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PROFESSOR: Welcome
back, and today I'm
00:00:23.590 --> 00:00:28.110
going to do for you three
problems all related
00:00:28.110 --> 00:00:31.310
to driven harmonic oscillators.
00:00:31.310 --> 00:00:32.930
Now one of the
things you'll notice
00:00:32.930 --> 00:00:36.080
as I'm doing these
problems, that it seems
00:00:36.080 --> 00:00:39.010
I'm doing the same thing
over and over again.
00:00:39.010 --> 00:00:41.850
You may have also noticed
that in the other problems
00:00:41.850 --> 00:00:45.280
I did earlier for you
in harmonic oscillators,
00:00:45.280 --> 00:00:47.440
et cetera.
00:00:47.440 --> 00:00:50.350
So at this stage, I have to
remind you about something.
00:00:50.350 --> 00:00:54.090
This is no accident
that it is like that.
00:00:54.090 --> 00:00:56.510
It's to do with the
scientific method,
00:00:56.510 --> 00:00:59.690
and I would want to
briefly remind you
00:00:59.690 --> 00:01:03.590
about how we are
solving problems.
00:01:03.590 --> 00:01:07.270
Here is a diagram of
a generic problem.
00:01:07.270 --> 00:01:08.440
What does it consists of?
00:01:08.440 --> 00:01:14.940
A typical problem is we are
told of a physical situation
00:01:14.940 --> 00:01:17.710
in words, ordinary language.
00:01:17.710 --> 00:01:23.090
And in ordinary language, we
are asked questions about given
00:01:23.090 --> 00:01:25.330
that physical
situation, what will
00:01:25.330 --> 00:01:27.470
happen as a function of time?
00:01:27.470 --> 00:01:33.870
What kind of emotion it
has, questions of that kind.
00:01:33.870 --> 00:01:38.680
What problem solving is,
doing the scientific method,
00:01:38.680 --> 00:01:39.970
is the following.
00:01:39.970 --> 00:01:50.300
The first thing we have to do
is, by studying this situation
00:01:50.300 --> 00:01:57.270
and using the laws
of nature, describing
00:01:57.270 --> 00:02:01.421
the physical situation in terms
of mathematical equations.
00:02:01.421 --> 00:02:02.420
This is always possible.
00:02:05.610 --> 00:02:09.820
So starting with as I
say this description,
00:02:09.820 --> 00:02:13.150
using laws of nature, we
end up with a description
00:02:13.150 --> 00:02:16.260
in terms of
mathematical equations.
00:02:16.260 --> 00:02:19.310
The next thing we
do is, we now forget
00:02:19.310 --> 00:02:21.650
that this has anything
to do with physics.
00:02:21.650 --> 00:02:24.260
We have a set of
mathematical equations,
00:02:24.260 --> 00:02:25.750
which we have to solve.
00:02:25.750 --> 00:02:27.180
This is mathematics.
00:02:27.180 --> 00:02:29.560
We use mathematics to
solve this problem.
00:02:29.560 --> 00:02:30.700
All right?
00:02:30.700 --> 00:02:34.810
Once we have found the solution
to this mathematical equation,
00:02:34.810 --> 00:02:41.380
there's only one thing
left, we identify the answer
00:02:41.380 --> 00:02:46.360
with the questions
posed in the problem.
00:02:46.360 --> 00:02:50.930
So understanding
of physics comes
00:02:50.930 --> 00:02:57.720
in this step, which I call
step 1, and in the last one.
00:02:57.720 --> 00:03:01.580
The middle one, step
2, is pure mathematics.
00:03:01.580 --> 00:03:14.050
Now, it so happens that if
you take all driven harmonic
00:03:14.050 --> 00:03:18.800
oscillators, the mathematics
that you end up with
00:03:18.800 --> 00:03:19.710
is identical.
00:03:19.710 --> 00:03:24.820
In other words, if I look at
the description of the problem
00:03:24.820 --> 00:03:30.820
in mathematical terms, I cannot
tell you which problem that
00:03:30.820 --> 00:03:32.230
description belongs.
00:03:32.230 --> 00:03:38.330
That's a generic description
of many, many, many situations.
00:03:38.330 --> 00:03:42.570
Today I'll take three of them,
but all three situations,
00:03:42.570 --> 00:03:45.130
all three problems--
although to you, they
00:03:45.130 --> 00:03:49.600
may look completely different--
after I've translated them
00:03:49.600 --> 00:03:53.020
in terms of mathematical
equations, are identical.
00:03:53.020 --> 00:03:56.460
And so it's not surprising
that from here on when we then
00:03:56.460 --> 00:03:59.580
try to solve them, et cetera,
we are doing the same problem,
00:03:59.580 --> 00:04:03.450
or at least it appears so.
00:04:03.450 --> 00:04:05.600
Having said that,
let's immediately
00:04:05.600 --> 00:04:10.090
dive in and take
one of the problems.
00:04:10.090 --> 00:04:12.810
So the first problem
I'm going to consider
00:04:12.810 --> 00:04:16.089
is an idealized-driven
RLC circuit.
00:04:18.750 --> 00:04:21.000
Why do I see idealized?
00:04:21.000 --> 00:04:22.690
Because as I've
mentioned before,
00:04:22.690 --> 00:04:27.390
in real life situations,
things extremely complicated,
00:04:27.390 --> 00:04:29.570
and if you take
everything into account,
00:04:29.570 --> 00:04:32.770
it will take you
essentially forever
00:04:32.770 --> 00:04:35.010
to get an exact solution.
00:04:35.010 --> 00:04:37.680
But we're often not most
of the time not interested
00:04:37.680 --> 00:04:39.290
in the very exact one.
00:04:39.290 --> 00:04:42.740
So we simplify a
physical situation
00:04:42.740 --> 00:04:48.480
to the bear essential parts
and solve that problem.
00:04:48.480 --> 00:04:54.690
So this is a problem where
you have a circuit consisting
00:04:54.690 --> 00:04:58.050
all a capacitance,
inductance, and a resistance,
00:04:58.050 --> 00:05:03.360
in series, which is driven by
an alternating voltage source V0
00:05:03.360 --> 00:05:04.670
cosine omega t.
00:05:07.330 --> 00:05:12.220
Before time t equals 0, I'm
assuming the circuit is open.
00:05:12.220 --> 00:05:15.330
In other words, there
is no charges anywhere,
00:05:15.330 --> 00:05:16.750
no current flowing.
00:05:16.750 --> 00:05:20.700
At t equals 0, we
close the circuit
00:05:20.700 --> 00:05:25.690
so the charge on the
capacitor t equals 0, is 0.
00:05:25.690 --> 00:05:27.790
The current flowing
in the circuit is 0.
00:05:27.790 --> 00:05:28.750
All right?
00:05:28.750 --> 00:05:31.750
And the question
we want to answer
00:05:31.750 --> 00:05:36.640
is, what will the current be
in the circuit at some time t
00:05:36.640 --> 00:05:42.570
a specific one, in 17 seconds
or whatever, some specific time.
00:05:42.570 --> 00:05:45.550
Since this is
completely specified,
00:05:45.550 --> 00:05:48.480
I can predict what will happen.
00:05:48.480 --> 00:05:51.730
That's what it means in this
case solving this problem.
00:05:51.730 --> 00:05:52.540
So let's do it.
00:06:01.270 --> 00:06:05.670
So we are now entering step 1.
00:06:05.670 --> 00:06:09.330
We're going to take
the physical situation
00:06:09.330 --> 00:06:16.050
and convert it to a description
in terms of mathematics.
00:06:16.050 --> 00:06:22.980
This is the conceptually
hardest step of any problem.
00:06:22.980 --> 00:06:28.240
It is the one where you need
to understand the physics.
00:06:28.240 --> 00:06:31.660
It may not necessarily be
the hardest or the longest.
00:06:31.660 --> 00:06:35.870
Often the solution
of the equations
00:06:35.870 --> 00:06:40.900
are incredibly tedious and
long and take all your time.
00:06:40.900 --> 00:06:43.860
But they are
routine mathematics.
00:06:43.860 --> 00:06:49.370
Here you've got to
conceptualize what's going on
00:06:49.370 --> 00:06:54.430
and translate it as I say to
the language of mathematics.
00:06:54.430 --> 00:07:00.230
So in the case of I've
had have to define
00:07:00.230 --> 00:07:02.240
coordinates for this problem.
00:07:02.240 --> 00:07:08.530
So what I have is, I'll assume
that at some instant of time
00:07:08.530 --> 00:07:10.590
that in this
circuit, it's closed
00:07:10.590 --> 00:07:12.770
because it's past
equal to 0, there
00:07:12.770 --> 00:07:18.230
is a current I flowing that
on this capacitor on the left
00:07:18.230 --> 00:07:20.830
there is plus charge
Q and minus charge
00:07:20.830 --> 00:07:23.240
Q on the right side
that is this inductance.
00:07:26.010 --> 00:07:31.300
And over here, we have an
alternating potential applied.
00:07:31.300 --> 00:07:34.000
Now I have to be very specific
at any instant of time which
00:07:34.000 --> 00:07:35.540
side is positive and negative.
00:07:35.540 --> 00:07:37.700
I'll assume that t equals 0.
00:07:37.700 --> 00:07:40.710
It's 0 over here,
and at this point
00:07:40.710 --> 00:07:45.340
here, it's plus
V0 cosine omega t.
00:07:45.340 --> 00:07:50.930
So these are the
coordinates of this problem.
00:07:50.930 --> 00:07:54.350
And now what drives the current?
00:07:54.350 --> 00:07:58.320
What decides what happens here?
00:07:58.320 --> 00:08:00.560
What are the law of
natures applicable?
00:08:00.560 --> 00:08:04.750
In this case, are the Maxwell's
equations or to be specific,
00:08:04.750 --> 00:08:06.810
the Faraday equation.
00:08:06.810 --> 00:08:08.340
All right?
00:08:08.340 --> 00:08:15.490
One more thing I should add
that I is a current here flowing
00:08:15.490 --> 00:08:18.430
charge by conservation
of charge.
00:08:18.430 --> 00:08:20.630
It's law of physics.
00:08:20.630 --> 00:08:23.930
The current at the
instant of time,
00:08:23.930 --> 00:08:28.700
will be equal to the rate
of change of the charge.
