 00:14:56,800
object is going to--
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angular frequency is going to
move, that omega is always
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lower than omega 0.
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And that shouldn't surprise
you, because when there is
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damping, there is something that
is opposing the motion.
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And so it shouldn't surprise
you that when you solve for
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omega that omega squared is
omega 0 squared minus gamma
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squared divided by 4.
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That is also something that we
would probably give you on the
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formula sheet, because you will
only find that if you
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00:15:26,400 --> 00:15:28,850
substitute this back into the
differential equation.
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00:15:28,850 --> 00:15:32,110
256
00:15:32,110 --> 00:15:35,090
We often-- particularly
French--
257
00:15:35,090 --> 00:15:39,630
like to write down q equals
omega divided by gamma.
258
00:15:39,630 --> 00:15:40,740
This is the quality--
259
00:15:40,740 --> 00:15:43,090
omega 0 divided by gamma.
260
00:15:43,090 --> 00:15:47,070
And if you do that, then omega
squared can also be written as
261
00:15:47,070 --> 00:15:55,360
omega 0 squared times 1 minus
1 over 4Q squared.
262
00:15:55,360 --> 00:15:59,445
So you see that if Q is, for
instance, 10, which is by no
263
00:15:59,445 --> 00:16:04,430
means absurdly high, that omega
is only 1/8 of a percent
264
00:16:04,430 --> 00:16:05,610
lower than omega 0.
265
00:16:05,610 --> 00:16:07,750
They are that close.
266
00:16:07,750 --> 00:16:11,770
And even when Q is two, the
difference is only 3% between
267
00:16:11,770 --> 00:16:15,160
omega and omega 0.
268
00:16:15,160 --> 00:16:21,520
So what happens here is that the
amplitude is decaying at a
269
00:16:21,520 --> 00:16:22,760
time constant--
270
00:16:22,760 --> 00:16:24,950
the 1 over e time constant,
which is 2
271
00:16:24,950 --> 00:16:27,760
divided by gamma seconds.
272
00:16:27,760 --> 00:16:31,430
If you put in t, 2 divided by
gamma, the amplitude goes down
273
00:16:31,430 --> 00:16:33,000
by a factor of e.
274
00:16:33,000 --> 00:16:35,690
Since energy is always
proportional to amplitude
275
00:16:35,690 --> 00:16:41,150
squared, the decay time of the
energy is not 2 over gamma but
276
00:16:41,150 --> 00:16:43,230
is 1 divided by gamma.
277
00:16:43,230 --> 00:16:47,230
278
00:16:47,230 --> 00:16:50,700
Now I'm going to
make a change.
279
00:16:50,700 --> 00:16:53,080
I'm going to drive
this system.
280
00:16:53,080 --> 00:16:57,690
This is no longer the
equilibrium position, but the
281
00:16:57,690 --> 00:17:00,280
equilibrium position
was really here.
282
00:17:00,280 --> 00:17:04,900
And I am driving the top of this
pendulum with a function
283
00:17:04,900 --> 00:17:13,020
eta is eta 0 times the
cosine of omega t.
284
00:17:13,020 --> 00:17:14,300
So I am driving it now.
285
00:17:14,300 --> 00:17:17,440
286
00:17:17,440 --> 00:17:20,720
So this eta is in terms of
inches-- millimeters.
287
00:17:20,720 --> 00:17:23,220
This is the motion of my hand.
288
00:17:23,220 --> 00:17:27,170
And this omega is no
longer negotiable.
289
00:17:27,170 --> 00:17:31,110
This omega has nothing to
do with this omega.
290
00:17:31,110 --> 00:17:32,760
This is the frequency
with which the
291
00:17:32,760 --> 00:17:35,210
system likes to oscillate.
292
00:17:35,210 --> 00:17:38,520
That, now, is the frequency with
which I want the system
293
00:17:38,520 --> 00:17:39,910
to oscillate.
294
00:17:39,910 --> 00:17:41,850
They're totally unrelated.
295
00:17:41,850 --> 00:17:43,560
This is my will.
296
00:17:43,560 --> 00:17:44,750
This is non-negotiable.
297
00:17:44,750 --> 00:17:46,490
I can make these anything
I want there.
298
00:17:46,490 --> 00:17:47,760
I can make it 0.
299
00:17:47,760 --> 00:17:49,020
I can make it large.
300
00:17:49,020 --> 00:17:52,640
I can make it, also, that
value, of course.
301
00:17:52,640 --> 00:17:55,280
So now the equilibrium
position is not here.
302
00:17:55,280 --> 00:18:01,390
303
00:18:01,390 --> 00:18:04,880
And I will call, now, this
distance from the equilibrium
304
00:18:04,880 --> 00:18:05,700
position x.
305
00:18:05,700 --> 00:18:09,090
You always use in your
coordinate system the
306
00:18:09,090 --> 00:18:12,590
equilibrium position,
as your 0.
307
00:18:12,590 --> 00:18:18,410
So what is the only thing that
is going to change now is that
308
00:18:18,410 --> 00:18:23,660
the sine of theta is no longer
x divided by l, but the sine
309
00:18:23,660 --> 00:18:31,940
of theta is x minus eta divided
by l, because notice,
310
00:18:31,940 --> 00:18:36,290
if this is eta, which is the
position of my hands, then the
311
00:18:36,290 --> 00:18:42,860
sine of this angle is now x
minus eta divided by l.
312
00:18:42,860 --> 00:18:46,190
And this, now, you have
to carry through your
313
00:18:46,190 --> 00:18:48,450
differential equations.
314
00:18:48,450 --> 00:18:53,270
And what you will see, then, if
you do that, that this now
315
00:18:53,270 --> 00:19:04,190
is not 0, but this now becomes
eta 0 times omega 0 squared
316
00:19:04,190 --> 00:19:09,725
times the cosine of
omega t, my omega.
317
00:19:09,725 --> 00:19:12,610
318
00:19:12,610 --> 00:19:17,230
So instead of sine theta x
divided by l, all you have to
319
00:19:17,230 --> 00:19:19,160
do is this.
320
00:19:19,160 --> 00:19:21,910
You know that eta is eta
0 cosine omega t.
321
00:19:21,910 --> 00:19:24,600
You carry that through, and
you will see that your
322
00:19:24,600 --> 00:19:29,560
differential equation
now changes.
323
00:19:29,560 --> 00:19:36,600
Notice that x double dot
is an acceleration.
324
00:19:36,600 --> 00:19:40,910
Notice that gamma times x
dot is an acceleration.
325
00:19:40,910 --> 00:19:45,100
Gamma is 1 over second, and x
dot is meters per second.
326
00:19:45,100 --> 00:19:47,300
This is an acceleration.
327
00:19:47,300 --> 00:19:49,550
Notice that this is
an acceleration.
328
00:19:49,550 --> 00:19:52,750
It has a dimension omega
squared, which is 1 second
329
00:19:52,750 --> 00:19:54,650
squared times meters.
330
00:19:54,650 --> 00:19:56,410
That is an acceleration.
331
00:19:56,410 --> 00:19:58,940
Notice that this is
an acceleration.
332
00:19:58,940 --> 00:20:02,550
Eta 0 is meters, and omega
squared is 1 divided by
333
00:20:02,550 --> 00:20:03,690
seconds squared.
334
00:20:03,690 --> 00:20:07,660
So these are apples, these are
apples, these are apples, and
335
00:20:07,660 --> 00:20:09,360
these are apples.
336
00:20:09,360 --> 00:20:14,290
So the equation looks
very kosher to me.
337
00:20:14,290 --> 00:20:17,750
Now, the solutions, that's
a different story.
338
00:20:17,750 --> 00:20:20,750
That is something that I
wouldn't want you to derive
339
00:20:20,750 --> 00:20:26,160
either, but the solutions
now become as follows.
340
00:20:26,160 --> 00:20:30,630
341
00:20:30,630 --> 00:20:35,240
I now get x as a
function of t.
342
00:20:35,240 --> 00:20:38,970
There's an amplitude, which is
a very strong function of
343
00:20:38,970 --> 00:20:43,640
omega, but I will simply write
it down as an amplitude A
344
00:20:43,640 --> 00:20:50,460
times the cosine of my omega t
minus some phase angle delta.
345
00:20:50,460 --> 00:20:53,770
346
00:20:53,770 --> 00:20:56,775
And this is what we call the
steady-state solution--
347
00:20:56,775 --> 00:20:59,770
348
00:20:59,770 --> 00:21:01,020
steady state.
349
00:21:01,020 --> 00:21:02,990
350
00:21:02,990 --> 00:21:07,670
And this A becomes, then, a
rather complicated function.
351
00:21:07,670 --> 00:21:10,550
352
00:21:10,550 --> 00:21:13,880
Upstairs, I get the eta
0 omega 0 squared.
353
00:21:13,880 --> 00:21:15,130
I get this part.
354
00:21:15,130 --> 00:21:19,180
355
00:21:19,180 --> 00:21:26,290
And downstairs, I get the square
root of omega 0 squared
356
00:21:26,290 --> 00:21:35,160
minus omega squared squared
plus omega gamma squared.
357
00:21:35,160 --> 00:21:38,590
And that is the amplitude
as a function of omega.
358
00:21:38,590 --> 00:21:44,780
And the tangent of delta becomes
omega gamma divided by
359
00:21:44,780 --> 00:21:48,180
omega 0 squared minus
omega squared.
360
00:21:48,180 --> 00:21:54,180
361
00:21:54,180 --> 00:22:02,980
If we plot that function A
as a function of omega--
362
00:22:02,980 --> 00:22:07,700
363
00:22:07,700 --> 00:22:12,530
so here is omega, and here,
I'm going to plot the A--
364
00:22:12,530 --> 00:22:18,970
then I can recognize that if
omega goes to 0, that means if
365
00:22:18,970 --> 00:22:26,580
I move this very slowly, that
for sure the amplitude of this
366
00:22:26,580 --> 00:22:30,230
object must be the same as my
hand-- must be at eta 0.
367
00:22:30,230 --> 00:22:33,160
If it takes me one week to go
from here to here, then of
368
00:22:33,160 --> 00:22:36,130
course, the pendulum is always
hanging below my hand.
369
00:22:36,130 --> 00:22:43,220
So when omega goes to 0, I can
always check the result here
370
00:22:43,220 --> 00:22:46,000
that A must go to eta 0.
371
00:22:46,000 --> 00:22:49,450
And indeed, if you put in omega
equals 0, you will see
372
00:22:49,450 --> 00:22:50,700
that is the case.
373
00:22:50,700 --> 00:22:53,080
374
00:22:53,080 --> 00:23:01,240
If omega goes to infinity,
then A goes to 0.
375
00:23:01,240 --> 00:23:04,260
And then there is a very special
case that is when
376
00:23:04,260 --> 00:23:12,790
omega happens to be omega 0,
which is the frequency of the
377
00:23:12,790 --> 00:23:15,140
system in the absence
of damping.
378
00:23:15,140 --> 00:23:16,380
This is my omega 0.
379
00:23:16,380 --> 00:23:19,160
It is not this one, but
it is the omega 0.
380
00:23:19,160 --> 00:23:26,140
Then you will get an amplitude
which is Q times eta 0.
381
00:23:26,140 --> 00:23:28,720
And again, I just happened
to remember that.
382
00:23:28,720 --> 00:23:31,370
If you substitute this back
in here-- this omega
383
00:23:31,370 --> 00:23:32,700
equals omega 0--
384
00:23:32,700 --> 00:23:34,230
you will see that.
385
00:23:34,230 --> 00:23:37,820
And that's always very nice to
remember, at that omega 0, you
386
00:23:37,820 --> 00:23:40,140
always get Q times more
than your driver.
