I'm Walter Lewin. I will be your lecturer this
term. Make sure you have a handout,
and make sure you read it. It tells you everything you
want to know about the course. The course is about waves and
vibrations, about oscillations, periodic, and not so periodic
events. And, when you look around in
the world, you see them everywhere.
For one thing: your heartbeat.
That's a periodic oscillation, at least I hope that most of
you it is periodic. Your breathing is some kind of
a periodic motion. The blinking of your eyes,
your daily routines, and your habits,
your eating, your sleeping,
taking a shower, your classes,
and occasionally doing some work: all those are periodic
actions. When you drink,
I drink some orange juice. Notice as I tried to move the
liquid down into my stomach that it's not a steady stream.
But it's a periodic motion. Look at my throat.
In fact, even if I don't want to swallow the liquid,
but simply have a bottle of liquid and I turn it over,
and we all know that the water doesn't come out like a steady
stream, but it goes glop, glop, glop, glop.
That's some kind of a periodic motion.
I have here a toy which I use to entertain my dinner guests.
Particularly, physicists is interesting.
And, this liquid here, the idea is to get the liquid
there. And then, the problem is,
how can you do it in the fastest possible way?
Well, if you turn it over, you'll see the phenomenon I
just mentioned, which is the glop,
glop, glop. It's not a steady stream.
It's almost pathetic the way that it runs from one side to the other.
It will take minutes before it's there.
But, it can be done and 17 seconds.
And, during the five minute intermission that we have,
you may give that a try. And, I hope you won't break it,
and see if any of you can think of a way that you can transfer
the liquid in 17 seconds. You have breakfast in the
morning, and you casually put your breakfast plate on the
table. What do you hear?
Some kind of a periodic motion. And, two things can happen to
this plate. It can move as a musilly, I call this as a musilly
because I'm an astronomer. But, it can also wobble.
In fact, something can wobble without moving as a musilly,
and something can move as a musilly without wobbling.
In this case, it does both.
And, a fabulous example of that is what's called the Euler's
disk, which is a metal disc, you'll see it shortly there,
and this metal disk we are going to wobble in the similar
way that I wobble plates. And then, we will follow its
motion as a musilly, and the wobbling frequency.
And, what is interesting, as you will see,
that as [a musile?] motion, which has a certain period,
that period gets longer and time, but the wobble motion,
the frequency goes up. So, I'll start it here and then
I'll show it to you in a way that is more appealing.
And you can follow that. It's an amazing toy to work out
to physics. It's very, very difficult.
I was told that Professor Wilczek at MIT once gave a
one-hour lecture exclusively on the explanation of this Euler's
disk. So, try to see this as a musile
motion. It will become clearer as it
slows down further. You may be able to hear the
wobble motion. I will hold my microphone close
up. Can you hear it?
Very high frequency already. Did you hear it?
It's quite amazing, isn't it, when you look at
this? The wobbling frequency
increases quite rapidly. Look as the musile motion slows
down, and how the frequency of the wobble goes up.
And now it comes to a stop. That's a very difficult piece
of physics right there.
If you take a tennis ball, this is a super ball,
and you bounce it. This is called student
involvement, thank you. Then you also get some kind of
the periodic motion whereby, again, the frequency increases
just like in the case of the Euler desk at the breakfast
plate. Here's an object that is
floating in a liquid in water. And if I push that a little
farther in and I let it go, it wobbles.
And, that is a very unique frequency that you'll be able to
calculate in 8.03, a very unique period of one
complete oscillation as this object goes up and down.
Even winds, steady winds, can generate periodic or almost
periodic motions, which all of you have
experienced. You walk outside,
it's windy, and your hair goes like this.
Your hair doesn't go flat like this.
It always has this tendency, just like a flag does the same
thing. If I generate wind here,
and I have here some aluminum, now, you will see that this
wind doesn't make the aluminum just go straight out,
but it wobbles. There is a certain period to
that. After work, if you want to have
some fun, what is more fun than riding your own rocking horse?
That's a periodic motion. Falling in love can be a
periodic event. Now, don't do it too often
because as most of you know it's quite exhausting.
The motion of electrons, atoms, molecules,
periodic and oscillatory, the motion of the moon,
the planets, and the stars:
periodic, oscillatory. Sound is a beautiful example.
I produce sound by oscillating my vocal chords.
I produce there by pressure waves.
My vocal chords push on the air, suck on the air,
push on the air, which produces a pressure wave.
And, the pressure wave propagates out in the lecture
hall, reaches your eardrum. Your eardrum starts to move
back and forth, and your brains tell you
that you hear a sound. I have here a tuning fork,
which is designed so that if I give it a hit at the prongs move
256 times per second. Recall that 256 Hz.
A hertz is one oscillation per second.
[SOUND PLAYS] And all of you can hear that.
Pressure waves are generated. We will deal with them in 8.03.
They travel through the air, reach your eardrum,
and your eardrum starts to shake.