00:08:28.700 --> 00:08:32.450
So at the instant time,
I is, in my language,
00:08:32.450 --> 00:08:35.240
Q with a dot on top of it.
00:08:35.240 --> 00:08:38.559
This is plus because
of the way in which I
00:08:38.559 --> 00:08:40.270
define my coordinates.
00:08:40.270 --> 00:08:43.640
Beware if you by
mistake say suppose
00:08:43.640 --> 00:08:47.300
I define my coordinates this
being minus and this plus,
00:08:47.300 --> 00:08:49.890
then this equation
would be a minus here,
00:08:49.890 --> 00:08:51.920
conservation of charge.
00:08:51.920 --> 00:08:54.350
But with the way I've
defined these coordinates,
00:08:54.350 --> 00:08:56.720
this is plus.
00:08:56.720 --> 00:08:57.850
So that's one.
00:08:57.850 --> 00:09:00.670
Next I said what
the law of nature
00:09:00.670 --> 00:09:03.120
which governs the
motion of the charges
00:09:03.120 --> 00:09:07.390
is the Faraday's
law, which states
00:09:07.390 --> 00:09:12.350
that if I take the line
integral of the electric field
00:09:12.350 --> 00:09:17.320
around a closed
loop, that's this,
00:09:17.320 --> 00:09:23.480
it equals to minus
the rate of change
00:09:23.480 --> 00:09:27.160
of total magnetic flux
linking this circuit,
00:09:27.160 --> 00:09:29.980
going through this circuit.
00:09:29.980 --> 00:09:34.580
Now, you know that the
magnetic flux through a circuit
00:09:34.580 --> 00:09:40.520
is equal to the current
times the self-inductance
00:09:40.520 --> 00:09:45.130
of that circuit or the total
inductance in that circuit.
00:09:45.130 --> 00:09:49.770
So taking this law of
nature, this law of nature,
00:09:49.770 --> 00:09:53.240
I can now translate
this into mathematics,
00:09:53.240 --> 00:09:54.970
and that's what I'm going to do.
00:09:54.970 --> 00:09:58.450
So first I'm going to take
the line integral of E dl
00:09:58.450 --> 00:10:01.700
around this circuit.
00:10:01.700 --> 00:10:06.490
If I go around
from here to here,
00:10:06.490 --> 00:10:10.300
this is positive
relative to this point,
00:10:10.300 --> 00:10:13.105
so the electric field
from here to here
00:10:13.105 --> 00:10:19.300
will be in this direction, and
so the line integral of E dl
00:10:19.300 --> 00:10:24.250
along here will be minus, and
by definition of a voltage,
00:10:24.250 --> 00:10:27.100
it will be just minus
the voltage drop
00:10:27.100 --> 00:10:30.360
across here, which
is V0 cosine omega t.
00:10:30.360 --> 00:10:33.725
So this is minus
V0 cosine omega t.
00:10:33.725 --> 00:10:35.910
I continue going around.
00:10:35.910 --> 00:10:39.530
From here to here, the electric
field is in this direction.
00:10:39.530 --> 00:10:42.117
And so that will be
the integral V dl
00:10:42.117 --> 00:10:46.210
is Q over C, that's from the
definition of a capacitance,
00:10:46.210 --> 00:10:47.840
of course.
00:10:47.840 --> 00:10:55.730
Then as we go around in
these idealized diagrams,
00:10:55.730 --> 00:11:00.250
all wires have 0 resistance and
therefore there cannot be field
00:11:00.250 --> 00:11:00.750
in them.
00:11:00.750 --> 00:11:03.400
That's why there's no
contribution to this
00:11:03.400 --> 00:11:06.060
from the field inside the wires.
00:11:06.060 --> 00:11:07.800
I continue.
00:11:07.800 --> 00:11:11.370
This inductance is
just coiled wire,
00:11:11.370 --> 00:11:15.320
so there's still no contribution
to this in the L. I continue.
00:11:15.320 --> 00:11:19.690
Here I know there is a current
flowing like this means
00:11:19.690 --> 00:11:22.590
the reason the electric
field in this direction
00:11:22.590 --> 00:11:27.170
and once again knowing the
definition of our Ohms law,
00:11:27.170 --> 00:11:30.400
I get that the integral of
the electric fields from here
00:11:30.400 --> 00:11:33.300
to here is I times R.
00:11:33.300 --> 00:11:37.160
So this is this
quantity, the integral
00:11:37.160 --> 00:11:39.800
of the electric field
around this closed loop.
00:11:39.800 --> 00:11:45.930
That must equal by Faraday's
law to minus the rate of change
00:11:45.930 --> 00:11:49.260
of magnetic flux,
which is minus L dI dt.
00:11:53.080 --> 00:11:55.170
We're almost finished.
00:11:55.170 --> 00:11:56.690
Now we do algebra.
00:11:56.690 --> 00:11:57.660
We're playing around.
00:11:57.660 --> 00:12:02.270
I can rewrite this like this
knowing that the current is
00:12:02.270 --> 00:12:07.490
the rate of change of Q, and
the rate of change of current
00:12:07.490 --> 00:12:09.200
is the second derivative.
00:12:09.200 --> 00:12:12.100
So this is just
straightforward algebra.
00:12:12.100 --> 00:12:15.510
Now, I'm going to rewrite
just moving terms around
00:12:15.510 --> 00:12:18.830
to make it look in
the way I prefer.
00:12:18.830 --> 00:12:24.190
And so I've rewritten this
equation now in this form.
00:12:24.190 --> 00:12:29.980
And now, I will just
redefine some constants
00:12:29.980 --> 00:12:32.720
so that I recognize
the equation better.
00:12:32.720 --> 00:12:38.350
So I will define R over
L by a constant gamma.
00:12:38.350 --> 00:12:43.430
And I will define 1 over LC by
the constant omega 0 squared.
00:12:43.430 --> 00:12:47.390
And I define V0 L
by the constant f.
00:12:47.390 --> 00:12:53.270
And so this equation
rewritten looks like that.
00:12:53.270 --> 00:12:54.740
Eureka.
00:12:54.740 --> 00:12:58.340
This is an equation I've
seen millions of times,
00:12:58.340 --> 00:12:59.840
and I know how to solve it.
00:13:03.950 --> 00:13:08.540
So what we've done is, we've
taken this physical situation,
00:13:08.540 --> 00:13:13.440
translated it into a
mathematical equation.
00:13:13.440 --> 00:13:16.530
This is the equation
of motion, this just
00:13:16.530 --> 00:13:20.490
defines the constants, and this
is the boundary conditions.
00:13:20.490 --> 00:13:23.870
It tells us that in
our particular problem,
00:13:23.870 --> 00:13:27.630
we switch that
circuit at t equals 0.
00:13:27.630 --> 00:13:29.680
At that time, what
was the condition
00:13:29.680 --> 00:13:31.570
of the current and charge?
00:13:31.570 --> 00:13:34.650
So Q at 0 we said was 0.
00:13:34.650 --> 00:13:37.090
Q dot at 0 was 0.
00:13:37.090 --> 00:13:41.530
This completely, mathematically
defines that problem.
00:13:41.530 --> 00:13:48.270
This is the mathematical
description of this problem.
00:13:48.270 --> 00:13:51.740
From now on, I don't
have to even remember
00:13:51.740 --> 00:13:55.380
that this has anything
to do with physics.
00:13:55.380 --> 00:13:58.790
This is now a problem
in mathematics.
00:13:58.790 --> 00:14:02.610
What is the solution
of this equation
00:14:02.610 --> 00:14:05.250
satisfying these
boundary conditions?
00:14:05.250 --> 00:14:07.380
We'll do that later.
00:14:07.380 --> 00:14:11.150
I'm now going to immediately
go to the next problem.
00:14:11.150 --> 00:14:12.990
See in each case,
I've done the physics.
00:14:12.990 --> 00:14:15.780
This is the end of the
physics in essence.
00:14:15.780 --> 00:14:18.220
From now on, I just
have to do mathematics.
00:14:18.220 --> 00:14:22.010
So let me take the next
problem, a different one.
00:14:22.010 --> 00:14:23.450
It seems completely different.
00:14:23.450 --> 00:14:27.240
Here we were dealing with
charges, currents, circuits, et
00:14:27.240 --> 00:14:31.290
cetera, and now I'm going
to look at the physical one.
00:14:31.290 --> 00:14:34.610
I could even bring you
a model of that problem.
00:14:34.610 --> 00:14:39.670
What I'm going to consider
here is an ideal pendulum,
00:14:39.670 --> 00:14:44.560
simple pendulum, ideal, which
is a heavy mass, considered
00:14:44.560 --> 00:14:49.400
to be a point mass, a
string, which is considered
00:14:49.400 --> 00:14:54.860
to be always taught
but yet massless.
00:14:54.860 --> 00:14:57.970
So it's an ideal pendulum.
00:14:57.970 --> 00:15:01.360
What I have here is an
approximation to an ideal one,
00:15:01.360 --> 00:15:05.970
and if I understand
what the ideal one does,
00:15:05.970 --> 00:15:11.070
this in reality, will you do
something pretty close to it.
00:15:11.070 --> 00:15:14.010
And so what I'm
going to consider,
00:15:14.010 --> 00:15:19.290
I am going to be oscillating
my hand backwards and forwards
00:15:19.290 --> 00:15:24.790
sinusoidally starting this
at some definite time,
00:15:24.790 --> 00:15:28.480
and I'll try to predict
the motion of this.
00:15:28.480 --> 00:15:31.770
And notice I can make all
sorts of motion out of this.
00:15:31.770 --> 00:15:35.580
If I do it slowly, look,
my hand and the ball
00:15:35.580 --> 00:15:37.820
are going in the same direction.
00:15:37.820 --> 00:15:42.720
If I go fast, they're going
in opposite direction.
00:15:42.720 --> 00:15:48.470
And the way this is
oscillating is different,
00:15:48.470 --> 00:15:52.870
depends on how I am moving
my hand backward and forward.
00:15:52.870 --> 00:15:56.450
So let me tell you
very precisely what is
00:15:56.450 --> 00:16:02.270
this idealized situation which
I am trying to solve and predict
00:16:02.270 --> 00:16:03.310
what will happen.
00:16:03.310 --> 00:16:07.120
So I have this ideal pendulum.
00:16:07.120 --> 00:16:11.270
It has a length, L, mass, m.