387
00:23:40,140 --> 00:23:42,490
Is Q is 100, you'll
get 100 times the
388
00:23:42,490 --> 00:23:45,940
amplitude of the driver.
389
00:23:45,940 --> 00:23:51,140
So here, you get eta
0 if omega is 0.
390
00:23:51,140 --> 00:23:56,190
And then, when you reach that
omega 0 frequency, you come
391
00:23:56,190 --> 00:24:03,280
out high, and then it goes back
to 0, and this point here
392
00:24:03,280 --> 00:24:07,450
is Q times eta 0.
393
00:24:07,450 --> 00:24:11,440
And we discussed before, I'll
never make too much of a deal
394
00:24:11,440 --> 00:24:15,490
out of that, that the actual
maximum of this curve is not
395
00:24:15,490 --> 00:24:18,930
at omega 0 but is always
a little lower.
396
00:24:18,930 --> 00:24:24,250
That is really more of an
algebraic interest than it is
397
00:24:24,250 --> 00:24:28,270
of physical interest, because if
gamma is low, if Q is high,
398
00:24:28,270 --> 00:24:29,850
the two almost coincide.
399
00:24:29,850 --> 00:24:33,270
400
00:24:33,270 --> 00:24:38,840
So this steady-state solution
has no adjustable constant
401
00:24:38,840 --> 00:24:42,070
under which the system has
lost its memory of what
402
00:24:42,070 --> 00:24:46,870
happened when we started
driving it.
403
00:24:46,870 --> 00:24:51,980
So at t equals 0, if we know
what x is, and if we know what
404
00:24:51,980 --> 00:24:54,970
x dot is, and if we know what
the driving term is at t
405
00:24:54,970 --> 00:24:59,560
equals 0, then the general
solution for all times larger
406
00:24:59,560 --> 00:25:03,600
than 0 is the sum of the
transient solution and the
407
00:25:03,600 --> 00:25:05,530
steady-state solution.
408
00:25:05,530 --> 00:25:08,230
And so here, you see the
transient solution, you can
409
00:25:08,230 --> 00:25:11,150
write it in this form, or you
can write it in this form.
410
00:25:11,150 --> 00:25:13,340
And here, you see the
steady-state solution.
411
00:25:13,340 --> 00:25:18,850
This A has nothing to do with
that A. This A follows from
412
00:25:18,850 --> 00:25:20,210
the initial conditions.
413
00:25:20,210 --> 00:25:23,460
Just like that alpha, this A
does not follow from the
414
00:25:23,460 --> 00:25:24,510
initial conditions.
415
00:25:24,510 --> 00:25:28,920
This A follows from eta 0, what
my amplitude is of my
416
00:25:28,920 --> 00:25:31,530
driver, and it follows
from omega.
417
00:25:31,530 --> 00:25:35,420
So the two As don't
confuse the two.
418
00:25:35,420 --> 00:25:43,800
So the general solution is
then the sum of the two.
419
00:25:43,800 --> 00:25:46,660
The transient one
will die out.
420
00:25:46,660 --> 00:25:50,760
If you wait a few times 2 over
gamma, the transient is gone,
421
00:25:50,760 --> 00:25:52,960
and so you end up, then,
only with the
422
00:25:52,960 --> 00:25:54,730
steady-state solution.
423
00:25:54,730 --> 00:25:58,430
And you had some chance
in your homework
424
00:25:58,430 --> 00:26:00,140
to work with that.
425
00:26:00,140 --> 00:26:03,490
426
00:26:03,490 --> 00:26:07,750
If we are not the driving the
system, and if we have one
427
00:26:07,750 --> 00:26:11,680
object on a spring or on a
pendulum or on a floating
428
00:26:11,680 --> 00:26:18,340
object in liquid, then there is
one frequency, one normal
429
00:26:18,340 --> 00:26:22,470
mode frequency, one resonance
frequency, we call it also the
430
00:26:22,470 --> 00:26:24,200
natural frequency.
431
00:26:24,200 --> 00:26:27,240
The moment that you couple
oscillators, if you couple
432
00:26:27,240 --> 00:26:30,800
two, you get two normal
mode frequencies.
433
00:26:30,800 --> 00:26:33,040
You get two resonance
frequencies.
434
00:26:33,040 --> 00:26:37,790
And if you have three objects,
you get three normal mode
435
00:26:37,790 --> 00:26:39,370
frequencies.
436
00:26:39,370 --> 00:26:44,650
So now I would like to discuss
with you a case whereby I
437
00:26:44,650 --> 00:26:48,880
couple two oscillators.
438
00:26:48,880 --> 00:26:53,540
If I gave you on an exam three
coupled oscillators, that
439
00:26:53,540 --> 00:26:57,190
would be very nasty, because
it's extremely time consuming.
440
00:26:57,190 --> 00:27:00,380
If I gave you four coupled
oscillators, that would be
441
00:27:00,380 --> 00:27:04,990
criminal, because you cannot
finish that in 85 minutes.
442
00:27:04,990 --> 00:27:07,450
So two is certainly
within reason.
443
00:27:07,450 --> 00:27:10,000
Three is marginally
within reason.
444
00:27:10,000 --> 00:27:13,620
Four is out of the question.
445
00:27:13,620 --> 00:27:16,240
When we have coupled
oscillators, we always leave
446
00:27:16,240 --> 00:27:17,640
damping out.
447
00:27:17,640 --> 00:27:19,490
And yet, we will learn
a lot from it.
448
00:27:19,490 --> 00:27:23,070
Even without damping,
we will learn a lot.
449
00:27:23,070 --> 00:27:28,750
So what I have chosen to do with
you is a spring system
450
00:27:28,750 --> 00:27:36,410
with two objects, two masses,
and the springs have no mass,
451
00:27:36,410 --> 00:27:37,330
negligible mass.
452
00:27:37,330 --> 00:27:39,950
The spring constants are
k, and the masses of
453
00:27:39,950 --> 00:27:42,960
the objects are m.
454
00:27:42,960 --> 00:27:48,220
So this is the equilibrium
position of these two objects,
455
00:27:48,220 --> 00:27:53,430
and the ends are fixed
of the string.
456
00:27:53,430 --> 00:27:58,830
I will introduce a shorthand
notation that omega 0 squared
457
00:27:58,830 --> 00:28:02,430
equals k over m.
458
00:28:02,430 --> 00:28:07,150
Now what I do, I offset both
these objects from equilibrium
459
00:28:07,150 --> 00:28:10,530
just as I did that here.
460
00:28:10,530 --> 00:28:13,480
I always offset them in the
same direction, and I call
461
00:28:13,480 --> 00:28:16,210
that positive.
462
00:28:16,210 --> 00:28:17,790
Is that a must?
463
00:28:17,790 --> 00:28:19,530
No.
464
00:28:19,530 --> 00:28:20,990
Is that useful?
465
00:28:20,990 --> 00:28:22,590
Yes.
466
00:28:22,590 --> 00:28:25,800
So I offset them.
467
00:28:25,800 --> 00:28:29,610
So this position, now-- this is
object number one, and this
468
00:28:29,610 --> 00:28:30,980
object number two--
469
00:28:30,980 --> 00:28:35,880
this is now a distance x1 away
from equilibrium, and this one
470
00:28:35,880 --> 00:28:40,530
is now at a position x2 away
from its equilibrium.
471
00:28:40,530 --> 00:28:45,310
You always want to know what the
displacement is relative
472
00:28:45,310 --> 00:28:48,200
to its own equilibrium.
473
00:28:48,200 --> 00:28:51,360
And its own equilibrium for 1
is here, and the equilibrium
474
00:28:51,360 --> 00:28:54,590
for 2 is there.
475
00:28:54,590 --> 00:28:59,060
I now make the following
assumption that x2
476
00:28:59,060 --> 00:29:01,730
is larger than x1.
477
00:29:01,730 --> 00:29:03,230
Is that a must?
478
00:29:03,230 --> 00:29:04,250
No.
479
00:29:04,250 --> 00:29:05,770
Is it useful?
480
00:29:05,770 --> 00:29:06,380
Yes.
481
00:29:06,380 --> 00:29:09,130
But if you want to assume
that x2 is smaller
482
00:29:09,130 --> 00:29:11,780
than x1, be my guest.
483
00:29:11,780 --> 00:29:15,750
I will assume that x2
is larger than x1.
484
00:29:15,750 --> 00:29:18,490
Now, follow me closely now.
485
00:29:18,490 --> 00:29:23,210
So we have an object here, and
we have an object here.
486
00:29:23,210 --> 00:29:26,850
And this is now one spring.
487
00:29:26,850 --> 00:29:32,220
This is the other spring, and
this is the third spring.
488
00:29:32,220 --> 00:29:35,010
It is immediately obvious that
this spring is longer than it
489
00:29:35,010 --> 00:29:35,950
wants to be.
490
00:29:35,950 --> 00:29:40,320
So there is a force that drives
it back to equilibrium.
491
00:29:40,320 --> 00:29:43,980
If x2 is larger than x1, this
spring is also longer than it
492
00:29:43,980 --> 00:29:44,840
wants to be.
493
00:29:44,840 --> 00:29:46,920
So it wants to contract,
so there is a
494
00:29:46,920 --> 00:29:48,170
force in that direction.
495
00:29:48,170 --> 00:29:50,670
496
00:29:50,670 --> 00:29:53,840
If this spring is longer than
it wants to be, it wants to
497
00:29:53,840 --> 00:29:58,210
contract, so there is a force
on x2 in this direction.
498
00:29:58,210 --> 00:30:01,310
This spring is clearly shorter
than it wants to be, so it's
499
00:30:01,310 --> 00:30:04,800
pushing, so there is a force
in this direction.
500
00:30:04,800 --> 00:30:08,840
So now, I'm going to write down
the differential equation
501
00:30:08,840 --> 00:30:12,350
first for object number one.
502
00:30:12,350 --> 00:30:17,030
So I've got n x1 double dot.
503
00:30:17,030 --> 00:30:25,850
So this one is minus k times
x1, and this one is plus k
504
00:30:25,850 --> 00:30:28,860
times x2 minus x1.
505
00:30:28,860 --> 00:30:32,410
506
00:30:32,410 --> 00:30:35,210
Is it a disaster if it
turns out that x2 is
507
00:30:35,210 --> 00:30:36,820
not larger than x1?
508
00:30:36,820 --> 00:30:37,980
Not at all.
509
00:30:37,980 --> 00:30:42,070
This equation now is correct
for all situations.
510
00:30:42,070 --> 00:30:45,970
The fact that I have assumed
that x2 is larger than x1 gave
511
00:30:45,970 --> 00:30:48,090
me a plus sign here.
512
00:30:48,090 --> 00:30:53,230
So my differential equation is
safe no matter what x1 is
513
00:30:53,230 --> 00:30:54,610
relative to x2.
514
00:30:54,610 --> 00:30:57,240
So you can always make that
assumption, and you don't have
515
00:30:57,240 --> 00:30:59,555
to worry later anymore
about signs.
516
00:30:59,555 --> 00:31:02,720
517
00:31:02,720 --> 00:31:09,800
I can divide, now, m out, and
then I get that x1 double dot
518
00:31:09,800 --> 00:31:13,290
plus 2 omega 0 squared
times x1.
519
00:31:13,290 --> 00:31:16,030
Well, you notice I have one
here, and I have one here that
520
00:31:16,030 --> 00:31:18,080
have both the minus sign.
521
00:31:18,080 --> 00:31:21,680
And then I have here plus sort
of becomes minus omega 0
522
00:31:21,680 --> 00:31:24,040
squared times x2.
