WEBVTT
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Professor Ramamurti
Shankar: Stable equilibrium,
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and if you disturb them,
they rock back and forth and
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there are two simple examples.
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The standard textbook example
is this mass on spring.
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This spring has a certain
natural length,
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which I'm showing you here.
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It likes to sit there.
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But if you pull it so the mass
comes here, you move it from
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x = 0 to a new location
x;
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then, there is a restoring
force F,
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which is -kx,
and that force will be equal to
00:41.193 --> 00:45.503
mass times acceleration by
Newton's law,
00:45.500 --> 00:48.730
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and so you're trying to solve
the equation d^(2)x/dt^(2) =
00:57.102 --> 01:02.692
-k/m times x,
and we use the symbol,
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ω^(2) = k/m,
or ω square root of
01:06.870 --> 01:08.740
k/m.
01:08.739 --> 01:10.419
And what's the solution to this
equation?
01:10.420 --> 01:12.760
I think we did a lot of talking
and said look,
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we are looking for a function
which,
01:14.730 --> 01:17.990
when differentiated twice,
looks like the same function up
01:17.994 --> 01:20.154
to some constants,
and we know they are
01:20.149 --> 01:21.359
trigonometry functions.
01:21.360 --> 01:24.340
Then, we finally found the
answer.
01:24.340 --> 01:30.380
The most general answer looks
like A cos ωt - φ.
01:30.379 --> 01:32.849
A is the amplitude,
φ is called the phase.
01:32.849 --> 01:36.059
That tells you what your clock
is doing when x is a
01:36.059 --> 01:39.099
maximum, and we can always
choose φ to be 0.
01:39.099 --> 01:43.649
That means when the time is 0,
x will have the biggest
01:43.650 --> 01:46.580
value A.
This is an example of simple
01:46.582 --> 01:48.852
harmonic motion.
But it is a very generic
01:48.848 --> 01:51.278
situation, so I'll give you a
second example.
01:51.280 --> 01:54.860
Now, we don't have a mass but
we have a bar,
01:54.862 --> 01:59.612
let's say, suspended by a
cable, hanging from the ceiling,
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and it's happy the way it is.
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But if you come and twist it by
angle θ,
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give it a little twist,
then it will try to untwist
02:08.880 --> 02:10.910
itself.
So, now we don't have a
02:10.909 --> 02:13.599
restoring force but we have a
restoring torque.
02:13.599 --> 02:18.269
What can be the expression for
the restoring torque?
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When you don't do anything,
it doesn't do anything,
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so it's a function of θ
that vanishes when θ is
02:24.411 --> 02:25.971
0.
If θ is not 0,
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it'll begin as some function of
θ, but the leading term
02:29.454 --> 02:30.974
would be just θ.
02:30.970 --> 02:34.480
But now, you can put a constant
here.
02:34.479 --> 02:37.479
It'll still be proportional to
θ, and you put a minus
02:37.484 --> 02:40.544
sign for the same reason you put
a minus sign here to tell you
02:40.538 --> 02:41.938
it's a restoring torque.
02:41.940 --> 02:43.810
That means if you make
θ positive,
02:43.812 --> 02:46.052
the torque will try to twist
you the other way.
02:46.050 --> 02:48.540
If you make θ negative
it'll try to bring you back.
02:48.539 --> 02:50.349
So, you have to find this
κ.
02:50.349 --> 02:54.589
If you find this κ then
you can say -κ times
02:54.585 --> 02:57.885
θ is I times
d^(2)θ over
02:57.887 --> 03:00.177
dt^(2).
Mathematically,
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that equation's identical to
that, and θ then will
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look like some constant.
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You can call it A cos (ωt-
φ, and ω [squared]
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now will be the ratio of this
κ at the moment of
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inertia,
because mathematically that's
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the role played by κ and
I.
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They are like k and
m.
03:20.780 --> 03:23.990
03:23.990 --> 03:26.320
Now, if someone tells you this
is κ,
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then you are done.
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You just stick it in and
mindlessly calculate all the
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formulas.
You can find θ,
03:30.974 --> 03:32.494
you can derivatives and so on.
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Sometimes, they may not tell
you what κ is.
03:35.210 --> 03:38.490
In the case of a spring,
they would have to give you
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k or they may tell you
indirectly if I pull this spring
03:42.411 --> 03:44.721
by 9 inches,
I exert a certain force,
03:44.717 --> 03:48.017
they're giving you F and
they're giving you x and
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you can find k.
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But in rotation problems,
the typical situation is the
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following.
Let us take the easiest problem
03:57.003 --> 04:00.443
of a pendulum.
The pendulum is hanging,
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say a massless rod,
with some mass m at the
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bottom.
So, it's very happy to stay
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this way.
If you leave it like this,
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it would stay this way forever.
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No torque, no motion.
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Now, you come along and you
disturb that by turning it by an
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angle θ.
If you do, then the force is
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like this, the separation vector
r from the pivot point is
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this,
and r cross F
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will be non-zero because the
angle between them is not 0,
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and you remember that torque is
equal to rF sin θ,
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and I told you for small
angles,
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this can be rF θ,
and r happens to be,
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in this problem,
the length of the pendulum,
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so it's really -mgl θ.
04:50.880 --> 04:54.300
04:54.300 --> 04:56.820
So, you have to do some work to
find θ [correction:
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should have said κ].
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It wasn't just given to you on
a plate.
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You did this piece of thinking,
and namely, you disturbed the
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system from equilibrium and
found the restoring torque.
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Then, you stared at the formula
and say hey, this guy must be
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the κ I'm looking for
because that number times
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θ is the restoring
torque.
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Then the ω here will
be--this κ is a generic
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κ that goes into
whatever it is for that problem.
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For our problem,
κ happens to be
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mgl.
That's something you should
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understand.
It's not a universal number
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that's known to everybody like
the spring constant of anything.
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In the case of mass and this
pendulum, it depends on how long
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the pendulum is,
how big the mass is,
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but you can work it out and
extract κ.
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Downstairs, you need the
momentum of inertia for a point
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mass m at a distance
l from the point of
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rotation,
which is ml^(2).
05:51.550 --> 05:54.360
So, you can cancel the
l, you can cancel the
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m, and you get ω
is root of g over
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l.
Let me remind you guys that
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ω is connected to what
you and I would call the
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frequency by this 2π,
or what you and I would call
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the time period by 2π
over T.
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06:17.230 --> 06:20.330
Okay, so this is the situation.
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Let me give one other example.
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This is a homework problem,
but I want to give you a hint
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on how to do this.
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Don't take a pendulum with all
the mass concentrated there.
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Take some irregularly shaped
object, a flat planar object.
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You drive a nail through it
there, hang it on the wall.
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It'll come to rest in a certain
configuration and you should
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think about where will the
center of mass be if I hang it
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like this on the wall?
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Center of mass is somewhere in
the body here.
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So, I want you to think about
it.
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I'll tell you in a second.
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The center of mass,
they claim, will lie somewhere
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on this vertical line going
through that point.
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I know that by default because
if the center of mass is here,
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for example,
we know all the force of
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gravity can be imagined to be
acting here [pointing to
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drawing].
Then if you do that separation
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and do the torque,
then you will find this is able
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to exert a torque on this point,
but that cannot be in
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equilibrium.
So, center of mass will align
07:24.935 --> 07:27.005
itself.
The body would swing a little
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bit and settle down;
the center of mass somewhere
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there.
If you now disturb the body,
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I don't want to draw another
picture, but I'll probably fail.
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This is the rotated body,
off from its ideal position,
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then there will be a torque.
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What will the torque be?
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It'll be the same thing minus
mgl sin θ,
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which I'm going to replace by
θ.
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l is now the distance
between the pivot point and the
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center of mass.
That's your torque.
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So, you can read the κ.
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In fact, it's just like the
pendulum.
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In other words,
as far as the torque is
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concerned, it's as if all the
mass were sitting here.
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The difference will be in a
moment of inertia.
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Moment of inertia of this is
not ml^(2),
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so don't make the mistake.
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All the mass is not sitting
here.
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All the mass is sitting all
over the place.
08:19.290 --> 08:23.100
So, you should know that if you
want the moment of inertia,
08:23.104 --> 08:26.664
it is I with respect to
center of mass plus this
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ml^(2).
That's the old parallel axis
08:29.966 --> 08:32.716
theorem.
So, take that moment of inertia
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and put that into this formula
here;
08:35.179 --> 08:37.789
take this κ and put it
there, you'll get some
08:37.788 --> 08:39.868
frequency.
That's the frequency with which
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it will oscillate.
08:40.769 --> 08:44.949
So, every problem that you will
ever get will look like one of
08:44.948 --> 08:47.028
these two.
Either there is something
08:47.025 --> 08:49.535
moving linearly with a
coordinate that you can call
08:49.537 --> 08:52.337
x,
or it's rotating or twisting by
08:52.339 --> 08:54.679
an angle you can call θ.
08:54.679 --> 08:57.749
And if you want to find out the
frequency of vibration,
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you have to disturb it from
equilibrium,
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either by pulling the mass or
by twisting the cable or
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displacing this pendulum from
its ideal position here,
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then finding the restoring
torque.
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Yep?
Student: What can we say
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[inaudible]
its restoring torque in the
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twisting example?
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How do we describe that?
09:14.580 --> 09:17.020
Professor Ramamurti
Shankar: That's a very good
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point.
His question is,
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if I gave you a cable and I
twist the cable by some angle,
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how are we going to calculate
restoring torque?
09:25.470 --> 09:26.800
Here is the good news.
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This problem is so hard we'll
give it to you.
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In other words,
there is, of course,
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an underlying answer.
09:34.620 --> 09:37.050
Given a cable,
made of some material and its
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torsional properties,
how much of a torque you'll get
09:39.998 --> 09:43.358
if you twist it by some amount,
but it's not something you can
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calculate from first principle,
so they'll simply have to give
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it to you,
okay, in such problems.
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The only time you'll have to
find the torque on your own is
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in a problem like this where I
believe you know enough to
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figure out the torque.
09:55.889 --> 09:58.779
Okay, what you will find is,
if you leave it alone,
09:58.779 --> 10:01.669
it'll go to a position where
there is no torque.
10:01.669 --> 10:03.969
If you move it off the
position, there will be a
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torque, and the torque will
always be proportional to the
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angle by which you've displaced
it.
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You read off the
proportionality constant and
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that's your κ.
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You should look at the forces
too.
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It is pretty interesting.
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This body, when it is hanging
in its rest position,
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has two forces on it.
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The nail, which is pushing up
and the weight of the body which
10:26.480 --> 10:28.780
is pushing down,
and they cancel each other.
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The nail will keep it from
falling.
10:31.090 --> 10:35.130
The nail will not keep it from
swinging, because the force of
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the nail acting as it does at
the pivot point is unable to
10:38.970 --> 10:41.970
exert a torque,
whereas the minute you rotate
10:41.967 --> 10:44.747
the body, mg is able to
exert a torque.
10:44.750 --> 10:46.950
That's why if you twist it,
it'll start rattling back and
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forth.
10:47.260 --> 10:50.590
10:50.590 --> 10:54.630
Alright.
So, what I'm going to do now is
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to go over more complicated
oscillations using some of the
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techniques we learned last time.
11:04.490 --> 11:06.550
So, here is the mega formula.
11:06.549 --> 11:07.899
I'm going to give it to you
guys.
11:07.899 --> 11:11.139
You're allowed to tattoo it on
your face.
11:11.140 --> 11:12.710
You can carry it with you.
11:12.710 --> 11:15.360
I will allow you to bring it
but you cannot forget this
11:15.359 --> 11:17.289
formula.
I think you know the formula
11:17.288 --> 11:19.308
I'm driving at.
Ready?
11:19.309 --> 11:23.269
e to the ix,
or θ,
11:23.274 --> 11:27.714
I don't care.
Let me call it θ here,
11:27.708 --> 11:30.518
is cos θ + I sin θ.
11:30.519 --> 11:33.979
That's a great formula from
Euler.
11:33.980 --> 11:37.100
From this formula,
if you take the complex
11:37.098 --> 11:40.668
conjugate of both sides,
that means change every
11:40.672 --> 11:45.852
i to minus i,
you will get e to the
11:45.848 --> 11:50.478
minus iθ is cos θ -
i sin θ.
11:50.480 --> 11:52.600
That also you should know.
11:52.600 --> 11:56.430
11:56.429 --> 11:58.499
If you got this in your
head--By the way,
11:58.502 --> 12:01.562
this is something you're not
going to derive on the spot.
12:01.560 --> 12:02.450
It takes a lot of work.
12:02.450 --> 12:06.110
If you tell me what are the few
things you really have to cram,
12:06.109 --> 12:08.879
that you don't want to carry in
your head, well,
12:08.883 --> 12:10.303
this is one of them.
12:10.299 --> 12:14.379
Okay, this is a formula worth
memorizing, unlike formula No.
12:14.380 --> 12:17.010
92 of your text.
It's not worth memorizing.
12:17.010 --> 12:18.340
This is worth memorizing.