This is a higher frequency, 440 Hz [SOUND PLAYS].
Most human beings can hear in the range from 20 Hz to 20 kHz.
There are animals you can go way beyond 20 kHz.
And, to be nice to you for the first time, this first lecture,
I would like to test your hearing.
And that will be free of charge.
I'm not so much interested in knowing what your lowest
frequency is, but what your highest frequency
is. So, I'm going to generate here
sounds. I will start with 100 Hz,
and then we'll go up higher and higher, and we'll see where your
hearing stops. So, let's start with 100 Hz.
I'm not going to ask you who can hear it because clearly all
of you can. Let's now go to kilohertz,
1,000 hertz. Piece of cake,
right? 2,000, no problem.
I have to change, now, my scale.
4,000, I didn't say that this was going to be a pleasant test.
5,000, this is where the violins come in.
6,000. Anyone in my audience who
cannot hear 6,000? 7,000.
Anyone in my audience who cannot hear 7,000?
I cannot hear 7,000. I hear nothing.
With age, you lose ability to hear higher frequencies.
You all will experience that in your lifetime.
You won't escape that. Now, for some people lose more
than that. I cannot hear above 6,000 Hz.
I hear nothing. OK, 10,000, 12,000,
14,000. Now, I want to see hands if you
cannot hear it any longer. Who cannot hear 14,000?
Don't be ashamed of it. It's not your fault.
14,000. All right, we're slowly going
up. 15,000, who cannot hear it?
Raise your hands. Ah, Professor [UNINTELLIGIBLE],
you're also getting older. 16,000, who cannot hear 16,000?
Of course, the ones who have already raised their hands,
you don't have to raise your hands again.
16,000, who cannot? 17,000?
18,000. OK, so now we're going to
change it. Now I want you to raise your
hands if you can hear it. And so, I first go to 20,000,
19,000. Sorry.
I was only off by a factor of 10.
19,000, who can hear it? Fantastic.
20,000. 21,000.
You see how it cut off? It's very sharp.
22,000. Very good.
23,000. 25,000.
27,000. Some of you have amazing ears
because I already turned it off at 21,000.
[LAUGHTER]
All right, key absolutely key in this course will be simple
harmonic oscillations because they are extremely common in
nature. A simple harmonic oscillation,
and you've seen this, of course, in 8.01, can be
written as follows: x equals x zero,
cosine (omega T plus phi). You can write a sign here if
you want to. X zero is the amplitude.
That's the largest displacement from equilibrium.
Omega is the angular frequency. Omega, which we expressed in
terms of radians per second, the period, T,
2 pi divided by omega is an expressed in terms of seconds,
and the frequency, F, which is one over T is what
we call hertz: number of cycles per second.
Do not confuse omega with F. There is a factor of 2 pi
difference. If I have a uniform circular
motion and I project that uniform circular motion on to
any line on the blackboard, then I get a simple harmonic
motion. So, I take for simplicity just
this horizontal line. But I could take any other
line. And let's call this the x
direction. And let this point be X zero.
And, I take an object which is rotating around.
Here is the object. It's going around,
uniform circular motion. If I project this onto the
x-axis, and this angle is theta, then this position here is x
zero cosine theta. And, if I make theta a function
of prime, theta equals omega T. This omega is what we call not
angular frequency, but we call it angular
velocity. It's very awkward in physics
that we have the same symbol for angular velocity and for angular
frequency. In this case,
it happens to be the same numerically because it's a
uniform circular motion. That's excellent.
So, now you see that x zero then becomes cosine omega T
because the two are the same. I do not have to call the
position T equals zero here. I can choose T equals zero
anywhere along the circumference,
and that introduces, then, phase angle phi.
We call that the initial condition.
So, x zero is the amplitude. Omega is the angular frequency,
and phi has to be adjusted so that at time T equals zero,
you get the right angle. You get the right position.
An easy example of a simple harmonic motion is a spring
system. If I have here a spring,
and this is in the relaxed position, a spring constant is
K, the mass is M, and x equals zero here,
and I bring it further out, I bring it to a position,
X. Then there is a spring force
that wants to drive it back to equilibrium.
It's a restoring force. That's the spring force.
That's arbitrarily called this direction force.
The spring force we call minus KX, minus because if X is
positive, then this force is in the opposite direction.
If the mass of the spring can be ignored, if it is negligibly
small compared to the mass of the object, I can write down
Newton's Second Law: F equals MA.
You may remember that from the good old days.
And so, MA is MX double dot. It is now minus KX.
It's really a vector notation, but since it's a
one-dimensional problem, the minus takes care of the
directions. And so, I can massage this a
little further and I can write this as x double dot plus K over
M times X equals zero. And, what is the solution to
this differential equation? This is a differential
equation: x double dot and x. This is the solution.