523
00:31:24,040 --> 00:31:25,510
That is 0.
524
00:31:25,510 --> 00:31:29,400
So that is my first differential
equation.
525
00:31:29,400 --> 00:31:30,650
So I put a 1 here.
526
00:31:30,650 --> 00:31:33,800
527
00:31:33,800 --> 00:31:37,370
So now I go to the second one.
528
00:31:37,370 --> 00:31:41,950
I get mx 2 double dot.
529
00:31:41,950 --> 00:31:45,240
Now I have two forces, both
in the negative direction.
530
00:31:45,240 --> 00:31:48,290
First, I have the one that
drives this one away from
531
00:31:48,290 --> 00:31:49,930
equilibrium.
532
00:31:49,930 --> 00:31:55,190
So I get minus k times
x2 minus x1.
533
00:31:55,190 --> 00:31:59,200
And then I have the force due
to the fact that this one is
534
00:31:59,200 --> 00:32:02,610
shorter than it wants to be
by an amount k2, so I get
535
00:32:02,610 --> 00:32:06,460
minus k times x2.
536
00:32:06,460 --> 00:32:10,260
It's shorter by an amount
x2 than it wants to be.
537
00:32:10,260 --> 00:32:15,730
So I can divide m out, so I get
mx2 double dot, and then I
538
00:32:15,730 --> 00:32:20,240
get plus omega 0 squared
times x2.
539
00:32:20,240 --> 00:32:26,165
And then I get minus omega 0
squared times x1 equals 0.
540
00:32:26,165 --> 00:32:29,400
541
00:32:29,400 --> 00:32:33,280
Now, compare this one
with this one.
542
00:32:33,280 --> 00:32:36,760
Notice the incredible
symmetry.
543
00:32:36,760 --> 00:32:41,610
I could have found this one by
changing a 1 to 2 here and by
544
00:32:41,610 --> 00:32:43,215
changing a 2 to 1.
545
00:32:43,215 --> 00:32:45,480
Of course, the system
is so symmetric--
546
00:32:45,480 --> 00:32:45,790
did I--
547
00:32:45,790 --> 00:32:46,575
AUDIENCE: [INAUDIBLE].
548
00:32:46,575 --> 00:32:47,870
PROFESSOR: Yes, thank
you very much.
549
00:32:47,870 --> 00:32:49,370
For the 2, right?
550
00:32:49,370 --> 00:32:51,180
Yeah, thank you.
551
00:32:51,180 --> 00:32:54,810
The system is so symmetric that
clearly, nature cannot
552
00:32:54,810 --> 00:32:57,430
make any distinction
between 1 and 2.
553
00:32:57,430 --> 00:33:00,550
So it is in this particular
case, because of the beautiful
554
00:33:00,550 --> 00:33:03,620
symmetry, it is obvious that
these two differential
555
00:33:03,620 --> 00:33:06,180
equations look very,
very similar.
556
00:33:06,180 --> 00:33:08,930
557
00:33:08,930 --> 00:33:12,320
Now I'm going to make
an important step.
558
00:33:12,320 --> 00:33:16,780
I'm going to substitute in
here x1 is C times cosine
559
00:33:16,780 --> 00:33:20,072
omega t, and x2--
560
00:33:20,072 --> 00:33:21,210
this is C1--
561
00:33:21,210 --> 00:33:23,290
is C2 times cosine omega t.
562
00:33:23,290 --> 00:33:26,210
I want to know what the normal
mode frequencies
563
00:33:26,210 --> 00:33:29,130
are for this system.
564
00:33:29,130 --> 00:33:31,250
I want to solve for omega.
565
00:33:31,250 --> 00:33:32,620
I want to find omega.
566
00:33:32,620 --> 00:33:34,030
I'm not driving the system.
567
00:33:34,030 --> 00:33:37,470
568
00:33:37,470 --> 00:33:41,050
Since I have no damping, in
which case at normal mode
569
00:33:41,050 --> 00:33:44,760
solutions, either of the two
objects are in phase with each
570
00:33:44,760 --> 00:33:47,040
other, or they're out of phase
with each other, but if
571
00:33:47,040 --> 00:33:49,530
they're out of phase, we can
always take care of that with
572
00:33:49,530 --> 00:33:50,720
a minus sign.
573
00:33:50,720 --> 00:33:53,010
So I'm now going to substitute
that in here.
574
00:33:53,010 --> 00:33:56,330
575
00:33:56,330 --> 00:34:00,910
And I may want to continue
working on the center board,
576
00:34:00,910 --> 00:34:04,770
in which case I might as well
erase this so that it stays a
577
00:34:04,770 --> 00:34:06,020
little compact.
578
00:34:06,020 --> 00:34:14,080
579
00:34:14,080 --> 00:34:18,739
So I'm going to substitute this
trial function into my
580
00:34:18,739 --> 00:34:22,179
differential equations.
581
00:34:22,179 --> 00:34:26,380
And every term will have a
cosine omega t, so I dump all
582
00:34:26,380 --> 00:34:28,040
the cosine omega t's.
583
00:34:28,040 --> 00:34:31,880
So I go to the equation, which
has a big 1 there, and so that
584
00:34:31,880 --> 00:34:34,770
1 becomes C1.
585
00:34:34,770 --> 00:34:38,500
Now, x1 double dot gives me a
minus omega squared, right?
586
00:34:38,500 --> 00:34:41,400
If I take cosine omega t, and I
do the second derivative, I
587
00:34:41,400 --> 00:34:44,120
get minus omega squared
in front of it.
588
00:34:44,120 --> 00:34:48,570
So I get C1 times 2
omega 0 squared.
589
00:34:48,570 --> 00:34:51,090
That is that 2 omega
0 squared.
590
00:34:51,090 --> 00:34:54,699
And then I get minus
omega squared.
591
00:34:54,699 --> 00:35:05,170
And then I get minus omega
0 squared times C2,
592
00:35:05,170 --> 00:35:06,590
and that equals 0.
593
00:35:06,590 --> 00:35:09,270
594
00:35:09,270 --> 00:35:11,900
And then I go to my second
differential equation, which
595
00:35:11,900 --> 00:35:13,300
is this one.
596
00:35:13,300 --> 00:35:16,720
But I'm going to rearrange
the C1s and the C2s.
597
00:35:16,720 --> 00:35:21,060
So I'm going to put the C1s
here, so I'm going to get
598
00:35:21,060 --> 00:35:25,350
minus omega 0 squared
times C1.
599
00:35:25,350 --> 00:35:30,060
And here, I'm going to get plus
2 omega 0 squared minus
600
00:35:30,060 --> 00:35:38,790
omega 0 squared times C2,
and that equals 0.
601
00:35:38,790 --> 00:35:43,310
So I can leave this plus out.
602
00:35:43,310 --> 00:35:45,530
So what do I have here now?
603
00:35:45,530 --> 00:35:49,930
I have here two equations
with three unknowns.
604
00:35:49,930 --> 00:35:53,430
I want to know what omega is
in the normal modes, but I
605
00:35:53,430 --> 00:35:56,520
also would like to know
what C1 and C2 is.
606
00:35:56,520 --> 00:35:58,100
Well, that is tough luck.
607
00:35:58,100 --> 00:36:00,110
You can't have it both ways.
608
00:36:00,110 --> 00:36:06,570
We do not know at all what
C1 and C2 is separately.
609
00:36:06,570 --> 00:36:09,500
But you can always find, in
the normal modes, without
610
00:36:09,500 --> 00:36:14,810
knowing anything else what
the ratios are of those
611
00:36:14,810 --> 00:36:16,840
amplitudes C1 and C2.
612
00:36:16,840 --> 00:36:19,190
And you can always solve for
omega, and you will see how
613
00:36:19,190 --> 00:36:20,450
that works.
614
00:36:20,450 --> 00:36:23,840
So with two equations and three
unknowns, you can only
615
00:36:23,840 --> 00:36:27,930
solve for omega and for
the ratio C1 over C2.
616
00:36:27,930 --> 00:36:30,560
617
00:36:30,560 --> 00:36:32,770
So now you may do that
any way you want to.
618
00:36:32,770 --> 00:36:35,380
This is not so difficult
to solve this.
619
00:36:35,380 --> 00:36:40,260
I will, however, use Cramer's
rule, and I will introduce a
620
00:36:40,260 --> 00:36:44,460
D, which is the determinant of
the following: 2 omega 0
621
00:36:44,460 --> 00:36:48,340
squared minus omega squared.
622
00:36:48,340 --> 00:36:50,675
And then I have here minus
omega 0 squared.
623
00:36:50,675 --> 00:36:53,370
624
00:36:53,370 --> 00:36:56,960
And then I have here minus
omega 0 squared.
625
00:36:56,960 --> 00:37:03,890
And here, I have 2 omega 0
squared minus omega 0 squared.
626
00:37:03,890 --> 00:37:10,620
And this, now, must be 0 if I am
looking for solutions which
627
00:37:10,620 --> 00:37:13,450
are not C1 is 0 and C2 is 0.
628
00:37:13,450 --> 00:37:17,310
It's clear that if C1 and
C2 is 0 that your
629
00:37:17,310 --> 00:37:19,240
equations are satisfied.
630
00:37:19,240 --> 00:37:20,040
0 is 0.
631
00:37:20,040 --> 00:37:22,560
But that's not an interesting
case.
632
00:37:22,560 --> 00:37:25,400
And the only way that you can
get a solution which is
633
00:37:25,400 --> 00:37:30,030
interesting, with values for C1
and C2 which are not 0, is
634
00:37:30,030 --> 00:37:34,690
by making this determinant 0.
635
00:37:34,690 --> 00:37:42,660
So that means that D, which now
becomes 2 omega 0 squared
636
00:37:42,660 --> 00:37:48,940
minus omega squared squared
minus omega 0 to the 4th, you
637
00:37:48,940 --> 00:37:51,880
have to make that 0.
638
00:37:51,880 --> 00:37:55,380
And when you do that, you find
two values for omega.
639
00:37:55,380 --> 00:38:01,060
You find that omega squared is
2 omega 0 squared plus or
640
00:38:01,060 --> 00:38:03,710
minus omega 0 squared.
641
00:38:03,710 --> 00:38:06,390
Those are the two solutions.
642
00:38:06,390 --> 00:38:10,960
And when we evaluate those two
solutions, you find the
643
00:38:10,960 --> 00:38:13,370
result, which is so
embarrassingly simple that you
644
00:38:13,370 --> 00:38:15,670
could almost have said that
without any work-- almost.
645
00:38:15,670 --> 00:38:18,330
646
00:38:18,330 --> 00:38:25,020
Notice that omega minus, which
is the lowest one of the two,
647
00:38:25,020 --> 00:38:30,140
is the same as omega 0, because
that's the minus sign.
648
00:38:30,140 --> 00:38:33,480
What does it mean if omega
minus is omega 0?
649
00:38:33,480 --> 00:38:38,040
Well, it means, of course, that
the two objects are just
650
00:38:38,040 --> 00:38:40,880
oscillating like this.
651
00:38:40,880 --> 00:38:43,780
The inner spring is never
stretched, is never longer
652
00:38:43,780 --> 00:38:46,830
than it wants to be, is never
shorter than it wants to be.
653
00:38:46,830 --> 00:38:51,390
So each one is only driven, so
to speak, by the outer spring.
654
00:38:51,390 --> 00:38:54,290
So that's immediately obvious
that that's the normal mode.
655
00:38:54,290 --> 00:38:57,830
And you can make the prediction
that C1/C2
656
00:38:57,830 --> 00:38:59,550
must be plus 1.