12:18.340 --> 12:21.060
Once you got this,
you should realize that in any
12:21.059 --> 12:23.269
expression involving complex
numbers,
12:23.269 --> 12:26.269
you can get another equation
where every i is changed
12:26.272 --> 12:27.292
to minus i.
12:27.290 --> 12:29.190
That'll give you this one.
12:29.190 --> 12:31.480
That's called complex
conjugation.
12:31.480 --> 12:34.180
I generally said,
if you have a complex number
12:34.175 --> 12:37.705
z, z star is equal
to x - iy,
12:37.710 --> 12:40.580
if z is equal to
x plus Iy.
12:40.580 --> 12:43.790
So, in our simple example,
this fellow here is z.
12:43.789 --> 12:47.539
This is the x and this
is the y.
12:47.539 --> 12:52.599
Now, you should be able to
invert this formula.
12:52.600 --> 12:56.970
To invert the formula,
you add the 2 and divide by 2.
12:56.970 --> 13:01.680
Then, you will find cos
θ is e^(iθ) +
13:01.683 --> 13:03.953
e^(-iθ) over 2.
13:03.950 --> 13:08.300
If you subtract and divide by
2i, you will find sin
13:08.296 --> 13:12.566
θ is e^(iθ) -
e^(iθ) over 2i.
13:12.570 --> 13:15.110
In other words,
my claim is that this funny
13:15.114 --> 13:18.694
exponential, sum of exponential,
is the family of cosine.
13:18.690 --> 13:20.670
It'll do everything your cosine
will do.
13:20.669 --> 13:22.939
It'll oscillate,
sine squared plus cosine
13:22.944 --> 13:25.624
squared would be 1,
it'll obey all trigonometric
13:25.617 --> 13:28.047
identities,
like sine 2 θ is 2
13:28.051 --> 13:31.461
sin θ cos θ,
everything comes out of this
13:31.458 --> 13:33.198
expression.
So, you should know,
13:33.201 --> 13:35.031
here is the main point you
should learn.
13:35.029 --> 13:38.929
We don't need trigonometric
functions anymore.
13:38.929 --> 13:43.129
Once you got the exponential
function, provided you let the
13:43.126 --> 13:47.466
exponent be complex or imaginary
you don't need trigonometric
13:47.467 --> 13:50.277
functions.
This is one example of grand
13:50.282 --> 13:52.462
unification.
People always say Maxwell
13:52.455 --> 13:54.585
unified this,
and Einstein tried to unify
13:54.592 --> 13:56.942
this.
Unification means things that
13:56.942 --> 14:00.122
you thought were unrelated are,
in fact, related,
14:00.123 --> 14:03.373
and there are different
manifestations of the same
14:03.370 --> 14:05.460
thing.
When we first discovered
14:05.456 --> 14:08.336
trigonometric functions,
we were drawing right-angle
14:08.339 --> 14:11.109
triangles and opposite side and
adjacent side.
14:11.110 --> 14:13.580
Then, we discovered the
exponential function which,
14:13.579 --> 14:16.199
by the way, was computed by
bankers who were trying to
14:16.197 --> 14:18.417
calculate compound interest
every second.
14:18.419 --> 14:21.279
The fact that those functions
are related is a marvelous
14:21.278 --> 14:24.238
result, but it happens only if
you got complex numbers.
14:24.240 --> 14:25.960
So, this is another thing you
should know.
14:25.960 --> 14:31.230
Okay, now, armed with this we
are ready to do anything we
14:31.231 --> 14:33.871
want.
Let me just tell you that every
14:33.865 --> 14:37.625
complex number z can be
written as its absolute value
14:37.625 --> 14:38.895
times some phase.
14:38.899 --> 14:41.499
In other words,
if my complex number z
14:41.497 --> 14:43.737
is here, it is x +
iy;
14:43.740 --> 14:45.050
that's one way to write it.
14:45.049 --> 14:47.599
You can also write it as its
absolute value.
14:47.600 --> 14:50.380
It'll be e to the
iφ φr,
14:50.378 --> 14:52.188
re to the iφ.
14:52.190 --> 14:55.290
r is the radial length
which also happens to be the
14:55.286 --> 14:56.966
length of the complex number.
14:56.970 --> 15:00.250
That's called the polar form of
the complex number.
15:00.250 --> 15:03.100
x + iy is the Cartesian
form of the complex number.
15:03.100 --> 15:06.320
15:06.320 --> 15:07.830
I may have called them r
and θ,
15:07.830 --> 15:08.670
r and φ.
15:08.670 --> 15:09.760
I just don't remember.
15:09.759 --> 15:13.569
I'm going to use these symbols
back and forth because people do
15:13.567 --> 15:16.427
use both the symbols,
which is why I'm sometimes
15:16.430 --> 15:19.660
calling this angle as θ
and sometimes this angle as
15:19.664 --> 15:21.914
φ,
but the basic idea is the same.
15:21.909 --> 15:26.509
This entity here can do it
either in this form or in this
15:26.514 --> 15:29.034
form.
This is something you should
15:29.033 --> 15:30.263
know.
If you don't know,
15:30.259 --> 15:32.509
or you don't understand,
you should stop and ask.
15:32.510 --> 15:34.310
Everything is built on this.
15:34.309 --> 15:43.339
So, what I'm going to do now is
to go back to this rather simple
15:43.336 --> 15:49.636
equation, d^(2)x/dt^(2) =
-ω^(2)x.
15:49.639 --> 15:52.549
By the way, today I'm going to
call this ω as
15:52.554 --> 15:53.904
ω_0.
15:53.900 --> 15:56.300
It's the same guy.
15:56.299 --> 15:58.579
I'm going to call it
ω_0 because we
15:58.575 --> 16:00.805
will find as the hour
progresses, there are going to
16:00.808 --> 16:02.338
be many ωs in the game.
16:02.340 --> 16:04.080
Two more ωs are going
to come in.
16:04.080 --> 16:06.290
So, we've got to be able to
tell them apart.
16:06.289 --> 16:07.909
As long as there's only one
ω in town,
16:07.913 --> 16:08.963
you just call it ω.
16:08.960 --> 16:10.890
If there are many,
you call this
16:10.890 --> 16:14.630
ω_0 to mean
it's the frequency of vibration
16:14.626 --> 16:17.176
of the system left to its own
design,
16:17.180 --> 16:18.240
and you pull it and let it go.
16:18.240 --> 16:19.010
What's the frequency?
16:19.009 --> 16:20.869
An angular frequency,
that's what we call
16:20.866 --> 16:21.976
ω_0.
16:21.980 --> 16:23.960
Now, how did we solve this
equation?
16:23.960 --> 16:27.620
The way I tried to solve it for
you is to say,
16:27.620 --> 16:30.060
turn it into a word problem.
16:30.059 --> 16:33.179
I'm looking for a function,
x(t) with a property
16:33.178 --> 16:36.758
that two derivatives of the
function look like the function,
16:36.759 --> 16:37.929
and we rake our brains
[inaudible]
16:37.929 --> 16:39.629
and we remembered,
hey, sines and cosines are the
16:39.629 --> 16:42.299
property.
One derivative is no good.
16:42.300 --> 16:43.410
Turn sign into cosine.
16:43.409 --> 16:47.079
Two derivatives bring back the
function you started with,
16:47.081 --> 16:50.491
which is why the answer could
be sines or cosines.
16:50.490 --> 16:53.320
But now I'm going to solve with
a different way.
16:53.320 --> 16:57.890
My thinking is going to be:
I know a function that repeats
16:57.891 --> 17:01.341
itself, even when it
differentiated once.
17:01.340 --> 17:04.720
Namely, the function has a
property;
17:04.720 --> 17:06.990
its first derivative looks like
the function itself.
17:06.990 --> 17:09.980
It's obvious that his 90-second
derivative will also look like
17:09.975 --> 17:12.955
the function because taking the
derivative leaves the function
17:12.960 --> 17:14.660
alone,
except for pulling out some
17:14.657 --> 17:16.447
numbers.
So, why not say,
17:16.450 --> 17:21.580
"I want an answer that looks
like this: x equals some
17:21.576 --> 17:24.266
A to the αt."
17:24.269 --> 17:26.919
That certainly has a property
that if you take two derivatives
17:26.916 --> 17:29.346
it's going to look like e
to the αt on the
17:29.345 --> 17:31.735
left-hand side,
and e to the αt
17:31.738 --> 17:34.398
on the right-hand side,
and then you can cancel them
17:34.403 --> 17:36.393
and you've got yourself a
solution.
17:36.390 --> 17:42.100
Do you guys remember why I
didn't follow this solution for
17:42.097 --> 17:45.257
the oscillator?
Why was it rejected?
17:45.260 --> 17:47.530
Yes?
Student: [inaudible]
17:47.527 --> 17:49.757
Professor Ramamurti
Shankar: Ah.
17:49.760 --> 17:50.830
That's a very good point.
17:50.829 --> 17:53.479
So, maybe I will take e
to the minus αt.
17:53.480 --> 17:56.080
How about that?
Will that give me a second
17:56.076 --> 17:57.696
derivative which is negative?
17:57.700 --> 18:00.980
You didn't fall for that, right?
18:00.980 --> 18:03.520
Because the first time you'll
get -α but second time
18:03.523 --> 18:04.733
you'll get +α^(2).
18:04.730 --> 18:07.420
That's correct.
So, this function is no good.
18:07.420 --> 18:09.320
Also, it doesn't look like what
I want.
18:09.319 --> 18:11.669
Even without doing much
mathematical physics,
18:11.669 --> 18:15.139
I know if you pull this spring
it's going to go back and forth,
18:15.140 --> 18:17.250
whereas these functions are
exponentially growing;
18:17.250 --> 18:20.330
they're exponentially falling,
they just don't do the trick.
18:20.329 --> 18:23.519
But now, we'll find that if
you're willing to work with
18:23.520 --> 18:26.180
complex exponentials,
this will do the job.
18:26.180 --> 18:29.910
So, we're just going to take
this guess and put it in the
18:29.914 --> 18:33.254
equation and see if I can get an
answer that works,
18:33.249 --> 18:36.069
okay?
This is called an ansatz.
18:36.069 --> 18:37.719
Ansatz are something we
use all the time.
18:37.720 --> 18:38.730
It's a German word.
18:38.730 --> 18:43.420
I'm not sure exactly what it
means, but we all use it to mean
18:43.417 --> 18:44.977
a tentative guess.
18:44.980 --> 18:46.160
That's what it is.
18:46.160 --> 18:47.770
If you're lucky, it'll work.
18:47.769 --> 18:50.059
If you're not lucky,
that's fine.
18:50.059 --> 18:51.459
You move on and try another
solution.
18:51.460 --> 18:54.700
So, we're going to say,
"Can a solution of this form be
18:54.704 --> 18:57.954
found, with A and
α completely free?"
18:57.950 --> 19:00.400
Maybe the equation will tell me
what A and α
19:00.402 --> 19:03.312
should be.
So, let's take it, put it here.
19:03.309 --> 19:07.019
By the way, I'm going to write
this equation in a form that's a
19:07.022 --> 19:09.832
little easier.
x double dot is minus
19:09.831 --> 19:13.601
ω_0^(2)x but
each dot is a derivative.
19:13.599 --> 19:18.059
It's just more compact that way
rather than to write all of this
19:18.063 --> 19:21.093
stuff.
So, let's take that guess and
19:21.090 --> 19:23.890
put it here and see what we get.
19:23.890 --> 19:27.630
I get, when I take two
derivatives, you can say,
19:27.634 --> 19:32.024
"Do I want e to the
αt or e to the
19:32.015 --> 19:34.985
-αt,"
you'll find it doesn't matter
19:34.993 --> 19:38.523
so I'm going to take e to
the αt and put it in.
19:38.519 --> 19:42.159
Now, the beauty of the x
function now is to take one
19:42.163 --> 19:43.863
derivative, x dot.
19:43.859 --> 19:46.549
I hope everybody knows the
x dot of this is,
19:46.551 --> 19:49.081
A [α]
e to the αt,
19:49.082 --> 19:51.722
and x double dot is
another α.
19:51.720 --> 19:55.070
Aα^(2)e to the
αt.
19:55.069 --> 19:56.649
That's the beauty of the
exponential.
19:56.650 --> 19:59.750
The act of differentiation is
trivial.
19:59.750 --> 20:02.340
You just multiply it by the
exponent, α.
20:02.339 --> 20:06.589
See, this being the result of
taking two derivatives,
20:06.592 --> 20:08.802
let me write it here now.
20:08.800 --> 20:09.560
What do I get?
20:09.560 --> 20:13.200
20:13.200 --> 20:16.440
So, I'm going to go to x
double dot equal to minus
20:16.438 --> 20:18.808
ω_0 squared
x.
20:18.809 --> 20:22.669
Into that, I'm going to put in
the guess, x equals
20:22.671 --> 20:26.121
Ae to the αt,
and what do I find?
20:26.120 --> 20:29.350
20:29.349 --> 20:32.499
In fact, let me write the
equation in a nicer way:
20:32.495 --> 20:35.895
x double dot plus
ω_0 squared
20:35.896 --> 20:37.626
x has to vanish.