00:16:11.270 --> 00:16:12.590
It's in the vertical plane.
00:16:12.590 --> 00:16:15.500
Gravity is down there.
00:16:15.500 --> 00:16:21.750
At one end of the string,
I will be moving my hand
00:16:21.750 --> 00:16:25.740
in perfect sinusoidal fashion.
00:16:25.740 --> 00:16:30.340
It will have an amplitude of
x0 and angular frequency omega,
00:16:30.340 --> 00:16:32.896
so it's x0 sine omega t.
00:16:32.896 --> 00:16:36.160
I am making the
following assumptions,
00:16:36.160 --> 00:16:37.610
that's why it's idealized.
00:16:37.610 --> 00:16:40.730
I'm assuming it's
a massless string.
00:16:40.730 --> 00:16:42.960
I'm assuming it's always taught.
00:16:42.960 --> 00:16:48.420
I'm going to assume that it's
all making small oscillations
00:16:48.420 --> 00:16:53.400
so that if in any
calculation, I can take a sine
00:16:53.400 --> 00:16:55.360
to be an angle to be
equal to the angle.
00:16:59.940 --> 00:17:04.210
Furthermore I'm going to assume
that there is a bit of drag,
00:17:04.210 --> 00:17:09.010
there is friction,
and I will assume
00:17:09.010 --> 00:17:14.210
that the frictional
force on this that mass
00:17:14.210 --> 00:17:18.160
is proportional to the
velocity, and the constant
00:17:18.160 --> 00:17:24.349
of that velocity is B. Now
you ask me, why choose that?
00:17:24.349 --> 00:17:28.300
Well, the simple
answer is, I want
00:17:28.300 --> 00:17:31.800
to find this situation
when I know I can solve it.
00:17:31.800 --> 00:17:34.710
And I know how to
solve this problem
00:17:34.710 --> 00:17:37.160
if it's proportional
to velocity.
00:17:37.160 --> 00:17:40.090
If it's something
more complicated,
00:17:40.090 --> 00:17:43.160
then the equation may
be correct at the end,
00:17:43.160 --> 00:17:47.300
but I will not be able to
just like that on the board,
00:17:47.300 --> 00:17:48.290
solve it for you.
00:17:48.290 --> 00:17:52.260
I could do it numerically,
use a computer, et cetera,
00:17:52.260 --> 00:17:56.620
but at the same time, I know
that if I have something which
00:17:56.620 --> 00:18:00.680
is very close to being what
the actual situation is,
00:18:00.680 --> 00:18:05.790
the prediction of the real world
compared to this idealized one
00:18:05.790 --> 00:18:07.490
will not be very different.
00:18:07.490 --> 00:18:10.410
Qualitatively, I will
get a good understanding
00:18:10.410 --> 00:18:11.220
of what's going on.
00:18:15.680 --> 00:18:19.870
So this is the assumption,
and what is the question?
00:18:19.870 --> 00:18:21.690
What do I want to predict?
00:18:21.690 --> 00:18:24.590
I want to now take a
very specific situation.
00:18:24.590 --> 00:18:28.980
I want to take a situation
that at t equals 0 initially,
00:18:28.980 --> 00:18:36.750
the mass is stationary, and
it's exactly below my hand,
00:18:36.750 --> 00:18:39.880
but my hand is not
stationary, it's moving.
00:18:39.880 --> 00:18:45.790
So I am moving this backwards
and forwards sinusoidally,
00:18:45.790 --> 00:18:46.790
all right?
00:18:46.790 --> 00:18:52.450
And in the instant, when
my hand is over the ball,
00:18:52.450 --> 00:18:55.980
I let go of the ball and
then I continue moving it.
00:18:55.980 --> 00:19:03.010
So I have specified this exactly
what this ball is doing at t
00:19:03.010 --> 00:19:05.760
equals 0, what my
hand is doing, and I
00:19:05.760 --> 00:19:12.480
want to predict what will happen
at some arbitrary later time.
00:19:12.480 --> 00:19:15.820
So my question is
what will happen,
00:19:15.820 --> 00:19:18.860
and this time to make it a
little more interesting also,
00:19:18.860 --> 00:19:26.870
I want to know in qualitative
terms what kind of motion
00:19:26.870 --> 00:19:31.450
will it have if I move
this slowly and quickly
00:19:31.450 --> 00:19:34.850
and check against
reality at the end.
00:19:34.850 --> 00:19:36.760
And [? originally ?]
I could tell you
00:19:36.760 --> 00:19:39.640
that things are different
when I'm doing it slowly.
00:19:39.640 --> 00:19:41.620
As I told you before,
if I do this slowly
00:19:41.620 --> 00:19:45.560
like this, my hand and the ball
going in the same direction.
00:19:45.560 --> 00:19:49.310
If I go fast, I can make them
go in opposite directions.
00:19:49.310 --> 00:19:53.830
All that should
come out of this.
00:19:53.830 --> 00:19:57.670
So the part of the question
is to qualitatively understand
00:19:57.670 --> 00:20:00.060
this phenomenon.
00:20:00.060 --> 00:20:02.780
So that's my problem number 2.
00:20:02.780 --> 00:20:07.100
The third problem
that I'm going to do.
00:20:07.100 --> 00:20:08.930
Sorry.
00:20:08.930 --> 00:20:09.860
I haven't done it.
00:20:09.860 --> 00:20:13.420
That's what I want to
do, and I have to do it.
00:20:13.420 --> 00:20:18.660
I wanted to get the exact things
over quicker, but no such luck.
00:20:21.229 --> 00:20:22.770
Now, I'm going to
solve this problem.
00:20:22.770 --> 00:20:24.360
So back again.
00:20:24.360 --> 00:20:27.370
Look at that diagram and
what's the first thing to do.
00:20:27.370 --> 00:20:38.150
Step 1 use the laws of nature
to restate this problem
00:20:38.150 --> 00:20:41.270
in terms of math and
language of mathematics.
00:20:41.270 --> 00:20:42.740
That's what we have to do.
00:20:42.740 --> 00:20:43.240
All right.
00:20:43.240 --> 00:20:45.430
Let's go a little fast.
00:20:45.430 --> 00:20:48.660
So I have to define
some coordinate system
00:20:48.660 --> 00:20:52.460
so I redraw my
picture and now I'm
00:20:52.460 --> 00:20:56.640
going to tell you that I'll take
some vertical stationary axis
00:20:56.640 --> 00:21:01.630
here, and I'll say it
instant t, this mass
00:21:01.630 --> 00:21:03.700
is at some position, y.
00:21:07.090 --> 00:21:11.600
My hand holding
here is at position
00:21:11.600 --> 00:21:15.700
is at x0 sine omega t, time t.
00:21:15.700 --> 00:21:16.370
Fine.
00:21:16.370 --> 00:21:18.420
That this string is taught.
00:21:18.420 --> 00:21:21.760
I said the assumption
is it's massless
00:21:21.760 --> 00:21:24.900
so therefore it
has to be taught.
00:21:24.900 --> 00:21:30.235
That's one of the reasons why
we have to make a massless.
00:21:30.235 --> 00:21:32.860
If it wasn't massless, it could
start getting kinks and I'll be
00:21:32.860 --> 00:21:34.957
doing this, but the
massless one cannot,
00:21:34.957 --> 00:21:37.540
and that's a good question for
you to think why that statement
00:21:37.540 --> 00:21:38.340
is correct.
00:21:38.340 --> 00:21:39.515
It's nontrivial.
00:21:39.515 --> 00:21:40.890
All right.
00:21:40.890 --> 00:21:46.230
Then at some instant of time,
this angle I'll called theta.
00:21:46.230 --> 00:21:51.030
And now this is a
problem in dynamics.
00:21:51.030 --> 00:21:55.490
Basically, I can draw my free
body diagram, a force diagram.
00:21:55.490 --> 00:21:57.990
I have this force, mass m.
00:21:57.990 --> 00:21:59.470
What forces are acting on it?
00:21:59.470 --> 00:22:03.510
Well, there is gravity
pulling it down, the force mg.
00:22:03.510 --> 00:22:06.170
There is a frictional
force which
00:22:06.170 --> 00:22:12.360
we assume will be
proportional to the velocity,
00:22:12.360 --> 00:22:13.920
so this is the
frictional force which
00:22:13.920 --> 00:22:18.200
will be minus b
times the velocity.
00:22:18.200 --> 00:22:21.230
There will be a tension
in this string pulling it
00:22:21.230 --> 00:22:22.670
in that direction.
00:22:22.670 --> 00:22:26.760
And the sum of forces, by
Newton's laws of motion,
00:22:26.760 --> 00:22:29.440
will give rise to
the acceleration.
00:22:29.440 --> 00:22:32.450
The acceleration
will be net force
00:22:32.450 --> 00:22:38.280
divided by C inertia, the
system which is the mass.
00:22:38.280 --> 00:22:39.270
We have now experience.
00:22:39.270 --> 00:22:41.390
I'm not going to go slowly.
00:22:41.390 --> 00:22:45.380
I'll immediately take
the components of forces.
00:22:45.380 --> 00:22:47.710
First let me consider
vertical motion.
00:22:47.710 --> 00:22:50.610
I told you the angle is small.
00:22:50.610 --> 00:22:52.800
If I make the angle
sufficiently small
00:22:52.800 --> 00:22:57.420
and that ball is oscillating,
it is not moving up and down.
00:22:57.420 --> 00:23:00.770
The distance is tiny
insignificant in this problem.
00:23:00.770 --> 00:23:03.790
So I can ignore the
vertical motion.
00:23:03.790 --> 00:23:07.400
So the vertical
acceleration is 0.
00:23:07.400 --> 00:23:11.880
The net vertical force
here is the component
00:23:11.880 --> 00:23:14.680
of t in the up direction
which is t cosine
00:23:14.680 --> 00:23:18.300
theta minus the gravity
down, and that's
00:23:18.300 --> 00:23:20.250
equal to 0, no acceleration.
00:23:20.250 --> 00:23:22.120
Therefore t is mg.
00:23:22.120 --> 00:23:24.080
T is the tension in this string.
00:23:24.080 --> 00:23:25.650
How about horizontal?
00:23:25.650 --> 00:23:29.100
For horizontal motion, I
have the horizontal component
00:23:29.100 --> 00:23:32.820
of this tension here
which is t sine theta.