657
00:38:59,550 --> 00:39:02,720
They must go in unison
like this.
658
00:39:02,720 --> 00:39:08,010
And you can confirm that by
substituting this omega 1.
659
00:39:08,010 --> 00:39:11,790
If you substitute that in this
equation, you will find the
660
00:39:11,790 --> 00:39:13,490
ratio C1 over C2.
661
00:39:13,490 --> 00:39:15,540
If you prefer this equation,
be my guest.
662
00:39:15,540 --> 00:39:16,940
You can do that, too.
663
00:39:16,940 --> 00:39:20,020
And you will find no matter
which one of the two you take
664
00:39:20,020 --> 00:39:22,420
that it is plus 1.
665
00:39:22,420 --> 00:39:26,110
Now, omega plus is
less obvious.
666
00:39:26,110 --> 00:39:29,480
It's the square root
of 3 times omega 0.
667
00:39:29,480 --> 00:39:32,250
That's when you take the plus.
668
00:39:32,250 --> 00:39:34,750
Even though it's not obvious
that it is the square root of
669
00:39:34,750 --> 00:39:37,160
3, it is clear what
is happening.
670
00:39:37,160 --> 00:39:40,260
You added two objects, and
now they're doing this.
671
00:39:40,260 --> 00:39:43,420
They are exactly 180 degrees out
of phase with each other.
672
00:39:43,420 --> 00:39:46,030
So now you can make comfortably
the prediction
673
00:39:46,030 --> 00:39:49,720
that C1 over C2 is minus 1.
674
00:39:49,720 --> 00:39:54,220
And indeed, if you substitute
the square root of 3 omega 0
675
00:39:54,220 --> 00:39:56,760
in either this or in that
equation, this,
676
00:39:56,760 --> 00:39:58,010
indeed, comes out.
677
00:39:58,010 --> 00:40:02,020
678
00:40:02,020 --> 00:40:06,570
So the general solution, now,
for a given initial
679
00:40:06,570 --> 00:40:09,170
condition--
680
00:40:09,170 --> 00:40:12,150
so if now I give you the
initial condition--
681
00:40:12,150 --> 00:40:19,110
then the general solution for
x1 would be x0 minus--
682
00:40:19,110 --> 00:40:23,940
the minus makes reference to
that lowest frequency--
683
00:40:23,940 --> 00:40:26,190
times the cosine.
684
00:40:26,190 --> 00:40:30,250
Or we got minus t plus some
phase angle minus.
685
00:40:30,250 --> 00:40:35,660
All these minus signs make
reference to this mode.
686
00:40:35,660 --> 00:40:42,860
And then I have plus x0 plus
times the cosine of omega plus
687
00:40:42,860 --> 00:40:46,420
t plus some phase angle plus.
688
00:40:46,420 --> 00:40:48,870
Of course, if you prefer,
instead of cosine, sines,
689
00:40:48,870 --> 00:40:52,050
that's fine, of course.
690
00:40:52,050 --> 00:40:54,500
And you have four adjustable
constants,
691
00:40:54,500 --> 00:40:57,510
one, two, three, four.
692
00:40:57,510 --> 00:41:00,230
And if you know the initial
condition, if you know what x1
693
00:41:00,230 --> 00:41:05,290
is and what x1 dot is and what
x2 is and what x2 dot is at
694
00:41:05,290 --> 00:41:08,200
time t equals 0, you can solve
for all four, in principle.
695
00:41:08,200 --> 00:41:10,950
696
00:41:10,950 --> 00:41:14,310
There is no longer any freedom
for number 2, because all four
697
00:41:14,310 --> 00:41:18,360
adjustable constants have
now been consumed.
698
00:41:18,360 --> 00:41:23,120
So therefore, you can write down
out a solution for x2 by
699
00:41:23,120 --> 00:41:26,370
simply making this one a 2.
700
00:41:26,370 --> 00:41:28,770
And this whole term
here is the same.
701
00:41:28,770 --> 00:41:31,740
The only difference is that this
plus sign now becomes a
702
00:41:31,740 --> 00:41:32,760
minus sign.
703
00:41:32,760 --> 00:41:35,690
Nothing else is different.
704
00:41:35,690 --> 00:41:38,640
The frequencies are the same,
otherwise they wouldn't be
705
00:41:38,640 --> 00:41:40,390
normal mode frequencies.
706
00:41:40,390 --> 00:41:45,740
And it is the ratio here of the
C1/C2 that is plus 1, and
707
00:41:45,740 --> 00:41:48,230
it is the ratio here that
is the minus 1.
708
00:41:48,230 --> 00:41:50,710
That's the reason why this
becomes a minus and why the
709
00:41:50,710 --> 00:41:54,040
plus here remains a plus.
710
00:41:54,040 --> 00:41:58,380
So that, now, is the general
solution if you also know the
711
00:41:58,380 --> 00:41:59,630
initial conditions.
712
00:41:59,630 --> 00:42:01,660
713
00:42:01,660 --> 00:42:06,190
So now we're going to
drive this system.
714
00:42:06,190 --> 00:42:15,960
So now we go back to where
we were, and I'm going
715
00:42:15,960 --> 00:42:19,300
to drive this end.
716
00:42:19,300 --> 00:42:24,350
And I'm going to drive this
end with my hand.
717
00:42:24,350 --> 00:42:35,260
This is eta, and eta equals eta
0 times cosine omega t.
718
00:42:35,260 --> 00:42:38,510
719
00:42:38,510 --> 00:42:44,100
There is only one term in all
of these on the blackboard
720
00:42:44,100 --> 00:42:46,040
that is going to change.
721
00:42:46,040 --> 00:42:47,290
And that is this term.
722
00:42:47,290 --> 00:42:50,470
723
00:42:50,470 --> 00:42:55,430
This spring here on the left
is no longer shorter by the
724
00:42:55,430 --> 00:43:02,130
amount x1 but is shorter by
the amount x1 minus eta.
725
00:43:02,130 --> 00:43:05,070
So it's only that term--
726
00:43:05,070 --> 00:43:09,330
it's only this minus kx1--
727
00:43:09,330 --> 00:43:13,100
that now have to be changed
into minus k
728
00:43:13,100 --> 00:43:15,960
times x1 minus eta.
729
00:43:15,960 --> 00:43:18,370
Nothing else changes.
730
00:43:18,370 --> 00:43:22,950
But that has major consequences,
of course.
731
00:43:22,950 --> 00:43:27,140
For one thing, if you're going
to substitute now these trial
732
00:43:27,140 --> 00:43:31,160
functions, omega is no
longer negotiable.
733
00:43:31,160 --> 00:43:34,160
Omega is my omega now.
734
00:43:34,160 --> 00:43:35,350
I set this omega.
735
00:43:35,350 --> 00:43:37,540
You're not going to solve
for that omega.
736
00:43:37,540 --> 00:43:39,180
That would be an insult to me.
737
00:43:39,180 --> 00:43:41,920
I dictate what omega is.
738
00:43:41,920 --> 00:43:44,390
So that means I now get
two equations with two
739
00:43:44,390 --> 00:43:46,430
unknowns, C1 and C2.
740
00:43:46,430 --> 00:43:48,720
Omega is no longer unknown.
741
00:43:48,720 --> 00:43:55,840
So now I get solutions for C1,
and I get solutions for C2.
742
00:43:55,840 --> 00:44:00,320
And of all the equations on the
blackboards that you have,
743
00:44:00,320 --> 00:44:07,030
the one that is going to change
is the first one.
744
00:44:07,030 --> 00:44:12,880
And when you carry through
that eta, which is eta 0
745
00:44:12,880 --> 00:44:21,450
cosine omega t, this 0 here
changes into eta 0
746
00:44:21,450 --> 00:44:22,725
times omega 0 squared.
747
00:44:22,725 --> 00:44:26,040
748
00:44:26,040 --> 00:44:29,770
The cosine omega t is gone,
because I've divided cosine
749
00:44:29,770 --> 00:44:31,690
omega t out in all those
other terms.
750
00:44:31,690 --> 00:44:34,340
751
00:44:34,340 --> 00:44:38,190
So now I use Cramer's rule, and
now I can actually come up
752
00:44:38,190 --> 00:44:44,320
with a solution for C1,
no longer ratio C1/C2.
753
00:44:44,320 --> 00:44:46,820
No, I can actually come up,
now, with solution.
754
00:44:46,820 --> 00:44:54,040
So C1, now, using Cramer's Rule,
this now is my first
755
00:44:54,040 --> 00:45:01,010
column, eta 0 omega
0 squared 0.
756
00:45:01,010 --> 00:45:07,140
And this now becomes my second
column, minus omega 0 squared,
757
00:45:07,140 --> 00:45:12,500
and I get here 2 omega 0 squared
minus omega squared
758
00:45:12,500 --> 00:45:16,010
divided by D.
759
00:45:16,010 --> 00:45:20,030
If I pick a random value
for omega, D is not 0.
760
00:45:20,030 --> 00:45:23,100
D is only 0 at those two
resonance frequencies.
761
00:45:23,100 --> 00:45:26,680
And then C2 becomes--
762
00:45:26,680 --> 00:45:30,170
the first column is like this.
763
00:45:30,170 --> 00:45:36,170
That is 2 omega 0 squared minus
omega squared minus
764
00:45:36,170 --> 00:45:38,350
omega 0 squared.
765
00:45:38,350 --> 00:45:44,290
And now the second column
becomes eta 0 omega 0 squared
766
00:45:44,290 --> 00:45:52,830
and a 0 divided by D.
767
00:45:52,830 --> 00:45:58,650
If I work C1 out a little
further, then I get C1 equals
768
00:45:58,650 --> 00:46:07,530
eta 0 omega 0 squared times 2
omega 0 squared minus omega 0
769
00:46:07,530 --> 00:46:15,210
squared divided by D. And
here, I get that C2.
770
00:46:15,210 --> 00:46:22,220
This is 0, so I get plus omega
0 to the 4th times eta 0
771
00:46:22,220 --> 00:46:28,980
divided by D. And
D, now, is this
772
00:46:28,980 --> 00:46:31,370
determinant that is not 0.
773
00:46:31,370 --> 00:46:34,105
It's only 0 for those two
special frequencies.
774
00:46:34,105 --> 00:46:37,980
775
00:46:37,980 --> 00:46:41,610
If you want to see what the
amplitudes now are as a
776
00:46:41,610 --> 00:46:44,060
function of omega--
it's a very,
777
00:46:44,060 --> 00:46:46,320
very interesting behavior--
778
00:46:46,320 --> 00:46:51,060
then what helps is you go omega
first to 0, and then you
779
00:46:51,060 --> 00:46:52,670
see what happens.
780
00:46:52,670 --> 00:46:55,070
And it's not so intuitive
what happens.
781
00:46:55,070 --> 00:46:59,290
If you put in omega equals
0, you'll find that C1 is
782
00:46:59,290 --> 00:47:02,860
plus 2/3 eta 0.
783
00:47:02,860 --> 00:47:06,020
You can confirm that by
substituting that into C1.
784
00:47:06,020 --> 00:47:07,900
You will see that that's
what happens.
785
00:47:07,900 --> 00:47:14,200
And you'll find that C2
is plus 1/3 eta 0.
786
00:47:14,200 --> 00:47:16,710
So that is at omega equals 0,
787
00:47:16,710 --> 00:47:20,220
And when you go to infinity
with frequencies, then it
788
00:47:20,220 --> 00:47:26,920
should not surprise you that
C1 is 0 and that C2 is 0.
789
00:47:26,920 --> 00:47:32,050
And then there is one case
which is pathetic.
