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LEWIN: You did very well
on your first exam.

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I was hoping
for an average of about 75;

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the class average was 89.

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So that leaves us
with two possibilities:

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either you are very smart,
this is an exceptional class,

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or the exam was too easy.

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Now, this exam was taken
by three instructors

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way before you took it.

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None of them thought
it was too easy,

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so I'd like to think that you
are really an exceptional class

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and I'd like to congratulate you
that you did so well.

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Here is a histogram
of the scores.

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If we had to decide
on this test alone--

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forgetting your quizzes,
forgetting your homework,

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on this test alone--

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the dividing line between
pass and fail would be 65.

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That means that five percent
of the class would fail,

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which is unusually low.

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Normally that is around 15%.

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But time will tell

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whether you are indeed
exceptionally smart

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or whether the exam was
too easy.

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The good news also is--
two pieces of good news--

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that we promise that the books
will arrive at the Coop today.

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Today we're going to talk
about springs, about pendulums

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and about simple
harmonic oscillators--

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one of the key topics in 801.

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If I have a spring...

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and this is the relaxed length
of the string... spring,

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I call that x equals zero.

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And I extend the string...
the spring, with a "p,"

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then there is a force
that wants to drive this spring

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back to equilibrium.

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And it is an experimental fact
that many springs--

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we call them ideal springs--
for many springs,

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this force is proportional
to the displacement, x.

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So if this is x, if you make x
three times larger,

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that restoring force is
three times larger.

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This is
a one-dimensional problem,

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so to avoid the vector notation,
we can simply say

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that the force, therefore,
is minus a certain constant,

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which we call
the spring constant--

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this is called
the spring constant--

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and the spring constant has
units newtons per meter.

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So the minus sign takes care
of the direction.

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When x is positive,

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then the force is
in the negative direction; 

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when F is negative, the force is
in the positive direction.

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It is a restoring force.

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Whenever this linear relation
between F and x holds,

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that is referred to
as Hooke's Law.

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How can we measure
the spring constant?

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That's actually
not too difficult.

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I can use gravity.

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Here is the spring
in its relaxed situation.

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I hang on the spring a mass, m,
and I make use of the fact

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that gravity now exerts
a force on the spring,

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and when you find
your new equilibrium--

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this is the new
equilibrium position--

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then the spring force,
of course,

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must be exactly the same as mg.

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No acceleration
when the thing is at rest.

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And so I could now make a plot

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whereby I could have here x and
I could have here this force F,

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which I know because
I know the masses.

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I can change the masses.

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I can go through
a whole lot of them.

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And you will see data points

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which scatter
around a straight line.

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And the spring constant follows,
then, if you take...

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if you call this delta F
and you call this delta x,

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then the spring constant, k,
is delta F divided by delta x.

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So you can even measure it.

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You don't have
to start necessarily

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at this point
where the spring is relaxed.

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You could already start when the
spring is already under tension.

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That is not a problem.

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You'll be surprised
how many springs

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really behave very nicely
according to Hooke's Law.

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Uh, I have one here.

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It's not
a very expensive spring.

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You see it here.

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And there is here
a holder on to it

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so it's already a little bit
under stress.

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That doesn't make
any difference.

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These marks here are
13 centimeters apart,

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and every time
that I put one kilogram on,

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you will see that it goes down
by roughly 13 centimeters.

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It goes down to this mark.

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I put another kilogram on,
it goes down to this mark.

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I put another kilogram on
and it goes back to this mark,

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all the way down.

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And if I take them all off...
so what I've done is

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I effectively went along this
curve, and if I take them off,

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if it is an ideal spring,

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then it goes back to its
original length, which it does.

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That's a requirement, of course,
for an ideal spring

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if it behaves according
to Hooke's Law.

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Now, I can, of course,
overdo things.

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I can take a spring
like this one and stretch it

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to the point that it no longer
behaves like Hooke's Law.

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I can damage it.

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Uh, I can do
permanent deformation.

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Look, that's easy.

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For sure, Hooke's Law is
no longer active.

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Look how much longer this spring
is than it was before.

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So there comes, of course,
a limit how far you can go

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before you permanently
deform your spring.

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What I have done now
with that spring,

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probably in the beginning
I went up along a straight line

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and then something like this
must have happened.

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I got a huge extension.

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My force did not increase
very much.

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And then when I relaxed,
when I took my force off,

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the spring was longer
at the ends

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than it was at the beginning.

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So I have a net extension
which will always be there,

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and that's not very nice, of
course, to do that to a spring.

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So Hooke's Law holds
only within certain limitations.

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You have to, uh, obey
a certain amount of discipline.

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There are ways that you can also
measure the spring constant

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in a dynamic way, which is
actually very interesting.

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Um, I have here a spring,
and this spring...

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this is x equals 0, and I attach
now to the spring a mass, m.

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This has to be
on a frictionless surface,

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and you will see, when
I extend it over a distance x,

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that you get your force,

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your spring force
that drives it back.

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We have, of course, gravity, mg,

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and we have the normal force
from the surface.

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So there is in the...
in the y direction,

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there is no acceleration,
so I don't have to worry

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about the forces
in the y direction at all.

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If I let this thing oscillate,
I let... I release it,

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it will start to oscillate about
this point, back and forth,

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then as I will show you now,

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you will find
that the period of oscillation,

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the time for
one whole oscillation

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is 2 pi times the square root
of the mass m

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divided by
the spring constant k.

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I will derive that--
you will see that shortly.

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In other words,
if you measured the period

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and you knew the math,
then you can calculate k.

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Alternatively, if you knew k
and you measure the period,

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you can calculate the mass,
even in the absence of gravity.

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I don't use gravity here.

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So a spring always allows you
to measure, uh, a mass

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even in the absence of gravity.

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The period that you see,
the time that it takes

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for this object to oscillate
once back and forth,

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is completely independent
of how far I move it out,

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which is very nonintuitive,

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but you will see that that
comes out of the derivation.

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There is no dependence
on how far I move it out.

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So whether I oscillate it
like this

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or whether I oscillate it
like this,

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as long as Hooke's Law holds,
you will see

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that the period is independent
of what we call that amplitude.

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So I'm going to derive the
situation now for an ideal case.

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Ideal case means
Hooke's Law must hold.

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There's no friction,
and the spring itself
 

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has negligible mass
compared to this one.

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Let's call it a massless spring.

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So now I'm going to write down
Newton's Second Law:

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ma, which is all in the
x direction, equals minus kx.

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a is the second derivative
of position,

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for which I will write
x double-dot, mx double-dot--

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one dot is the first derivative,
that's the velocity; 

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two dots is the acceleration--
plus kx equals zero.

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I divide by m

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and I get x double-dot plus
k over m times x equals zero.

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And this is arguably

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the most important equation
in all of physics.

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It's a differential equation.

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Some of you may already have
solved differential equations.

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The outcome of this,
you will see, is very simple.

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x is, of course, changing in
some way as a function of time,

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and when you have
the correct solution

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for x as a function of time

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and you substitute that back
into that differential equation,

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that equation will have
to be satisfied.

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What would a solution be
to this differential equation?

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I'm going to make you see
this oscillation first.

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I'm going to make you see x
as a function,

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and I'm going to do that
in the following way.

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I have here a spray paint can

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which is suspended
between two springs,

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and I can oscillate it
vertically,

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which is your x direction,
like that.