20:37.630 --> 20:41.210
I just brought everything to
one side.
20:41.210 --> 20:44.340
Now, put e^(αt),
A to the αt as
20:44.336 --> 20:47.106
your guess, then you find that
α^(2),
20:47.109 --> 20:52.299
Ae^(α)t,
plus ω_0
20:52.297 --> 20:56.987
squared Ae^(αt) has to
be 0.
20:56.990 --> 21:00.220
So, we know what to do.
21:00.220 --> 21:01.600
Let's group a few terms.
21:01.599 --> 21:07.399
So, this means I get A
times α^(2) + 1,
21:07.404 --> 21:10.804
e^(αt) has to vanish.
21:10.799 --> 21:14.809
If you can make that happen,
you got a solution because
21:14.812 --> 21:17.192
you've satisfied the equation.
21:17.190 --> 21:20.240
No one's going to say,
well, you got it by guesswork.
21:20.240 --> 21:23.060
Well, it turns out solving
differential equation is only
21:23.055 --> 21:25.085
guesswork.
There's no other way to solve
21:25.091 --> 21:27.641
the equation other than to make
a guess, stick it in,
21:27.640 --> 21:30.140
fiddle with the parameters,
see if it will work.
21:30.140 --> 21:32.990
So, we're not done yet because
we want this to vanish.
21:32.990 --> 21:36.830
How many ways are there to kill
this answer?
21:36.830 --> 21:38.790
You can say maybe A is 0.
21:38.790 --> 21:40.700
A is 0;
you've got what you call a
21:40.696 --> 21:41.396
trivial solution.
21:41.400 --> 21:43.640
Yes?
Student: [inaudible]
21:43.638 --> 21:45.518
Professor Ramamurti
Shankar: Ah,
21:45.522 --> 21:46.972
yes.
Thank you.
21:46.970 --> 21:51.140
Yep.
Thank you very much.
21:51.140 --> 21:56.280
So, this one is α^(2) +
ω_0^(2).
21:56.280 --> 21:59.690
Does everybody follow that?
21:59.690 --> 22:02.220
Yes, from here to here it's
very clear what to do.
22:02.220 --> 22:03.630
So, we cannot kill A.
22:03.630 --> 22:08.310
Yep?
Student: [inaudible]
22:08.309 --> 22:12.549
Professor Ramamurti
Shankar: Yeah,
22:12.547 --> 22:15.617
yeah, yeah.
I'm trying to rule out certain
22:15.622 --> 22:16.792
options I don't like.
22:16.789 --> 22:21.149
The option I don't like to get
this to be 0 is to say that is
22:21.149 --> 22:22.729
0.
That's not happening.
22:22.730 --> 22:25.240
e^(α)t is not going to
vanish.
22:25.240 --> 22:28.630
Maybe it'll vanish for some
negative infinity time,
22:28.625 --> 22:31.465
but I want this to be true at
all times.
22:31.470 --> 22:33.410
The equation has to be obeyed
at all times,
22:33.413 --> 22:35.683
so this guy certainly is not
zero at all times,
22:35.680 --> 22:39.760
so the only way to fix that is
to have α^(2) +
22:39.757 --> 22:42.167
ω_0^(2) = 0.
22:42.170 --> 22:44.310
That's the only solution,
which is not a trivial
22:44.309 --> 22:44.809
solution.
22:44.810 --> 22:48.510
22:48.509 --> 22:50.309
Well, that's a simple equation,
right?
22:50.309 --> 22:53.729
It's a quadratic equation that
means α^(2) =
22:53.726 --> 22:57.446
ω_0^(2),
or α is plus or minus
22:57.453 --> 22:59.553
iω_0.
22:59.550 --> 23:05.810
23:05.809 --> 23:08.339
So, when we were young and we
didn't know about complex
23:08.337 --> 23:11.377
numbers, we would come to this
stage and we would quit and say,
23:11.380 --> 23:14.200
"Look, exponentials won't work,
but now we are not afraid of
23:14.200 --> 23:17.320
complex exponentials,
so we embrace the solution."
23:17.320 --> 23:19.410
So now, I've got two solutions.
23:19.410 --> 23:21.520
By the way, what's the value of
A?
23:21.520 --> 23:23.710
Think about this.
23:23.710 --> 23:28.160
If this condition is satisfied,
you realize A can be
23:28.156 --> 23:29.686
whatever you like.
23:29.690 --> 23:32.420
A is completely
arbitrary.
23:32.420 --> 23:36.730
So, what this equation has told
me is the following.
23:36.730 --> 23:39.380
Yes, there are solutions of
this form.
23:39.380 --> 23:41.780
For A you can pick any
number you like,
23:41.778 --> 23:44.228
in fact, real or complex,
it doesn't matter.
23:44.230 --> 23:45.580
The equation is satisfied.
23:45.579 --> 23:48.729
But your α is not an
arbitrary number.
23:48.730 --> 23:52.420
It can be only one of two
numbers: +iω_0
23:52.415 --> 23:54.285
or -iω_0.
23:54.289 --> 23:58.149
So, we've got a problem now
that I look at one answer,
23:58.146 --> 24:00.346
I got two.
So, I will write them both.
24:00.349 --> 24:03.579
So, let's say that the solution
x_1 of
24:03.584 --> 24:06.104
t, which is some number
A,
24:06.099 --> 24:09.199
e to the
iω_0t,
24:09.200 --> 24:13.180
then I got another solution
x_2 of
24:13.176 --> 24:16.356
t,
which is equal to any other
24:16.363 --> 24:18.843
number B,
e to the
24:18.836 --> 24:21.066
-iω_0t.
24:21.070 --> 24:25.610
24:25.609 --> 24:29.599
Now, I think you guys can see
in your head that if you take
24:29.597 --> 24:32.207
this and you put it in the
equation,
24:32.210 --> 24:35.810
it works, and if you take this
and put it in the equation,
24:35.812 --> 24:38.312
that also works.
Because when you take two
24:38.313 --> 24:41.583
derivatives, iω whole
square would be minus ω
24:41.584 --> 24:43.324
square,
minus iω_0
24:43.324 --> 24:45.674
whole square would also be minus
ω not square,
24:45.670 --> 24:49.100
so both would work.
24:49.099 --> 24:52.669
Now, we have to ask,
"How do I decide between that
24:52.672 --> 24:54.862
solution and that solution?"
24:54.859 --> 24:58.719
It turns out that we don't have
to pick.
24:58.720 --> 25:02.450
We can pick both,
and I'll tell you what I mean
25:02.445 --> 25:04.385
by "we can pick both."
25:04.390 --> 25:06.670
Now, this is a very,
very important property for all
25:06.665 --> 25:08.985
of you who are going into
economics or engineering or
25:08.985 --> 25:11.195
chemistry.
This or other disciplines,
25:11.202 --> 25:14.142
which are mathematical,
the property I'm going to
25:14.141 --> 25:17.571
mention of this equation is very
important, so please pay
25:17.571 --> 25:19.961
attention.
This is called a linear
25:19.960 --> 25:20.600
equation.
25:20.600 --> 25:24.590
25:24.589 --> 25:28.079
A linear equation will obey a
certain property called
25:28.077 --> 25:30.987
superposition.
I'll have to tell you what it
25:30.986 --> 25:32.956
is.
If I give you differential
25:32.955 --> 25:35.925
equation, you know,
96 derivative of x,
25:35.925 --> 25:40.075
plus 52 times the 37 derivative
of x, blah blah blah adds
25:40.082 --> 25:43.542
up to 0;
what makes it a linear equation
25:43.541 --> 25:48.581
is that throughout the equation,
either you find the function
25:48.578 --> 25:52.638
x or its derivatives,
but never the squares of cubes
25:52.642 --> 25:55.032
at higher powers of x or
the derivatives.
25:55.030 --> 25:58.620
Okay?
The function appears to first
25:58.623 --> 26:00.793
power, not to second power.
26:00.789 --> 26:03.249
For example,
if this were the equation you
26:03.254 --> 26:06.504
were trying to solve,
that is not a linear equation.
26:06.500 --> 26:08.120
That's called a non-linear
equation.
26:08.119 --> 26:10.639
If you have a linear equation,
there's a very,
26:10.644 --> 26:13.904
very profound consequence and
I'm going to tell you what it
26:13.897 --> 26:15.867
is,
and that lies at the heart of
26:15.865 --> 26:17.055
so many things we do.
26:17.059 --> 26:21.379
So, let me write down two
solutions.
26:21.380 --> 26:24.330
One of them will be
x_1 double dot
26:24.328 --> 26:27.158
+ ω_0
^(2)x_1 = 0.
26:27.160 --> 26:29.910
Second one obeys
x_2 double dot
26:29.905 --> 26:32.705
+ ω_0^(2)x
_2 = 0.
26:32.710 --> 26:35.490
Don't even have to look at
these two solutions.
26:35.490 --> 26:38.140
Take a general problem where
you have a linear equation that
26:38.141 --> 26:39.041
are two solutions.
26:39.040 --> 26:41.660
Add the two equations.
26:41.660 --> 26:44.860
On the left-hand side,
I get a second derivative of
26:44.864 --> 26:47.884
x_1 plus a
second derivative of
26:47.877 --> 26:50.607
x_2,
and remember that is,
26:50.609 --> 26:52.329
in fact, the second derivative.
26:52.329 --> 26:55.419
Now, I go back to the old
notation of x_1 +
26:55.424 --> 26:56.554
x_2.
26:56.549 --> 26:59.419
This has to do with the fact
the derivative of a sum is the
26:59.417 --> 27:01.987
sum of the derivatives,
and the sum of the derivative
27:01.988 --> 27:03.618
is the derivative of the sum.
27:03.620 --> 27:05.420
Yep?
Student: So,
27:05.418 --> 27:07.388
because you have these two
linear solutions,
27:07.385 --> 27:09.755
does that mean you can generate
an infinite number of
27:09.763 --> 27:11.273
[inaudible]
Professor Ramamurti
27:11.272 --> 27:12.372
Shankar: Yes.
27:12.370 --> 27:13.870
I'm coming to that in a minute.
27:13.869 --> 27:17.419
I'm first doing a more modest
goal of showing that their sum
27:17.417 --> 27:18.677
is also a solution.
27:18.680 --> 27:22.190
In the second term,
I get ω_0
27:22.190 --> 27:26.540
squared times x_1 +
x_2 = 0.
27:26.540 --> 27:29.260
You stare at this equation.
27:29.259 --> 27:33.009
Look, this plus this implies
what I've written down simply by
27:33.007 --> 27:35.707
adding.
But say in words what you found
27:35.714 --> 27:36.974
out.
What you found out,
27:36.967 --> 27:39.547
is that if x_1
satisfies the equation,
27:39.547 --> 27:42.457
x_2 satisfies
the equation x_1 +
27:42.461 --> 27:45.161
x_2,
let's call it x plus,
27:45.161 --> 27:48.021
is x_1 +
x_2 also
27:48.022 --> 27:51.152
satisfies the equation,
and the proof is right in front
27:51.147 --> 27:53.577
of you.
And the key was the derivative
27:53.584 --> 27:56.274
of a sum is the sum of the
derivatives.
27:56.269 --> 27:59.739
Now, I'm going to generalize it
more and say,
27:59.739 --> 28:04.309
suppose I multiplied all of the
first equation by A,
28:04.313 --> 28:08.023
and all of the second equation
by B.
28:08.019 --> 28:10.099
Well, that's certainly still
true, right?
28:10.099 --> 28:12.959
You take something equal to 0,
multiply both sides by A,
28:12.957 --> 28:14.357
it's still going to be 0.
28:14.359 --> 28:18.079
Now, imagine adding this to
this [adding the two equations
28:18.077 --> 28:20.077
together].
Add this to this,
28:20.081 --> 28:21.591
and what do you get?
28:21.589 --> 28:24.769
You will find
d^(2)/dt^(2) of
28:24.765 --> 28:28.935
Ax_1 +
Bx_2 +
28:28.938 --> 28:34.508
ω_0^(2),
Ax_1 +
28:34.510 --> 28:37.980
Bx_2 is 0.
28:37.980 --> 28:40.870
So, here is the punch line.
28:40.869 --> 28:43.579
If you give me two solutions to
this equation,
28:43.580 --> 28:47.070
I can manufacture another one
by taking the first one times
28:47.074 --> 28:50.024
any number I like,
plus the second one times any
28:50.023 --> 28:51.203
other number I like.
28:51.200 --> 28:55.650
So, the story is a lot more
complicated than it looked.
28:55.650 --> 28:57.580
It looked like there are two
solutions;
28:57.579 --> 29:00.249
actually, there's infinite
number of solutions because you
29:00.248 --> 29:02.588
can pick A and B
any way you like.
29:02.589 --> 29:05.789
So, linear equations typically
have infinite number of
29:05.787 --> 29:09.767
solutions and you build them up
by taking a few building blocks.
29:09.769 --> 29:12.679
They're like unit vectors
I and J.