790
00:47:32,050 --> 00:47:34,730
And that is the case
that I make omega
791
00:47:34,730 --> 00:47:37,310
squared, 2 omega 0 squared.
792
00:47:37,310 --> 00:47:41,770
At that frequency, C1 becomes
0, but C2 is not 0.
793
00:47:41,770 --> 00:47:44,590
And I spent almost the whole
lecture with you on that
794
00:47:44,590 --> 00:47:48,010
demonstrating that in three
different ways that indeed,
795
00:47:48,010 --> 00:47:50,590
there is this bizarre
solution.
796
00:47:50,590 --> 00:47:56,680
So when omega squared equals
2 omega 0 squared, then C1
797
00:47:56,680 --> 00:48:00,210
becomes 0, but C2 is not 0.
798
00:48:00,210 --> 00:48:03,920
799
00:48:03,920 --> 00:48:09,830
So now you can make a plot of
C's as a function of omega.
800
00:48:09,830 --> 00:48:12,110
And during that lecture, I
showed you three of those
801
00:48:12,110 --> 00:48:19,020
plots, and I used the
convention, then, that when
802
00:48:19,020 --> 00:48:23,590
the object is moving in phase
with the driver, I put it
803
00:48:23,590 --> 00:48:25,950
positive and out of
phase, negative.
804
00:48:25,950 --> 00:48:29,710
And I plot here C divided
by eta 0.
805
00:48:29,710 --> 00:48:32,260
And I will use a color code.
806
00:48:32,260 --> 00:48:37,930
I will do C1 in red,
and I will do--
807
00:48:37,930 --> 00:48:45,220
C1 is red, and C2 I will
do in white chalk.
808
00:48:45,220 --> 00:48:46,470
So here is 0 omega.
809
00:48:46,470 --> 00:48:49,590
810
00:48:49,590 --> 00:48:54,580
Here is omega 0, which
is my omega minus.
811
00:48:54,580 --> 00:49:00,190
And then my omega
plus was at 3--
812
00:49:00,190 --> 00:49:01,250
the square root of 3.
813
00:49:01,250 --> 00:49:03,616
So that's about 1.7.
814
00:49:03,616 --> 00:49:06,060
So here is my omega plus.
815
00:49:06,060 --> 00:49:07,550
And then things go nuts.
816
00:49:07,550 --> 00:49:11,630
That is when D goes to 0.
817
00:49:11,630 --> 00:49:14,545
So C1 is 2/3 here.
818
00:49:14,545 --> 00:49:17,220
819
00:49:17,220 --> 00:49:21,770
And then it goes to
infinity there.
820
00:49:21,770 --> 00:49:28,240
And then here, when it is 1.4
times omega 0, the square root
821
00:49:28,240 --> 00:49:34,680
of 2, then it goes through 0,
and it comes up here and then
822
00:49:34,680 --> 00:49:38,190
I get a curve here, without
being too precise.
823
00:49:38,190 --> 00:49:45,200
And for C2, I get 1/3,
and then it goes up.
824
00:49:45,200 --> 00:49:49,070
And then C2, something like
this, and then I get
825
00:49:49,070 --> 00:49:50,320
something like that.
826
00:49:50,320 --> 00:49:58,960
827
00:49:58,960 --> 00:50:01,810
Yeah, I can live with that.
828
00:50:01,810 --> 00:50:03,870
Now, we ignored damping.
829
00:50:03,870 --> 00:50:06,800
And because we ignored
damping, we get these
830
00:50:06,800 --> 00:50:13,020
un-physical infinities when you
hit the frequency omega
831
00:50:13,020 --> 00:50:17,090
minus and omega plus, these
resonance frequencies.
832
00:50:17,090 --> 00:50:20,630
Now, if you include damping,
then the solutions become
833
00:50:20,630 --> 00:50:21,910
extremely complicated.
834
00:50:21,910 --> 00:50:24,850
But of course, you avoid
the infinity values.
835
00:50:24,850 --> 00:50:28,900
But the resonance amplitude
can still be very high.
836
00:50:28,900 --> 00:50:32,800
So these plots are still very
useful provided that you don't
837
00:50:32,800 --> 00:50:36,850
interpret infinities is as being
real but something that
838
00:50:36,850 --> 00:50:37,750
is very large.
839
00:50:37,750 --> 00:50:40,310
If Q is high, then of course,
the amplitudes
840
00:50:40,310 --> 00:50:42,210
are enormously high.
841
00:50:42,210 --> 00:50:44,000
And this can lead
to destruction.
842
00:50:44,000 --> 00:50:46,710
We've seen the movie of
the Tacoma Bridge.
843
00:50:46,710 --> 00:50:49,020
And then we have seen the
dramatic experiment of the
844
00:50:49,020 --> 00:50:50,890
breaking wine glass.
845
00:50:50,890 --> 00:50:54,690
You remember that wine
glass demonstration.
846
00:50:54,690 --> 00:51:01,210
Because of the catastrophic
success, to use Bush's words,
847
00:51:01,210 --> 00:51:06,100
of that demonstration,
students have
848
00:51:06,100 --> 00:51:08,310
asked me for an encore.
849
00:51:08,310 --> 00:51:11,150
They would like to
see it again.
850
00:51:11,150 --> 00:51:13,735
And that's a very reasonable
thing to do.
851
00:51:13,735 --> 00:51:17,170
It fits very nicely into
this concept of coupled
852
00:51:17,170 --> 00:51:18,290
oscillators.
853
00:51:18,290 --> 00:51:24,480
A wine glass would have a huge
number of oscillators coupled.
854
00:51:24,480 --> 00:51:26,010
And a wine glass--
855
00:51:26,010 --> 00:51:27,940
I have one here--
856
00:51:27,940 --> 00:51:32,040
can be made to oscillate
easily in its
857
00:51:32,040 --> 00:51:34,210
lowest normal mode.
858
00:51:34,210 --> 00:51:39,180
I have to wash my hands to make
you listen to it, because
859
00:51:39,180 --> 00:51:43,080
my hands are now a little
greasy from the chalk.
860
00:51:43,080 --> 00:51:46,075
And this is the frequency.
861
00:51:46,075 --> 00:51:47,340
That is the lowest mode.
862
00:51:47,340 --> 00:51:51,130
863
00:51:51,130 --> 00:51:53,710
If it is a circle, like this,
looking from above, it becomes
864
00:51:53,710 --> 00:51:56,090
an ellipse, and then it becomes
an ellipse like this,
865
00:51:56,090 --> 00:51:58,600
and the ellipse like this, and
the ellipse like that.
866
00:51:58,600 --> 00:52:02,620
And if now you drive that with
sound at exactly that
867
00:52:02,620 --> 00:52:05,360
frequency, and you put
in enough power--
868
00:52:05,360 --> 00:52:09,000
another way of saying is if you
make eta 0 large enough--
869
00:52:09,000 --> 00:52:11,220
then the system can break.
870
00:52:11,220 --> 00:52:14,310
871
00:52:14,310 --> 00:52:15,620
It's making a lot of noise.
872
00:52:15,620 --> 00:52:16,910
I warn you.
873
00:52:16,910 --> 00:52:19,230
Those who are sitting here,
I really think you should
874
00:52:19,230 --> 00:52:20,785
protect your ears.
875
00:52:20,785 --> 00:52:24,850
The sound can be deafening,
so be careful.
876
00:52:24,850 --> 00:52:27,010
And those who are a little bit
further away, make sure that
877
00:52:27,010 --> 00:52:28,630
you close your years.
878
00:52:28,630 --> 00:52:32,250
I have the luxury that I can
turn my hearing aids off.
879
00:52:32,250 --> 00:52:35,230
But without my hearing aids,
believe me, I can still hear a
880
00:52:35,230 --> 00:52:39,640
lot, so I also will have
to protect myself.
881
00:52:39,640 --> 00:52:44,770
So one hearing aid is off, the
other is off, and I'm going to
882
00:52:44,770 --> 00:52:48,040
put this on anyhow.
883
00:52:48,040 --> 00:52:52,440
I'm going to stroke this glass
with a frequency which is only
884
00:52:52,440 --> 00:52:57,700
slightly different from this
frequency of the sound.
885
00:52:57,700 --> 00:53:03,110
That is about 427 hertz, which
you can very easily hear.
886
00:53:03,110 --> 00:53:06,070
And then you will see the glass
in slow motion because
887
00:53:06,070 --> 00:53:10,580
of the fact that the
stroboscopic light has a
888
00:53:10,580 --> 00:53:13,865
slightly different frequency
than the--
889
00:53:13,865 --> 00:53:18,480
890
00:53:18,480 --> 00:53:20,850
that should be coming up now.
891
00:53:20,850 --> 00:53:22,185
AUDIENCE: Yeah, it will.
892
00:53:22,185 --> 00:53:23,895
PROFESSOR: And you
say it will.
893
00:53:23,895 --> 00:53:35,010
894
00:53:35,010 --> 00:53:35,740
Thank you very much.
895
00:53:35,740 --> 00:53:39,590
I'm going to make it very dark
now, because this is important
896
00:53:39,590 --> 00:53:42,030
that you can see
this very well.
897
00:53:42,030 --> 00:53:46,350
So we'll make it completely
dark in the room.
898
00:53:46,350 --> 00:53:48,430
I'm going to protect my ears.
899
00:53:48,430 --> 00:53:50,220
I hope you can still hear me.
900
00:53:50,220 --> 00:53:53,190
I can hardly hear myself.
901
00:53:53,190 --> 00:53:57,410
And I'm going to drive the
system now at very low
902
00:53:57,410 --> 00:54:01,080
amplitude at a frequency
close to its resonance.
903
00:54:01,080 --> 00:54:06,200
904
00:54:06,200 --> 00:54:09,090
You can already see that
the glass is moving.
905
00:54:09,090 --> 00:54:11,060
I'm going to increase
the amplitude.
906
00:54:11,060 --> 00:54:14,030
907
00:54:14,030 --> 00:54:16,670
You see it moving?
908
00:54:16,670 --> 00:54:19,330
I can go a little bit
over the resonance
909
00:54:19,330 --> 00:54:20,590
and under the resonance.
910
00:54:20,590 --> 00:54:23,220
You will see it stop
moving then.
911
00:54:23,220 --> 00:54:24,470
So I do that purposely now.
912
00:54:24,470 --> 00:54:28,800
913
00:54:28,800 --> 00:54:29,970
Now I'm off resonance.
914
00:54:29,970 --> 00:54:31,220
I'm over it.
915
00:54:31,220 --> 00:54:33,980
916
00:54:33,980 --> 00:54:36,160
And now I'm off resonance.
917
00:54:36,160 --> 00:54:37,385
I'm under resonance.
918
00:54:37,385 --> 00:54:37,750
I'm below.
919
00:54:37,750 --> 00:54:44,050
And now it's 423 hertz, but
the resonance is at 427.
920
00:54:44,050 --> 00:54:50,140
So I'm going to put
it back at 427.
921
00:54:50,140 --> 00:54:52,605
And now I'm going to
increase the sound.
922
00:54:52,605 --> 00:54:53,855
So I warn you.
923
00:54:53,855 --> 00:54:59,850
924
00:54:59,850 --> 00:55:01,740
And that's it.
925
00:55:01,740 --> 00:55:02,990
It broke.
926
00:55:02,990 --> 00:55:06,910
927
00:55:06,910 --> 00:55:08,172
And it broke very fast.
928
00:55:08,172 --> 00:55:15,870
929
00:55:15,870 --> 00:55:19,290
So this is an ideal moment
for a break.