It's a simple harmonic motion, provided that omega is the
square root of K over M. So, I advise you to take this
function, substitute it in here, and you will see that out pops,
yes, you can satisfy this equation, provided that omega is
the square root of K over M. Notice, which is not so
intuitive, but this angular frequency of omega,
and therefore also the period of oscillation,
2 pi divided by omega, is independent of x zero.
So, it's independent of how far and move it away from
equilibrium. If I move it far out,
it will take the same amount of time for one oscillation than if
I move it out a teeny-weeny little bit.
Not so intuitive. So, omega is independent of my
initial conditions. It's independent on how I start
the experiment off. It's independent of phi.
It's independent of what I call T equals zero.
Nature doesn't give a damn what I call T equals zero.
Nature has one answer for the frequency.
That's only determined by K and by M, not by my initial
conditions, not so intuitive. If I take the same spring,
and if I hang the spring vertically, there is the spring,
due to gravity the object will come to a halt equilibrium a
little lower, obviously.
If now I displace it from this equilibrium position and I let
it oscillate, I get exactly the same
frequency. Maybe that's not so intuitive
either. And, you can work that out for
yourself. It's an 8.01 problem.
What it means is that you can define this as X equals zero,
ignore gravity completely, and set up your differential
equation as if there was no gravity.
And this x equals zero. So, you offset it over a
distance, X, from that equilibrium position.
You only allow for a spring force minus KX,
and everything works. And, of course,
you should be able to prove that that is correct.
If a spring oscillates in the simple harmonic fashion,
and we have such a spring here, Marcos, if you can do me a
favor and get it up here, then I should be able to
demonstrate that a uniform circular motion, projected on
the wall, we call shadow projection,
should be able, thank you Marcos,
to be able to have the same motion as my spring,
provided, of course, that we very carefully make a
period of oscillation of the spring exactly the same as the
time for this object to go around.
We then shadow projected on there, and then I will even try
to release this one, this is very difficult,
at the same time that this one is here.
And what you'll see, then, you will see the uniform
circular motion projected, becomes the simple harmonic
motion, and you'll see the spring simple harmonic.
And so, we'll try to do that, and shadow projection will make
it a little darker.
And for that, I need some light here.
OK, and somebody already turned it off.
So here, you see the spring. In there, you see this object
which is rotating in a circle. But you think it's simple
harmonic motion. And that's, of course,
my objective. And so, now that is difficult.
I will have to block you for a few seconds.
I will try now to release this at the same time,
and also the same amplitude. Boy, that wasn't my best day,
was it? No.
Oh, this is perhaps the best I can do today.
So, they don't go exactly next to each other,
but you see they have the same periods, and they both represent
simple harmonic oscillation, the spring, because we just
calculated that. And, the projection of the
uniform circular motion.
So, if we return to the spring, and maybe we should remove
this, if we return to the spring, then we have a period
for the spring system which is 2 pi times the square root of M
over K. And, I want to bring this to a
test, to a quantitative test. How accurate is this?
I'm going to double the mass that I'm going to hang on that
spring. We're going to measure the
periods. And then, I want the mass,
which is twice as high. I want that period to be the
square root of two times higher because that's what this
equation predicts. Now, whenever you want to do a
measurement in physics whereby you want to compare numbers,
so you have a certain goal in mind, a measurement without the
uncertainty in the measurements is completely meaningless.
You must know the accuracy of your measurement.
So, M1 is 500 plus or minus 0.2 g.
And, M2 is thousands plus or minus 0.2 g.
That's the best we can do. That's an extremely small
error. This is an error of only 0.04%.
And this is only half as large. So, now comes the question,
if I measure the periods of oscillation with the 500 g
hanging on the spring, how accurately can I do that?
On a good day, I can do it to 0.1 seconds
accuracy. I have to start it,
and I had to stop it. And if I do that ten times,
obviously you'll get different answers.
And, they vary by about a tenth of a second on a good day.
Now, on a bad day, 2/10 of a second.
I don't know whether today is a good day or whether it's a bad
day, but lets it's in-between. So, let's say I can do to 0.15
seconds, which I cannot guarantee you,
but I'll try. So, I can measure the periods
to plus or minus 0.15 seconds. This is with M1.
However, I can get a very accurate measurement for time,
for the period, if I oscillate if I make 10
oscillations because if I make 10 oscillations,
the error in T goes down by a factor of ten because the 0.15
is 0.15. That's not going to change.
So, I'm going to oscillate it ten times.
And then, we're going to make a prediction about what we should
measure for the higher mass. So, let's first measure the
period of, this is the spring, and here is the 500 g plus or
minus 0.2. I'm going to oscillate it.
We already know that it's independent of amplitude.
And then, I'm going to start it when it's down.
That's the easiest for me. And, I will count to ten.
You will count to ten, and then we'll stop.
So, let's give it an oscillation.
Yeah, one, two, three, four,
five, six, seven, eight, nine,
ten. 14.96.
Let's write this down. We have 14.96.