29:12.680 --> 29:15.230
You can take any multiple of
this unit, vector any multiple
29:15.225 --> 29:17.415
of that unit vector,
and this combination will also
29:17.420 --> 29:18.430
solve the equation.
29:18.430 --> 29:21.860
29:21.859 --> 29:24.339
Just so you don't think this is
going to happen all the time,
29:24.342 --> 29:25.172
let me remind you.
29:25.170 --> 29:27.450
You don't have to write this
down but you guys follow what
29:27.450 --> 29:29.610
I'm saying.
Suppose this quantity here was
29:29.605 --> 29:32.985
not x_1 but the
square of x_1
29:32.985 --> 29:34.285
and here.
Okay.
29:34.289 --> 29:36.129
You don't have to write this
because this is not a linear
29:36.128 --> 29:37.458
equation.
We're not interested in this.
29:37.460 --> 29:40.250
But notice this is non-linear,
and you guys should be able to
29:40.246 --> 29:41.956
pick it up.
It's non-linear because I've
29:41.960 --> 29:43.190
got the square of the unknown.
29:43.190 --> 29:45.980
Now, can you see if I add these
two equations,
29:45.983 --> 29:49.153
forget A and B,
let them both be 1.
29:49.150 --> 29:51.820
I get x_1
squared plus
29:51.820 --> 29:55.690
x_2 squared,
but that does not equal to
29:55.686 --> 29:59.056
x_1 +
x_2,
29:59.060 --> 30:00.700
the whole thing square.
30:00.700 --> 30:03.710
So, even though the second
derivative of the sum is the sum
30:03.709 --> 30:06.509
of the second derivative,
the sum of the squares is not
30:06.511 --> 30:07.861
the square of the sum.
30:07.859 --> 30:10.839
So, for a non-linear equation
you cannot add solutions,
30:10.836 --> 30:13.806
but for linear equations you
can combine solutions with
30:13.811 --> 30:15.301
arbitrary coefficients.
30:15.299 --> 30:19.139
That's the lesson you have
learned today.
30:19.140 --> 30:23.520
So, harmonic oscillating
equation is a linear equation,
30:23.515 --> 30:28.295
so feel free to combine them
and get the following solution:
30:28.296 --> 30:34.426
x(t) is equal to Ae^(iω0) ,
plus B e^(-iω0t).
30:34.430 --> 30:38.750
A and B are whatever you like,
but ω_0 is
30:38.754 --> 30:41.294
the original root of k/m.
30:41.290 --> 30:48.460
30:48.460 --> 30:51.650
But if I gave you this
solution, will you be happy to
30:51.645 --> 30:54.585
take it as a solution for the
mass spring system,
30:54.586 --> 30:57.156
and if not, what is it you
don't like?
30:57.160 --> 31:02.260
31:02.260 --> 31:04.280
Yes?
Anybody have a view?
31:04.279 --> 31:06.179
I mean, would you take that as
a good solution?
31:06.180 --> 31:08.820
Yes?
Student: Wouldn't you
31:08.816 --> 31:12.836
need to figure out how to choose
the right A and B?
31:12.839 --> 31:14.049
Professor Ramamurti
Shankar: In order to achieve
31:14.052 --> 31:16.272
what?
Student: Because your
31:16.266 --> 31:19.346
system starts at some amplitude
[inaudible]
31:19.351 --> 31:22.291
Professor Ramamurti
Shankar: Well,
31:22.290 --> 31:24.200
he's raising a point.
31:24.200 --> 31:25.530
Let me repeat his point.
31:25.529 --> 31:28.669
A and B in
general--they are arbitrary.
31:28.670 --> 31:31.000
For this mass and this spring
you can pick any A and
31:31.004 --> 31:31.814
B you like.
31:31.809 --> 31:34.229
But on a given day,
when you pull it by 9
31:34.226 --> 31:37.666
centimeters and release it,
the answer has to be chosen so
31:37.669 --> 31:40.289
that our t = 0,
x becomes 9,
31:40.287 --> 31:42.597
and the velocity,
let's say, was 0 when you
31:42.603 --> 31:45.393
released it,
so when you take the derivative
31:45.393 --> 31:48.753
of this, the velocity should
vanish at t = 0.
31:48.750 --> 31:50.480
Or maybe it will have some
other velocity.
31:50.480 --> 31:53.780
But you can fit initial
coordinate, initial velocity,
31:53.776 --> 31:56.816
the two numbers,
by picking these two numbers.
31:56.820 --> 31:58.030
You understand that.
31:58.030 --> 32:00.650
But that's not enough.
32:00.650 --> 32:02.360
I've got another problem.
32:02.360 --> 32:05.210
Yes?
Student: It has
32:05.211 --> 32:05.941
imaginary numbers in it?
32:05.940 --> 32:06.560
Professor Ramamurti
Shankar: Yes.
32:06.560 --> 32:08.160
That should bother you.
32:08.160 --> 32:13.020
The answer is manifestly not
real, okay, and we know x
32:13.017 --> 32:14.877
is the real function.
32:14.880 --> 32:18.700
That is not a mathematical
requirement of the equation,
32:18.697 --> 32:22.657
but a physical requirement that
when you pull a mass by 9
32:22.655 --> 32:26.465
centimeters and you release it,
it's going to oscillate with
32:26.470 --> 32:28.720
some real x and not a
complex x,
32:28.720 --> 32:29.930
so you've got to fix that.
32:29.930 --> 32:34.750
So, to say that x is
real really means the following.
32:34.750 --> 32:38.010
You remember that in a
complex--world of complex
32:38.013 --> 32:42.323
numbers, a complex number x +
iy has a complex conjugate
32:42.318 --> 32:45.858
x - iy and the property
of real numbers,
32:45.859 --> 32:48.749
is that when you take the
complex conjugate,
32:48.746 --> 32:52.636
nothing happens because
i goes to minus i.
32:52.640 --> 32:56.700
If the number was purely real,
it satisfies the condition
32:56.701 --> 32:58.661
z = z star.
32:58.660 --> 33:03.110
So, real numbers are their own
complex conjugates.
33:03.109 --> 33:05.509
In general, a complex number is
not its own conjugate,
33:05.505 --> 33:07.535
but if you wanted me to draw
you a picture,
33:07.539 --> 33:10.109
that's where the z is,
and that's where the z
33:10.111 --> 33:12.281
star is, but fellows on the
x axis have the
33:12.284 --> 33:14.254
properties,
z is the same as
33:14.245 --> 33:16.755
z star.
z star is a reflection
33:16.763 --> 33:19.483
of the x axis of
z, and therefore,
33:19.475 --> 33:22.355
if the number is real,
it's its own reflection.
33:22.359 --> 33:25.389
So, I'm going to demand that
this solution,
33:25.389 --> 33:28.559
in addition to satisfying the
basic equation,
33:28.563 --> 33:31.563
also is real.
To do that, I'm going to find
33:31.563 --> 33:32.973
the complex conjugate.
33:32.970 --> 33:35.390
It's denoted by the x
star of t.
33:35.390 --> 33:38.550
I want to conjugate everything
in sight.
33:38.549 --> 33:41.719
The complex conjugate of
A I'm going to call
33:41.724 --> 33:45.664
A star.
Complex conjugate of e^(iωt)
33:45.659 --> 33:47.979
is e^(-iω0t).
Why?
33:47.980 --> 33:49.470
Because the i goes to
minus i,
33:49.473 --> 33:51.353
ω_0 and
t are real numbers.
33:51.350 --> 33:52.800
Nothing happens to them.
33:52.799 --> 33:56.429
And B becomes B
star and this becomes e
33:56.425 --> 33:58.755
to the plus
iω_0t.
33:58.759 --> 34:01.569
And I demand that these two
fellows are equal.
34:01.570 --> 34:05.480
34:05.480 --> 34:10.460
If they are both equal,
then I take this exponential
34:10.456 --> 34:13.966
and demand that those be the
same.
34:13.969 --> 34:17.489
That means I demand that
A be the same as B
34:17.488 --> 34:21.068
star, because if A is the
same as B star,
34:21.070 --> 34:24.420
this will go into this,
and then I will demand that
34:24.420 --> 34:27.100
B is the same as A
star.
34:27.099 --> 34:29.649
If these conditions are
satisfied, x star will
34:29.648 --> 34:30.528
become x.
34:30.530 --> 34:35.540
34:35.539 --> 34:38.139
So, the trick is to take the
function, take its complex
34:38.137 --> 34:41.117
conjugate, equate it to itself,
and look at the consequence.
34:41.119 --> 34:43.409
Now, I want to tell you that if
A = B star,
34:43.405 --> 34:45.735
then you don't have to worry
about this as an extra
34:45.737 --> 34:47.707
condition,
because I'm your complex
34:47.710 --> 34:51.010
conjugate, you are my complex
conjugate, but the way it works
34:51.010 --> 34:54.200
is if you took A and
changed every i to minus
34:54.199 --> 34:56.569
i,
right, or if you took B,
34:56.571 --> 34:59.361
whatever its complex form was,
change i to minus
34:59.361 --> 35:00.861
i you get A.
35:00.860 --> 35:03.880
But if you do it one more time,
you come back to where you are.
35:03.880 --> 35:06.020
In other words,
for any complex number,
35:06.023 --> 35:08.513
if you star it and you star it
one more time,
35:08.505 --> 35:10.475
you come back to where you are.
35:10.480 --> 35:13.230
Therefore, if A = B
star, take the complex conjugate
35:13.231 --> 35:15.951
of the left-hand side,
you'll find A star is
35:15.952 --> 35:18.612
equal to B star star
which is B.
35:18.610 --> 35:20.380
So, you don't need these two
conditions.
35:20.380 --> 35:22.430
You do need A = B
star.
35:22.430 --> 35:26.570
35:26.570 --> 35:29.310
Okay, so you have to put that
extra condition now.
35:29.309 --> 35:34.379
So, in the real world,
for people who want a real
35:34.377 --> 35:40.607
answer, you want to write it as
Ae^(iω0t) + A
35:40.605 --> 35:43.345
star e^(-iωt).
35:43.349 --> 35:44.739
In other words,
B is not an independent
35:44.741 --> 35:46.181
number.
B has to be the complex
35:46.180 --> 35:47.070
conjugate of A.
35:47.070 --> 35:50.230
Then, all of you can see at a
glance that this is now real,
35:50.234 --> 35:52.094
because whatever this animal
is,
35:52.090 --> 35:54.650
this is the complex conjugate
of that, which got all the
35:54.652 --> 35:56.472
is turned into minus
i.
35:56.469 --> 35:58.699
When you add them,
all the is will cancel;
35:58.700 --> 35:59.780
answer will be real.
35:59.780 --> 36:03.610
But A is not necessarily
real.
36:03.610 --> 36:08.020
The complex number A,
which is some complex number,
36:08.015 --> 36:11.025
has a length and it can have a
phase,
36:11.030 --> 36:13.520
because every complex number
can be written in this form.
36:13.519 --> 36:19.749
So, if you remember that,
then you find x looks like
36:19.748 --> 36:25.488
absolute value of
Ae^(iω0t) p+ φ
36:25.488 --> 36:32.448
plus absolute value of
Ae^(-iωt) + φ.
36:32.450 --> 36:40.660
36:40.659 --> 36:43.189
Well, maybe that's too fast for
you, so let me repeat what I
36:43.192 --> 36:45.012
did.
In the place of A,
36:45.012 --> 36:47.952
write the absolute value times
e^(iφ).
36:47.949 --> 36:51.219
e^(iφ) will combine it
e^(iωt) and form this
36:51.215 --> 36:53.475
exponential.
Second term will be just a
36:53.480 --> 36:55.200
conjugate of the whole thing.
36:55.200 --> 36:56.570
You don't even have to think.
36:56.570 --> 37:00.720
Now, what is this function I
have?
37:00.720 --> 37:04.200
I want you to think about it.
37:04.199 --> 37:08.869
Do you recognize this creature
here as something familiar?
37:08.870 --> 37:13.140
Yes?
Do you have any idea?
37:13.140 --> 37:17.930
Yeah?
Yep?
37:17.929 --> 37:18.699
Student: [inaudible]
Professor Ramamurti
37:18.696 --> 37:19.256
Shankar: Without the
A.
37:19.260 --> 37:23.390
Student: [inaudible]
Professor Ramamurti
37:23.392 --> 37:27.442
Shankar: You remember this
great identity.
37:27.440 --> 37:29.010
Where is it?
Here?
37:29.010 --> 37:34.400
This one.
e^(iθ) + e^(-iθ) is 2
37:34.400 --> 37:36.970
times cos θ.
37:36.969 --> 37:43.289
So, this becomes 2 times
absolute value of A cos ωt +
37:43.289 --> 37:49.719
φ, and if you want you can
call 2 A as another
37:49.721 --> 37:54.801
number C cos ω_0t
+ φ.
37:54.800 --> 38:02.100
Student: [inaudible]
Professor Ramamurti
38:02.098 --> 38:08.948
Shankar: Oh,
but for any complex number.
38:08.949 --> 38:11.039
You can pull certainly this
number out of both.