930
00:55:19,290 --> 00:55:22,540
931
00:55:22,540 --> 00:55:25,250
Since you had such a good time
with the breaking glass, let's
932
00:55:25,250 --> 00:55:26,310
settle for four minutes.
933
00:55:26,310 --> 00:55:27,710
We'll reconvene in
four minutes.
934
00:55:27,710 --> 00:55:33,140
935
00:55:33,140 --> 00:55:35,830
[BLOWS WHISTLE]
936
00:55:35,830 --> 00:55:37,080
OK.
937
00:55:37,080 --> 00:55:39,480
938
00:55:39,480 --> 00:55:44,670
I now want to discuss with
you continuous media.
939
00:55:44,670 --> 00:55:48,850
If you have n coupled
oscillators, then you get n
940
00:55:48,850 --> 00:55:50,530
normal modes.
941
00:55:50,530 --> 00:55:53,300
When you make n infinitely
high, then you
942
00:55:53,300 --> 00:55:56,460
get continuous media.
943
00:55:56,460 --> 00:56:00,770
And if we have one-dimensional
continuous media, like a
944
00:56:00,770 --> 00:56:04,810
string or a pipe with air--
945
00:56:04,810 --> 00:56:06,310
sound--
946
00:56:06,310 --> 00:56:11,580
then we have studied the
transverse motion of a string.
947
00:56:11,580 --> 00:56:14,570
948
00:56:14,570 --> 00:56:16,940
And then we derived that
now you have to
949
00:56:16,940 --> 00:56:18,430
use the wave equation.
950
00:56:18,430 --> 00:56:20,990
We derive the wave equation.
951
00:56:20,990 --> 00:56:27,930
And then you get that d2y / dx
squared is 1 over v squared
952
00:56:27,930 --> 00:56:32,930
times d2y / dt squared, y being
now the displacement
953
00:56:32,930 --> 00:56:35,310
away from equilibrium
in this position if
954
00:56:35,310 --> 00:56:36,350
x is in this direction.
955
00:56:36,350 --> 00:56:38,498
AUDIENCE: [INAUDIBLE].
956
00:56:38,498 --> 00:56:42,850
PROFESSOR: And yeah--
957
00:56:42,850 --> 00:56:43,370
that is fine.
958
00:56:43,370 --> 00:56:44,096
AUDIENCE: [INAUDIBLE].
959
00:56:44,096 --> 00:56:46,425
PROFESSOR: And so-- excuse me?
960
00:56:46,425 --> 00:56:49,770
Ah, it's amazing how you can
look at it and think that it's
961
00:56:49,770 --> 00:56:52,210
fine, and it is not.
962
00:56:52,210 --> 00:56:56,950
All right, so v, in the case
of a string, is the square
963
00:56:56,950 --> 00:56:58,100
root of t over mu.
964
00:56:58,100 --> 00:57:01,640
That was just a special
case for the string.
965
00:57:01,640 --> 00:57:08,450
We also examined longitudinal
motion, again 1D.
966
00:57:08,450 --> 00:57:12,600
Sound is a longitudinal wave.
967
00:57:12,600 --> 00:57:14,700
And if you deal with
pressure--
968
00:57:14,700 --> 00:57:17,160
you think of sound as being
a pressure wave--
969
00:57:17,160 --> 00:57:23,820
then you get d2p dx squared
equals 1 over v squared times
970
00:57:23,820 --> 00:57:26,760
d2p dt squared.
971
00:57:26,760 --> 00:57:30,250
P, then, being positive, would
be overpressure, over and
972
00:57:30,250 --> 00:57:32,200
above the ambient
one atmosphere.
973
00:57:32,200 --> 00:57:34,710
And if it's negative,
then it is below.
974
00:57:34,710 --> 00:57:36,950
So it's not the total pressure,
but it is the
975
00:57:36,950 --> 00:57:39,140
overpressure.
976
00:57:39,140 --> 00:57:42,760
And v, then, is the speed of
sound, which in air, room
977
00:57:42,760 --> 00:57:46,790
temperature, is about 340
meters per second.
978
00:57:46,790 --> 00:57:50,480
If you prefer not to work in
pressure but in terms of the
979
00:57:50,480 --> 00:57:55,930
actual position of the air
molecules in analogy with the
980
00:57:55,930 --> 00:57:59,030
position of the string, then
you can write down the same
981
00:57:59,030 --> 00:58:03,640
equation in terms of xi, or d2
xi dx squared 1 over v squared
982
00:58:03,640 --> 00:58:06,950
d2 xi dt squared, that, then,
gives you the position, the
983
00:58:06,950 --> 00:58:09,990
actual motion, of the
air molecules.
984
00:58:09,990 --> 00:58:13,600
I often prefer the pressure,
and I will
985
00:58:13,600 --> 00:58:16,500
follow that also today.
986
00:58:16,500 --> 00:58:19,425
So there are an infinite
number of normal modes.
987
00:58:19,425 --> 00:58:21,940
988
00:58:21,940 --> 00:58:26,670
And the ratios of the amplitudes
of two adjacent
989
00:58:26,670 --> 00:58:27,480
oscillators--
990
00:58:27,480 --> 00:58:29,530
they are coupled now--
991
00:58:29,530 --> 00:58:33,590
reflects itself, of course, in
terms of the overall shape,
992
00:58:33,590 --> 00:58:37,050
which you can best see when
you deal with a string.
993
00:58:37,050 --> 00:58:45,420
994
00:58:45,420 --> 00:58:50,410
As I said, it's easiest to see
with transverse oscillations
995
00:58:50,410 --> 00:58:54,210
what the displacements
look like.
996
00:58:54,210 --> 00:58:56,065
It's harder to see
that with sound.
997
00:58:56,065 --> 00:58:59,210
998
00:58:59,210 --> 00:59:05,610
I also discussed with you a
special situation that I
999
00:59:05,610 --> 00:59:09,020
connected to media--
1000
00:59:09,020 --> 00:59:13,330
medium one and medium two.
1001
00:59:13,330 --> 00:59:17,820
And I gave these two
media different
1002
00:59:17,820 --> 00:59:20,500
masses per unit length.
1003
00:59:20,500 --> 00:59:23,760
When you set up a traveling
wave on a string,
1004
00:59:23,760 --> 00:59:25,480
or you set it up--
1005
00:59:25,480 --> 00:59:27,410
you could set up a pulse--
1006
00:59:27,410 --> 00:59:29,480
it reflects at the end.
1007
00:59:29,480 --> 00:59:32,970
And how it reflects depends
on the boundary
1008
00:59:32,970 --> 00:59:34,820
conditions at the end.
1009
00:59:34,820 --> 00:59:38,870
And when you connect them, two
media, then you get not only a
1010
00:59:38,870 --> 00:59:41,970
reflection, but you also
get some of the pulse.
1011
00:59:41,970 --> 00:59:45,720
Some of the wave goes
into medium two.
1012
00:59:45,720 --> 00:59:49,110
So let's assume that we have
an incident wave coming in
1013
00:59:49,110 --> 00:59:56,340
like this, and we have here mu
1, and we have here v1 given
1014
00:59:56,340 --> 00:59:57,420
by this equation.
1015
00:59:57,420 --> 01:00:00,550
They both have the same tension,
T, and here we have
1016
01:00:00,550 --> 01:00:02,980
mu 2, and you have v2.
1017
01:00:02,980 --> 01:00:05,790
1018
01:00:05,790 --> 01:00:07,740
We discussed that.
1019
01:00:07,740 --> 01:00:13,370
We used the wave equation to
solve what happens when we
1020
01:00:13,370 --> 01:00:18,020
have an incident harmonic
wave coming in.
1021
01:00:18,020 --> 01:00:22,070
We derived even this speed
of propagation
1022
01:00:22,070 --> 01:00:23,650
using the wave equation.
1023
01:00:23,650 --> 01:00:25,305
None of this came
out of the blue.
1024
01:00:25,305 --> 01:00:28,230
We always derived that.
1025
01:00:28,230 --> 01:00:33,040
And then we found by using the
boundary conditions at the
1026
01:00:33,040 --> 01:00:36,610
junction, namely that the string
is not breaking, and
1027
01:00:36,610 --> 01:00:42,730
that dy dx on the left side is
the same as dy dx on the right
1028
01:00:42,730 --> 01:00:45,560
side with no boundary
conditions, we found that the
1029
01:00:45,560 --> 01:00:48,380
amplitude of the reflected
wave--
1030
01:00:48,380 --> 01:00:50,670
and the same would hold
for a pulse--
1031
01:00:50,670 --> 01:00:57,020
divided by the amplitude of the
incident wave was v2 minus
1032
01:00:57,020 --> 01:01:00,360
v1 divided by v1 plus v2.
1033
01:01:00,360 --> 01:01:02,040
And I called that R--
reflectivity.
1034
01:01:02,040 --> 01:01:05,690
1035
01:01:05,690 --> 01:01:09,520
And we found that the amplitude
of the transmitted
1036
01:01:09,520 --> 01:01:14,870
wave, or pulse, divided by the
amplitude of the incident 1
1037
01:01:14,870 --> 01:01:19,630
was 2v2 divided by v1 plus v2.
1038
01:01:19,630 --> 01:01:22,845
And I called that shorthand
notation transmitivity.
1039
01:01:22,845 --> 01:01:27,880
1040
01:01:27,880 --> 01:01:35,100
And when we had done that, I put
in some simple test cases
1041
01:01:35,100 --> 01:01:38,630
where our intuition
is very good.
1042
01:01:38,630 --> 01:01:41,260
The first thing I did
was, suppose mu 2
1043
01:01:41,260 --> 01:01:43,610
is infinitely high.
1044
01:01:43,610 --> 01:01:46,550
So mu 2 is infinitely high.
1045
01:01:46,550 --> 01:01:51,590
In other words, that medium
two is a wall.
1046
01:01:51,590 --> 01:01:55,480
That means the string number
one cannot move.
1047
01:01:55,480 --> 01:01:57,850
It's fixed at the end.
1048
01:01:57,850 --> 01:02:03,100
So that means v2 is 0.
1049
01:02:03,100 --> 01:02:06,870
And we go to this equation,
and we find that R
1050
01:02:06,870 --> 01:02:09,300
equals minus 1.
1051
01:02:09,300 --> 01:02:10,110
v2 is 0.
1052
01:02:10,110 --> 01:02:12,190
You get minus v1 over v1.
1053
01:02:12,190 --> 01:02:15,770
And we like that because what
it means is that when a
1054
01:02:15,770 --> 01:02:19,610
mountain rolls in, it comes
back as a valley.
1055
01:02:19,610 --> 01:02:21,840
And when a valley rolls in, it
comes back as a mountain.
1056
01:02:21,840 --> 01:02:25,300
And I demonstrated that
with strings.
1057
01:02:25,300 --> 01:02:29,150
It was very pleasing that
T of R is then 0.
1058
01:02:29,150 --> 01:02:30,750
Well, it better be 0, right?
1059
01:02:30,750 --> 01:02:33,530
If everything comes back at
you-- upside down, but
1060
01:02:33,530 --> 01:02:36,770
nevertheless, everything comes
back at you, you expect that
1061
01:02:36,770 --> 01:02:38,730
nothing goes into the wall.
1062
01:02:38,730 --> 01:02:43,630
And you see when v2
is 0 that TR is 0.
1063
01:02:43,630 --> 01:02:49,920
And we were all very happy,
and we could all sleep.
1064
01:02:49,920 --> 01:02:51,005
But then--
1065
01:02:51,005 --> 01:02:52,680
and you guessed it--
1066
01:02:52,680 --> 01:02:56,890
then I said, let's suppose
mu 2 becomes 0.