00:23:32.820 --> 00:23:36.430
This is theta, so t sine theta
is the horizontal component.
00:23:36.430 --> 00:23:41.300
Now, this is moving
here, therefore,
00:23:41.300 --> 00:23:44.580
there will be a frictional
force proportional to that.
00:23:44.580 --> 00:23:49.520
So that's minus, this is
the velocity, times b,
00:23:49.520 --> 00:23:50.860
so this is the frictional force.
00:23:50.860 --> 00:23:53.330
It's always minus, it's
always in opposition
00:23:53.330 --> 00:23:55.280
to the actual velocity.
00:23:55.280 --> 00:23:58.800
And that by Newton's
law is equal to mass
00:23:58.800 --> 00:24:01.900
times the horizontal
acceleration.
00:24:01.900 --> 00:24:06.480
Beware of my symbols.
00:24:06.480 --> 00:24:12.060
This whole problem for me
is really in one dimension.
00:24:12.060 --> 00:24:18.320
So I called x the horizontal
motion of my hand,
00:24:18.320 --> 00:24:23.620
and I'm calling y the
horizontal motion of the mass.
00:24:23.620 --> 00:24:26.400
This has nothing to do
with x, y, z, one being
00:24:26.400 --> 00:24:27.930
up and the other horizontal.
00:24:27.930 --> 00:24:29.650
It's all horizontal.
00:24:29.650 --> 00:24:30.440
I had a choice.
00:24:30.440 --> 00:24:33.700
I could have called
this x1 and this x2.
00:24:33.700 --> 00:24:35.940
I chose to call them x and y.
00:24:35.940 --> 00:24:38.822
You're smart enough to
figure the difference,
00:24:38.822 --> 00:24:40.280
to follow that kind
of terminology.
00:24:43.590 --> 00:24:47.660
So now I can rewrite this.
00:24:47.660 --> 00:24:53.570
I know what sine theta is
if I look at this diagram.
00:24:53.570 --> 00:25:03.650
Sine theta is simply
equal to the distance.
00:25:03.650 --> 00:25:05.740
It's x minus y.
00:25:05.740 --> 00:25:06.875
It's sine theta.
00:25:09.570 --> 00:25:13.290
This angle equals, I
should draw this here.
00:25:13.290 --> 00:25:16.180
This angle equals that
one, so sine theta
00:25:16.180 --> 00:25:19.750
is this distance
divided by that length,
00:25:19.750 --> 00:25:23.740
and this distance
is x minus y over l.
00:25:23.740 --> 00:25:27.530
So this is sine theta
minus by equals this.
00:25:27.530 --> 00:25:30.140
Quickly, I can play
with the algebra,
00:25:30.140 --> 00:25:32.040
and I end up with this equation.
00:25:32.040 --> 00:25:35.610
And lo and behold,
Eureka, once again I
00:25:35.610 --> 00:25:40.230
get almost the same equation.
00:25:40.230 --> 00:25:43.860
I purposely chose it so it
didn't come out exactly right.
00:25:43.860 --> 00:25:46.480
Before I ended up
with a cosine, now I
00:25:46.480 --> 00:25:48.690
have ended up with a sine.
00:25:48.690 --> 00:25:49.470
Intention.
00:25:49.470 --> 00:25:50.190
All right.
00:25:50.190 --> 00:25:59.650
So to show that something which
seems different often is not.
00:25:59.650 --> 00:26:04.020
So rewriting this in terms
of constant, which I like,
00:26:04.020 --> 00:26:07.030
I do the same as before.
00:26:07.030 --> 00:26:12.310
I define this b over m as
gamma, this g over l as omega
00:26:12.310 --> 00:26:17.720
squared, this thing
here as f, and I end up
00:26:17.720 --> 00:26:23.150
with this equation where
the constants are here
00:26:23.150 --> 00:26:27.940
and the initial conditions
are that the position
00:26:27.940 --> 00:26:32.934
of this mass at time 0 is 0,
and that the velocity was 0.
00:26:32.934 --> 00:26:35.100
You remember I considered
the mass being stationary.
00:26:38.170 --> 00:26:43.010
This is now a set of
mathematical equations.
00:26:43.010 --> 00:26:50.210
I have described this problem
in terms of mathematics.
00:26:50.210 --> 00:26:53.720
From now on, I can
forget that this
00:26:53.720 --> 00:26:57.210
has anything to do
with that problem.
00:26:57.210 --> 00:26:59.650
I have to solve now a
mathematical problem.
00:27:03.530 --> 00:27:05.360
Finally.
00:27:05.360 --> 00:27:07.070
One more.
00:27:07.070 --> 00:27:10.840
Let me consider
one more problem,
00:27:10.840 --> 00:27:18.010
and once we've done all
three, at least part
00:27:18.010 --> 00:27:25.820
of step 1 of all three we'll go
and solve them all in one go.
00:27:25.820 --> 00:27:33.380
And I repeat what I've done
here I've done all the physics.
00:27:33.380 --> 00:27:38.620
This is the part which
causes most difficulty,
00:27:38.620 --> 00:27:42.740
needs most understanding of the
physical situation, et cetera.
00:27:42.740 --> 00:27:46.720
It may or may not
take a very long time.
00:27:46.720 --> 00:27:50.930
The next step is
routine mathematics,
00:27:50.930 --> 00:27:54.044
which can be the part which
takes you a whole night solving
00:27:54.044 --> 00:27:55.460
because you make
algebraic errors,
00:27:55.460 --> 00:27:58.480
and you sweat and you
go back, et cetera.
00:27:58.480 --> 00:28:02.080
That's the part you may be
cursing when you're doing it,
00:28:02.080 --> 00:28:05.530
but it is routine from then on.
00:28:05.530 --> 00:28:09.890
It doesn't need an
understanding of the physics
00:28:09.890 --> 00:28:13.590
of the situation, and what I'm
trying to help you understand
00:28:13.590 --> 00:28:17.050
is physics, not
mathematics at this stage.
00:28:17.050 --> 00:28:18.690
So let's take another
problem, which
00:28:18.690 --> 00:28:22.100
again, seems
completely different.
00:28:22.100 --> 00:28:23.270
All right.
00:28:23.270 --> 00:28:26.810
Here it's a seismograph,
in other words,
00:28:26.810 --> 00:28:32.470
a device for detecting
tremors of the Earth.
00:28:32.470 --> 00:28:34.640
And the following would work.
00:28:34.640 --> 00:28:39.910
Take attached to
the floor a spring.
00:28:39.910 --> 00:28:46.470
Put the mass, m, on it,
and just look at the mass
00:28:46.470 --> 00:28:48.380
how it's behaving.
00:28:48.380 --> 00:28:53.550
If the Earth starts
oscillating, an earthquake,
00:28:53.550 --> 00:28:55.540
the mass will oscillate.
00:28:55.540 --> 00:28:58.130
How much it oscillates
will give you
00:28:58.130 --> 00:29:01.630
a measure of how badly
the Earth is oscillating.
00:29:01.630 --> 00:29:08.800
So I'm going to here do
an idealized seismograph.
00:29:08.800 --> 00:29:17.280
I will model it by a mass, m,
sitting on top of a massless,
00:29:17.280 --> 00:29:25.830
ideal spring, which is
attached to the floor.
00:29:25.830 --> 00:29:29.190
I will idealize the
earthquake and saying
00:29:29.190 --> 00:29:31.510
it's oscillating with
a single frequency
00:29:31.510 --> 00:29:32.710
with a single amplitude.
00:29:32.710 --> 00:29:39.092
The floor is oscillating by
an amplitude y0 sine omega t.
00:29:39.092 --> 00:29:39.800
It's oscillating.
00:29:42.550 --> 00:29:46.440
So my assumptions are this
spring is ideal massless.
00:29:46.440 --> 00:29:49.480
The base, Hooke's law,
that's the ideal part.
00:29:49.480 --> 00:29:53.570
I will again assume that
there is a damping, which
00:29:53.570 --> 00:29:57.130
will be proportional to the
velocity, for the same reasons
00:29:57.130 --> 00:30:00.560
as I took the other one to
be proportional to velocity.
00:30:00.560 --> 00:30:03.050
And the question
this time will be
00:30:03.050 --> 00:30:07.510
not to predict where that mass
is at a given interval of time,
00:30:07.510 --> 00:30:12.330
but the question is, for
what values of the mass
00:30:12.330 --> 00:30:14.960
and the spring
constant, k, will I
00:30:14.960 --> 00:30:18.360
get the most
sensitive instrument?
00:30:18.360 --> 00:30:20.560
So this seems
completely different
00:30:20.560 --> 00:30:23.780
and the question seems different
the other ones et cetera,
00:30:23.780 --> 00:30:25.880
but as you will see
it all boils down
00:30:25.880 --> 00:30:29.170
to one and the same question.
00:30:29.170 --> 00:30:31.750
So how do we do this?
00:30:31.750 --> 00:30:35.860
So now let me come
down to step 1,
00:30:35.860 --> 00:30:38.930
understanding the
situation, the problem
00:30:38.930 --> 00:30:41.310
in terms of mathematics.
00:30:41.310 --> 00:30:42.890
All right.
00:30:42.890 --> 00:30:47.520
So I've re-drawn
here the situation.
00:30:47.520 --> 00:30:49.940
This is my coordinate
system of the problem.
00:30:52.930 --> 00:30:56.580
The surface of the floor,
I'll assume it's here.
00:30:56.580 --> 00:30:58.080
It's moving.
00:30:58.080 --> 00:30:59.670
It is not a good
reference place.
00:30:59.670 --> 00:31:00.570
It's moving.
00:31:00.570 --> 00:31:01.430
All right.
00:31:01.430 --> 00:31:03.590
So it's here.
00:31:03.590 --> 00:31:06.230
The mass, m, is there
attached to the spring
00:31:06.230 --> 00:31:09.450
of natural length, l0,
spring constant, k.
00:31:09.450 --> 00:31:12.420
There is the mass sitting on it.
00:31:12.420 --> 00:31:17.420
Now, we've got to take a
coordinate system which
00:31:17.420 --> 00:31:18.090
isn't moving.
00:31:18.090 --> 00:31:20.430
Got to have an inertial one.
00:31:20.430 --> 00:31:22.690
The surface of the Earth
is no good for this.
00:31:22.690 --> 00:31:24.500
It's vibrating.