38:11.039 --> 38:13.599
You pull it out and you got
e to the i
38:13.595 --> 38:16.145
something plus either the minus
i something--
38:16.151 --> 38:18.611
Student: [inaudible]
Professor Ramamurti
38:18.607 --> 38:21.187
Shankar: So,
this is a long and difficult
38:21.185 --> 38:23.025
way to get back to the old
answer.
38:23.030 --> 38:26.070
You got this by filling it out
and guessing and arguing
38:26.068 --> 38:27.888
physically.
But I want to show you that
38:27.893 --> 38:30.163
with an exponential function,
you would have come to this
38:30.161 --> 38:32.801
answer anyway.
So, your point of view is,
38:32.795 --> 38:35.035
you know, we don't need this.
38:35.039 --> 38:36.839
We've got enough problems in
life;
38:36.840 --> 38:40.170
we're doing well with cosines,
thank you.
38:40.170 --> 38:42.050
Why do you bring this
exponential on us?
38:42.050 --> 38:45.120
Well, now I'm going to give you
a problem where you cannot talk
38:45.119 --> 38:48.189
your way out of this by just
turning it into a word problem.
38:48.190 --> 38:51.140
The word problem I meant,
we asked a word question,
38:51.138 --> 38:54.438
find me a function where the
second derivative looks like
38:54.441 --> 38:56.501
itself,
and you can either start at
38:56.498 --> 38:59.628
exponentials and differentiate
twice, or sines and cosines.
38:59.630 --> 39:00.790
So, you don't need exponentials.
39:00.789 --> 39:02.169
But look at the following
problem.
39:02.170 --> 39:06.620
This is a problem of a mass
m, force constant
39:06.623 --> 39:10.643
k, moving on a surface
with friction.
39:10.640 --> 39:14.130
39:14.130 --> 39:16.730
The minute you got friction,
you have an extra force.
39:16.730 --> 39:19.520
So you find m,
x double dot equal to
39:19.520 --> 39:22.030
minus kx,
which would be all the force
39:22.026 --> 39:23.276
due to the spring.
39:23.280 --> 39:27.480
Friction also exerts a force
which has got some coefficient
39:27.482 --> 39:30.962
b, but it's multiplied by
the velocity.
39:30.960 --> 39:32.830
We know that if you're moving
to the right,
39:32.834 --> 39:35.514
the force of the left you are
moving to the left the force is
39:35.511 --> 39:37.691
to the right,
frictional force is velocity
39:37.691 --> 39:41.221
dependent.
So, the equation you want to
39:41.220 --> 39:47.800
solve, when you've got friction,
is really mx double dot
39:47.797 --> 39:52.887
plus bx dot plus
kx equal to 0.
39:52.889 --> 39:56.249
I'm going to divide everything
by m.
39:56.250 --> 39:59.500
Take the m,
put it here,
39:59.500 --> 40:03.880
and put it here.
Then, I'm going to rewrite for
40:03.880 --> 40:07.770
us the equation we want to solve
with friction.
40:07.769 --> 40:11.619
x double dot,
plus γx dot,
40:11.623 --> 40:16.153
plus ω_0
squared x = 0,
40:16.151 --> 40:20.391
where γ is just b over
x.
40:20.389 --> 40:22.129
So, this is the equation you
want to solve.
40:22.130 --> 40:26.820
40:26.820 --> 40:30.990
Can you solve this by your
usual word problem?
40:30.989 --> 40:33.899
It's going to be difficult,
because you want a function
40:33.897 --> 40:35.887
which, when I take two
derivatives,
40:35.889 --> 40:40.089
it looks like the function plus
some amount of its own
40:40.089 --> 40:43.169
derivative.
If you take cos ωt,
40:43.166 --> 40:45.566
it won't do it.
If you take sin ωt,
40:45.573 --> 40:46.283
it won't do it.
40:46.280 --> 40:50.350
40:50.349 --> 40:55.159
So, we can still solve it
because even this equation,
40:55.162 --> 40:59.052
you solve by the same mindless
approach,
40:59.050 --> 41:01.320
which is to say,
let x look like
41:01.315 --> 41:03.695
A, Ae to the
αt,
41:03.700 --> 41:07.860
and it'll work.
Let's see how it'll work.
41:07.860 --> 41:10.720
It has to work.
You can see why because when I
41:10.719 --> 41:12.919
take two derivatives of
x, I'll get
41:12.918 --> 41:16.258
α^(2).
Let me pull the A out of
41:16.264 --> 41:20.824
everything plus αγ plus
ω_0^(2),
41:20.819 --> 41:23.489
e to the αt = 0.
41:23.490 --> 41:30.470
41:30.469 --> 41:35.879
So, what we learn is yes,
there are solutions of this
41:35.875 --> 41:38.365
form to this equation.
41:38.369 --> 41:41.059
Once again, A can be
whatever number you like,
41:41.055 --> 41:43.685
because if A vanishes
you've killed the whole
41:43.688 --> 41:46.088
solution.
e to the αt is
41:46.089 --> 41:48.709
not going to vanish at every
instant in time,
41:48.714 --> 41:51.164
so the only way is for this to
vanish.
41:51.159 --> 41:54.849
That means α that you
put into this guess is not any
41:54.851 --> 41:58.481
old number, but the solution to
this quadratic equation,
41:58.480 --> 42:01.700
α^(2) + αγ +
ω_0^(2) = 0.
42:01.700 --> 42:05.650
42:05.650 --> 42:09.390
So, we want α^(2) +
αγ not square is
42:09.390 --> 42:12.050
ω_0^(2) = 0.
42:12.050 --> 42:14.920
So, α must be the root
of this equation,
42:14.918 --> 42:18.518
and we all know a quadratic
equation will have two roots,
42:18.519 --> 42:22.329
and we will get two solutions,
and we can add them with any
42:22.326 --> 42:23.766
coefficient we like.
42:23.770 --> 42:25.930
That'll also be a solution.
42:25.930 --> 42:27.520
So, let me write it down.
42:27.519 --> 42:30.589
So, in other words,
I'm going to explicitly solve
42:30.594 --> 42:32.264
this quadratic equation.
42:32.260 --> 42:37.600
So, go back to the old days and
remember that the solution of
42:37.598 --> 42:41.778
that equation will be α
= -γ,
42:41.780 --> 42:45.680
plus or minus square root of
γ^(2) -
42:45.681 --> 42:49.221
4ω_0 square
over 2.
42:49.219 --> 42:54.339
I would like to rewrite this as
minus γ over 2,
42:54.340 --> 42:59.270
plus or minus square root of
γ over 2 square,
42:59.271 --> 43:02.971
minus ω_0
square.
43:02.969 --> 43:07.839
And let's call the two roots,
one with a plus sign and one
43:07.841 --> 43:12.031
with a minus sign,
as α plus and minus.
43:12.030 --> 43:14.450
This is a shorthand for this
whole combination.
43:14.450 --> 43:17.590
43:17.590 --> 43:20.580
You understand that if I give
you the mass and the spring
43:20.576 --> 43:23.506
constant, and I give you the
coefficient of friction,
43:23.510 --> 43:26.060
the number b,
α plus and α
43:26.057 --> 43:28.987
minus are uniquely known,
so you'll get two precise
43:28.988 --> 43:32.638
numbers coming out of this whole
game, and your answer then to
43:32.643 --> 43:36.423
this problem will be x of
t is any number A
43:36.419 --> 43:39.959
times e to the minus
α plus t with any
43:39.955 --> 43:43.425
number B times e
to the minus α minus
43:43.431 --> 43:45.561
t,
where α plus or minus
43:45.564 --> 43:46.404
are these two numbers.
43:46.400 --> 43:51.300
43:51.300 --> 43:55.380
I want you to notice both
numbers are falling
43:55.379 --> 43:57.879
exponentials.
Very important,
43:57.882 --> 44:02.482
because now I'm talking about a
mass where I pull the spring,
44:02.480 --> 44:05.100
there's a lot of friction,
and I let it go,
44:05.103 --> 44:08.043
I don't want an exponentially
growing answer.
44:08.040 --> 44:09.390
That makes no sense.
44:09.389 --> 44:12.689
The x should eventually
vanish and that will,
44:12.686 --> 44:15.336
in fact, happen.
Let's make sure you understand
44:15.341 --> 44:17.121
that.
Look at these two roots.
44:17.119 --> 44:20.569
Minus γ over 2,
plus and minus,
44:20.571 --> 44:25.751
well, that is this solution
minus γ over 2 square,
44:25.747 --> 44:29.287
minus ω_0
square.
44:29.289 --> 44:32.989
I'm going to assume in this
calculation that γ over
44:32.986 --> 44:35.596
2 is bigger than
ω_0.
44:35.600 --> 44:36.430
Let's do that first.
44:36.430 --> 44:39.510
44:39.510 --> 44:41.440
Then, you can see that that's a
negative number,
44:41.444 --> 44:43.754
that's a negative number,
the whole thing is negative,
44:43.750 --> 44:51.370
so α minus is
definitely, let me see how I
44:51.369 --> 44:53.769
wrote it.
Oh, I'm sorry,
44:53.772 --> 44:54.912
I made a mistake here.
44:54.909 --> 44:59.569
I think these are called just
plus and minus α,
44:59.566 --> 45:04.996
α has two roots and I
have to write them as they are.
45:05.000 --> 45:06.510
Okay.
I'm sorry.
45:06.510 --> 45:09.170
So please remove this,
I put unwanted sign there.
45:09.170 --> 45:11.610
Because they are the two
answers for α.
45:11.610 --> 45:13.860
So, I put e to the
α plus t.
45:13.860 --> 45:17.200
So again, I should change my
answer here.
45:17.200 --> 45:18.900
I'm really sorry about this one.
45:18.900 --> 45:22.590
Now, I'm going to write it
below so you have a second
45:22.591 --> 45:24.251
chance.
So, here is what it looks like.
45:24.250 --> 45:28.590
Ae to the minus
γ over 2,
45:28.585 --> 45:33.695
plus square root of γ
over 2 square,
45:33.699 --> 45:37.709
minus ω_0
square t,
45:37.706 --> 45:43.196
plus Be to the minus
γ over 2 minus γ
45:43.203 --> 45:47.583
over 2 square minus
ω_0 square
45:47.582 --> 45:50.582
t.
Student: You forgot the
45:50.577 --> 45:51.267
minus.
Professor Ramamurti
45:51.270 --> 45:51.620
Shankar: Pardon me?
45:51.619 --> 45:54.369
Student:
You forgot the minus in.
45:54.369 --> 45:58.239
Professor Ramamurti
Shankar: I think I did mean
45:58.239 --> 46:00.469
this.
Or did I get it wrong?
46:00.469 --> 46:05.359
These are the two values of
α.
46:05.360 --> 46:08.540
So, they both have the first
number to be minus γ
46:08.540 --> 46:11.830
over 2, and the square root
comes for the plus sign for one
46:11.834 --> 46:14.394
of them and minus sign for the
other one.
46:14.389 --> 46:16.079
I probably have A and
B mixed up.
46:16.080 --> 46:17.200
Let me check that.
46:17.200 --> 46:20.970
No, I think even that's okay.
46:20.970 --> 46:27.080
46:27.080 --> 46:30.150
Yes?
I'm willing to be corrected if
46:30.149 --> 46:31.719
I got A and B
mixed up.
46:31.720 --> 46:32.930
I know these are correct.
46:32.929 --> 46:35.219
I want to make sure this is
what I call B.
46:35.219 --> 46:37.439
B has a coefficient
α minus,
46:37.443 --> 46:38.823
which would be this one.
46:38.820 --> 46:42.000
Student: [inaudible]
Professor Ramamurti
46:41.998 --> 46:43.488
Shankar: Oh,
yes.
46:43.490 --> 46:46.190
Yes.
So, what should I do,
46:46.187 --> 46:47.497
call this A?
46:47.500 --> 46:49.490
Call this B?
46:49.490 --> 46:52.890
How is that now?
Let's see if that works out.
46:52.889 --> 46:56.879
This should be Be to the
α minus t.
46:56.880 --> 47:03.440
B over the minus
γ over 2 minus that.
47:03.440 --> 47:06.660
Yes.
Student: [inaudible]
47:06.659 --> 47:10.609
Professor Ramamurti
Shankar: Pardon me?
47:10.610 --> 47:16.870
Student: [inaudible]
Professor Ramamurti
47:16.866 --> 47:19.416
Shankar: Okay.
47:19.420 --> 47:22.380
The point of writing it this
way is to show you that both
47:22.383 --> 47:24.453
these powers,
this is clearly a positive
47:24.447 --> 47:27.567
number so it's e to the
minus a negative number.
47:27.570 --> 47:30.440
I just want to show you that
this object--that's all I wanted
47:30.442 --> 47:33.642
to emphasize.
Inside the square brackets is
47:33.644 --> 47:36.504
also positive.
Okay, this is γ over 2.
47:36.500 --> 47:39.400
You can see this is a number
smaller than γ over 2
47:39.401 --> 47:41.181
square.
So, if you take the square root
47:41.176 --> 47:43.126
of that number,
it'll be smaller than γ
47:43.131 --> 47:45.341
over 2, so this cannot overturn
the sign of this.