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So x changes with time.

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The time axis I will introduce
by pulling on the string.

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When the spray paint
is going to spray,

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I'm going to pull
on that string,
 

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and if I can do that
at a constant speed,

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then you get horizontally a time
axis and of course vertally...

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vertically you will get
the position of x.

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So I want you
to just see qualitatively

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what kind of a weird curve
x as a function of time is,

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which then will have to satisfy
that differential equation.

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All right.

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It's always a messy experiment
because the paint is dripping,

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but I will try to get
the spray paint going.

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There we go.

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Okay.

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Now I'll pull...

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All right.

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Will you give me a hand?

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00:11:31 --> 00:11:37

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Yeah, could you, please?

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I will cut it here
and then you...

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Be very careful,
because it's... it is messy.

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Oh, let's put it...
let's take it out this way.

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Aah...

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Okay, just walk back.

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Just walk-- yeah, great.

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I'll hold up the top
so that we can see it fine.

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Okay.

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What is... what does it
remind you of?

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(student responds )

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LEWIN:
Sinusoids-- reminds me
of a cosinusoid, by the way.

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Sinusoid or a cosine--
same thing.

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All right.

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Let's try to substitute
in that equation

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a sinusoid's
or a cosine solution,

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whichever one you prefer.

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Makes no difference.

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So I'm going to substitute
in this equation--

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this is my trial function--

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that x as a function of time
is a constant, A--

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I will get back to that
in a minute--

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times cosine omega t plus phi.

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This A we call the amplitude.

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Notice the cosine function is...
the highest value is plus one

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and the lowest value is minus
one, so the amplitude indicates

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that is... the farthest
displacement from zero

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on this side would be plus A and
on this side would be minus A.

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So that's in meters.

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This omega we call
the angular frequency.

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Don't confuse it
with angular velocity.

247
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We call it angular frequency,
and the units are the same.

248
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The units are
in radians per second,

249
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the same as angular velocity.

250
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If I advance
this time little t--

251
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if I advance that
by 2 pi divided by omega...

252
00:13:31 --> 00:13:37
if I advance this time
by 2 pi divided by omega,

253
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then this angle here increases
here by 2 pi radians,

254
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which is 360 degrees, and so
that's the time that it takes

255
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for the oscillation
to repeat itself.

256
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So this is the period
of the oscillation,

257
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and that is in seconds.

258
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And you can determine...
if you want to,

259
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you can define the frequency
of the oscillations

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which is one over T,

261
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which we express always
in terms of hertz.

262
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And then here we have
what we call the phase angle,

263
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and I will return to that.

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That's in radians.

265
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And this trial function
I'm going to substitute now

266
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into this equation.

267
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So the first thing I have to do,
I have to find

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what the second derivative is
of x as a function of time.

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Well, that's my function.

270
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I have here first
the first derivative, x dot.

271
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That becomes minus A omega.

272
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I get an omega out because
there's a time here, and now

273
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I have to take the derivative
of the function itself

274
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so I get the sine
of omega t plus phi.

275
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Of course I could have started
off here with a sine curve; 

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I hope you realize that.

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I just picked the cosine one.

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x double-dot.

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Now I get another omega out,
so I get minus A omega squared.

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The derivative of the sine
is the cosine.

281
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Cosine omega t plus phi.

282
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And that is also
minus omega squared times x,

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because notice I have
A cosine omega t plus phi,

284
00:15:15 --> 00:15:21
which itself is x.

285
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So now I'm ready
to substitute this result

286
00:15:20 --> 00:15:26
into that differential equation.

287
00:15:23 --> 00:15:29
This must always hold
for any value of x,

288
00:15:27 --> 00:15:33
for any moment in time.

289
00:15:29 --> 00:15:35
And therefore,

290
00:15:30 --> 00:15:36
the only way that this can work
is if omega squared is k over m.

291
00:15:34 --> 00:15:40
So omega squared must be
k over m.

292
00:15:38 --> 00:15:44
And therefore, we now have
the solution to this problem.

293
00:15:43 --> 00:15:49
So we have omega equals
the square root of k over m

294
00:15:49 --> 00:15:55
and the period is 2 pi times
the square root of m over k.

295
00:15:55 --> 00:16:01
And what is striking,
really remarkable,

296
00:15:59 --> 00:16:05
that this is independent
of the amplitude,

297
00:16:02 --> 00:16:08
and it's also independent

298
00:16:04 --> 00:16:10
of this angle phi,
this phase angle.

299
00:16:06 --> 00:16:12
What is this business
of this phase angle?

300
00:16:09 --> 00:16:15
It's a peculiar thing
that we have there.

301
00:16:11 --> 00:16:17
Well, you can think
about the physics, actually.

302
00:16:14 --> 00:16:20
When I start this oscillation,
I have a choice of two things.

303
00:16:19 --> 00:16:25
I can start it off at a certain
position which I can choose.

304
00:16:23 --> 00:16:29
I can give it a certain
displacement from zero

305
00:16:26 --> 00:16:32
and simply let it go.

306
00:16:27 --> 00:16:33
But I can also, when I let it
go, give it a certain velocity.

307
00:16:31 --> 00:16:37
That's my choice.

308
00:16:32 --> 00:16:38
So I have two choices:

309
00:16:33 --> 00:16:39
where I let it go
and what velocity I give it.

310
00:16:36 --> 00:16:42
And that is reflected
in my solution:

311
00:16:39 --> 00:16:45
namely, that ultimately
in the solution

312
00:16:41 --> 00:16:47
I got the result of A
and the result of phi,

313
00:16:44 --> 00:16:50
which doesn't determine
the period,

314
00:16:46 --> 00:16:52
but it results from what we call
my initial conditions.

315
00:16:51 --> 00:16:57
And I want to do an example
whereby you see how A and phi

316
00:16:56 --> 00:17:02
immediately follow
from the initial conditions.

317
00:16:59 --> 00:17:05
So in this example,

318
00:17:01 --> 00:17:07
I release the object at
x equals zero and t equals zero.

319
00:17:06 --> 00:17:12
So I release it
at the equilibrium.

320
00:17:09 --> 00:17:15
At that moment in time,

321
00:17:11 --> 00:17:17
I give it a velocity, which is
minus three meters per second.

322
00:17:15 --> 00:17:21
My units are always
in MKS units.

323
00:17:18 --> 00:17:24
The spring constant k equals
ten newtons per meter,

324
00:17:22 --> 00:17:28
and the mass of the object is
0.1 kilograms.

325
00:17:26 --> 00:17:32
And now I can ask you what now
is x as a function of time,

326
00:17:30 --> 00:17:36
including the amplitude A,
including the phase angle phi?

327
00:17:34 --> 00:17:40
Well, let's first
calculate omega.

328
00:17:38 --> 00:17:44
That is
the square root of k over m.

329
00:17:42 --> 00:17:48
That would be
ten radians per second.

330
00:17:46 --> 00:17:52
The period T, which is
2 pi divided by omega,

331
00:17:53 --> 00:17:59
would be roughly 6.28 seconds.

332
00:17:57 --> 00:18:03
And the frequency f would be
about 0.16 hertz,

333
00:18:04 --> 00:18:10
just to get some numbers.

334
00:18:07 --> 00:18:13
1.6 hertz-- sorry.