00:31:24.500 --> 00:31:28.220
So the best thing we
have are the fixed stars.
00:31:28.220 --> 00:31:32.750
So let's take the fixed
stars as a reference.
00:31:32.750 --> 00:31:37.640
So this is a height
in that room, which
00:31:37.640 --> 00:31:43.370
is fixed in position
relative to the fixed stars.
00:31:43.370 --> 00:31:47.350
Now, the surface of the Earth
can be moving relative to that,
00:31:47.350 --> 00:31:51.280
and I am making the assumption
it's moving sinusoidally
00:31:51.280 --> 00:31:52.660
with amplitude y0.
00:31:52.660 --> 00:31:56.990
So this is the floor, position
of the floor at time, t.
00:31:56.990 --> 00:32:05.430
The mass relative to this,
which I defined as my y equal 0,
00:32:05.430 --> 00:32:07.025
is at position y of m.
00:32:09.970 --> 00:32:15.440
So this now is again
a problem in dynamics,
00:32:15.440 --> 00:32:21.320
a mass with forces acting on
it, what is the acceleration,
00:32:21.320 --> 00:32:24.280
and what is the solution
of the dynamic equation,
00:32:24.280 --> 00:32:25.820
the equation of motion.
00:32:25.820 --> 00:32:30.040
So the force diagram
or free body diagram
00:32:30.040 --> 00:32:36.120
is a point mass, m, with a
force fg due to gravity on it,
00:32:36.120 --> 00:32:43.050
a force, f, due to the
spring acting on it,
00:32:43.050 --> 00:32:45.740
and I forgot to put this.
00:32:45.740 --> 00:32:54.560
Of course, there will be a
force here due to friction
00:32:54.560 --> 00:32:55.795
acting on it.
00:32:55.795 --> 00:32:57.670
This is probably not a
good place to draw it.
00:32:57.670 --> 00:33:00.480
Let me draw it so it
doesn't confuse you.
00:33:00.480 --> 00:33:06.570
I'll draw it next
to this, friction.
00:33:06.570 --> 00:33:10.692
So these are in same
direction et cetera.
00:33:10.692 --> 00:33:11.460
All right.
00:33:11.460 --> 00:33:16.110
So now let's use
Newton's laws of motion,
00:33:16.110 --> 00:33:21.260
f equals ma, and
relate the forces
00:33:21.260 --> 00:33:24.310
to the resulting acceleration.
00:33:24.310 --> 00:33:29.560
So first of all what is
the net up-going force?
00:33:29.560 --> 00:33:32.980
I'm taking a coordinate system
where this is a one dimensional
00:33:32.980 --> 00:33:35.320
problem, so I don't
have to use vectors.
00:33:35.320 --> 00:33:37.380
I'll consider just
the magnitude.
00:33:37.380 --> 00:33:42.430
My positive force is upwards.
00:33:42.430 --> 00:33:46.800
So at any instant
of time, this mass
00:33:46.800 --> 00:33:50.410
will have a force due to the
spring and it will depend.
00:33:50.410 --> 00:33:55.790
If this spring is compressed,
the force will be up.
00:33:55.790 --> 00:34:00.530
If it's elongated, stretched,
the force will be down.
00:34:00.530 --> 00:34:05.780
And the magnitude would be
k times the change of length
00:34:05.780 --> 00:34:09.570
from the natural length,
so the upward force
00:34:09.570 --> 00:34:14.190
will be the difference
between those two subtracted
00:34:14.190 --> 00:34:16.909
from the natural length.
00:34:16.909 --> 00:34:21.449
So k times the natural length
minus this quantity, which
00:34:21.449 --> 00:34:23.170
is the difference
between this and this,
00:34:23.170 --> 00:34:28.900
which therefore this is how
much the string is compressed,
00:34:28.900 --> 00:34:31.139
how much is compressed
and is forced upwards.
00:34:31.139 --> 00:34:34.230
If this is a positive
number, it's a negative one,
00:34:34.230 --> 00:34:36.580
it's stretched, it's
putting it down.
00:34:36.580 --> 00:34:39.810
So this is the force
up due to the spring,
00:34:39.810 --> 00:34:42.330
this is the force
down due to gravity,
00:34:42.330 --> 00:34:45.723
and this is the force in
opposition to the velocity.
00:34:45.723 --> 00:34:47.264
I don't know whether
it's up or down.
00:34:47.264 --> 00:34:52.190
It depends which way it's
moving due to the friction.
00:34:52.190 --> 00:34:54.239
This is the net
force, and that's
00:34:54.239 --> 00:34:56.429
got to be equal to the
inertia of the system, which
00:34:56.429 --> 00:34:59.220
is the mass times the
acceleration of it
00:34:59.220 --> 00:35:01.570
to the upward direction.
00:35:01.570 --> 00:35:05.500
And I told you this spring
is massless, et cetera,
00:35:05.500 --> 00:35:07.850
so I don't have to
take into account
00:35:07.850 --> 00:35:09.020
the motion of the spring.
00:35:09.020 --> 00:35:11.395
And that's why I have to
take an idealized situation.
00:35:13.910 --> 00:35:15.590
Back now to algebra.
00:35:15.590 --> 00:35:19.810
From here, I manipulate
this, and I end up
00:35:19.810 --> 00:35:20.970
with this equation.
00:35:20.970 --> 00:35:25.920
You can do the same,
and I've replaced
00:35:25.920 --> 00:35:30.790
for example, the motion of
the floor by y0 sine omega t,
00:35:30.790 --> 00:35:33.690
et cetera, from here.
00:35:33.690 --> 00:35:35.850
And at first I say, oh, hell.
00:35:39.390 --> 00:35:45.220
This looks different than the
one equation I've seen before.
00:35:45.220 --> 00:35:50.450
But then I don't have to be an
Einstein to realize, hold on.
00:35:50.450 --> 00:36:00.140
If I redefine this as my
variable, in other words,
00:36:00.140 --> 00:36:08.460
if I take define y as ym plus mg
over k minus l0, this quantity,
00:36:08.460 --> 00:36:13.310
the rate of change of y is
the rate of change of ym
00:36:13.310 --> 00:36:17.350
because these are constants,
and the second derivative of y
00:36:17.350 --> 00:36:20.710
is the second derivative
of this, is a constant.
00:36:20.710 --> 00:36:23.360
And therefore, I can
rewrite the equation
00:36:23.360 --> 00:36:28.350
as well y double dot plus
gamma y dot plus omega 0
00:36:28.350 --> 00:36:33.750
squared y is equal to a
constant and sine omega t
00:36:33.750 --> 00:36:40.700
where these constants are
written here from this equation
00:36:40.700 --> 00:36:43.590
and I redefine my variable.
00:36:43.590 --> 00:36:47.720
And now, Eureka, I'm back
again to the same problem.
00:36:47.720 --> 00:36:50.100
So once, again, what we've done.
00:36:50.100 --> 00:36:54.470
We've taken a physical situation
and a physical question
00:36:54.470 --> 00:36:57.210
from our understanding
of the laws of nature
00:36:57.210 --> 00:36:59.840
which apply to that situation.
00:36:59.840 --> 00:37:04.580
We have translated it to
a problem in mathematics.
00:37:04.580 --> 00:37:06.200
And here it is.
00:37:06.200 --> 00:37:10.110
That's the problem,
and I can now,
00:37:10.110 --> 00:37:12.580
at least for the
next few minutes,
00:37:12.580 --> 00:37:16.040
forget that this is
physics, treat it
00:37:16.040 --> 00:37:18.060
as a problem in mathematics.
00:37:18.060 --> 00:37:20.190
And that's what I'll do
it now, but first I've
00:37:20.190 --> 00:37:22.120
got to make some
room for myself.
00:37:22.120 --> 00:37:25.570
I have nowhere to write, so
I have to erase some things,
00:37:25.570 --> 00:37:29.440
and we'll continue
in a few minutes.
00:37:29.440 --> 00:37:30.270
All right.
00:37:30.270 --> 00:37:34.720
We've erased the boards, and
we can therefore continue.
00:37:34.720 --> 00:37:37.450
Let me remind you
what we have shown
00:37:37.450 --> 00:37:48.140
is we've taken three completely
different systems, described
00:37:48.140 --> 00:37:51.440
them in terms of mathematics.
00:37:51.440 --> 00:37:53.800
The three systems with
an electrical circuit,
00:37:53.800 --> 00:37:56.600
an RLC circuit.
00:37:56.600 --> 00:38:03.240
We took a pendulum, driven on
a pendulum, and a seismograph.
00:38:03.240 --> 00:38:11.175
Each one of these ended up
with almost identical equation.
00:38:11.175 --> 00:38:15.110
In fact, the only
difference is one of them
00:38:15.110 --> 00:38:20.380
ended up with a cosine here,
the other two with a sine.
00:38:20.380 --> 00:38:24.520
I will comment about
that in a second.
00:38:24.520 --> 00:38:28.720
We are now, forgetting
about the physics,
00:38:28.720 --> 00:38:31.690
we are now in the
world of mathematics.
00:38:31.690 --> 00:38:36.110
These equations of
motion need to be
00:38:36.110 --> 00:38:38.460
solved if we are to
predict what will
00:38:38.460 --> 00:38:41.270
happen in those situations.
00:38:41.270 --> 00:38:46.300
Now, there are many ways of
solving differential equations.
00:38:46.300 --> 00:38:51.810
It is not my job
to teach you how
00:38:51.810 --> 00:38:55.360
to solve differential equations.
00:38:55.360 --> 00:38:58.470
For an example, you can go
to the lectures of Professor
00:38:58.470 --> 00:39:02.350
Walter Lewin, and
he does illustrate
00:39:02.350 --> 00:39:05.790
how one can solve these
using complex amplitudes.
00:39:08.740 --> 00:39:14.060
I am satisfied if I
find the solutions.
00:39:14.060 --> 00:39:15.530
I don't care how.
00:39:15.530 --> 00:39:17.880
I can take them
from what he did,
00:39:17.880 --> 00:39:21.190
I can take them from a book
of mathematics, et cetera,
00:39:21.190 --> 00:39:25.070
and what you will
find is that if you
00:39:25.070 --> 00:39:29.490
have a generic second
ordered differential
00:39:29.490 --> 00:39:35.960
equation of this
form, it has solution
00:39:35.960 --> 00:39:40.030
which looks like this, the
most general solution that
00:39:40.030 --> 00:39:41.800
looks like that.