If you want to see whether this is a good day or whether this is
a bad day, measure it again.
One, two, three, four, five, six,
seven, eight, nine, ten.
14.98. So, this is not a bad day.
But, it's luck that I comes out so close, of course.
So, now, we can make a prediction that ten times TM2
must be 1.414, which is the square root of two
times 14.97. OK, so I take the 1.414.
I multiply that by 14.97, and I find 21.17.
And of course, process to be multiplied also
by 1.4, so that becomes plus or minus 0.2 seconds.
That is a prediction. This is predict.
And now comes the observation. This is a thrilling moment for
you because what is at stake is the integrity of physics.
And, this is going to be measured plus or minus 0.15,
right? Every time I make a
measurement, plus or minus 0.15. I'm nervous.
So, I'm going to add 500 g. Period will indeed increase.
I'm going to oscillate it.
Yeah. One, two, three,
four, five, six, seven, eight,
nine, yeah. Oh, boy.
Oh, boy, what have I done? [LAUGHTER]
What have I done? What have I done? What is it?
20.52. We have a problem.
Physics is not working. Do any one of you have an
idea whether there's something wrong with the equation,
or whether there's something wrong with Walter Lewin?
Any idea? Come on, give it a try,
the worst case your suggestion is not correct,
yeah? Ah, you accuse me,
right? [LAUGHTER] What's your name?
[LAUGHTER] Questioning my 0.15. You say, man,
you couldn't even do better than 0.4 seconds,
maybe. And then, of course,
the two would be consistent with each other.
Thank you, very nice of you. Yeah?
OK, now, that's a very good suggestion.
In other words, you begin to think like a
physicist. Now, you also thought like a
physicist because, indeed, if my uncertainty is
higher than 0.15, you could be right.
Friction in this case, we will deal with friction
later in the course, has such a negligibly small
effect that it couldn't be measured to either one of these
two. In any case,
it's almost the same for both because the shape has not
changed much. It's a very good suggestion;
friction doesn't come near the proper explanation,
but you tried. And that's good.
One more try. Mass of the spring:
we have said earlier; you may not have heard it,
but I did say it, we can replay the tape,
I can prove it to you; I said if the mass of the
spring is negligible, then that is the equation.
Now, what do we do when the mass cannot be ignored?
That's not so easy. But, I requested some mandatory
reading, and I'm sure all of you have done that before this
lecture, and the mandatory reading was French,
p 60-61 among others. And, French says that the mass
of the spring itself is capital M, and if capital M divided by
3 is substantially less than the mass at the end of the
spring, then a very, very good approximation is that
the period of oscillation is, then, this.
And, he actually derives it [SOUND OFF/THEN ON] period is
higher. So, we can bring this to the
test now. In other words,
the mass of the spring, we have weighed that.
In our case it's 175.6 plus or minus 0.2 g.
And so, M divided by three is 58.5 plus or minus 0.07 g.
That's a very small error, by the way.
It's a 0.1% error. And so, we can now do the
following test. We can now take the ratio of
these two and eliminate, thereby, K.
So, we can write ten T. M2 divided by ten T M1 is now
the square root of M2 plus M divided by three divided by M1
plus M divided by three. And, that number is easy to
calculate because M2, you know M1,
you know these numbers, and I have calculated it for
you. And, it is 1.377.
And, your uncertainty is so small compared to my tiny
uncertainty that I don't even have to allow for any
uncertainty in that number because, remember,
the uncertainties in these masses was of the order of 0.1%.
Compare that with the uncertainty in the observations
of the time which were closer to 1%.
So, we can bring this now to a test.
And all I must do now is multiply if I want to find out
ten, TM2. And, I take 1.377,
and I multiply it by TM1, times ten TM1.
So, I take this number. And now, I'm really getting
nervous, not joking. 14.97 multiplied by 1.377,
that is 20.61. And, the uncertainty would be
the same uncertainty as in there which is a 1% uncertainty.
So, that is 0.2 seconds. This number you can now compare
with this number on the button. Within the error of
measurements, they now agree.
This is what we observed, and this is what we predict if
we apply the proper relation and take the mass of the spring into
account. So, you see that physics works,
except that this equation was too simple to be used for our
observation. Notice, by the way,
that this 1.414, in our case,
is lower. All right, in 8.03,
we will often, though not always,
use complex notation. And, the reason why we do that
is that it can at times simplify your life.
And you are completely free to choose when you want to use it
and when you don't want to use it.
You can be the judge. So, let's talk a little bit
about complex numbers. I start with a circle.
And this is the complex plane. The blackboard is a complex
plane. That's quite a promotion for
the blackboard. And here, I call this axis the
real axis. So, all the real numbers are on
this axis. And, let this be plus one,
let this be minus one, and I call this axis the
imaginary axis. So, this one is plus J,
and this one is minus J. And, J is the square root of
minus one. We don't call it I in general
because I is, stands for the current, so we pick J.