335
00:18:08 --> 00:18:14

336
00:18:10 --> 00:18:16
This is not my day.

337
00:18:12 --> 00:18:18
This is 0.628,
and this is 1.6 hertz.

338
00:18:20 --> 00:18:26
2 pi divided by omega; 
you can see this is ten.

339
00:18:24 --> 00:18:30
Six divided by ten is about 0.6.

340
00:18:27 --> 00:18:33
All right, so now I know that
at t equals zero, x equals zero.

341
00:18:33 --> 00:18:39
So I see my solution
right there.

342
00:18:36 --> 00:18:42
Right here I put in
t equals zero,

343
00:18:39 --> 00:18:45
and I know that x is zero.

344
00:18:41 --> 00:18:47
So I get zero equals
A times the cosine of phi.

345
00:18:47 --> 00:18:53
Well, A is not zero.

346
00:18:48 --> 00:18:54
If I release that thing
at equilibrium

347
00:18:50 --> 00:18:56
and I give it a velocity
of three meters per second,

348
00:18:53 --> 00:18:59
it's going to oscillate.

349
00:18:55 --> 00:19:01
So A is not zero.

350
00:18:56 --> 00:19:02
So the only solution is
that cosine phi is zero,

351
00:19:00 --> 00:19:06
and so that leaves me
with phi is pi over two,

352
00:19:03 --> 00:19:09
or phi is 3 pi...
phi is 3 pi over two.

353
00:19:06 --> 00:19:12
That's the only
two possibilities.

354
00:19:09 --> 00:19:15
Now I go to my
next initial condition,

355
00:19:11 --> 00:19:17
that the velocity is
minus 3.

356
00:19:14 --> 00:19:20
Now, here you see
the equation for the velocity.

357
00:19:17 --> 00:19:23
This is minus 3
at t equals zero.

358
00:19:21 --> 00:19:27
So minus 3 equals minus A,
and A is... we don't know yet.

359
00:19:27 --> 00:19:33
Minus A, and then
we have omega squared.

360
00:19:31 --> 00:19:37
Omega-- sorry-- which is ten.

361
00:19:34 --> 00:19:40
t is zero.

362
00:19:36 --> 00:19:42
I get the sine of phi.

363
00:19:38 --> 00:19:44

364
00:19:40 --> 00:19:46
If I pick pi over 2,
then the sine of phi is 1.

365
00:19:45 --> 00:19:51
And so you find immediately
that A equals plus 0.3

366
00:19:50 --> 00:19:56
And so the solution now,
which includes now phi and A,

367
00:19:56 --> 00:20:02
is that x equals plus 0.3 times
the cosine of omega,

368
00:20:02 --> 00:20:08
which is 10t plus pi over 2.

369
00:20:06 --> 00:20:12
So you see that
the initial conditions...

370
00:20:09 --> 00:20:15
what the conditions are
at t equals zero,

371
00:20:11 --> 00:20:17
they determine my A and
they determine my phase angle.

372
00:20:15 --> 00:20:21
If you had chosen this as
the phase angle-- 3 pi over 2--

373
00:20:20 --> 00:20:26
that would have been fine.

374
00:20:22 --> 00:20:28
You would have found
a minus sign here,

375
00:20:25 --> 00:20:31
and that's exactly the same.

376
00:20:26 --> 00:20:32
So you would have found
nothing different.

377
00:20:29 --> 00:20:35

378
00:20:31 --> 00:20:37
I want to demonstrate to you
that the period of oscillations,

379
00:20:36 --> 00:20:42
nonintuitive as that may be,

380
00:20:39 --> 00:20:45
is independent of the amplitude
that I give the object.

381
00:20:44 --> 00:20:50
And I want to do that here
with this air track.

382
00:20:48 --> 00:20:54
I have a... an object here.

383
00:20:50 --> 00:20:56
This object has a mass--
186 plus or minus 1 gram.

384
00:20:58 --> 00:21:04

385
00:21:03 --> 00:21:09
Call it m1.

386
00:21:04 --> 00:21:10

387
00:21:07 --> 00:21:13
I'm going to oscillate it

388
00:21:09 --> 00:21:15
and we're going to measure
the periods.

389
00:21:11 --> 00:21:17
But instead of measuring
one period,

390
00:21:13 --> 00:21:19
I'm going to measure
ten periods,

391
00:21:15 --> 00:21:21
because that gives me
a smaller uncertainty,

392
00:21:18 --> 00:21:24
a smaller relative error
in my measurements.

393
00:21:21 --> 00:21:27
So I'm going to do it
as an amplitude,

394
00:21:24 --> 00:21:30
which is 15 centimeters.

395
00:21:26 --> 00:21:32
Let's make it 20 centimeters.

396
00:21:28 --> 00:21:34
So I get 10T,
I get a certain number,

397
00:21:31 --> 00:21:37
and I get an error
which is my reaction error,

398
00:21:34 --> 00:21:40
which is about
a tenth of a second.

399
00:21:37 --> 00:21:43
That's about the reaction error
that we all have, roughly.

400
00:21:40 --> 00:21:46
Then I will do it
at 40 centimeters.

401
00:21:41 --> 00:21:47
We get a 10T, and we get, again,
plus or minus 0.1 seconds.

402
00:21:48 --> 00:21:54

403
00:21:49 --> 00:21:55
And we'll see
how much they differ.

404
00:21:52 --> 00:21:58
They should be the same,
if this is an ideal spring,

405
00:21:55 --> 00:22:01
within the uncertainty
of my measurements.

406
00:21:58 --> 00:22:04
You see the timing there.

407
00:22:00 --> 00:22:06
I'm going to give it
a 20-centimeter offset,

408
00:22:03 --> 00:22:09
which is here,

409
00:22:04 --> 00:22:10
and then I will start it
when it comes back here.

410
00:22:07 --> 00:22:13
So I will allow it
one oscillation first.

411
00:22:09 --> 00:22:15
That's easier for me to see it
stand still when I start it.

412
00:22:12 --> 00:22:18
There we go.

413
00:22:13 --> 00:22:19

414
00:22:15 --> 00:22:21
One... two... three...
four... five...

415
00:22:23 --> 00:22:29
six... seven...
eight... nine... ten.

416
00:22:31 --> 00:22:37
What do we see?

417
00:22:33 --> 00:22:39
15.16.

418
00:22:38 --> 00:22:44
15.16 seconds.

419
00:22:39 --> 00:22:45
By the way, you can derive the
spring constant from this now

420
00:22:43 --> 00:22:49
because you know the mass
and you know the time.

421
00:22:46 --> 00:22:52
Now I'm going to give it
a displacement, an amplitude

422
00:22:50 --> 00:22:56
which is twice as high.

423
00:22:52 --> 00:22:58
So I make it 40 centimeters.

424
00:22:54 --> 00:23:00
So this is ten.

425
00:22:56 --> 00:23:02
40 centimeters--
a huge displacement.

426
00:23:00 --> 00:23:06
Now... one... two...
three... four...

427
00:23:08 --> 00:23:14
five... six... seven...
eight... nine... ten.

428
00:23:17 --> 00:23:23
15.13.

429
00:23:19 --> 00:23:25

430
00:23:24 --> 00:23:30
Fantastic agreement within the
uncertainty of my measurements.

431
00:23:28 --> 00:23:34
They're within
3/100 of a second.