47:45.340 --> 47:47.740
It'll still be a positive over
all signs to be negative;
47:47.739 --> 47:50.739
so, if you draw the picture
here, it's a sum of two
47:50.744 --> 47:53.154
exponentials,
it'll just die down after a
47:53.148 --> 47:58.118
while.
And how do I find A and
47:58.122 --> 48:01.602
B?
To find A and B,
48:01.603 --> 48:05.523
what I have to do is to take
extra data, one of them may be
48:05.516 --> 48:08.886
initial position,
x of 0 is given to me.
48:08.889 --> 48:12.499
If x of 0 is given to me
as some number,
48:12.495 --> 48:16.405
I take the solution,
I put t = 0 and have it
48:16.414 --> 48:19.054
match this.
If I put t = 0,
48:19.048 --> 48:22.888
well all this exponential will
vanish and I just get A +
48:22.892 --> 48:25.612
B, because e to the 0
is 1.
48:25.610 --> 48:28.760
So, I already have one
condition on A and
48:28.762 --> 48:31.052
B.
Their sum must be the initial
48:31.054 --> 48:34.484
position.
How about their initial
48:34.482 --> 48:38.442
velocity?
Go back to this x and
48:38.444 --> 48:40.284
take dx/dt.
48:40.280 --> 48:43.150
What do you get?
dx/dt is α plus
48:43.150 --> 48:46.160
A, e to the
α plus t,
48:46.159 --> 48:49.819
plus α minus B,
e to the α minus
48:49.820 --> 48:53.360
t.
The derivative of my answer is
48:53.363 --> 48:56.763
this.
Evaluate it at t = 0.
48:56.760 --> 49:01.700
At t = 0,
you put t = 0 here,
49:01.701 --> 49:07.581
you find x dot of 0,
equal to α plus
49:07.583 --> 49:12.763
A, plus α minus
B.
49:12.760 --> 49:16.150
So, here are the two equations
you need to find A and
49:16.148 --> 49:18.548
B.
Solve these two simultaneous
49:18.553 --> 49:20.113
equations.
In other words,
49:20.114 --> 49:23.524
I will tell you the initial
position and I will tell you the
49:23.516 --> 49:24.666
initial velocity.
49:24.670 --> 49:30.150
You will take then this number,
known, and that number known.
49:30.150 --> 49:32.350
This is a linear simultaneous
equation for A and
49:32.351 --> 49:34.021
B.
You will fiddle around and
49:34.020 --> 49:34.960
solve for A.
49:34.960 --> 49:38.310
Yes?
Student: Are A
49:38.313 --> 49:39.633
and B still complex
numbers?
49:39.630 --> 49:40.280
Professor Ramamurti
Shankar: Ah.
49:40.280 --> 49:41.810
Now, we have to ask the
following question.
49:41.809 --> 49:45.479
In this problem,
if everything is real--See
49:45.478 --> 49:49.928
previously, what happened is
when I took the complex
49:49.932 --> 49:54.282
conjugate of this function,
the conjugate turned into the
49:54.282 --> 49:56.552
other function,
and the conjugate of that
49:56.547 --> 49:58.697
function turned back into this
one.
49:58.699 --> 50:00.239
That's why A and
B were related,
50:00.242 --> 50:01.102
by complex conjugation.
50:01.099 --> 50:03.799
Here, if I take the complex
conjugate of x,
50:03.802 --> 50:06.732
e to the α plus
t, if it's real,
50:06.726 --> 50:07.936
it remains itself.
50:07.940 --> 50:09.540
So, A just goes into
A star,
50:09.541 --> 50:11.851
B goes into B
star, and you want the answer to
50:11.845 --> 50:13.625
be unchanged.
That requires A equal to
50:13.625 --> 50:15.455
A star and B equal
to B star.
50:15.460 --> 50:18.390
Thanks for pointing out that
here.
50:18.389 --> 50:23.279
We do want A equal to
A star and B equal
50:23.281 --> 50:24.831
to B star.
50:24.829 --> 50:29.259
So, what I'm telling you is go
take the solution,
50:29.260 --> 50:32.860
conjugate it and equate it to
itself,
50:32.860 --> 50:36.000
and remember that now this
exponential remains real to
50:35.996 --> 50:38.596
begin with so it doesn't go into
anything.
50:38.599 --> 50:41.149
It remains itself,
and I compare this exponential
50:41.147 --> 50:44.217
before and after its coefficient
that went from A to
50:44.224 --> 50:46.034
A star,
they have to be equal so
50:46.027 --> 50:48.037
A is A star and
B is B star.
50:48.040 --> 50:50.150
Yes?
Student: [inaudible]
50:50.152 --> 50:52.072
Professor Ramamurti
Shankar: Well,
50:52.067 --> 50:53.357
look at these functions.
50:53.360 --> 50:55.770
There are no complex numbers
here.
50:55.770 --> 50:57.320
That's why everything is real.
50:57.320 --> 51:03.130
51:03.130 --> 51:05.780
So, what problems have I solved
now?
51:05.780 --> 51:08.770
First, I put γ = 0.
51:08.770 --> 51:11.540
No friction.
And I just re-derived the
51:11.539 --> 51:15.789
harmonic oscillator because when
I got the cos (ω_0
51:15.788 --> 51:19.238
t - φ)--Then I took a
problem with friction.
51:19.239 --> 51:22.319
Then, I got a solution that was
just exponentially falling
51:22.316 --> 51:25.066
because even though it looks
like α plus,
51:25.070 --> 51:26.770
that's a negative number,
that's a negative number,
51:26.769 --> 51:27.789
in the end both are falling.
51:27.789 --> 51:32.009
It says you pull the mass and
let it go, it'll relax to its
51:32.012 --> 51:34.272
initial position,
it'll stop.
51:34.269 --> 51:37.999
But all of you must know that
there's got to be something in
51:37.998 --> 51:39.648
between.
In the one case,
51:39.654 --> 51:43.564
I have a mass that oscillates
forever, the other case I have a
51:43.560 --> 51:45.930
mass when I pull it and let it
go,
51:45.929 --> 51:50.879
it goes back to equilibrium and
reaches 0 at infinite time.
51:50.880 --> 51:53.990
But we all know that the real
situation you run into all the
51:53.988 --> 51:56.358
time when you pull something and
let it go;
51:56.360 --> 51:59.480
it vibrates and it vibrates
less and less and less and less
51:59.475 --> 52:01.405
and then eventually comes to
rest.
52:01.410 --> 52:03.010
Where is that solution?
52:03.010 --> 52:06.680
It's got to come out of this
thing.
52:06.680 --> 52:08.760
Yes?
Student: [inaudible]
52:08.756 --> 52:09.806
Professor Ramamurti
Shankar: That's correct.
52:09.809 --> 52:12.049
So, you've got to go back to
the roots I found.
52:12.050 --> 52:15.250
For α,
it's minus γ over 2,
52:15.248 --> 52:19.208
plus or minus square root of
γ over 2 square,
52:19.209 --> 52:22.179
minus ω_0
square.
52:22.179 --> 52:24.089
I've taken γ over 2 to
be bigger than
52:24.088 --> 52:25.128
ω_0.
52:25.130 --> 52:29.070
So, what I did was I turned on
friction but I didn't turn on a
52:29.067 --> 52:30.807
small amount of friction.
52:30.809 --> 52:33.939
I turned on a rather hefty
amount of friction so that the
52:33.944 --> 52:37.474
friction term was bigger than
the ω_0 term.
52:37.469 --> 52:40.239
But if you imagine the other
limit where you have no friction
52:40.244 --> 52:42.284
and you turn on a tiny amount of
friction,
52:42.280 --> 52:44.860
then the tiny γ over 2
will be smaller than this
52:44.862 --> 52:45.992
ω_0.
52:45.989 --> 52:48.339
Of course, then you've got the
square root of a negative
52:48.337 --> 52:52.057
number, right?
So, you should really write it.
52:52.059 --> 52:54.889
So, let's take γ over 2 less
than ω_0,
52:54.890 --> 52:57.570
then you will write this as
minus γ over 2,
52:57.570 --> 53:02.190
plus or minus i times
ω_0 square,
53:02.194 --> 53:04.794
minus γ over 2 square.
53:04.789 --> 53:08.279
In other words,
write this stuff inside as
53:08.281 --> 53:12.711
minus of this number now,
and take the square root of
53:12.709 --> 53:15.859
minus 1 and write it as
i.
53:15.860 --> 53:21.020
53:21.019 --> 53:23.319
So now, what do the solutions
look like?
53:23.320 --> 53:28.810
They look like x equal
to Ae to the minus
53:28.808 --> 53:30.738
γ over 2.
53:30.739 --> 53:35.189
Let me write the exponentials
if you like.
53:35.190 --> 53:38.780
Ae to the minus
γ over 2,
53:38.783 --> 53:44.403
t plus iω prime
t, but I'm going to call
53:44.404 --> 53:48.094
this combination as ω
prime.
53:48.090 --> 53:50.590
That's why we have
ω_0 and
53:50.592 --> 53:54.352
ω prime.
Plus Be to the minus
53:54.350 --> 53:59.370
γ over 2 minus
iω prime t.
53:59.370 --> 54:05.230
54:05.230 --> 54:07.860
Now e to the minus
γ over γt over 2
54:07.855 --> 54:09.325
is common to both the factors.
54:09.330 --> 54:10.540
You can pull it out.
54:10.539 --> 54:14.229
Imagine pulling this factor out
common to the whole expression.
54:14.230 --> 54:17.020
Then, you just got A to
the iω prime,
54:17.016 --> 54:19.526
plus B to the minus
iω prime.
54:19.530 --> 54:20.990
That's a very familiar problem.
54:20.989 --> 54:24.169
If you want that to be real,
you want A to be equal
54:24.172 --> 54:26.352
to the complex conjugate of
B.
54:26.349 --> 54:28.449
Then, you repeat everything I
did before, so I don't want to
54:28.446 --> 54:29.296
do that one more time.
54:29.300 --> 54:32.150
You will get some number C,
times e to the minus
54:32.146 --> 54:36.906
γ over 2t,
times cos ω prime
54:36.905 --> 54:42.105
t, plus,
you can add a φ,
54:42.108 --> 54:45.248
if you like.
There.
54:45.250 --> 54:50.480
54:50.480 --> 54:51.940
So, what does this look like?
54:51.940 --> 54:53.570
What does this graph look like?
54:53.570 --> 54:56.670
It looks like cos ωt
but it's got this thing in the
54:56.672 --> 54:58.602
front.
If γ vanishes,
54:58.600 --> 55:01.330
then forget the exponential
completely.
55:01.330 --> 55:03.470
This is our oscillating mass.
55:03.469 --> 55:07.009
If γ is not 0,
imagine it's one part in
55:07.012 --> 55:11.632
10,000, then for the first one
second this number will hardly
55:11.633 --> 55:14.493
change from 1.
Meanwhile, this would have
55:14.491 --> 55:16.361
oscillated some number of times.
55:16.360 --> 55:19.620
But if you wait long enough
this exponential will start
55:19.615 --> 55:22.745
coming into play and the way to
think about it is,
55:22.750 --> 55:27.450
it's an oscillation whose
amplitude itself falls with
55:27.447 --> 55:29.847
time.
And if you draw a picture of
55:29.847 --> 55:33.567
that, I think you will not be
surprised that if you draw the
55:33.573 --> 55:35.723
picture it will look like this.
55:35.719 --> 55:36.859
It's called a damped
oscillation.
55:36.860 --> 55:40.900
55:40.900 --> 55:46.740
So, that's the method by which
you describe the three cases.
55:46.739 --> 55:51.129
One is no friction and one is
small friction or friction
55:51.128 --> 55:54.958
obeying this condition,
when you have oscillatory
55:54.959 --> 55:58.709
motion whose amplitude is damped
with time.
55:58.710 --> 56:01.690
Then you have the other case
where the gammas has crossed
56:01.686 --> 56:04.396
some threshold when
ω_0 is smaller
56:04.397 --> 56:07.427
than γ over 2,
then you get two falling
56:07.434 --> 56:10.524
exponentials.
So, what do you imagine the
56:10.521 --> 56:12.911
mass is doing?
There's no friction if you pull
56:12.908 --> 56:14.868
it and let it go;
it vibrates forever.
56:14.870 --> 56:16.640
That's not very realistic.
56:16.639 --> 56:19.329
If you turn a tiny amount of
friction, it will do this,
56:19.333 --> 56:20.683
which is very ubiquitous.
56:20.680 --> 56:21.790
We see it all the time.
56:21.789 --> 56:23.759
You pull a spring and you let
it go.
56:23.760 --> 56:26.000
After a while, it comes to rest.
56:26.000 --> 56:29.190
This tells you it never quite
comes to rest because as long as
56:29.188 --> 56:31.538
t is finite,
this number is going to be
56:31.540 --> 56:33.370
non-0, but it'll be very small.
56:33.369 --> 56:35.499
At some point you just cannot
see it.
56:35.500 --> 56:37.550
Third option is overdamped.