I now pick a position here, which now represents a complex
number called this angle, theta.
And I protect this. This is position Z,
complex number. This is the real part of that
complex number, and this is the imaginary part
of a complex number. So, you can see that,
indeed, Z can be written since this length is one,
is the cosine of theta equals J times the sine of theta.
So, does this part, which is real.
And, this is the sine of theta because this is one.
I have to multiply that by J. And this now,
according to Euler, great mathematician Euler,
after whom this disk was also mentioned, already in 1748,
he proved that this was the same as E to the power of Jphi,
sorry, theta. This equality is mind-boggling.
And, when I saw this equality for the first time,
I didn't believe it, number one, and I could hardly
sleep at night because I couldn't prove it.
See, I hadn't had any Taylor expansion yet.
So, I couldn't prove it. Because my teacher in high
school said, this is the case and I said, why?
He said this is the way it is. But we now can prove this.
You can do the Taylor expansion of the cosine theta,
Taylor expansion of the sine theta, and the Taylor expansion
of E to the power J theta, and it's exactly correct,
not an approximation. So, why would we ever want to
use this? Well, if you make this thing go
around, going back to my uniform circular motion here,
and if I make that point go around, and I only look at the
real part, I have a simple harmonic motion.
And so, if I change theta into omega T, then I get that Z
equals the cosine of omega T plus J times the sine of omega
T, the real part of which is a simple harmonic motion.
And, of course, I'm not stuck to an amplitude
of one. I can easily make the amplitude
A times larger. And, of course,
there's nothing wrong depending upon my initial conditions to
have here a phase angle, phi.
And, this, then, is A times E to the power J
omega T plus phi according to Euler.
So, what that means is that if you use this as your trial
function to solve a differential equation, and you can manipulate
this very easily, you could take first
derivative, second derivative of exponentials extremely easy,
and then when you're done, you take the real part of Z,
and out pops X as a function of time.
And, you're done. As I said, it's up to you when
you want to use it next lecture. I will give you an example that
it's clearly the way to go. I wouldn't even know how to do
it any other way. But, often you do have a
choice. So, we are interested,
then, in the real part of that, which is then our acceptable
solution. So, if we have a complex
number, Z equals A plus JB, but we should always be able to
write it as an the torqueamplitude times E to the power J theta.
And then, the amplitude, A, is the square root of A
squared plus B squared and tangent of theta is B over A.
That follows immediately from that figure.
And so, in problem set one you will get some chance to
practice. It will give you a few
interesting cases. And a classic case that all of
you in your lifetime have to be able to do once is the very
non-intuitive problem J to the power J.
When I saw for the first time J to the power J,
I said to myself, well, what on Earth can be more
complex than J to the power J? But, it's real.
It is not complex. And, you will wrestle with
this. There is an infinite number of
solutions. Not one, all of them are
correct. And, I will help you little
because the first time I want to be nice to you.
But it's only the first time. I can also write J as E to the
power J times pi over two; do you agree?
Because it simply means that the angle theta is pi over two.
It's here, so I end up here. But that's J.
I'm not saying it is a very nice way of expressing J,
but it is J. But not only is this J,
I can also rotate an integer number times 360°
whereby N zero, one, two, three rotate either
clockwise or counterclockwise. And, it's again J because if I
rotate 90°, it's J. But, if I rotate another 360°,
it's again J, or if I rotate back 360°.
And so, you see that this is also a way to write J.
And, that will help you, believe me.
I will always have a five-minute break during this 85
minute lecture so that you can stretch your legs.
If you can manage to make it back and forth to the bathroom,
that's fine but that's your problem.
I will start exactly after five minutes.
However, every Tuesday, during part of these five
minutes we will have a mini quiz.
It's really mini, this small.
And we will collect it after the lecture, and you'll even get
some credit for that. Before, that's only on
Tuesdays, but not today. Before we go into this
five-minute break today, I want you to see something so
that you have something to think about.
Believe me, it's healthy for an MIT student to sleep,
but is also healthy sometimes to not sleep.
Sleepless nights and worry, just the way that I had
sleepless nights in high school about Euler's equation:
it's healthy. The reason why that's healthy
is because once you see the solution, you say,
ah, of course, and you never forget it,
whereas if someone tells you from the start you say,
yeah, of course. And, you forget it.
And the next day you don't remember.
So, what I want you to see is a remarkable example of an
oscillation that can be produced not by wind, as we have seen,
but by heat and by cooling. I have here a nice pipe,
and there is a grid here. I can touch it.
I'm touching it now. That's all there is.
It's an open pipe, and there's a grid here.
And, when I heat that grid and cool it, somehow it generates
110 Hz oscillation, a pressure wave,
which you will be able to hear. And, I'll give you until the
end of December, maybe mid-December,
to come up with a solution why it's doing that.
I'm heating the grid now.