432
00:23:30 --> 00:23:36
Of course if you try it
many times,

433
00:23:32 --> 00:23:38
you won't always get that close,
because my reaction time

434
00:23:35 --> 00:23:41
is really not much better
than a tenth of a second.

435
00:23:37 --> 00:23:43
Now I will show you something
else which is quite interesting,

436
00:23:42 --> 00:23:48
and that is how the behavior
of the period is on the...

437
00:23:46 --> 00:23:52
on the mass of the object.

438
00:23:49 --> 00:23:55
I have here another car
which weighs roughly the same.

439
00:23:54 --> 00:24:00
Uh, I'm going to add
the two together,

440
00:23:58 --> 00:24:04
and so we get m2 is about
372 plus or minus 1 gram.

441
00:24:02 --> 00:24:08
The plus or minus 1 comes in

442
00:24:04 --> 00:24:10
because our scale is no more
accurate than one gram.

443
00:24:07 --> 00:24:13
So we put them both on the scale

444
00:24:10 --> 00:24:16
and we find this to be
the uncertainty.

445
00:24:12 --> 00:24:18
So now I'm going to measure

446
00:24:15 --> 00:24:21
the ten periods of this object
with mass m2, so twice the mass.

447
00:24:22 --> 00:24:28
So that should be

448
00:24:24 --> 00:24:30
the square root of m2
divided by m1

449
00:24:28 --> 00:24:34
times 10 times the period of m1.

450
00:24:33 --> 00:24:39
And so I can make a prediction

451
00:24:35 --> 00:24:41
because this is the square root
of two, and I know what this is.

452
00:24:39 --> 00:24:45
So I will take my calculator

453
00:24:41 --> 00:24:47
and I will take the square root
of two, and I multiply that

454
00:24:46 --> 00:24:52
by, uh, let's take 15.15.

455
00:24:50 --> 00:24:56

456
00:24:51 --> 00:24:57
And so that comes out
to be 21.42.

457
00:24:57 --> 00:25:03
21.42.

458
00:24:58 --> 00:25:04
It's not clear that
this two is meaningful.

459
00:25:00 --> 00:25:06
And now comes the $64 question:
What is the uncertainty?

460
00:25:04 --> 00:25:10
This is a prediction.

461
00:25:06 --> 00:25:12

462
00:25:09 --> 00:25:15
And this now becomes
a little tricky.

463
00:25:11 --> 00:25:17
So what I'm telling you now
may confuse you a bit.

464
00:25:14 --> 00:25:20
It's not meant to be,

465
00:25:15 --> 00:25:21
but I really won't hold you
responsible for it.

466
00:25:18 --> 00:25:24
You may now think

467
00:25:19 --> 00:25:25
that the uncertainty
in these measurements follows

468
00:25:22 --> 00:25:28
from the uncertainty in this,
which is true,

469
00:25:24 --> 00:25:30
which is about 0.6%, and
from the uncertainty in this,

470
00:25:28 --> 00:25:34
so this has about
an uncertainty of 0.6%.

471
00:25:31 --> 00:25:37
I got it low because I measured
ten oscillations, you see?

472
00:25:35 --> 00:25:41
The uncertainty is only
one out of 150, which is low.

473
00:25:39 --> 00:25:45
You may think
that the uncertainty in there

474
00:25:43 --> 00:25:49
equals the square root
of 372 plus or minus 1

475
00:25:46 --> 00:25:52
divided by 186 plus or minus 1.

476
00:25:49 --> 00:25:55
And now you may argue,

477
00:25:50 --> 00:25:56
and it's completely reasonable
that you would argue that way--

478
00:25:54 --> 00:26:00
you would say,
"Well, this is roughly

479
00:25:56 --> 00:26:02
"a quarter of a percent error
here under the square root

480
00:25:59 --> 00:26:05
and this is roughly
half a percent error."

481
00:26:02 --> 00:26:08
One out of 200 is about half.

482
00:26:03 --> 00:26:09
So you would add up
the two errors--

483
00:26:05 --> 00:26:11
a quarter plus half,
that's about 0.7--

484
00:26:08 --> 00:26:14
and because of the square roots,
that becomes 0.35%,

485
00:26:12 --> 00:26:18
and that's wrong.

486
00:26:14 --> 00:26:20
And the reason why
that is completely wrong--

487
00:26:16 --> 00:26:22
that has to do with the fact

488
00:26:18 --> 00:26:24
that these two errors
are coupled to each other.

489
00:26:21 --> 00:26:27
See, we... the 186 is included
in the 372.

490
00:26:26 --> 00:26:32
The best way
I can show you this--

491
00:26:28 --> 00:26:34
suppose I measured
m1 divided by m1, which would be

492
00:26:31 --> 00:26:37
186 plus or minus 1
divided by 186 plus or minus 1.

493
00:26:37 --> 00:26:43
That number is one
with a hundred zeros.

494
00:26:39 --> 00:26:45
This number is one.

495
00:26:40 --> 00:26:46
You have the mass of one object; 

496
00:26:42 --> 00:26:48
you divide it
by the same object.

497
00:26:43 --> 00:26:49
Whereas if you would say, "Ah,
this is a half a percent error

498
00:26:46 --> 00:26:52
and this is a half a percent
error," you would say

499
00:26:49 --> 00:26:55
the ratio has an error of 1%,
and that's not the case.

500
00:26:51 --> 00:26:57
So I will not bother you
with that.

501
00:26:53 --> 00:26:59
I will not hold you
responsible for that,

502
00:26:55 --> 00:27:01
but it turns out
that if you do it correctly

503
00:26:58 --> 00:27:04
and you take the error of this
into account, of about 0.6%,

504
00:27:02 --> 00:27:08
that the error in this ratio is
really much less than .2%.

505
00:27:07 --> 00:27:13
You can almost forget about it.

506
00:27:09 --> 00:27:15
I will allow, generously, for
a 1% error in the final answer,

507
00:27:13 --> 00:27:19
and so I stick to my prediction

508
00:27:16 --> 00:27:22
that the 10T of double the mass
is going to be like this.

509
00:27:22 --> 00:27:28

510
00:27:24 --> 00:27:30
And now we're going to get
the observation:

511
00:27:28 --> 00:27:34
10T times m2, which is double
the mass, and that, of course,

512
00:27:33 --> 00:27:39
always has my uncertainty
of my reaction time.

513
00:27:38 --> 00:27:44
There's nothing I can do
about that.

514
00:27:40 --> 00:27:46
And we will compare
these two numbers.

515
00:27:42 --> 00:27:48
So I will put the other mass
on top of it.

516
00:27:45 --> 00:27:51
Goes here.

517
00:27:47 --> 00:27:53

518
00:27:48 --> 00:27:54
Tape them together
so that they won't fall off.

519
00:27:52 --> 00:27:58
There we go.

520
00:27:54 --> 00:28:00

521
00:27:56 --> 00:28:02
So, I hope I did that correctly.

522
00:28:00 --> 00:28:06
The square root of two
times 15.15.

523
00:28:03 --> 00:28:09
We'll give it a... amplitude,

524
00:28:05 --> 00:28:11
something like 30,
maybe 35 centimeters.

525
00:28:08 --> 00:28:14
There we go.

526
00:28:09 --> 00:28:15

527
00:28:12 --> 00:28:18
One... two-- much slower, eh?
You see that.

528
00:28:16 --> 00:28:22
Three... four... five... six...