56:37.550 --> 56:39.680
Overdamped is when γ is
bigger than
56:39.678 --> 56:42.518
ω_0 over 2,
then the answer has nothing
56:42.516 --> 56:44.876
that oscillates.
Nothing oscillates,
56:44.878 --> 56:47.858
everything falls exponentially
with time.
56:47.860 --> 56:51.140
That is, you should imagine you
pull something,
56:51.139 --> 56:53.919
let it go, it comes slowly and
stops.
56:53.920 --> 56:57.590
When you buy your shocks in
your car, the shock absorbers,
56:57.585 --> 57:00.925
they're supposed to damp the
vibration of the car.
57:00.929 --> 57:05.209
It's got a spring suspending
the tires, but it's immersed in
57:05.205 --> 57:06.795
some viscous medium.
57:06.800 --> 57:08.790
So that is, vibrations are
damped.
57:08.789 --> 57:11.389
So, when you hit a bump,
if your shocks are dead,
57:11.386 --> 57:14.196
you vibrate a lot of times and
slowly settle down.
57:14.199 --> 57:16.509
That's when your shocks are in
this regime.
57:16.510 --> 57:19.850
When you bought them,
they were in this regime.
57:19.849 --> 57:21.549
When you bought them they were
doing this.
57:21.550 --> 57:24.400
Once you hit a bump and you
overshoot from your position,
57:24.396 --> 57:26.426
you come down to 0 and you stay
there.
57:26.429 --> 57:29.509
That's the ideal situation for
damping.
57:29.510 --> 57:36.060
57:36.060 --> 57:40.120
All right.
So, I'm going to the last topic
57:40.124 --> 57:45.364
in oscillation but is there
anything that you need
57:45.362 --> 57:48.002
clarification?
Yes?
57:48.000 --> 57:50.020
Student: [inaudible]
Professor Ramamurti
57:50.021 --> 57:50.971
Shankar: Ah,
yes.
57:50.969 --> 57:55.129
So, when γ over 2
equals ω_0,
57:55.126 --> 57:58.916
this thing vanishes,
and you seem to have only one
57:58.915 --> 58:02.595
solution, e to the minus
γt over 2.
58:02.599 --> 58:05.639
You don't have the plus or
minus, right.
58:05.639 --> 58:07.969
And we sort of know in every
problem there must be two
58:07.972 --> 58:10.392
solutions, because we should be
able to pick the initial
58:10.393 --> 58:11.893
position and velocity at will.
58:11.889 --> 58:15.709
If a solution has only one free
parameter, you cannot pick two
58:15.712 --> 58:18.612
numbers at will.
Now, that's a piece of
58:18.609 --> 58:23.599
mathematics I don't want to do
now, but what you can show is in
58:23.597 --> 58:27.617
that case, the second solution
looks like this.
58:27.619 --> 58:30.619
It's a new function,
not an exponential but t
58:30.617 --> 58:33.377
times an exponential will solve
the equation.
58:33.380 --> 58:36.680
Those who want to see that this
is true can put that into the
58:36.683 --> 58:37.843
equation and check.
58:37.840 --> 58:40.030
In other words,
pick your γ carefully
58:40.034 --> 58:41.784
so that the square root
vanishes.
58:41.780 --> 58:44.560
Only a γ is left,
but that γ is not
58:44.559 --> 58:46.059
independent of ω.
58:46.059 --> 58:49.779
That γ is equal to
2ω_0.
58:49.780 --> 58:53.320
For that problem you can check
that that's a solution,
58:53.321 --> 58:56.201
so is that.
There are nice ways to motivate
58:56.197 --> 58:59.237
that but I think I don't have
time to do that.
58:59.240 --> 59:01.300
Yes?
Student: [inaudible]
59:01.300 --> 59:03.780
Professor Ramamurti
Shankar: Why was A
59:03.775 --> 59:05.275
equal to A star here?
59:05.280 --> 59:08.550
You mean here?
Okay, take this x.
59:08.550 --> 59:12.800
Demand that when you conjugate
it nothing should happen,
59:12.799 --> 59:15.089
okay.
In this particular example,
59:15.091 --> 59:18.531
when α plus an
α minus real numbers,
59:18.530 --> 59:20.870
e to the α plus
t, remains e to
59:20.874 --> 59:23.024
the α plus t when
you take conjugate.
59:23.019 --> 59:25.579
So therefore,
A has to remain A
59:25.584 --> 59:28.964
star, and B must also
remain invariant when you take
59:28.964 --> 59:30.994
the conjugation.
In other words,
59:30.988 --> 59:34.018
if this was an imaginary
exponential like in the other
59:34.020 --> 59:36.070
problem,
when you take the conjugate,
59:36.074 --> 59:39.184
this exponential becomes this
guy and that exponential becomes
59:39.179 --> 59:41.199
that guy.
Then, by matching the solution,
59:41.198 --> 59:43.968
you will be able to show this
must be the conjugate of that;
59:43.969 --> 59:45.949
that must be the conjugate of
this.
59:45.949 --> 59:48.439
But if no imaginary
exponentials appear,
59:48.440 --> 59:50.740
each term must separately be
real.
59:50.739 --> 59:54.409
Now, this is something you will
have to think about.
59:54.409 --> 59:57.039
I'm not saying my repeating it
makes it any clearer,
59:57.037 --> 59:59.507
so what I want you to do is
take the x,
59:59.510 --> 1:00:02.660
take the complex conjugate of
x, equate the two sides.
1:00:02.659 --> 1:00:06.029
There's a rule among functions
that if you got e to the
1:00:06.034 --> 1:00:08.724
plus αt,
its coefficient should match on
1:00:08.724 --> 1:00:11.894
both sides and e to the
minus αt should have
1:00:11.893 --> 1:00:13.263
matching coefficients.
1:00:13.260 --> 1:00:15.380
It's like saying when two
vectors are equal,
1:00:15.383 --> 1:00:18.303
the coefficient of I
should match and the coefficient
1:00:18.297 --> 1:00:19.727
of J should match.
1:00:19.730 --> 1:00:22.130
There's a similar theorem that
if you take a sum of two
1:00:22.125 --> 1:00:24.815
independent functions,
equate it to the sum of [same]
1:00:24.821 --> 1:00:27.921
two independent functions,
the coefficient must separately
1:00:27.920 --> 1:00:30.180
match.
If you impose that you will
1:00:30.184 --> 1:00:31.634
find what I told you.
1:00:31.630 --> 1:00:36.120
Okay, now for the really
interesting problem.
1:00:36.119 --> 1:00:39.049
The really interesting problem
is this.
1:00:39.050 --> 1:00:41.690
I got the mass,
I got the spring,
1:00:41.688 --> 1:00:45.808
I got the friction,
but I'm going to apply an extra
1:00:45.811 --> 1:00:49.111
force, F_0 cos
ωt.
1:00:49.110 --> 1:00:51.640
This is called a "driven
oscillator."
1:00:51.639 --> 1:00:54.549
So far, the oscillator was not
driven.
1:00:54.550 --> 1:00:57.750
In other words,
no one is pushing and pulling
1:00:57.746 --> 1:00:59.446
it.
Of course, you pulled it in the
1:00:59.454 --> 1:01:01.844
beginning and you released it,
but once you released it,
1:01:01.839 --> 1:01:03.399
no one's touching the
oscillator.
1:01:03.400 --> 1:01:06.720
The only forces on it are due
to internal frictional forces or
1:01:06.722 --> 1:01:09.882
spring force.
But now, I want to imagine a
1:01:09.881 --> 1:01:14.731
case where I am actively driving
the oscillator by my hand,
1:01:14.730 --> 1:01:17.490
exerting cos ωt force.
1:01:17.489 --> 1:01:24.489
So ω here is the
frequency of the driving force,
1:01:24.489 --> 1:01:26.749
okay.
That's why there are so many
1:01:26.750 --> 1:01:28.120
ωs in this problem.
1:01:28.119 --> 1:01:31.449
The ω prime that you
saw here is not going to appear
1:01:31.450 --> 1:01:33.840
too many times.
It's a matter of convenience to
1:01:33.842 --> 1:01:35.092
call this ω prime.
1:01:35.090 --> 1:01:37.100
But this ω will appear
all the time.
1:01:37.099 --> 1:01:39.139
When I write an ω with
no subscript,
1:01:39.138 --> 1:01:41.128
it's the frequency of the
driving force,
1:01:41.130 --> 1:01:46.400
and the equation we want to
solve is x double dot,
1:01:46.402 --> 1:01:50.662
plus γ plus dot,
plus ω_0
1:01:50.659 --> 1:01:54.069
square x equal to
F over m,
1:01:54.070 --> 1:01:56.670
because I divided everything by
m, cos ωt.
1:01:56.670 --> 1:02:00.410
1:02:00.409 --> 1:02:01.639
So, we've got to solve this
problem.
1:02:01.640 --> 1:02:07.930
1:02:07.929 --> 1:02:10.979
Now, this is really difficult
because you cannot guess the
1:02:10.984 --> 1:02:12.864
answer to this by a word
problem.
1:02:12.860 --> 1:02:16.160
Now, you can do the following.
1:02:16.159 --> 1:02:19.799
If the right-hand side had been
e to the iωt
1:02:19.801 --> 1:02:23.101
instead of cos ωt,
you would be fine because then
1:02:23.096 --> 1:02:25.906
you can pick an x that
looks like e to the
1:02:25.911 --> 1:02:27.391
iωt,
and when you take two
1:02:27.394 --> 1:02:29.304
derivatives that look like
e to the iωt,
1:02:29.300 --> 1:02:31.430
one derivative would look like
e to the iωt;
1:02:31.429 --> 1:02:33.569
x itself would look like
e to the iωt,
1:02:33.567 --> 1:02:34.797
you can cancel it on both
sides.
1:02:34.800 --> 1:02:37.450
But what you have is cos
ωt.
1:02:37.450 --> 1:02:40.190
So, here is what people do.
1:02:40.190 --> 1:02:41.500
It's a very clever trick.
1:02:41.500 --> 1:02:46.040
People say, let me manufacture
a second problem.
1:02:46.040 --> 1:02:47.630
Nobody gave me this problem.
1:02:47.630 --> 1:02:49.330
Okay, this is a problem you
gave me.
1:02:49.329 --> 1:02:52.299
I make up a new problem,
the answer to which is called
1:02:52.302 --> 1:02:52.922
u.
1:02:52.920 --> 1:02:57.980
1:02:57.980 --> 1:03:00.460
But y is the answer to
the following problem.
1:03:00.460 --> 1:03:08.790
The driving force is sin
ωt, but this is what you
1:03:08.792 --> 1:03:11.622
give me to solve.
1:03:11.620 --> 1:03:15.890
This is my artifact.
1:03:15.889 --> 1:03:18.519
I introduced a new problem you
did not give me,
1:03:18.522 --> 1:03:20.642
but I want to look at this
problem.
1:03:20.640 --> 1:03:24.450
Now, here is a trick.
1:03:24.449 --> 1:03:28.659
You multiply this equation by
any number;
1:03:28.659 --> 1:03:31.199
it's still going to satisfy the
equation.
1:03:31.200 --> 1:03:33.070
Let me multiply by i.
1:03:33.070 --> 1:03:35.190
Put an i here,
put an i here,
1:03:35.194 --> 1:03:37.474
put an i here,
put an i here.
1:03:37.469 --> 1:03:41.229
Multiply both sides by i
and add them.
1:03:41.230 --> 1:03:46.180
1:03:46.179 --> 1:03:49.509
Then I have got x plus
iy, double dot,
1:03:49.509 --> 1:03:51.859
plus γ times x
dot,
1:03:51.860 --> 1:03:53.770
plus iy dot is
ω_0 square
1:03:53.766 --> 1:03:55.046
times x,
plus iy.
1:03:55.050 --> 1:03:58.310
Let's introduce a number
z, which is x plus
1:03:58.311 --> 1:04:00.261
iy.
It varies with time.
1:04:00.260 --> 1:04:03.150
Then this equation,
by adding the two equations,
1:04:03.145 --> 1:04:07.035
look like z double dot,
plus γz dot,
1:04:07.044 --> 1:04:12.654
plus ω_0
square z is F over
1:04:12.654 --> 1:04:15.754
me to the Iωt.
1:04:15.750 --> 1:04:21.390
1:04:21.389 --> 1:04:25.909
So, I have manufactured a
problem in which the thing
1:04:25.907 --> 1:04:29.447
that's vibrating is not a real
number.
1:04:29.449 --> 1:04:33.219
The force driving it is also
not a real number.
1:04:33.219 --> 1:04:36.139
It's a cos ωt plus I
sin ωt.
1:04:36.139 --> 1:04:41.059
But if I can solve this problem
by some trick,
1:04:41.061 --> 1:04:44.781
at the end what do I have to
do?
1:04:44.780 --> 1:04:48.580
I have to take the real part of
the answer, because the answer
1:04:48.576 --> 1:04:51.746
will look like a real part and
an imaginary part;
1:04:51.750 --> 1:04:53.930
I'll have to dump the imaginary
part.
1:04:53.929 --> 1:04:56.579
That'll be the answer to the
question I posed.