110 Hz, roughly. If you want to play with this,
don't break it. Try to transfer the liquid in
17 seconds. I will start.
I will resume this lecture exactly 5 minutes from now.
[SOUND OFF/THEN ON] If you turn this in a tornado,
you rotate it; then you open up a funnel of
air. And so it is never the problem
that the liquid cannot go through.
As always, pressure equilibrium, and I don't
remember how long it takes, but I thought it was 17
seconds. But, if you want to,
we can time that. It may even be less.
I now want to address the issue of simple harmonic oscillation
of a pendulum. As you will remember from 8.01,
if you have a pendulum length L, mass M, and if the mass of
the string is negligibly small compared to the mass that is
hanging here, then a period of oscillations
is two pi times the square root of L over G.
G in the Boston area being to a high degree of accuracy 9.80 m
per second squared. If you simply take L
approximately 1 m, then you can see that you get a
period of about two seconds. And, if you make the length
about 25 cm, that is four times shorter than you would expect
this period, which is two times shorter, which is about one
second. And, without any pretense of
accuracy, just eyeballing, not really testing if I just
eyeball this to be about a meter.
And, if I oscillate this back and forth, it's about two
seconds for one oscillation: one, two, one,
two. If I, however,
make a 25 cm four times shorter, then it is very close
to one second. No interference here:
one, one, one, one, one, one.
Remarkable when you look at this equation is that just like
in the case of the spring, it is independent of the
amplitude. In other words,
whether I have a large amplitude or a small amplitude,
it would take the same amount of time to go back and forth.
Well, not quite for a pendulum. When we derive this period,
you remember that you have to assume what we call small angle
approximations. You'll see that again and again
with 8.03 called small angle approximations.
With small angle approximations,
the final theta is always the sames as theta in radians.
Now, if you ask me how small is small, that's a matter of taste.
In 26-100, we have the mother of all pendulums.
5.18 m long, quite impressive.
So, we have a pendulum whose L is 5.1 plus or minus 0.05 m.
We cannot measure it any better than 5 cm because it has to be
under stretch when we measure it.
And then, you have to go all the way to the ceiling,
and all the way down. Marcos does that,
risking his life, and he claims that the best he
can do is 5 cm. We have 31 pounds hanging under
there. We tried during the summer,
believe me. We tried with technicians of
MIT to have that pendulum here. And, one day it looked good.
But finally they said, no we can't do it.
We can't install it here as a safety issue.
So, unfortunately we don't have the mother of all pendulums
here. In 26-100, when I lectured
Newtonian mechanics, I demonstrated that.
That this pendulum produces is extremely close within the error
of measurement what you predict. In other words,
the mass of the string is indeed negligibly small compared
to the mass of the object. We read the string ones.
I don't remember what it was, it was such a small fraction of
M that indeed could be ignored. And so, the prediction,
then, is if you simply put this L in there, Tpredicted
purely on the basis of that simply equation equals 4.57 plus
or minus 0.02 seconds. And, this 0.02 is the result of
the 0.05. There is a 1% error in here,
right? Five out of 518 is 1%.
And so, the error in T is half a percent because it's the
square root. And so, you get a half a
percent error. And, I rounded that off.
So, that is the prediction. And then, I made two
measurements: one at a 5° angle,
and one at a 10° angle. And I did that ten times.
So, 10T at five degrees, and 10T at 10°.
Now, this was in 1999. Those were my good days.
They were my good times, right?
Past is always the good. And so, I then claimed that I
could do this to an accuracy of 0.1 seconds.
I had a lot of courage in those days.
And, I measured the 5°, and what did I find?
Unbelievable: truly unbelievable,
purely lucky. I found exactly that number,
which is, of course, is an accident because my
accuracy was no better than 0.1 seconds, and then I did it at a
10° angle. And then I found this.
And so, I demonstrated that, indeed, five and 10° I still
considered small angles for that approximation.
And, it is within the uncertainty of my measurement
what you expect. Then, I wanted to demonstrate,
which is not so intuitive, that the period is independent
of mass, which is not the case for the spring.
So now, if you change the mass, and you don't change L,
you expect no change in period. And that's what I really wanted
to show you here. But I can't.
And therefore, I have decided to show you what
I did in 1999. If you can show that two minute
version of my video lectures, then you can judge for yourself
to what extent the mass does not influence the [SOUND OFF/THEN
ON]. One of the most remarkable
things I just mentioned to you is that the period of the
oscillations is independent of the mass of the object.
That would mean, if I joined the ball and I
swing down with the ball that you should get that same period.
Or, should you not? I'm asking you a question
before we do this awful experiment.
Would the period come out to be the same or not?
Some of you think is the same. Have you thought about it that
I'm a little bit taller than the object and that therefore maybe
effectively the length of the string has become a little less
if I sit up like this? And, if the lengths of the
string is a little less, the period would be a little
shorter, yeah? Be prepared for that.