529
00:28:24 --> 00:28:30
seven... eight... nine--
I'm not looking-- ten.

530
00:28:33 --> 00:28:39
21.36.

531
00:28:35 --> 00:28:41

532
00:28:38 --> 00:28:44
21.36.

533
00:28:42 --> 00:28:48
You can round it off
if you want to,

534
00:28:44 --> 00:28:50
and you see that
the agreement is spectacular.

535
00:28:47 --> 00:28:53
Within the uncertainty
of my measurements,

536
00:28:49 --> 00:28:55
it comes out amazingly well.

537
00:28:51 --> 00:28:57
You could have removed
this two, of course,

538
00:28:53 --> 00:28:59
because if you have
an uncertainty of .2 here,

539
00:28:55 --> 00:29:01
it's a little silly to have
that little two hanging there.

540
00:28:58 --> 00:29:04
But you see that indeed

541
00:28:59 --> 00:29:05
this spring is very close
to an ideal spring.

542
00:29:02 --> 00:29:08
It obeys Hooke's Law,
and it is also nearly massless.

543
00:29:07 --> 00:29:13
Here is the pendulum.

544
00:29:10 --> 00:29:16
Here is the mass, and it's
offset at an angle, theta.

545
00:29:15 --> 00:29:21
The length of the pendulum is l,
the length of the string.

546
00:29:20 --> 00:29:26
There is gravity here, mg, and
the other force on the object,

547
00:29:25 --> 00:29:31
the only other force,
is the tension, T.

548
00:29:29 --> 00:29:35
Don't confuse that with
period T; this is tension T.

549
00:29:34 --> 00:29:40
It's in newtons.

550
00:29:35 --> 00:29:41
Those are the only two forces.

551
00:29:37 --> 00:29:43
There is nothing else.

552
00:29:39 --> 00:29:45
The thing is going
to arc around like this

553
00:29:44 --> 00:29:50
and it's going to oscillate.

554
00:29:47 --> 00:29:53
I call this the y direction
and I call this the x direction,

555
00:29:52 --> 00:29:58
and here x equals 0.

556
00:29:57 --> 00:30:03
Well, I'm going to decompose
the tension into the y

557
00:30:03 --> 00:30:09
and into the x direction
as we have done before.

558
00:30:08 --> 00:30:14
So this is going to be
the y component.

559
00:30:12 --> 00:30:18
This is the x component.

560
00:30:15 --> 00:30:21
So this y component equals
T cosine theta

561
00:30:21 --> 00:30:27
and the x component equals
T sine theta.

562
00:30:28 --> 00:30:34
And now I'm going to write down

563
00:30:30 --> 00:30:36
the differential equations
of motion,

564
00:30:33 --> 00:30:39
first in the x direction.

565
00:30:35 --> 00:30:41
Second... Newton's Second Law:
ma equals...

566
00:30:41 --> 00:30:47
this is the only force
in the x direction.

567
00:30:44 --> 00:30:50
It's a restoring force,
just like with the spring.

568
00:30:46 --> 00:30:52
I therefore have to give it
a minus sign.

569
00:30:49 --> 00:30:55
So equals minus T times
the sine of theta.

570
00:30:54 --> 00:31:00
T itself could easily be
a function of theta.

571
00:30:57 --> 00:31:03
So I have to allow for that.

572
00:30:59 --> 00:31:05
The sine of theta equals x

573
00:31:02 --> 00:31:08
if it's here
at position x divided by l,

574
00:31:06 --> 00:31:12
and so I can write for this
minus T--

575
00:31:08 --> 00:31:14
which may be a function of
theta-- times x divided by l.

576
00:31:13 --> 00:31:19
That is my differential equation
in the x direction,

577
00:31:17 --> 00:31:23
and I prefer always for this a
to write down x double-dot.

578
00:31:24 --> 00:31:30
Now the y direction.

579
00:31:26 --> 00:31:32
In the y direction, I have
m y double-dot equals...

580
00:31:32 --> 00:31:38
this is my plus direction, so
I have T cosine theta minus mg.

581
00:31:41 --> 00:31:47
This is equation one
and this is equation two.

582
00:31:44 --> 00:31:50
And so now we have to solve two
coupled differential equations,

583
00:31:49 --> 00:31:55
which is a hopeless task.

584
00:31:51 --> 00:31:57
It looks like a zoo,
and it is a zoo.

585
00:31:54 --> 00:32:00
And now we're going to make
some approximations,

586
00:31:58 --> 00:32:04
and the approximations
that we will make

587
00:32:02 --> 00:32:08
which we will often see
in physics

588
00:32:05 --> 00:32:11
when something oscillates--

589
00:32:07 --> 00:32:13
what we call
the small-angle approximations.

590
00:32:11 --> 00:32:17
Small-angle-- we will not allow
theta to become too large.

591
00:32:16 --> 00:32:22
I'll be quantitative,
what I mean by too large.

592
00:32:19 --> 00:32:25
When theta, which is in radians,
equals much, much less than one,

593
00:32:24 --> 00:32:30
we call that a small angle.

594
00:32:27 --> 00:32:33
If that's the case, the cosine
of theta is very close to 1.

595
00:32:32 --> 00:32:38
You will say, "Well, blah,
blah, blah-- how close to 1?"

596
00:32:37 --> 00:32:43
Okay, five degrees--
the cosine is 0.996.

597
00:32:43 --> 00:32:49
That's close to 1.

598
00:32:45 --> 00:32:51
Ten degrees--
the cosine is 0.985.

599
00:32:50 --> 00:32:56
That's only 1½% different
from one.

600
00:32:53 --> 00:32:59
So even at ten degrees,
you're doing extremely well.

601
00:32:57 --> 00:33:03
So, this is
consequence number one

602
00:33:00 --> 00:33:06
of the small-angle
approximation.

603
00:33:03 --> 00:33:09
But there is
a second consequence

604
00:33:05 --> 00:33:11
of the small-angle
approximation.

605
00:33:08 --> 00:33:14
Look at the excursion
that this object made

606
00:33:11 --> 00:33:17
from equilibrium
in the x direction.

607
00:33:13 --> 00:33:19
That's this big.

608
00:33:15 --> 00:33:21
Look at the excursion it makes
in the y direction.

609
00:33:18 --> 00:33:24
It's this small.

610
00:33:20 --> 00:33:26
It's way smaller than the
excursion in the x direction,

611
00:33:24 --> 00:33:30
provided that
your angle is small.

612
00:33:27 --> 00:33:33
I'll give you an example.

613
00:33:29 --> 00:33:35
At five degrees, this excursion
is only 4% of this excursion.

614
00:33:36 --> 00:33:42
At ten degrees, this excursion
is only 9% of this excursion.

615
00:33:40 --> 00:33:46
And since the excursion
in the y direction

616
00:33:43 --> 00:33:49
is so much smaller
than in the x direction,

617
00:33:46 --> 00:33:52
we say that the acceleration
in the y direction

618
00:33:50 --> 00:33:56
can be approximated
to be roughly zero.

619
00:33:53 --> 00:33:59
There is almost no acceleration
in the y direction.

620
00:33:58 --> 00:34:04
With these two conclusions,

621
00:34:00 --> 00:34:06
which follow from
the small-angle approximation,

622
00:34:03 --> 00:34:09
I go back to my equation
number two, and I find

623
00:34:08 --> 00:34:14
that zero equals T, which
could be a function of theta.