1067
01:02:56,890 --> 01:03:00,570
So I attach that string
to nothing--
1068
01:03:00,570 --> 01:03:03,190
to empty space.
1069
01:03:03,190 --> 01:03:05,030
Is that practical?
1070
01:03:05,030 --> 01:03:06,020
Yes.
1071
01:03:06,020 --> 01:03:07,300
I can do it.
1072
01:03:07,300 --> 01:03:10,730
I can take a nickel wire, and I
have here a magnetic field--
1073
01:03:10,730 --> 01:03:11,740
very strong--
1074
01:03:11,740 --> 01:03:14,420
and I can pull on that nickel
wire, and the end of the
1075
01:03:14,420 --> 01:03:17,630
nickel wire ends up in nothing,
in empty space, but I
1076
01:03:17,630 --> 01:03:18,870
keep the tension on.
1077
01:03:18,870 --> 01:03:20,130
So it's completely practical.
1078
01:03:20,130 --> 01:03:20,950
It can be done.
1079
01:03:20,950 --> 01:03:22,870
It's not just notes.
1080
01:03:22,870 --> 01:03:25,480
So mu 2 goes to 0.
1081
01:03:25,480 --> 01:03:29,690
That means that v2
goes to infinity.
1082
01:03:29,690 --> 01:03:36,260
And then we looked at R, and
we say, well, if v2 goes to
1083
01:03:36,260 --> 01:03:40,510
infinity, then R
equals plus 1.
1084
01:03:40,510 --> 01:03:42,110
And we were all very happy.
1085
01:03:42,110 --> 01:03:45,080
A mountain comes back as
a mountain, and I even
1086
01:03:45,080 --> 01:03:47,060
demonstrated that.
1087
01:03:47,060 --> 01:03:50,910
We were still able to
sleep at that point.
1088
01:03:50,910 --> 01:03:56,670
But then came the awful thing,
that if we substitute v2
1089
01:03:56,670 --> 01:04:00,440
equals infinity in this
equation that TR
1090
01:04:00,440 --> 01:04:03,800
becomes plus 2.
1091
01:04:03,800 --> 01:04:08,040
And now we can no longer sleep,
because this is absurd.
1092
01:04:08,040 --> 01:04:11,330
All the energy that rolls in
comes back, but there is
1093
01:04:11,330 --> 01:04:15,210
something in addition that goes
into that second medium.
1094
01:04:15,210 --> 01:04:16,850
Now, admit it.
1095
01:04:16,850 --> 01:04:18,890
Who could not sleep
that night?
1096
01:04:18,890 --> 01:04:21,780
1097
01:04:21,780 --> 01:04:24,440
You should all fail this
course, by the way.
1098
01:04:24,440 --> 01:04:27,240
1099
01:04:27,240 --> 01:04:30,640
But in any case, I don't have
to feel guilty, right?
1100
01:04:30,640 --> 01:04:33,740
Who thought about this
and said, there
1101
01:04:33,740 --> 01:04:35,180
is something weird.
1102
01:04:35,180 --> 01:04:37,970
I must find an explanation
for that.
1103
01:04:37,970 --> 01:04:42,080
In my case, I had to find an
explanation, because I
1104
01:04:42,080 --> 01:04:44,340
couldn't sleep.
1105
01:04:44,340 --> 01:04:45,740
Who found an explanation?
1106
01:04:45,740 --> 01:04:50,970
Who could say, oh yes,
don't worry about it.
1107
01:04:50,970 --> 01:04:52,412
What was your solution?
1108
01:04:52,412 --> 01:04:53,778
AUDIENCE: [INAUDIBLE].
1109
01:04:53,778 --> 01:04:54,756
PROFESSOR: Excuse me.
1110
01:04:54,756 --> 01:04:57,690
AUDIENCE: Mass two is 0, so
there's no energy being
1111
01:04:57,690 --> 01:04:58,180
transmitted.
1112
01:04:58,180 --> 01:05:00,340
PROFESSOR: Very good.
1113
01:05:00,340 --> 01:05:02,440
That's a very nice way
of looking at it.
1114
01:05:02,440 --> 01:05:05,840
So that's probably why
you could sleep.
1115
01:05:05,840 --> 01:05:08,420
Well actually, I'll tell
you why I could sleep.
1116
01:05:08,420 --> 01:05:11,810
But your solution
is even shorter.
1117
01:05:11,810 --> 01:05:17,070
But the reason why I want to
show you what I'm going to
1118
01:05:17,070 --> 01:05:20,420
show you is that I want to also
expose you to the idea,
1119
01:05:20,420 --> 01:05:24,730
which you already alluded to,
namely that there is energy
1120
01:05:24,730 --> 01:05:27,040
involved when we deal
with pulses and when
1121
01:05:27,040 --> 01:05:31,720
we deal with waves.
1122
01:05:31,720 --> 01:05:39,220
Do you remember when we have a
traveling wave that the total
1123
01:05:39,220 --> 01:05:43,780
energy per wavelength lambda--
1124
01:05:43,780 --> 01:05:46,035
I only did it per wavelength--
1125
01:05:46,035 --> 01:05:49,190
that that total energy-- perhaps
you remember that--
1126
01:05:49,190 --> 01:05:56,540
equals 2A squared times pi
squared times the tension T
1127
01:05:56,540 --> 01:05:58,190
divided by lambda.
1128
01:05:58,190 --> 01:05:59,530
A was the amplitude.
1129
01:05:59,530 --> 01:06:02,620
Energy is always proportional
to amplitude squared.
1130
01:06:02,620 --> 01:06:05,780
This was the tension, and
this was the wavelength.
1131
01:06:05,780 --> 01:06:10,160
And if v2 goes to infinity,
then the
1132
01:06:10,160 --> 01:06:11,960
wavelengths go to infinity.
1133
01:06:11,960 --> 01:06:14,000
That's obvious, right?
1134
01:06:14,000 --> 01:06:17,790
If something moves with the
speed of light, even faster--
1135
01:06:17,790 --> 01:06:20,470
infinity's even faster than
the speed of light--
1136
01:06:20,470 --> 01:06:24,870
then lambda goes to infinity,
so this goes to 0.
1137
01:06:24,870 --> 01:06:27,310
So we came to the
same conclusion.
1138
01:06:27,310 --> 01:06:32,950
Some decadent solution, TR
equals plus 2, has no meaning
1139
01:06:32,950 --> 01:06:35,020
because there's no
energy in there.
1140
01:06:35,020 --> 01:06:38,475
So I was able to sleep.
1141
01:06:38,475 --> 01:06:41,530
1142
01:06:41,530 --> 01:06:45,900
Let's now turn to normal modes
of continuous media.
1143
01:06:45,900 --> 01:06:50,462
And I suggested we
go longitudinal.
1144
01:06:50,462 --> 01:06:53,510
Because we did so much
transverse stuff, let's go
1145
01:06:53,510 --> 01:06:56,210
longitudinal.
1146
01:06:56,210 --> 01:07:04,220
I have here a pipe, has length
L, and it is open here, and it
1147
01:07:04,220 --> 01:07:07,100
is open there.
1148
01:07:07,100 --> 01:07:11,590
That means the overpressure
or underpressure here
1149
01:07:11,590 --> 01:07:12,550
can never build up.
1150
01:07:12,550 --> 01:07:14,530
It's connected to
the universe.
1151
01:07:14,530 --> 01:07:17,970
So at these two boundary
conditions, this p that I have
1152
01:07:17,970 --> 01:07:21,020
there must be 0.
1153
01:07:21,020 --> 01:07:24,900
If I write down the general
equation for a standing wave--
1154
01:07:24,900 --> 01:07:27,790
because normal mode solutions
are standing waves--
1155
01:07:27,790 --> 01:07:32,575
I can write down p equals some
amplitude times the sine or
1156
01:07:32,575 --> 01:07:33,460
the cosine--
1157
01:07:33,460 --> 01:07:35,430
let me take the sine--
1158
01:07:35,430 --> 01:07:39,590
2 pi divided by lambda
times x.
1159
01:07:39,590 --> 01:07:41,080
I'll be very general.
1160
01:07:41,080 --> 01:07:44,400
I will introduce some phase
angle alpha for which I will
1161
01:07:44,400 --> 01:07:46,330
find the solution
very shortly.
1162
01:07:46,330 --> 01:07:50,780
And then cosine omega
t or sine omega t,
1163
01:07:50,780 --> 01:07:51,420
if you prefer that.
1164
01:07:51,420 --> 01:07:56,350
This is a standing wave
in very general terms.
1165
01:07:56,350 --> 01:07:59,890
Everything here is in terms--
is the space.
1166
01:07:59,890 --> 01:08:00,590
x--
1167
01:08:00,590 --> 01:08:03,320
this is x.
1168
01:08:03,320 --> 01:08:08,130
And here, all the information
here deals with time, which is
1169
01:08:08,130 --> 01:08:11,370
typical for a standing wave.
1170
01:08:11,370 --> 01:08:18,729
I prefer always to write down
for 2 pi over lambda, k, and I
1171
01:08:18,729 --> 01:08:25,930
must observe the situation that
p must be 0 at x equals 0
1172
01:08:25,930 --> 01:08:31,500
and also at x equals L. If
p is 0 at x equals 0,
1173
01:08:31,500 --> 01:08:35,430
immediately you see
that alpha is 0.
1174
01:08:35,430 --> 01:08:36,910
So I'm going to rewrite
it now.
1175
01:08:36,910 --> 01:08:43,680
I'm going to write down p is p 0
times the sine of kx, So I'm
1176
01:08:43,680 --> 01:08:44,990
going to replace this by k.
1177
01:08:44,990 --> 01:08:50,859
I know that alpha as 0, times
the cosine of omega t, and now
1178
01:08:50,859 --> 01:08:55,250
I must meet the boundary
condition that when x equals L
1179
01:08:55,250 --> 01:08:58,000
that p is, again, 0.
1180
01:08:58,000 --> 01:09:01,520
And that, now, breaks open
a whole spectrum of
1181
01:09:01,520 --> 01:09:05,390
possibilities in which I
introduced this normal mode
1182
01:09:05,390 --> 01:09:13,979
number n as in Nancy whereby n
can be 1, 2, 3, et cetera.
1183
01:09:13,979 --> 01:09:19,899
And then I get solutions when k
of n equals n pi divided by
1184
01:09:19,899 --> 01:09:20,990
L.
1185
01:09:20,990 --> 01:09:24,930
You see that immediately,
because if I make x now L,
1186
01:09:24,930 --> 01:09:26,680
then I get the sine of n pi.
1187
01:09:26,680 --> 01:09:31,359
No matter what n is,
I always get 0.
1188
01:09:31,359 --> 01:09:38,140
So my lambda of n, which is 2
pi divided by k, is then 2L
1189
01:09:38,140 --> 01:09:41,290
divided by n.
1190
01:09:41,290 --> 01:09:47,880
So I can rewrite, now, this
equation as p0 times the sine
1191
01:09:47,880 --> 01:09:54,650
of n pi x divided by L. And now
I have cosine omega n t,
1192
01:09:54,650 --> 01:09:58,450
because omega n is now--
1193
01:09:58,450 --> 01:10:02,840
and now the frequencies which
I associated with the n-th
1194
01:10:02,840 --> 01:10:06,840
mode, n being n as in Nancy.
1195
01:10:06,840 --> 01:10:09,040
What, now, is the connection
between this
1196
01:10:09,040 --> 01:10:11,920
omega n and this k?
1197
01:10:11,920 --> 01:10:14,140
Well, that connection
you will find
1198
01:10:14,140 --> 01:10:16,760
through the wave equation.