00:39:41.800 --> 00:39:46.000
There is no other solution
in the universe which
00:39:46.000 --> 00:39:48.690
is not represented
by this equation.
00:39:48.690 --> 00:39:51.250
You can manipulate it,
make it look different,
00:39:51.250 --> 00:39:53.720
but it will boil down
to this equation.
00:39:53.720 --> 00:39:55.500
How do I know that?
00:39:55.500 --> 00:39:57.250
Because of something
I mentioned earlier,
00:39:57.250 --> 00:40:00.180
the so-called
uniqueness theorem.
00:40:00.180 --> 00:40:05.240
This is a second order
differential equation.
00:40:05.240 --> 00:40:09.110
This equation satisfies it.
00:40:09.110 --> 00:40:11.740
Try it, take it,
differentiate it twice this,
00:40:11.740 --> 00:40:16.590
et cetera, u
[? dissatisfied. ?] Furthermore,
00:40:16.590 --> 00:40:19.030
it has two arbitrary constants.
00:40:22.650 --> 00:40:27.530
This C can be anything and
this phi can be anything.
00:40:27.530 --> 00:40:29.210
They are arbitrary.
00:40:29.210 --> 00:40:34.560
Everything else in this equation
is completely determined.
00:40:34.560 --> 00:40:41.490
For example, a of
omega depends on omega.
00:40:41.490 --> 00:40:48.450
This amplitude, a,
is equal to f divided
00:40:48.450 --> 00:40:52.460
by the square root of omega
0 squared minus omega squared
00:40:52.460 --> 00:40:56.760
0 squared plus gamma
omega 0 squared.
00:40:56.760 --> 00:41:02.310
All these constants--
omega, gamma, f, omega 0--
00:41:02.310 --> 00:41:04.285
are known in each
of the problems.
00:41:07.270 --> 00:41:10.630
Let me just as an example
take the last problem we
00:41:10.630 --> 00:41:15.310
did When we derived
the equation of motion,
00:41:15.310 --> 00:41:20.090
we furthermore knew exactly
what each one of those constants
00:41:20.090 --> 00:41:20.630
meant.
00:41:20.630 --> 00:41:24.140
They are known, so I don't
need to solve for them.
00:41:24.140 --> 00:41:28.500
So a of omega is this quantity.
00:41:28.500 --> 00:41:33.330
The delta, the phase there
is, again, a known quantity.
00:41:33.330 --> 00:41:35.800
The time of delta is this.
00:41:35.800 --> 00:41:38.965
It is only under
these conditions
00:41:38.965 --> 00:41:43.090
that this satisfies
that equation.
00:41:43.090 --> 00:41:46.690
And find the omega prime here
also is a known quantity.
00:41:46.690 --> 00:41:48.400
Nothing here is unknown.
00:41:48.400 --> 00:41:51.660
The only unknowns are
the C, and the phi,
00:41:51.660 --> 00:41:54.220
and they are
arbitrary constants.
00:41:54.220 --> 00:41:57.790
This works for any value
of C and any value of phi.
00:42:02.340 --> 00:42:05.760
Now, one more comment.
00:42:05.760 --> 00:42:08.080
These look equations.
00:42:08.080 --> 00:42:11.000
One looks like a
cosine theta sine.
00:42:11.000 --> 00:42:13.310
If you stop and
think for a second,
00:42:13.310 --> 00:42:16.670
they're actually one
and the same equation
00:42:16.670 --> 00:42:20.840
where you've
redefined what is t 0.
00:42:20.840 --> 00:42:26.015
So by just changing the phase
of this, you'll get to this.
00:42:26.015 --> 00:42:28.040
You've changed it by 90 degrees.
00:42:28.040 --> 00:42:31.610
Just redefine what
t0 is, your clock,
00:42:31.610 --> 00:42:34.780
and you'll get
from this to that.
00:42:34.780 --> 00:42:39.150
So if this equation
I've written here
00:42:39.150 --> 00:42:44.380
is a solution to
the cosine part,
00:42:44.380 --> 00:42:46.850
if my driver has this
phase, in other words,
00:42:46.850 --> 00:42:51.400
sine omega t, then the answer
will be displaced with respect
00:42:51.400 --> 00:42:53.290
to this one by that same thing.
00:42:53.290 --> 00:42:56.740
Instead of having a cosine
here, you have a sign.
00:42:56.740 --> 00:43:01.570
So from my perspective, these
two are identical equations,
00:43:01.570 --> 00:43:06.140
and the solution for
them is identical.
00:43:06.140 --> 00:43:10.420
It's also worth looking at what
the mathematics of this looks
00:43:10.420 --> 00:43:11.070
like.
00:43:11.070 --> 00:43:15.840
What this looks like is the
solution to these equations
00:43:15.840 --> 00:43:22.460
is some function with a certain
amplitude, which is oscillatory
00:43:22.460 --> 00:43:28.350
with the same frequency
as the driver.
00:43:28.350 --> 00:43:30.760
So this is just an
oscillatory function.
00:43:30.760 --> 00:43:32.870
It's got a different
phase to that.
00:43:32.870 --> 00:43:35.330
They lag will lead
it, different phase,
00:43:35.330 --> 00:43:39.740
but it's just constant
amplitude oscillatory function.
00:43:39.740 --> 00:43:45.270
Added to it is another
oscillatory function
00:43:45.270 --> 00:43:49.870
with a slightly
different frequency,
00:43:49.870 --> 00:43:55.920
but which with time decays,
because of this e to the minus
00:43:55.920 --> 00:43:57.910
gamma over 2t.
00:43:57.910 --> 00:44:03.240
And by the way, one more
thing I should just emphasize.
00:44:03.240 --> 00:44:07.400
In writing this solution,
I've made the assumption
00:44:07.400 --> 00:44:09.510
that this oscillator does oscil.
00:44:09.510 --> 00:44:12.775
You're driving something
which oscillates and is
00:44:12.775 --> 00:44:18.230
an overdamped
system in which case
00:44:18.230 --> 00:44:22.560
this term here would be
different and not interesting
00:44:22.560 --> 00:44:25.730
when normally you wouldn't drive
things and be interested in how
00:44:25.730 --> 00:44:30.540
it responds if it's
incredibly damped system.
00:44:30.540 --> 00:44:33.820
So I am making the
assumption that this
00:44:33.820 --> 00:44:38.110
is an underdamped system, which
means that this quantity here
00:44:38.110 --> 00:44:42.420
is greater than
this quantity here.
00:44:42.420 --> 00:44:46.000
And if you're having
difficulty with this part,
00:44:46.000 --> 00:44:54.080
I urge you to review what
happens with just not driven
00:44:54.080 --> 00:44:57.860
oscillators under conditions
of being underdamped
00:44:57.860 --> 00:45:00.875
or overdamped.
00:45:00.875 --> 00:45:01.875
That's the same physics.
00:45:04.860 --> 00:45:08.000
So we found the most
general solution.
00:45:08.000 --> 00:45:12.100
This solution must apply to
every problem we've done.
00:45:12.100 --> 00:45:15.150
So I'll now take one of them.
00:45:15.150 --> 00:45:19.130
For example, let's
consider this problem.
00:45:19.130 --> 00:45:19.950
We did this.
00:45:19.950 --> 00:45:22.170
We have an RLC circuit.
00:45:22.170 --> 00:45:24.890
They are an equation of motion.
00:45:24.890 --> 00:45:28.300
From here, I can
immediately write
00:45:28.300 --> 00:45:30.930
what is the solution of this.
00:45:30.930 --> 00:45:37.060
As a function of
time, the charge Q
00:45:37.060 --> 00:45:47.650
will be equal to-- just look
at that-- a of omega times
00:45:47.650 --> 00:45:58.990
cosine omega t minus delta-- I'm
just copying from over there--
00:45:58.990 --> 00:46:09.640
plus some arbitrary constant
e to the minus gamma over 2t.
00:46:09.640 --> 00:46:16.710
And here we have a cosine, so I
took the cosine here, and here
00:46:16.710 --> 00:46:26.850
is cosine of omega prime t
plus this arbitrary phase phi.
00:46:26.850 --> 00:46:31.460
So this is a general
equation, which
00:46:31.460 --> 00:46:38.820
satisfies that one at all times.
00:46:38.820 --> 00:46:41.310
And I'm repeating myself.
00:46:41.310 --> 00:46:51.450
This, you know, it will
be given by A of omega
00:46:51.450 --> 00:46:55.030
will be equal to
f, and what is f?
00:46:55.030 --> 00:47:02.950
f is v0 over L-- I just
read it off there-- divided
00:47:02.950 --> 00:47:10.470
by square root of
omega 0 squared,
00:47:10.470 --> 00:47:22.490
which is 1 over LC
minus omega squared
00:47:22.490 --> 00:47:28.790
That's just the frequencies with
which the voltage is driving
00:47:28.790 --> 00:47:37.260
there all squared
plus gamma omega.
00:47:37.260 --> 00:47:39.277
Gamma is R over L
omega all squared.
00:47:44.840 --> 00:47:52.530
And time delta is
equal to gamma omega.
00:47:52.530 --> 00:47:59.780
Gamma is R over L
divided by omega 0
00:47:59.780 --> 00:48:06.070
squared, which is 1 over
LC minus omega squared.
00:48:11.560 --> 00:48:17.750
So far we know everything
except the C and the phi.
00:48:17.750 --> 00:48:22.710
The question actually
was, I've now erased,
00:48:22.710 --> 00:48:27.030
it but the question was
what is the current flowing?
00:48:27.030 --> 00:48:33.780
And the current
flowing I of t, is
00:48:33.780 --> 00:48:39.560
equal to dq dt with
the way we define
00:48:39.560 --> 00:48:48.100
the signs with a part plus here,
which is equal to Q dot, which
00:48:48.100 --> 00:48:52.520
I can get from this by
differentiating this equation,
00:48:52.520 --> 00:48:57.630
and I get differentiate
this minus omega
00:48:57.630 --> 00:49:08.930
A of omega sine of
omega t minus delta.
00:49:08.930 --> 00:49:13.720
Go to differentiate
this, I get minus gamma
00:49:13.720 --> 00:49:25.220
over 2 Ce to the minus gamma
over 2t times cosine omega
00:49:25.220 --> 00:49:29.090
prime t plus phi.