624
00:34:13 --> 00:34:19
The cosine of theta is
one minus mg.

625
00:34:17 --> 00:34:23
So I find that T equals mg.

626
00:34:20 --> 00:34:26
Notice it's no longer
even a function of theta.

627
00:34:24 --> 00:34:30
So I simply have,
in my small-angle approximation,

628
00:34:29 --> 00:34:35
that I can make T
the same as mg.

629
00:34:32 --> 00:34:38
It's approximately, but I still
put an equals sign there.

630
00:34:35 --> 00:34:41
I substitute this back
in my equation number one.

631
00:34:39 --> 00:34:45
And so now I get
that m times x double-dot--

632
00:34:43 --> 00:34:49
and now I bring this
on the other side--

633
00:34:47 --> 00:34:53
plus-- T is now mg--

634
00:34:50 --> 00:34:56
mg times x divided by l
equals zero.

635
00:34:57 --> 00:35:03
And now comes
the wonderful result:

636
00:35:01 --> 00:35:07
x double-dot plus g over l
times x equals zero.

637
00:35:06 --> 00:35:12
And this is
such a beautiful result

638
00:35:09 --> 00:35:15
that it almost makes me cry.

639
00:35:11 --> 00:35:17
This is a simple
harmonic oscillation.

640
00:35:16 --> 00:35:22
This equation looks

641
00:35:18 --> 00:35:24
like a carbon copy
of the one that we have there.

642
00:35:23 --> 00:35:29
Here we have k over m,
and there we have g over l.

643
00:35:28 --> 00:35:34
That's all.

644
00:35:29 --> 00:35:35
Other than that,
there is no difference.

645
00:35:31 --> 00:35:37
So you can write down
immediately

646
00:35:34 --> 00:35:40
the solution to this
differential equation.

647
00:35:37 --> 00:35:43
x will be some amplitude times
the cosine of omega t plus phi,

648
00:35:43 --> 00:35:49
just as we had before,

649
00:35:46 --> 00:35:52
and omega will now be
the square root of g over l.

650
00:35:51 --> 00:35:57
And so the period
of the pendulum will be

651
00:35:53 --> 00:35:59
2 pi times
the square root of l over g.

652
00:35:57 --> 00:36:03

653
00:35:59 --> 00:36:05
Just falls into our lap,
because we did all the work.

654
00:36:04 --> 00:36:10
I want you to realize that
these results for a pendulum

655
00:36:07 --> 00:36:13
have their restrictions.

656
00:36:09 --> 00:36:15
Small angles,
and we discussed quantitatively

657
00:36:11 --> 00:36:17
how small you would like
to allow,

658
00:36:13 --> 00:36:19
and also the mass has to be
exclusively in here

659
00:36:17 --> 00:36:23
and not in the string.

660
00:36:19 --> 00:36:25
We call that a massless string.

661
00:36:22 --> 00:36:28
To give you some rough idea
of what these periods will be,

662
00:36:28 --> 00:36:34
substitute for l, one meter.

663
00:36:30 --> 00:36:36
Now, you take for g 9.8,
take the square root

664
00:36:34 --> 00:36:40
and you multiply by 2 pi,
 

665
00:36:36 --> 00:36:42
and what you find is that the
period is about two seconds.

666
00:36:40 --> 00:36:46
So a pendulum one meter long has
a period of about two seconds.

667
00:36:45 --> 00:36:51

668
00:36:47 --> 00:36:53
One... two... three...
four... five... six.

669
00:36:53 --> 00:36:59
So to go from here to here
is about one second.

670
00:36:57 --> 00:37:03
If I make it four times
shorter-- l four times shorter--

671
00:37:01 --> 00:37:07
the square root of four is two.

672
00:37:03 --> 00:37:09
Then the period is ch...
the period is changing.

673
00:37:06 --> 00:37:12
Four times shorter, the period
must be two times shorter.

674
00:37:11 --> 00:37:17
To make roughly 25 centimeters.

675
00:37:13 --> 00:37:19
I'm not trying to be
very quantitative here.

676
00:37:15 --> 00:37:21
Now the whole period must be
about one second.

677
00:37:18 --> 00:37:24
One... two... three...
four... five... six.

678
00:37:24 --> 00:37:30
Roughly one second.

679
00:37:25 --> 00:37:31
So, you see that the period is
extremely sensitive

680
00:37:31 --> 00:37:37
to the length of the string.

681
00:37:34 --> 00:37:40

682
00:37:37 --> 00:37:43
I now want to compare with you

683
00:37:39 --> 00:37:45
the results that we have
for the spring

684
00:37:42 --> 00:37:48
with the results that
we have from the pendulum

685
00:37:46 --> 00:37:52
to give you
some further insight.

686
00:37:49 --> 00:37:55
We have the string,
and we have the pendulum.

687
00:37:53 --> 00:37:59
And I'm only going to look
at the period T,

688
00:37:56 --> 00:38:02
which here is 2 pi divided
by the square root of m over k,

689
00:38:01 --> 00:38:07
and here is 2 pi times
the square root of l over g.

690
00:38:06 --> 00:38:12

691
00:38:07 --> 00:38:13
If I look here,
there is a mass in here.

692
00:38:11 --> 00:38:17
If I look here,
it's independent of the mass.

693
00:38:15 --> 00:38:21
Why is there a mass in here?

694
00:38:18 --> 00:38:24
That is very easy to see.

695
00:38:20 --> 00:38:26
If I take a spring

696
00:38:22 --> 00:38:28
and I extend the spring
over a certain distance,

697
00:38:26 --> 00:38:32
then there is a certain force
that I feel.

698
00:38:29 --> 00:38:35
That force is independent

699
00:38:31 --> 00:38:37
of the mass that I put
at the end of the spring.

700
00:38:33 --> 00:38:39
The spring doesn't know what the
mass is you're going to put on.

701
00:38:36 --> 00:38:42
All it knows is "I am too long

702
00:38:38 --> 00:38:44
and I want to go back
to equilibrium."

703
00:38:40 --> 00:38:46
That force is a fixed force.

704
00:38:42 --> 00:38:48
If I double the mass,
that fixed force will give,

705
00:38:45 --> 00:38:51
on double the mass,
half the acceleration.

706
00:38:47 --> 00:38:53
If the acceleration goes down,

707
00:38:49 --> 00:38:55
the period of oscillation
goes up.

708
00:38:51 --> 00:38:57
It's very clear.

709
00:38:52 --> 00:38:58
So you can immediately see
that with the spring,

710
00:38:56 --> 00:39:02
the mass must enter
into the period.

711
00:38:58 --> 00:39:04
Now go to the pendulum.

712
00:39:00 --> 00:39:06
If I double the mass of my bob
at the end of a pendulum,

713
00:39:04 --> 00:39:10
then the vertical component
of the tension will also double.

714
00:39:10 --> 00:39:16
That means this restoring force,

715
00:39:12 --> 00:39:18
which is proportional with the
tension, will also double.

716
00:39:15 --> 00:39:21
So now the restoring force
doubles and the mass doubles,

717
00:39:18 --> 00:39:24
the acceleration remains
the same,

718
00:39:20 --> 00:39:26
the period remains the same.