On the other hand, well, I'm not quite prepared
for it. I will try to hold my body as
horizontal as I possibly can in order to be at the same level as
the bulb. I will start when I come to a
halt here. There we go.
Now. You count.
(CROWD: One.) This hurts. (CROWD: Two.
Three.) I want to hear you loud.
(CROWD: Four. Five.) Thank you.
(CROWD: Six. Seven.
Eight. [LAUGHTER] Nine.
Ten.) [APPLAUSE] Ten T with Walter Lewin.
45.6 plus or minus 0.1 seconds. Physics works,
I'm telling you. [SOUND OFF/THEN ON] All right,
so I think it was convincing at least for the freshman at
indeed, the period of the pendulum is independent of the
mass provided that you can ignore the mass of the string
itself, which is the case for that pendulum.
Many pendulums, as some we will see in 8.03 are
more complex, more complicated,
then simply a mass-less string with an object at the end.
And, those pendulums we call a physical pendulum.
For instance, I could have this pair of
compasses, and just let it oscillate like this.
That is not just a simple pendulum.
Or, I could have a ruler like this with a pole through here,
and a have a pin, and have it oscillate like
this. But, I can also have it
oscillate here. It's a different period.
If I oscillate it right in the middle, then it doesn't
oscillate at all. So, now comes the question:
how do we deal with that? And most of you must have seen
that in 8.01. But I do want to address that
in quite some detail. So, a physical pendulum,
then, looks like this. This is an object,
and I drill a hole in here, point P.
And, I put a pin in the wall. And it can, without friction it can
oscillate back and forth. And, let the center of mass be
here, position O, and the separation between P
and O is B. And, O is the center of mass.
You can choose P anywhere you want to.
There's no restriction on P. so, you can see that this
pendulum is offset over an angle, theta,
and it will start to oscillate back and forth.
And, the question is, what is the period?
So, clearly, we may put the entire
gravitational force at point O in the center of mass.
So, this is the force acting at that point.
And now comes the question: are there any other forces
acting on this object, or is this the only one?
Because when you study it, you've got to take all forces
into account. Who is happy that we have taken
all forces into account? Raise your hand.
Most of you are getting scared, right?
Who says, no, it has to be least one other
force? And which force is that?
Yeah, where does it act?
What location? Yeah, so there must be somehow
a force at P to hold it up. Otherwise, it would just start
to accelerate down. Now, I'm not even sure that it
is straight up. I doubt that.
It may simply be at a direction.
I don't want to think about that.
But surely, there has to be a force up.
Now, remember, F equals MA.
When you deal with rotation of objects, and this is going to be
rotational, then this equation changes into Tau,
the torque is moment of inertia times alpha whereby
alpha is theta double dot. It's the angular acceleration.
And so, if I pick P as my point of origin, then the torque,
due to this force, does not contribute to my
torque equation because the torque is R cross F.
It's a cross product between the position vector and the
force. And, this is the position
vector to a center of mass, and the position vector from P
to P is zero. So, if we deal with the torque
relative to point P, that force is of no
consequence. So, I'm going to take P as my
origin. And so, now is the question,
what is the torque relative to point P?
Well, it's R cross F. R is this distance which is B.
F is MG. But, I have a cross product,
so I have to take the sign of this angle into account.
So, that is the magnitude of the torque.
And, the magnitude of that torque, then,
according to my rotational equivalent of F equals MA equals
the moment of inertia for rotation about that point,
P, times theta double dot. However, it is a restoring
torque. The torque, you can do that
with your right hand, whatever way you've learned how
to do that. The torque is in the blackboard
perpendicular to the blackboard in the blackboard.
R cross F is in the blackboard. I have rotated it
counterclockwise, which is a vector out of the
blackboard. So, one is like this and the
others like this. That is like saying the torque
is restoring. Same reason why we wrote down F
equals minus KX with the spring is why we now write this equals
minus this. Take into account the direction
of the vectors. And so, this is the
differential equation that you would have to solve and,
if now we go to small angle approximation,
then the sine of theta goes to theta if theta is in radians.
And so, I can rewrite this theta double dot plus BMG
divided by the moment of inertia about point P times theta equals
zero. And, now, small angle
approximation, we have a differential equation
which is, again, a piece of cake,
simple harmonic oscillation. And so, the simple harmonic
oscillation, the solution must be that theta is some maximum
angle theta zero times the cosine of omega T plus phi.
But this omega is the square root of this number just like we
earlier had the square root of K over M, omega now must be this.
Omega is the square root of BMG divided by I of P.
That means T. The period of oscillation is
the moment of inertia about point P divided by BMG.
I want to repeat what I said earlier.
This omega is called angular frequency.
Angular frequency is a given. That's the angular frequency.
Do not confuse that with theta dot, which we also call omega,
which is called angular velocity.
The angular velocity in this case is a strong function of
time. When the object comes to a
halt, the angular velocity is zero because theta dot is zero.