1:04:56.579 --> 1:05:00.369
The imaginary part of it will
be the answer to the fictitious
1:05:00.369 --> 1:05:01.569
question I posed.
1:05:01.570 --> 1:05:03.140
Now, why am I doing this?
1:05:03.140 --> 1:05:05.100
The reason is the following.
1:05:05.099 --> 1:05:07.899
If the driving force is
e to the iωt,
1:05:07.895 --> 1:05:09.975
I can make the following
ansatz,
1:05:09.977 --> 1:05:13.237
or a guess.
z is some constant times
1:05:13.238 --> 1:05:15.518
e to the iωt.
1:05:15.519 --> 1:05:19.229
I will show you now solutions
of this form do exist,
1:05:19.233 --> 1:05:23.023
because let me take that
assumed form and put it into
1:05:23.019 --> 1:05:26.219
this equation.
Then, what do I get?
1:05:26.219 --> 1:05:30.679
I get minus ω square,
Z_0e to the
1:05:30.684 --> 1:05:34.844
iωt because two
derivatives of e to the
1:05:34.836 --> 1:05:38.906
iωt give me iω
times iω.
1:05:38.909 --> 1:05:42.859
Then, one derivative gives me
iω times γ,
1:05:42.861 --> 1:05:46.391
times z_0 e to
the iωt,
1:05:46.389 --> 1:05:49.469
and no derivatives,
just leaves it alone,
1:05:49.472 --> 1:05:54.022
e to the iωt
equals F_0 over
1:05:54.018 --> 1:05:56.698
m,
e to the iωt.
1:05:56.700 --> 1:06:00.380
1:06:00.380 --> 1:06:02.440
So, let me rewrite this as
follows.
1:06:02.440 --> 1:06:06.830
Let me rewrite this as minus
ω square,
1:06:06.827 --> 1:06:11.117
plus iωγ,
plus ω_0
1:06:11.117 --> 1:06:14.817
square,
times z_0e to
1:06:14.820 --> 1:06:18.500
the iωt,
is F over m,
1:06:18.502 --> 1:06:21.222
e to the iωt.
1:06:21.220 --> 1:06:23.530
This is what I want to be true.
1:06:23.530 --> 1:06:26.100
Well, e to the
iωt, whatever it is,
1:06:26.097 --> 1:06:28.767
can be canceled on two sides
because it is not 0.
1:06:28.769 --> 1:06:31.319
Anything that's not 0 you can
always cancel,
1:06:31.321 --> 1:06:34.111
and here is the interesting
result you learn.
1:06:34.110 --> 1:06:39.280
For this equation to be valid,
the z_0 that
1:06:39.279 --> 1:06:44.009
you pick here in your guess
satisfies the condition,
1:06:44.010 --> 1:06:48.850
z_0 equals
F over m divided
1:06:48.849 --> 1:06:53.689
by ω_0 square
minus ω square,
1:06:53.690 --> 1:06:55.450
plus iωγ.
1:06:55.450 --> 1:07:07.430
1:07:07.429 --> 1:07:10.429
Now, parts of it may be easy,
parts of it may be difficult.
1:07:10.429 --> 1:07:13.899
The easy part is to take the
guess I made and stick it into
1:07:13.904 --> 1:07:16.544
the equation,
cancel exponentials and get the
1:07:16.540 --> 1:07:19.280
answer.
What you should understand is,
1:07:19.282 --> 1:07:21.852
x was what I was looking
for.
1:07:21.849 --> 1:07:24.039
I brought in a partner
y, and I solved for
1:07:24.043 --> 1:07:26.103
z, which is x plus
iy,
1:07:26.099 --> 1:07:31.069
and z was assumed to
take this form with the
1:07:31.067 --> 1:07:35.337
amplitude, which itself could be
complex,
1:07:35.340 --> 1:07:37.060
times e to the
iωt.
1:07:37.059 --> 1:07:39.609
If z_0 looks
like this, then z,
1:07:39.612 --> 1:07:42.072
which is z_0e
to the iωt,
1:07:42.070 --> 1:07:45.670
looks like F over
m, e to the
1:07:45.672 --> 1:07:49.792
iωt divided by this
number in the denominator,
1:07:49.789 --> 1:07:52.409
I'm going to call I for
impedance.
1:07:52.409 --> 1:07:57.339
I is called impedance
and there's the following
1:07:57.343 --> 1:08:01.443
complex number,
ω_0 square,
1:08:01.438 --> 1:08:05.718
minus ω square,
plus iωγ.
1:08:05.720 --> 1:08:15.540
1:08:15.540 --> 1:08:16.420
Okay, we're almost done now.
1:08:16.420 --> 1:08:21.790
1:08:21.789 --> 1:08:25.979
The z looks like
F_0 over
1:08:25.975 --> 1:08:30.505
m, e to the
Iωt divided by this
1:08:30.508 --> 1:08:32.948
complex number I.
1:08:32.949 --> 1:08:34.769
Imagine this complex number
I.
1:08:34.770 --> 1:08:35.870
What does it look like?
1:08:35.869 --> 1:08:38.899
It's a got a real part and an
imaginary part.
1:08:38.899 --> 1:08:42.059
Imaginary part is iωγ,
real part is
1:08:42.056 --> 1:08:46.016
ω_0 square
minus ω square.
1:08:46.020 --> 1:08:49.450
1:08:49.449 --> 1:08:54.099
This is the complex number
I.
1:08:54.100 --> 1:08:57.790
So, we are told the answer to
our problem is to find this
1:08:57.788 --> 1:09:00.618
number z,
then take the real part.
1:09:00.619 --> 1:09:05.049
Does everybody agree that every
complex number I can be written
1:09:05.054 --> 1:09:09.064
as an absolute value times
e to the iφ.
1:09:09.060 --> 1:09:11.570
φ, which is going to be
here.
1:09:11.569 --> 1:09:16.359
And I'm going to write it that
way because now things become a
1:09:16.358 --> 1:09:19.418
lot simpler if you write it that
way.
1:09:19.420 --> 1:09:23.540
F_0 over
m, divided by absolute
1:09:23.538 --> 1:09:26.968
value of i,
e to the iφ,
1:09:26.969 --> 1:09:29.409
e to the iωt.
1:09:29.409 --> 1:09:33.209
Now, this e to the
iφ, I can delete it here
1:09:33.210 --> 1:09:36.010
and put it upstairs as minus
iφ.
1:09:36.010 --> 1:09:40.110
1:09:40.109 --> 1:09:41.609
So, let me write it in that
form.
1:09:41.610 --> 1:09:42.540
Then it's very easy.
1:09:42.539 --> 1:09:45.929
F_0 over
m, divided by the
1:09:45.926 --> 1:09:50.286
magnitude of this I,
times e to the
1:09:50.289 --> 1:09:57.029
Iωt minus φ,
where φ is this angle.
1:09:57.029 --> 1:10:00.209
Well, now that I got
z, I know how to find
1:10:00.213 --> 1:10:03.373
x.
x is the real part of
1:10:03.369 --> 1:10:05.929
z.
Now, when I look at real part,
1:10:05.931 --> 1:10:09.421
all of these are real numbers
so I will keep them as they are.
1:10:09.420 --> 1:10:12.160
F_0 over
m divided by i,
1:10:12.155 --> 1:10:14.835
and the real part of this
function, I hope you know by
1:10:14.841 --> 1:10:17.021
now, is cos ωt minus
φ.
1:10:17.020 --> 1:10:23.860
1:10:23.860 --> 1:10:27.120
And that's the answer.
1:10:27.119 --> 1:10:30.939
The answer to the problem that
was originally given to us is
1:10:30.935 --> 1:10:33.225
the following.
You know the magnitude of the
1:10:33.231 --> 1:10:35.061
applied force,
the amplitude of the applied
1:10:35.058 --> 1:10:36.658
force.
You know the mass of the
1:10:36.657 --> 1:10:38.737
particle.
You need to find the absolute
1:10:38.740 --> 1:10:40.540
value of I and φ.
1:10:40.539 --> 1:10:44.269
For that, you construct this
complex number whose real part
1:10:44.271 --> 1:10:47.361
is this, whose imaginary part is
iωγ.
1:10:47.359 --> 1:10:51.459
Then, the absolute value of
I is just by Pythagoras'
1:10:51.456 --> 1:10:55.616
theorem, ω_0
square minus ω square
1:10:55.623 --> 1:10:59.583
square,
plus ω square γ
1:10:59.579 --> 1:11:03.039
square.
And the phase φ obeys
1:11:03.039 --> 1:11:07.129
the condition tan
φ is equal to
1:11:07.134 --> 1:11:12.144
ωγ divided by
ω_0 square,
1:11:12.140 --> 1:11:18.230
minus ω square square
is ω square γ
1:11:18.234 --> 1:11:20.054
square.
I'm sorry.
1:11:20.050 --> 1:11:22.190
It's just the imaginary part
over the real part.
1:11:22.190 --> 1:11:28.340
1:11:28.340 --> 1:11:33.230
So, this is the answer to the
problem that was given to us.
1:11:33.229 --> 1:11:37.149
But there's one subtle point
you should notice,
1:11:37.149 --> 1:11:39.279
which is the following.
1:11:39.279 --> 1:11:41.849
Where are the three parameters
in this problem?
1:11:41.850 --> 1:11:46.660
Everything is determined in
this problem.
1:11:46.659 --> 1:11:48.729
φ, absolute I,
all these are known,
1:11:48.734 --> 1:11:51.194
but we know every equation
should have two free numbers.
1:11:51.190 --> 1:11:52.530
You know what they are?
1:11:52.530 --> 1:11:59.140
Anybody know?
Okay, let me tell you what the
1:11:59.140 --> 1:12:01.530
answer to that question is.
1:12:01.529 --> 1:12:07.499
I'm saying, I can add to this
the solution I got earlier on
1:12:07.498 --> 1:12:12.398
without the driving force,
because in this problem,
1:12:12.402 --> 1:12:16.702
when I had a right-hand side
with a driving force,
1:12:16.700 --> 1:12:18.770
let me add a 0 to that.
1:12:18.770 --> 1:12:22.020
That's harmless,
and by superposition principle,
1:12:22.019 --> 1:12:25.059
if that force will produce that
displacement,
1:12:25.060 --> 1:12:26.720
0 will produce what?
1:12:26.720 --> 1:12:30.010
Well, we have seen all morning
that even when the right-hand
1:12:30.005 --> 1:12:32.785
side is 0, they are the
solutions I got for you.
1:12:32.789 --> 1:12:36.109
e to the γ over
2 cos ω prime t
1:12:36.114 --> 1:12:38.224
etc., so you can always add to
this,
1:12:38.220 --> 1:12:41.480
what you call a complementary
function, which is the solution
1:12:41.484 --> 1:12:43.664
of the equation with no driving
force,
1:12:43.659 --> 1:12:46.599
which is what we were studying
earlier in the class.
1:12:46.600 --> 1:12:49.370
But usually,
people don't bother with this
1:12:49.372 --> 1:12:53.502
because they all have in them
e to the minus γt
1:12:53.496 --> 1:12:56.256
over 2.
So, if you wait long enough,
1:12:56.260 --> 1:12:59.800
this will die out,
and this is the only thing that
1:12:59.797 --> 1:13:02.587
will remain.
So at earlier times,
1:13:02.586 --> 1:13:05.056
this is not the full answer.
1:13:05.060 --> 1:13:08.950
You should add to this the
answer when there is no driving
1:13:08.945 --> 1:13:12.145
force and together they form the
full answer,
1:13:12.149 --> 1:13:15.339
and the numbers A and
B that you have there
1:13:15.340 --> 1:13:18.190
will be chosen to match some
initial conditions,
1:13:18.189 --> 1:13:22.929
like initial position,
initial velocity.
1:13:22.930 --> 1:13:24.910
Okay now, time is up.
1:13:24.909 --> 1:13:28.069
I'm going to stop,
so I have not been able to
1:13:28.067 --> 1:13:32.157
finish some parts of this,
so I'm trying to see what I can
1:13:32.157 --> 1:13:34.707
do.
I had asked for you to give
1:13:34.714 --> 1:13:38.294
this problem set to me on--what
is this today?
1:13:38.290 --> 1:13:39.480
On Wednesday, right.
1:13:39.480 --> 1:13:41.210
So, you don't have enough time.
1:13:41.210 --> 1:13:43.740
I mean, I haven't taught you a
few other things you need about
1:13:43.739 --> 1:13:44.899
five minutes more of work.
1:13:44.900 --> 1:13:46.640
I don't want to keep you back.
1:13:46.640 --> 1:13:47.910
So, here is what I will do.
1:13:47.909 --> 1:13:51.459
I will post on the website some
notes on just the missing part
1:13:51.464 --> 1:13:55.084
here, so you can read it and you
can do one or two problems you
1:13:55.077 --> 1:13:57.347
may not be able to do without
that.
1:13:57.350 --> 1:14:00.450
Then, I will come on Wednesday,
I will teach that to you again,
1:14:00.449 --> 1:14:02.999
but you will be able to hand in
your problem set.