719
00:39:22 --> 00:39:28
So you can simply argue

720
00:39:24 --> 00:39:30
that there should be no mass
in here, and there isn't.

721
00:39:28 --> 00:39:34
How about this k?

722
00:39:30 --> 00:39:36
If k is high,
then a spring is stiff.

723
00:39:33 --> 00:39:39
What does that mean,
a stiff spring?

724
00:39:35 --> 00:39:41
It means that if I give it
a small extension,

725
00:39:38 --> 00:39:44
that the spring force is huge.

726
00:39:40 --> 00:39:46
If I have a huge spring force,

727
00:39:42 --> 00:39:48
the acceleration on a given mass
will be high.

728
00:39:45 --> 00:39:51
If I have a high acceleration,
the period will be short,

729
00:39:48 --> 00:39:54
and that's exactly what you see.

730
00:39:50 --> 00:39:56
If k is high,
the period will be short.

731
00:39:54 --> 00:40:00
g.

732
00:39:56 --> 00:40:02
Imagine that you have
a pendulum in outer space,

733
00:39:58 --> 00:40:04
that there is
no gravity, nothing.

734
00:40:00 --> 00:40:06
The pendulum will not swing.

735
00:40:03 --> 00:40:09
The period of the pendulum
will be infinitely long.

736
00:40:07 --> 00:40:13
Going to the shuttle

737
00:40:09 --> 00:40:15
where the perceived gravity
in their frame of reference--

738
00:40:12 --> 00:40:18
perceived;
they're weightless, remember--

739
00:40:15 --> 00:40:21
their perceived gravity is zero.

740
00:40:17 --> 00:40:23
You take a pendulum
in the shuttle

741
00:40:18 --> 00:40:24
and you put it at this angle,
you let it go,

742
00:40:20 --> 00:40:26
it will stay there
forever and ever and ever.

743
00:40:22 --> 00:40:28
The period is infinitely long.

744
00:40:24 --> 00:40:30
But take a spring in the shuttle
and let the spring oscillate,

745
00:40:28 --> 00:40:34
and it does.

746
00:40:30 --> 00:40:36
So you can actually measure the
mass of an object using a spring

747
00:40:34 --> 00:40:40
on the shuttle

748
00:40:35 --> 00:40:41
and let it oscillate
if you know the spring constant,

749
00:40:37 --> 00:40:43
and that's the way
it's actually done.

750
00:40:38 --> 00:40:44
So, you see indeed
that these things make sense

751
00:40:44 --> 00:40:50
when you think about it
in a rational way.

752
00:40:49 --> 00:40:55
We have here in 26-100
the mother of all pendulums.

753
00:40:56 --> 00:41:02
It is a pendulum...

754
00:40:57 --> 00:41:03
(object clangs )

755
00:40:58 --> 00:41:04
Oops.

756
00:41:00 --> 00:41:06
It is a pendulum
which is 5.1 meters long,

757
00:41:06 --> 00:41:12
and there is a mass at the end
of it which is 15 kilograms.

758
00:41:12 --> 00:41:18

759
00:41:18 --> 00:41:24
The length is 5.18 meters

760
00:41:25 --> 00:41:31
and the uncertainty is
about five centimeters.

761
00:41:28 --> 00:41:34
We can't measure it any better.

762
00:41:31 --> 00:41:37
And the mass at the end of it,

763
00:41:34 --> 00:41:40
which doesn't enter into the
period, is about 15 kilograms.

764
00:41:38 --> 00:41:44

765
00:41:40 --> 00:41:46
The period, which is 2 pi times
the square root of l over g,
 

766
00:41:46 --> 00:41:52
if you substitute in
your length of 5.1 meters,

767
00:41:51 --> 00:41:57
you will find 4.57 seconds.

768
00:41:56 --> 00:42:02
4.5 second... seven.

769
00:41:59 --> 00:42:05
Since you have a 1% error in l,

770
00:42:01 --> 00:42:07
you're going to have a half
a percent error in your period,
 

771
00:42:04 --> 00:42:10
so that is about 0.02 seconds.

772
00:42:09 --> 00:42:15
So this is my prediction.

773
00:42:11 --> 00:42:17

774
00:42:19 --> 00:42:25
And now I'm going
to oscillate it for you

775
00:42:22 --> 00:42:28
and I'm going to do it
from two different angles.

776
00:42:27 --> 00:42:33
I'm going to do
at a five-degree angle

777
00:42:30 --> 00:42:36
and I'm going to do it
at a ten-degree angle.

778
00:42:33 --> 00:42:39
In order to get
my relative error down,

779
00:42:37 --> 00:42:43
I will oscillate ten times.

780
00:42:39 --> 00:42:45
So I'm going to get

781
00:42:41 --> 00:42:47
at an angle theta maximum
of roughly five degrees.

782
00:42:46 --> 00:42:52
I get ten T equals something
plus or minus my reaction time,

783
00:42:52 --> 00:42:58
which is 0.1 of a second.

784
00:42:54 --> 00:43:00
And then I will do it
from ten degrees

785
00:42:57 --> 00:43:03
and I will do again ten T,

786
00:42:59 --> 00:43:05
and again my reaction time is
not much better than 0.1 second.

787
00:43:04 --> 00:43:10
So, let's do that first.

788
00:43:06 --> 00:43:12
I will move this out of the way

789
00:43:09 --> 00:43:15
because if that 15-kilogram
object hits this,

790
00:43:13 --> 00:43:19
that is not funny.

791
00:43:16 --> 00:43:22
All right.

792
00:43:18 --> 00:43:24
Zero.

793
00:43:19 --> 00:43:25

794
00:43:22 --> 00:43:28
I have a mark here on the floor.

795
00:43:24 --> 00:43:30
This is about five degrees,
and this is about ten degrees.

796
00:43:28 --> 00:43:34
I will first do it
from five degrees.

797
00:43:30 --> 00:43:36
I will let it swing
one oscillation,

798
00:43:32 --> 00:43:38
and when it comes to a halt
here, I will start the timer.

799
00:43:35 --> 00:43:41
That's, for me, the easiest.

800
00:43:37 --> 00:43:43
But I count on you
when it comes to counting.

801
00:43:40 --> 00:43:46
You ready? You ready?

802
00:43:42 --> 00:43:48
You're sure?

803
00:43:43 --> 00:43:49
I'm ready, too.

804
00:43:44 --> 00:43:50

805
00:43:47 --> 00:43:53
Okay.

806
00:43:49 --> 00:43:55
Now, keep counting and
don't confuse me again, now.

807
00:43:52 --> 00:43:58
You're completely responsible
for the counting.

808
00:43:56 --> 00:44:02
So you only have to tell me
is when...

809
00:43:58 --> 00:44:04
when eight or nine is coming up.

810
00:44:00 --> 00:44:06
That's all I want to know.

811
00:44:01 --> 00:44:07
Don't even bother me with three.

812
00:44:03 --> 00:44:09
Don't even bother me with four.

813
00:44:05 --> 00:44:11
Just let me know

814
00:44:06 --> 00:44:12
when I have to get in position
for the final kill.

815
00:44:09 --> 00:44:15

816
00:44:15 --> 00:44:21
Notice there's almost no denting
on this pendulum.

817
00:44:18 --> 00:44:24
The amplitude remains almost
the same, whereas with the...

818
00:44:20 --> 00:44:26
with the air track
you could actually see

819
00:44:22 --> 00:44:28
that there was already some kind
of friction... Where are we now?