It is unfortunate that we give them the same symbol.
So, this is independent of time.
But, theta dot does depend on time.
And, theta dot is the angular velocity.
And, in the case of the uniform circular motion,
the two omegas are the same. So now, we have all the
ingredients in hand to calculate for absurd looking objects of
what the period of oscillation is, provided that we are able to
calculate the moment of inertia about the point of rotation.
And of course, we have to know B,
and the mass of the object. You have a wonderful example in
your problem set. I will solve that equation,
calculate this T, for a hoop.
This is the hoop. All the mass is that the
circumference, so it should be very easy to
calculate the moment of inertia. And, we have a hole in here.
And so, we are going to oscillate it right at the rim.
And so, our geometry is easy. But, we should be able to bring
this equation to a rigid test, provided that we take into
account the uncertainty of our measurements.
And so, let me put here this circle, this hoop.
So, all the mass to very good approximation is at the
circumference. And, the oscillation is about
an axis perpendicular to the blackboard, point P.
This is the center of mass O, and I'm going to offset this
hoop. So, this is when it is in
equilibrium. This is offset over an angle,
theta. So, point O is now here.
We'll call it O prime. And so, in analogy with what
we did there, we have here the force,
MG. And, the derivation is
identical. We don't have to go over that
again. And the radius is R.
M is given if you need it, and R is given.
I'll show you what these numbers are later.
So, all I have to do now is go through this equation and
calculate the moment of inertia for rotation of an axis like
this to a point, P.
Who remembers how to do that? 8.01.
Come on, in the worst case it's wrong.
I see one hand there. Who remembers?
Let me ask you this. Suppose you were rotating
through an axis right through the center of mass.
That's difficult because there's nothing to hold onto.
Would you know then what the moment of inertia is?
What is it then? Yeah?
You say yes, but now you're quiet.
OK, moment of inertia is never MR.
It's dimensionally wrong, but you tried which is better
than not trying. Yeah, MR squared is what the
moment of inertia would be if the axis were straight through
O. I'm slowly working you up now.
Now we move the axis from O to P.
What happens now? What do we call that theorem?
Parallel axis theorem. Now, we have to add the mass
times the distance between the center of mass and that point
squared. That's the parallel axis
theorem. And so, the moment of inertia
about point P is M R squared for rotation about this point.
We take the same axis. We move it to P,
and we have to add M distance squared.
So, we have to add plus MR squared.
So, we get 2MR squared. And then we have B.
What is B? What is the distance from P to
the center of mass? That's R.
So, we come now with the prediction that P is 2 pi times
the square root of 2MR squared divided by RMG.
M goes. M always goes with pendulums.
You never have to worry about M if you do it right.
M always goes, not with springs,
but with pendulums. One R goes, and so you get 2 pi
times the square root of 2R over G.
Before we bring these to a test, there is a remarkable
answer. What does it make you think of?
Excuse me? It makes you think of a single
pendulum whereby the length is 2R, which is by no means
obvious, is it? In other words,
if I had pendulum here, and I would hang here in
object, M, that would have the same period because it has a
length, 2R. So, T is 2 pi times the length
divided by G. By no means obvious,
absolutely not clear why that is.
But that's the way it is. So, now comes the acid test.
And so, we don't have to measure the mass.
But we did measure as accurately as we can the radius.
That's really all we have to do.
And, the measurement of the radius is a little uncertain
because it's not a perfect circle.
So, we measured it at various places and we find that R equals
40.0 plus or minus 0.5 cm. But, that's a 1% uncertainty.
And so, we make a prediction now, T.
We get the square root of R, so the 1% uncertainty becomes a
half a percent because of the square root.
You take 2R. You divide it by G,
and you will find that the prediction, this is a
prediction, is that T is 1.795 plus or minus,
that is, your half a percent, 0.01 seconds.
That's because of the square root.
So, this becomes half a percent error.
And now, we do the observation, and you guessed it,
of course. We're going to do 10T.
And, if this is a good day, 0.1.
But, we'll give myself a little bit extra leeway today,
0.15. I'm fairly sure I should be
able to do that. And so, we bring this now to a
test. If you're ready for this,
oh, it's still on. I always like to start the
timer when the object comes to a halt.
That is a better criteria than when it goes through
equilibrium. And I will not look at the,
even if I did look at it as no way I can stop that when I want
to. So, we'll give it an offset.
I first wanted to swing in a way that it's not wobbling
because I've make a very strong prediction.
I want to get that number: 17.95.
So, I better make sure that it's oscillating happily.
No, this is not happy. I don't want any wobbling like
this. Maybe a little more.
OK, I think this looks good. You're ready;
I am ready. One, two, three,
four, five, six, seven, eight,
nine: 17.80. Ah, man, 0.15.
[LAUGHTER] So, it wasn't such a bad day after all
for me OK, see you Tuesday, 981 1:18:04,669 and work on your problem sets.