00:49:29.090 --> 00:49:31.350
So I've differentiated
this times that.
00:49:31.350 --> 00:49:34.220
Now I've got to take
this differentiated
00:49:34.220 --> 00:49:41.570
by that, which minus Ce e to
the minus gamma over 2t times
00:49:41.570 --> 00:49:52.291
omega prime times sine
omega prime t plus phi.
00:49:52.291 --> 00:49:52.790
Sorry.
00:49:52.790 --> 00:49:55.430
That's so.
00:49:55.430 --> 00:50:00.660
So and again, we
know every quantity
00:50:00.660 --> 00:50:09.030
here except for C and phi.
00:50:09.030 --> 00:50:10.065
Those are unknowns.
00:50:14.340 --> 00:50:17.640
So this would be the
answer to my question
00:50:17.640 --> 00:50:20.970
if I knew C and phi,
then everything is known.
00:50:20.970 --> 00:50:22.960
Then I could
predict exactly what
00:50:22.960 --> 00:50:25.320
would happen with
any instant of time.
00:50:25.320 --> 00:50:27.390
How do I find those two?
00:50:27.390 --> 00:50:31.190
This is where the boundary
conditions come in.
00:50:31.190 --> 00:50:33.820
That's why you need
at least two boundary
00:50:33.820 --> 00:50:36.850
conditions or
initial conditions,
00:50:36.850 --> 00:50:41.080
two facts about this system
to be able to determine
00:50:41.080 --> 00:50:43.300
those constants,
and we're always
00:50:43.300 --> 00:50:47.610
in the physical
situation of this kind,
00:50:47.610 --> 00:50:49.340
you will have two
bits of information.
00:50:49.340 --> 00:50:53.490
For example, in this
case, I told you
00:50:53.490 --> 00:50:55.650
that we will assume
that t equals
00:50:55.650 --> 00:50:59.340
0 the charge and
the capacitance is 0
00:50:59.340 --> 00:51:01.972
and the current in
the circuit is 0.
00:51:01.972 --> 00:51:04.180
I could have told you
something else at the beginning
00:51:04.180 --> 00:51:06.350
and the answer
would be different.
00:51:06.350 --> 00:51:07.170
All right.
00:51:07.170 --> 00:51:09.310
And the thing that
would be different
00:51:09.310 --> 00:51:11.050
would be those constants.
00:51:11.050 --> 00:51:13.270
So how do we do it?
00:51:13.270 --> 00:51:22.520
Simply by taking this equation
and writing it at the time t
00:51:22.520 --> 00:51:23.380
equals 0.
00:51:23.380 --> 00:51:31.510
So at t equals 0, we
know what Q of 0 is,
00:51:31.510 --> 00:51:35.170
and we know what Q dot of 0 is.
00:51:39.290 --> 00:51:45.340
So I take this
equation, set t to 0,
00:51:45.340 --> 00:51:49.270
and write down this equal to 0.
00:51:53.000 --> 00:51:53.500
Sorry.
00:51:53.500 --> 00:51:55.965
I take this equation,
the Q. This one set
00:51:55.965 --> 00:52:01.740
t to 0 and write 0 equals
that with t equals 0.
00:52:01.740 --> 00:52:04.670
I write another
equation from here.
00:52:04.670 --> 00:52:11.790
At t equals 0, Q dot is 0, so
I take this one, set t to 0,
00:52:11.790 --> 00:52:13.510
and write another equation.
00:52:13.510 --> 00:52:17.680
I will end up with two
algebraic equations
00:52:17.680 --> 00:52:21.430
with two unknowns
which you can solve.
00:52:21.430 --> 00:52:26.660
Now remember I told you,
the hard part conceptually
00:52:26.660 --> 00:52:32.280
is getting the equations
of motion right.
00:52:32.280 --> 00:52:38.710
But the slog, the
sweat, and a hard labor
00:52:38.710 --> 00:52:44.320
is getting this solution
right, and I'm not
00:52:44.320 --> 00:52:48.760
going to go here now and
embarrass myself making error
00:52:48.760 --> 00:52:53.110
after algebraic error
solving those equations.
00:52:53.110 --> 00:52:55.920
I know you can solve them
if you keep your cool
00:52:55.920 --> 00:52:59.020
and follow all the
constants, et cetera.
00:52:59.020 --> 00:53:04.070
You take these two algebraic
equations, solve them,
00:53:04.070 --> 00:53:10.340
you come out with a value for C
and a value for the phase phi,
00:53:10.340 --> 00:53:12.250
and then I have
the final answer.
00:53:12.250 --> 00:53:19.350
I then have this equation I
of t and Q of t for any T.
00:53:19.350 --> 00:53:24.310
And so I can predict at
any time with this will do.
00:53:24.310 --> 00:53:27.200
End of story.
00:53:27.200 --> 00:53:33.204
Now, let me take another case
and to look at this one here,
00:53:33.204 --> 00:53:34.370
and I'll go a little faster.
00:53:38.110 --> 00:53:45.380
This is the mathematical
description of the pendulum.
00:53:45.380 --> 00:53:49.880
So again, I'll take
this, initially I'll
00:53:49.880 --> 00:53:59.110
do the same as I did before, so
I will write that y at time t
00:53:59.110 --> 00:54:03.650
is equal to-- and I have
to once again just follow
00:54:03.650 --> 00:54:12.320
the thing-- it's f, which
is gx over L divided
00:54:12.320 --> 00:54:21.140
by the square root
of omega squared,
00:54:21.140 --> 00:54:35.160
which is g over L minus omega
squared all squared plus gamma,
00:54:35.160 --> 00:54:40.460
which is b over m this time
gamma omega all squared.
00:54:46.100 --> 00:54:49.010
So times.
00:54:49.010 --> 00:54:52.210
Now, here I have a
sign so it's going
00:54:52.210 --> 00:55:08.650
to be the sine omega
t minus delta plus--
00:55:08.650 --> 00:55:16.067
and to save time, not
rewrite it, the full thing,
00:55:16.067 --> 00:55:17.400
but I'll start at the beginning.
00:55:17.400 --> 00:55:23.190
There's the Ce to the
minus gamma over 2t
00:55:23.190 --> 00:55:30.220
here at cosine et cetera.
00:55:30.220 --> 00:55:36.550
So what we see
here as before, we
00:55:36.550 --> 00:55:42.920
can get what this mass will
do as a function of time,
00:55:42.920 --> 00:55:48.600
and as before, I don't want
to just go around and around
00:55:48.600 --> 00:55:50.370
repeating the thing is.
00:55:50.370 --> 00:55:55.590
In order to solve, everything
here is known except the C
00:55:55.590 --> 00:56:02.070
and this phase here-- maybe I
should at least write that in.
00:56:02.070 --> 00:56:07.780
That is omega prime t plus phi.
00:56:07.780 --> 00:56:11.350
And in principle,
this and that you
00:56:11.350 --> 00:56:16.570
can get, again, by using
the initial conditions.
00:56:16.570 --> 00:56:17.930
I don't want to focus on that.
00:56:17.930 --> 00:56:21.630
In this case, I want to
focus what the problem said.
00:56:21.630 --> 00:56:25.410
What it said it
was, discuss what
00:56:25.410 --> 00:56:29.030
you would expect the
motion to be for that.
00:56:29.030 --> 00:56:32.030
In other words, when
we took this special,
00:56:32.030 --> 00:56:40.530
we took this and took in
given initial conditions
00:56:40.530 --> 00:56:42.720
and let's go.
00:56:42.720 --> 00:56:46.280
This will describe what goes on.
00:56:46.280 --> 00:56:51.850
And let's look at this solution.
00:56:51.850 --> 00:56:58.660
What we see is that there
are two terms here as before.
00:56:58.660 --> 00:57:05.890
This is an oscillation of the
same frequency as a driver.
00:57:05.890 --> 00:57:09.850
So other words, that
term is something
00:57:09.850 --> 00:57:17.050
which has this moving with
the same frequency as my hand.
00:57:17.050 --> 00:57:21.910
This is a slightly
different frequency,
00:57:21.910 --> 00:57:26.200
which is exponentially decaying.
00:57:26.200 --> 00:57:30.090
When you add two
oscillating functions,
00:57:30.090 --> 00:57:33.150
you've learned from Professor
Walter Lewin's class,
00:57:33.150 --> 00:57:35.130
you get beats.
00:57:35.130 --> 00:57:42.820
So at first when this is
large, these two wave motions,
00:57:42.820 --> 00:57:44.720
these oscillatory
motions will beat
00:57:44.720 --> 00:57:47.365
minutes one against
the other and you
00:57:47.365 --> 00:57:50.450
will get some kind
of a crazy motion.
00:57:50.450 --> 00:57:52.850
You can't make sense of it.
00:57:52.850 --> 00:57:57.840
As a function of time, as t
becomes bigger and bigger,
00:57:57.840 --> 00:58:03.410
this exponential kills this
term, and after a long time,
00:58:03.410 --> 00:58:07.670
you'll end up with
only this term, which
00:58:07.670 --> 00:58:09.970
is simple sinusoidal motion.
00:58:09.970 --> 00:58:12.550
Try it for yourself.
00:58:12.550 --> 00:58:15.900
If I start any
old way and I then
00:58:15.900 --> 00:58:19.000
move with a given
frequency, you'll
00:58:19.000 --> 00:58:25.340
find this has a constant
frequency, some amplitude given
00:58:25.340 --> 00:58:29.860
by this, and it continues
doing this forever
00:58:29.860 --> 00:58:32.740
consistent with it.
00:58:32.740 --> 00:58:35.810
But there is another
interesting feature,
00:58:35.810 --> 00:58:38.350
which I would like
you to focus on.
00:58:38.350 --> 00:58:44.630
If the friction is very
small, like this, this
00:58:44.630 --> 00:58:49.980
will have very little
damping, after a long time,
00:58:49.980 --> 00:58:54.470
this has gone and this
term I can neglect.
00:58:54.470 --> 00:59:01.040
And what I see
here is that this,
00:59:01.040 --> 00:59:07.050
the amplitude of the
oscillations of this not only
00:59:07.050 --> 00:59:12.680
depends on the amplitude-- and
again, I notice the time here;
00:59:12.680 --> 00:59:18.150
this is the node quantity
here-- not only depends
00:59:18.150 --> 00:59:24.140
on the amplitude of how
far my head is moving,
00:59:24.140 --> 00:59:28.900
but it depends critically
on the difference between g
00:59:28.900 --> 00:59:31.920
over l and omega squared.