820
00:44:25 --> 00:44:31
STUDENTS:
Nine.

821
00:44:27 --> 00:44:33
LEWIN:
Nine? Nine, right?

822
00:44:30 --> 00:44:36

823
00:44:32 --> 00:44:38
STUDENTS:
Ten.

824
00:44:34 --> 00:44:40

825
00:44:36 --> 00:44:42
STUDENT:
Oh, my God!

826
00:44:37 --> 00:44:43
LEWIN:
45.70.

827
00:44:39 --> 00:44:45

828
00:44:43 --> 00:44:49
45.70.

829
00:44:44 --> 00:44:50
Where is my chalk?

830
00:44:47 --> 00:44:53
45.70.

831
00:44:52 --> 00:44:58
What was my prediction?

832
00:44:54 --> 00:45:00
(students responding )

833
00:44:57 --> 00:45:03
(applause )

834
00:44:59 --> 00:45:05
LEWIN:
Yeah!

835
00:45:00 --> 00:45:06
Yeah!

836
00:45:01 --> 00:45:07
Yeah, exactly.

837
00:45:03 --> 00:45:09
You get the picture.

838
00:45:05 --> 00:45:11
That is pure luck,

839
00:45:07 --> 00:45:13
because my accuracy is no better
than a tenth of a second.

840
00:45:10 --> 00:45:16
Now we do from ten...

841
00:45:11 --> 00:45:17
ten degrees, and I want
to show you now

842
00:45:14 --> 00:45:20
that the effect on the angle--
you go from five to ten--

843
00:45:17 --> 00:45:23
is small, so small
that you cannot measure it

844
00:45:20 --> 00:45:26
within the accuracy
of your measurement.

845
00:45:23 --> 00:45:29

846
00:45:27 --> 00:45:33
Yeah!

847
00:45:29 --> 00:45:35
Okay.

848
00:45:31 --> 00:45:37
Again, relax and count.

849
00:45:34 --> 00:45:40

850
00:45:47 --> 00:45:53
Aah, nerve-wracking!

851
00:45:50 --> 00:45:56
Ooh!

852
00:45:52 --> 00:45:58

853
00:45:56 --> 00:46:02
Where are we now?

854
00:45:57 --> 00:46:03
STUDENTS:
Seven.

855
00:45:59 --> 00:46:05
LEWIN:
Seven.

856
00:46:00 --> 00:46:06

857
00:46:03 --> 00:46:09
STUDENTS:
Eight.

858
00:46:05 --> 00:46:11

859
00:46:07 --> 00:46:13
Nine.

860
00:46:09 --> 00:46:15

861
00:46:12 --> 00:46:18
Ten.

862
00:46:14 --> 00:46:20
Oh!

863
00:46:15 --> 00:46:21
(applause )

864
00:46:16 --> 00:46:22

865
00:46:19 --> 00:46:25
LEWIN:
Did you expect anything else?

866
00:46:21 --> 00:46:27
(students laugh )

867
00:46:22 --> 00:46:28
LEWIN:
45.75.

868
00:46:27 --> 00:46:33
One of the most remarkable
things I just mentioned to you

869
00:46:32 --> 00:46:38
is that the period
of the oscillations

870
00:46:35 --> 00:46:41
is independent
of the mass of the object.

871
00:46:38 --> 00:46:44
That would mean

872
00:46:40 --> 00:46:46
if I joined the bob
and I swing down with the bob

873
00:46:44 --> 00:46:50
that you should get
that same period.

874
00:46:47 --> 00:46:53
Or should you not?

875
00:46:49 --> 00:46:55
I'm asking you a question before
we do this awful experiment.

876
00:46:55 --> 00:47:01
Would the period come out
to be the same or not?

877
00:47:00 --> 00:47:06
(students respond )

878
00:47:01 --> 00:47:07
LEWIN:
Some of you think it's the same.

879
00:47:04 --> 00:47:10
Have you thought about it,

880
00:47:05 --> 00:47:11
that I'm a little bit taller
than this object

881
00:47:08 --> 00:47:14
and that therefore
maybe effectively

882
00:47:11 --> 00:47:17
the length of the string
has become a little less

883
00:47:14 --> 00:47:20
if I sit up like this?

884
00:47:16 --> 00:47:22
And if the length of the string
is a little less,

885
00:47:19 --> 00:47:25
the period would be
a little shorter.

886
00:47:22 --> 00:47:28
Yeah?

887
00:47:23 --> 00:47:29
Be prepared for that.

888
00:47:24 --> 00:47:30
On the other hand,
I'm also pre...

889
00:47:25 --> 00:47:31
well, I'm not quite
prepared for it.

890
00:47:27 --> 00:47:33
(students laughing )

891
00:47:29 --> 00:47:35
LEWIN:
I will try to hold my body
as horizontal as I possibly can

892
00:47:34 --> 00:47:40
in order to be
at the same level as the bob.

893
00:47:37 --> 00:47:43
I will start
when I come to a halt here.

894
00:47:41 --> 00:47:47
There we go.

895
00:47:43 --> 00:47:49
(students laughing )

896
00:47:44 --> 00:47:50

897
00:47:46 --> 00:47:52
LEWIN:
Now!

898
00:47:47 --> 00:47:53

899
00:47:49 --> 00:47:55
You count!

900
00:47:52 --> 00:47:58
This hurts!

901
00:47:54 --> 00:48:00
Aah!

902
00:47:55 --> 00:48:01
(students counting
in background )

903
00:47:58 --> 00:48:04
(laughter continues )

904
00:47:59 --> 00:48:05
LEWIN:
I want to hear you loud!

905
00:48:04 --> 00:48:10
STUDENTS:
Four!

906
00:48:06 --> 00:48:12
LEWIN (groaning ):
Oh...

907
00:48:08 --> 00:48:14
STUDENTS:
Five!

908
00:48:10 --> 00:48:16
YOUNG MAN:
Halfway!

909
00:48:12 --> 00:48:18
LEWIN:
Thank you!

910
00:48:13 --> 00:48:19
STUDENTS:
Six...

911
00:48:14 --> 00:48:20
LEWIN (groaning ):
Oh...

912
00:48:17 --> 00:48:23
STUDENTS:
Seven...

913
00:48:22 --> 00:48:28
Eight.

914
00:48:23 --> 00:48:29
LEWIN (groaning ):
Aah...

915
00:48:26 --> 00:48:32
STUDENTS:
Nine...

916
00:48:28 --> 00:48:34

917
00:48:31 --> 00:48:37
Ten!

918
00:48:32 --> 00:48:38
LEWIN:
Ah!

919
00:48:33 --> 00:48:39
STUDENTS:
Oh!

920
00:48:34 --> 00:48:40
(cheering and applause )

921
00:48:37 --> 00:48:43

922
00:48:42 --> 00:48:48
LEWIN:
Ten T with Walter Lewin.

923
00:48:47 --> 00:48:53
45.6 plus or minus 0.1 second.

924
00:48:53 --> 00:48:59
Physics works, I'm telling you!

925
00:48:56 --> 00:49:02
I'll see you Monday.

926
00:48:57 --> 00:49:03
Have a good weekend.

927
00:48:59 --> 00:49:05
(applause )

928
00:49:01 --> 00:49:07

929
00:49:28 --> 00:49:34.000