1199
01:10:16,760 --> 01:10:21,590
You now have to substitute this
result back into the wave
1200
01:10:21,590 --> 01:10:26,000
equation, which is this one,
to solve for omega.
1201
01:10:26,000 --> 01:10:27,840
And I want to do
that with you.
1202
01:10:27,840 --> 01:10:29,120
It is not that much work.
1203
01:10:29,120 --> 01:10:31,770
1204
01:10:31,770 --> 01:10:36,700
I go to d2p dx squared.
1205
01:10:36,700 --> 01:10:42,930
So here it is, d2p dx squared.
1206
01:10:42,930 --> 01:10:47,250
So all I get is I get this out
twice, so I get n squared pi
1207
01:10:47,250 --> 01:10:49,800
squared over L squared--
1208
01:10:49,800 --> 01:10:53,180
n squared pi squared
over L squared.
1209
01:10:53,180 --> 01:10:56,840
I get a minus sign, because if I
take twice the derivative, I
1210
01:10:56,840 --> 01:10:58,460
always end up with
a minus sign.
1211
01:10:58,460 --> 01:11:01,310
But the sine comes
back as a sine.
1212
01:11:01,310 --> 01:11:03,550
And not only does the sine come
back, but all the rest
1213
01:11:03,550 --> 01:11:04,800
comes back.
1214
01:11:04,800 --> 01:11:07,020
So I will just write p here.
1215
01:11:07,020 --> 01:11:09,550
So this p is exactly that p.
1216
01:11:09,550 --> 01:11:12,950
So this is the only thing that
was added by taking the second
1217
01:11:12,950 --> 01:11:16,060
derivative against x.
1218
01:11:16,060 --> 01:11:20,420
Now, what is d2p dt squared?
1219
01:11:20,420 --> 01:11:22,570
Now I have to go to this
function, my partial
1220
01:11:22,570 --> 01:11:23,440
derivatives.
1221
01:11:23,440 --> 01:11:26,520
x is constant here, but t is
the one that is changing,
1222
01:11:26,520 --> 01:11:29,790
whereas here, we had that x was
1223
01:11:29,790 --> 01:11:31,520
changing but t was constant.
1224
01:11:31,520 --> 01:11:35,590
So now we're going to get
minus omega n squared,
1225
01:11:35,590 --> 01:11:36,950
again, times p.
1226
01:11:36,950 --> 01:11:40,090
The whole function comes back.
1227
01:11:40,090 --> 01:11:44,120
And now we are in business,
because now the wave equation
1228
01:11:44,120 --> 01:11:46,345
will tell us the connection
between the two.
1229
01:11:46,345 --> 01:11:50,630
It tells us that n squared times
pi squared divided by L
1230
01:11:50,630 --> 01:11:54,000
squared is now 1 over
v squared--
1231
01:11:54,000 --> 01:11:56,800
that is the 1 over v squared
that I have there.
1232
01:11:56,800 --> 01:11:59,350
And there is a minus sign here,
and there's a minus sign
1233
01:11:59,350 --> 01:12:01,370
here times omega n squared.
1234
01:12:01,370 --> 01:12:03,930
1235
01:12:03,930 --> 01:12:07,690
So this is not connected.
1236
01:12:07,690 --> 01:12:12,330
So you see the solution for
omega n just is being
1237
01:12:12,330 --> 01:12:15,230
presented to you on
a silver plate.
1238
01:12:15,230 --> 01:12:24,560
Omega n is now n pi times v
divided by L. That follows
1239
01:12:24,560 --> 01:12:27,020
from the wave equation.
1240
01:12:27,020 --> 01:12:31,500
So if you prefer the frequency
in hertz, then you have to
1241
01:12:31,500 --> 01:12:33,280
divide this by 2 pi.
1242
01:12:33,280 --> 01:12:35,640
So you get nv divided by 2L.
1243
01:12:35,640 --> 01:12:39,800
1244
01:12:39,800 --> 01:12:46,375
So if I want to plot now, so
this is x equals 0, and this
1245
01:12:46,375 --> 01:12:51,730
is x equals L, I can plot now
here the pressure in terms of
1246
01:12:51,730 --> 01:12:56,110
this overpressure or
underpressure p, and you get a
1247
01:12:56,110 --> 01:13:01,300
curve which looks very similar
to a transverse solution.
1248
01:13:01,300 --> 01:13:02,980
But of course, it is
not transverse.
1249
01:13:02,980 --> 01:13:05,020
It's really longitudinal.
1250
01:13:05,020 --> 01:13:09,940
But you would get, then, that
for n equals 1, you
1251
01:13:09,940 --> 01:13:11,190
would get this mode.
1252
01:13:11,190 --> 01:13:13,270
1253
01:13:13,270 --> 01:13:16,260
There must be a pressure node
here and there, because we can
1254
01:13:16,260 --> 01:13:17,420
never build up pressure.
1255
01:13:17,420 --> 01:13:18,785
It's connected with
the universe.
1256
01:13:18,785 --> 01:13:21,310
You can never build
up overpressure.
1257
01:13:21,310 --> 01:13:25,360
So the pressure builds up here
positive, and then later in
1258
01:13:25,360 --> 01:13:26,830
time, it will be negative,
and then
1259
01:13:26,830 --> 01:13:29,400
positive, and then negative.
1260
01:13:29,400 --> 01:13:36,040
And when you go to n equals 2,
then you get another node in
1261
01:13:36,040 --> 01:13:37,650
pressure here.
1262
01:13:37,650 --> 01:13:41,190
So now, you get this.
1263
01:13:41,190 --> 01:13:42,620
Always a pressure node here.
1264
01:13:42,620 --> 01:13:44,180
Always a pressure node there.
1265
01:13:44,180 --> 01:13:48,280
But now you end up with another
pressure node there.
1266
01:13:48,280 --> 01:13:50,940
And if you have any difficulties
to see what the
1267
01:13:50,940 --> 01:13:55,310
air molecules are doing, I would
recommend you go back to
1268
01:13:55,310 --> 01:14:00,490
xi space, which is the actual
position of the molecules.
1269
01:14:00,490 --> 01:14:03,850
And when you do that, you will
always find that where the
1270
01:14:03,850 --> 01:14:09,370
pressure has an anti-node, which
is here, xi always has a
1271
01:14:09,370 --> 01:14:13,690
node, and where the pressure
has a node, xi
1272
01:14:13,690 --> 01:14:15,100
always has an anti-node.
1273
01:14:15,100 --> 01:14:19,100
Of course, the molecules can
freely move in and out.
1274
01:14:19,100 --> 01:14:20,890
There is no problem.
1275
01:14:20,890 --> 01:14:23,630
So the molecules can freely
move in and out here.
1276
01:14:23,630 --> 01:14:28,210
So xi has its largest amplitude
is anti-nodes.
1277
01:14:28,210 --> 01:14:30,580
That is where the pressure
cannot build up.
1278
01:14:30,580 --> 01:14:34,300
But going to xi space, actually,
often helps me to
1279
01:14:34,300 --> 01:14:37,360
see precisely what is
going on with the
1280
01:14:37,360 --> 01:14:38,630
motion of the molecules.
1281
01:14:38,630 --> 01:14:41,450
1282
01:14:41,450 --> 01:14:51,260
I have here a linear system,
which is a sound cavity.
1283
01:14:51,260 --> 01:14:53,230
It's made of magnesium.
1284
01:14:53,230 --> 01:14:57,240
It is not air, but
it is magnesium.
1285
01:14:57,240 --> 01:15:02,330
And it is open-open
on both sides.
1286
01:15:02,330 --> 01:15:04,300
You can't have it any better.
1287
01:15:04,300 --> 01:15:06,015
I'll make a drawing for you.
1288
01:15:06,015 --> 01:15:09,670
1289
01:15:09,670 --> 01:15:12,430
So here is my magnesium.
1290
01:15:12,430 --> 01:15:13,910
It is a rod.
1291
01:15:13,910 --> 01:15:19,160
It's one dimensional, and the
length is 122 centimeters.
1292
01:15:19,160 --> 01:15:24,920
And the speed of sound in
magnesium is about 5,000
1293
01:15:24,920 --> 01:15:26,170
meters per second.
1294
01:15:26,170 --> 01:15:29,430
1295
01:15:29,430 --> 01:15:36,460
When I hit that magnesium rod
on the side, it wants to go
1296
01:15:36,460 --> 01:15:39,020
into standing waves.
1297
01:15:39,020 --> 01:15:40,850
It prefers the lowest mode.
1298
01:15:40,850 --> 01:15:42,560
It almost always does.
1299
01:15:42,560 --> 01:15:46,080
But it may also create second
and third harmonics.
1300
01:15:46,080 --> 01:15:53,180
So the lowest mode F1 is
then v divided by 2L.
1301
01:15:53,180 --> 01:15:59,360
So the lowest frequency F1 is v
divided by 2L, which, when I
1302
01:15:59,360 --> 01:16:03,990
calculate it, is about
250 hertz.
1303
01:16:03,990 --> 01:16:09,430
And the actual value we measured
is about 2,044.
1304
01:16:09,430 --> 01:16:12,860
And there may also be, when I
hit it with a hammer, there
1305
01:16:12,860 --> 01:16:16,180
may also be some that
is twice as high.
1306
01:16:16,180 --> 01:16:20,280
So that may be 4,100 hertz.
1307
01:16:20,280 --> 01:16:22,550
That would be double
the frequency.
1308
01:16:22,550 --> 01:16:25,750
That would be the second mode.
1309
01:16:25,750 --> 01:16:27,770
And this is quite remarkable.
1310
01:16:27,770 --> 01:16:30,050
So this is not filled
with air, but this
1311
01:16:30,050 --> 01:16:31,830
is filled with magnesium.
1312
01:16:31,830 --> 01:16:33,710
But by exciting it here--
1313
01:16:33,710 --> 01:16:36,660
it's like blowing on
the flute there--
1314
01:16:36,660 --> 01:16:40,750
it goes into these normal mode
solutions, and it is this one
1315
01:16:40,750 --> 01:16:42,190
that you will hear
loud and clear.
1316
01:16:42,190 --> 01:16:43,740
It's a beautiful tone.
1317
01:16:43,740 --> 01:16:46,200
And this one, you may hear in
the beginning, but the high
1318
01:16:46,200 --> 01:16:51,185
harmonics often die out faster
than the lower harmonics.
1319
01:16:51,185 --> 01:16:53,820
So you ready for this?
1320
01:16:53,820 --> 01:16:54,845
Open-open.
1321
01:16:54,845 --> 01:16:56,800
It's open on both sides.
1322
01:16:56,800 --> 01:16:59,220
And I just bang the
hell out of it.
1323
01:16:59,220 --> 01:17:03,670
1324
01:17:03,670 --> 01:17:06,525
2,044.
1325
01:17:06,525 --> 01:17:07,860
Is that beautiful?
1326
01:17:07,860 --> 01:17:09,110
It's oscillating like this.
1327
01:17:09,110 --> 01:17:11,700
1328
01:17:11,700 --> 01:17:17,290
It's exactly oscillating the way
I derived for you for air,
1329
01:17:17,290 --> 01:17:19,240
and here, you see it that
it holds [INAUDIBLE].
1330
01:17:19,240 --> 01:17:24,900
1331
01:17:24,900 --> 01:17:27,210
Only the fundamental is there.
1332
01:17:27,210 --> 01:17:30,840
1333
01:17:30,840 --> 01:17:31,660
OK.
1334
01:17:31,660 --> 01:17:33,750
I wish you luck on Thursday.
1335
01:17:33,750 --> 01:17:35,000
I'll see you then.
1336
01:17:35,000 --> 01:17:45,360