00:59:31.920 --> 00:59:37.850
This, if you remember from
the simple pendulum-- that's
00:59:37.850 --> 00:59:44.150
this omega 0 squared--
gives the frequency
00:59:44.150 --> 00:59:49.400
of oscillation of a free
pendulum of this length.
00:59:49.400 --> 00:59:56.710
So what happens depends
critically how different
00:59:56.710 --> 00:59:59.950
and of which magnitude
is between the frequency
00:59:59.950 --> 01:00:02.840
of oscillation of
my hand compared
01:00:02.840 --> 01:00:06.820
to the natural frequency
of oscillation.
01:00:06.820 --> 01:00:11.110
If-- and here I
won't do the algebra.
01:00:11.110 --> 01:00:13.590
I'll let you do this
as an exercise-- play
01:00:13.590 --> 01:00:19.280
for yourself changing
the value of this making
01:00:19.280 --> 01:00:22.030
it bigger than that
or smaller than that.
01:00:24.710 --> 01:00:35.250
And what you will find is that
the motion of that and sorry,
01:00:35.250 --> 01:00:37.180
there's one more
thing I should have
01:00:37.180 --> 01:00:40.940
said I didn't, and let
me just say it now.
01:00:40.940 --> 01:00:44.590
There is also this question
of this phase delta,
01:00:44.590 --> 01:00:53.920
tan delta is equal
to gamma omega
01:00:53.920 --> 01:01:01.900
over omega 0 squared
minus omega squared.
01:01:01.900 --> 01:01:09.350
So the sine of delta depends on
whether it's bigger or smaller.
01:01:09.350 --> 01:01:13.670
Play for yourself
changing this value.
01:01:13.670 --> 01:01:16.960
See what happens to
the amplitude here
01:01:16.960 --> 01:01:21.730
and what happens to this
phase, and what you'll see
01:01:21.730 --> 01:01:28.880
is that if I go slowly,
the phase is such
01:01:28.880 --> 01:01:31.830
that the two go
in phase together
01:01:31.830 --> 01:01:36.570
if my oscillating
frequency is lower
01:01:36.570 --> 01:01:40.020
than the natural frequency
of oscillation of this,
01:01:40.020 --> 01:01:44.480
while if I go faster, in
other words, higher frequency,
01:01:44.480 --> 01:01:49.790
this becomes a negative
and the phase gets out
01:01:49.790 --> 01:01:53.610
of phase by 180
degrees, the two.
01:01:53.610 --> 01:01:57.710
Furthermore, if you
look at this amplitude,
01:01:57.710 --> 01:02:01.290
you can completely understand
it in the following way.
01:02:06.130 --> 01:02:10.490
Imagine that I had this
string and I move it slowly.
01:02:15.100 --> 01:02:18.910
I'm moving this slowly,
and what you'll see
01:02:18.910 --> 01:02:29.750
is it's going in phase with
that, and the angle of this
01:02:29.750 --> 01:02:34.890
will be such that
this string appears
01:02:34.890 --> 01:02:40.180
to be the continuation
of a single string, which
01:02:40.180 --> 01:02:41.275
has a longer length.
01:02:44.060 --> 01:02:48.760
And the string with the longer
length has a lower frequency.
01:02:48.760 --> 01:02:53.410
So if I move here with
the lower frequency
01:02:53.410 --> 01:02:56.360
than the natural
frequency, this will
01:02:56.360 --> 01:03:02.810
appear to be part of a string
which has just the right length
01:03:02.810 --> 01:03:07.970
to give me the
frequency of my hand.
01:03:07.970 --> 01:03:12.010
While if I move with
the higher frequency,
01:03:12.010 --> 01:03:16.700
what the string
will do is do this.
01:03:16.700 --> 01:03:19.580
Here is my hand moving
backwards and forwards.
01:03:19.580 --> 01:03:21.290
There is the mass,
and the mass is
01:03:21.290 --> 01:03:23.060
moving backwards and forwards.
01:03:23.060 --> 01:03:24.210
All right.
01:03:24.210 --> 01:03:30.200
This string is
oscillating as if it
01:03:30.200 --> 01:03:35.120
was a free string
of a shorter length.
01:03:35.120 --> 01:03:39.880
I urge you as an exercise,
to play with the numbers
01:03:39.880 --> 01:03:45.540
here and consider
these two cases.
01:03:45.540 --> 01:03:48.790
It is a beautiful,
educational tool
01:03:48.790 --> 01:03:53.720
to understand why if you
have a driven system,
01:03:53.720 --> 01:03:57.740
response can be in
phase or out of phase
01:03:57.740 --> 01:04:01.430
depending on the relative
frequency you're driving
01:04:01.430 --> 01:04:04.380
the system compared to
the natural frequency
01:04:04.380 --> 01:04:07.170
of oscillation of
the oscillator.
01:04:07.170 --> 01:04:11.810
And this example
brings everything out.
01:04:11.810 --> 01:04:15.010
Take this example,
consider very little
01:04:15.010 --> 01:04:19.120
damping so that you can do
the calculations easily.
01:04:19.120 --> 01:04:21.170
See what's the
relative amplitude
01:04:21.170 --> 01:04:25.010
of the driver compared
to the response
01:04:25.010 --> 01:04:28.350
and what's the relative
phase of the driver compared
01:04:28.350 --> 01:04:32.110
to the response, and all
this characteristics you'll
01:04:32.110 --> 01:04:35.720
see which you can play for
yourselves with a little model.
01:04:35.720 --> 01:04:38.490
I really urge you to do that.
01:04:38.490 --> 01:04:43.570
Finally today, I
will just comment.
01:04:43.570 --> 01:04:46.410
You see we're now in
a world of repeating.
01:04:46.410 --> 01:04:49.260
We're doing the same
thing over and over again.
01:04:49.260 --> 01:05:03.680
I would just like to comment
about the seismograph.
01:05:03.680 --> 01:05:11.580
Once again, this is an
equation of this form.
01:05:11.580 --> 01:05:13.620
It's the sine version.
01:05:13.620 --> 01:05:18.160
So this is the solution
to it with a sine,
01:05:18.160 --> 01:05:23.510
and the question is,
in this case was,
01:05:23.510 --> 01:05:30.290
what the value of k over
m makes it most sensitive?
01:05:30.290 --> 01:05:34.130
That's equivalent to
asking the question,
01:05:34.130 --> 01:05:39.460
after a long time--
this earthquake
01:05:39.460 --> 01:05:45.930
has lasted a long time--
this so-called transient,
01:05:45.930 --> 01:05:48.770
this term dies away.
01:05:48.770 --> 01:05:50.290
So this we can forget.
01:05:50.290 --> 01:05:56.360
This is the response of the
mass to the earthquake which
01:05:56.360 --> 01:06:05.540
we assumed it's a uniform
frequency, sinusoidal driver.
01:06:05.540 --> 01:06:10.640
And so the question is, what is
the amplitude of the response?
01:06:10.640 --> 01:06:14.080
How big is this term?
01:06:14.080 --> 01:06:19.920
This term we know how
big it is, and if you
01:06:19.920 --> 01:06:26.080
look at this function, you see
that this amplitude depends
01:06:26.080 --> 01:06:27.570
on omega, on the driver.
01:06:30.660 --> 01:06:38.860
Now let's assume the
damping is not very big.
01:06:38.860 --> 01:06:41.790
This will be small
compared to that.
01:06:41.790 --> 01:06:45.600
And so if you want
a big response,
01:06:45.600 --> 01:06:49.330
you want this quantity
to be very small
01:06:49.330 --> 01:06:53.680
so that f divided by
this is a big number.
01:06:53.680 --> 01:06:56.610
When will this be small?
01:06:56.610 --> 01:06:59.500
When these two are almost equal.
01:07:02.180 --> 01:07:07.090
So what you want, if you want
to have an instrument which
01:07:07.090 --> 01:07:16.930
is very sensitive, you want to
study so the dominant frequency
01:07:16.930 --> 01:07:20.630
of oscillations of
earthquakes in your area
01:07:20.630 --> 01:07:27.870
and adjust the natural frequency
of oscillation close to that.
01:07:27.870 --> 01:07:32.470
And the natural frequency is
of course given by k over m.
01:07:32.470 --> 01:07:38.490
So what matters is, it doesn't
matter what m is, what k is,
01:07:38.490 --> 01:07:41.870
but what's the
ratio of k over m?
01:07:41.870 --> 01:07:46.910
You want that to be close
to the natural frequency,
01:07:46.910 --> 01:07:49.920
that the natural frequency of
this oscillator, that value
01:07:49.920 --> 01:07:54.220
k over m, to be close to the
frequency of the oscillations
01:07:54.220 --> 01:07:56.390
of the earth in the earthquake.
01:08:01.030 --> 01:08:10.500
If this driver really
had a definite frequency,
01:08:10.500 --> 01:08:15.060
I could solve this problem
to find the maximum.
01:08:15.060 --> 01:08:18.960
If I plotted this as
a function of omega,
01:08:18.960 --> 01:08:23.439
I would get a curve
something like this.
01:08:23.439 --> 01:08:28.260
This is a function of
omega, the amplitude,
01:08:28.260 --> 01:08:31.810
the response of that oscillator.
01:08:31.810 --> 01:08:38.609
if you do that, you'll find
that if the natural frequency
01:08:38.609 --> 01:08:55.040
of oscillation is
over here, the peak
01:08:55.040 --> 01:08:58.000
is slightly at the
lower frequency
01:08:58.000 --> 01:09:01.199
than the natural frequency
of oscillation omega 0.
01:09:03.750 --> 01:09:09.819
And so you could do this
exactly by plotting it out
01:09:09.819 --> 01:09:16.330
to find out what is the
optimum value of the frequency
01:09:16.330 --> 01:09:20.620
that you want to adjust so
that you get the best response.
01:09:20.620 --> 01:09:23.910
So much for driven oscillators.
01:09:23.910 --> 01:09:30.130
We will now move to
couple oscillators,
01:09:30.130 --> 01:09:32.040
and that we'll do next
time, some problems.
01:09:32.040 --> 01:09:33.490
Thank you.