8.01x – Lect 10 – Hooke’s Law, Springs, Pendulums, Simple Harmonic Motion

Today we are going to talk
about springs about pendulums. and about simple
harmonic oscillators– one of the key topics in 801. If I have a spring… and this is the relaxed length
of the string… spring, I call that x equals zero. And I extend the string…
the spring, with a “p,” then there is a force
that wants to drive this spring back to equilibrium. And it is an experimental fact
that many springs– we call them ideal springs–
for many springs, this force is proportional
to the displacement, x. So if this is x, if you make x
three times larger, that restoring force is
three times larger. This is
a one-dimensional problem, so to avoid the vector notation,
we can simply say that the force, therefore,
is minus a certain constant, which we call
the spring constant– this is called
the spring constant– and the spring constant has
units newtons per meter. So the minus sign takes care
of the direction. When x is positive, then the force is
in the negative direction; when F is negative, the force is
in the positive direction. It is a restoring force. Whenever this linear relation
between F and x holds, that is referred to
as Hooke’s Law. How can we measure
the spring constant? That’s actually
not too difficult. I can use gravity. Here is the spring
in its relaxed situation. I hang on the spring a mass, m,
and I make use of the fact that gravity now exerts
a force on the spring, and when you find
your new equilibrium– this is the new
equilibrium position– then the spring force,
of course, must be exactly the same as mg. No acceleration
when the thing is at rest. And so I could now make a plot whereby I could have here x and
I could have here this force F, which I know because
I know the masses. I can change the masses. I can go through
a whole lot of them. And you will see data points which scatter
around a straight line. And the spring constant follows,
then, if you take… if you call this delta F
and you call this delta x, then the spring constant, k,
is delta F divided by delta x. So you can even measure it. You don’t have
to start necessarily at this point
where the spring is relaxed. You could already start when the
spring is already under tension. That is not a problem. You’ll be surprised
how many springs really behave very nicely
according to Hooke’s Law. Uh, I have one here. It’s not
a very expensive spring. You see it here. And there is here
a holder on to it so it’s already a little bit
under stress. That doesn’t make
any difference. These marks here are
13 centimeters apart, and every time
that I put one kilogram on, you will see that it goes down
by roughly 13 centimeters. It goes down to this mark. I put another kilogram on,
it goes down to this mark. I put another kilogram on
and it goes back to this mark, all the way down. And if I take them all off…
so what I’ve done is I effectively went along this
curve, and if I take them off, if it is an ideal spring, then it goes back to its
original length and which it does. That’s a requirement, of course,
for an ideal spring if it behaves according
to Hooke’s Law. Now, I can, of course,
overdo things. I can take a spring
like this one and stretch it to the point that it no longer
behaves like Hooke’s Law. I can damage it. Uh, I can do
permanent deformation. Look, that’s easy. For sure, Hooke’s Law is
no longer active. Look how much longer this spring
is than it was before. So there comes, of course,
a limit how far you can go before you permanently
deform your spring. What I have done now
with that spring, probably in the beginning
I went up along this straight line and then something like this
must have happened. I got a huge extension. My force did not increase
very much. And then when I relaxed,
when I took my force off, the spring was longer
at the ends than it was at the beginning. So I have a net extension
which will always be there, and that’s not very nice, of
course, to do that to a spring. So Hooke’s Law holds
only within certain limitations. You have to, uh, obey
a certain amount of discipline. There are ways that you can also
measure the spring constant in a dynamic way, which is
actually very interesting. Um, I have here a spring,
and this spring… this is x equals 0, and I attach
now to the spring a mass, m. This has to be
on a frictionless surface, and you will see, when
I extend it over a distance x, that you get your force, your spring force
that drives it back. We have, of course, gravity, mg, and we have the normal force
from the surface. So there is in the…
in the y direction, there is no acceleration,
so I don’t have to worry about the forces
in the y direction at all. If I let this thing oscillate,
I let… I release it, it will start to oscillate about
this point, back and forth, then as I will show you now, you will find
that the period of oscillation, the time for
one whole oscillation is 2 pi times the square root
of the mass m divided by
the spring constant k. I will derive that–
you will see that shortly. In other words,
if you measured the period and you knew the math,
then you can calculate k. Alternatively, if you knew k
and you measure the period, you can calculate the mass,
even in the absence of gravity. I don’t use gravity here. So a spring always allows you
to measure, uh, a mass even in the absence of gravity. The period that you see,
the time that it takes for this object to oscillate
once back and forth, is completely independent
of how far I move it out, which is very nonintuitive, but you will see that that
comes out of the derivation. There is no dependence
on how far I move it out. So whether I oscillate it
like this or whether I oscillate it
like this, as long as Hooke’s Law holds,
you will see that the period is independent
of what we call that amplitude. So I’m going to derive the
situation now for an ideal case. Ideal case means
Hooke’s Law must hold. There’s no friction,
and the spring itself has negligible mass
compared to this one. Let’s call it a massless spring. So now I’m going to write down
Newton’s Second Law: ma, which is all in the
x direction, equals minus kx. a is the second derivative
of position, for which I will write
x double-dot, mx double-dot– one dot is the first derivative,
that’s the velocity; two dots is the acceleration–
plus kx equals zero. I divide by m and I get x double-dot plus
k over m times x equals zero. And this is arguably the most important equation
in all of physics. It’s a differential equation. Some of you may already have
solved differential equations. The outcome of this,
you will see, is very simple. x is, of course, changing in
some way as a function of time, and when you have
the correct solution for x as a function of time and you substitute that back
into that differential equation, that equation will have
to be satisfied. What would a solution be
to this differential equation? I’m going to make you see
this oscillation first. I’m going to make you see x
as a function, and I’m going to do that
in the following way. I have here a spray paint can which is suspended
between two springs, and I can oscillate it
vertically, which is your x direction,
like that. So x changes with time. The time axis I will introduce
by pulling on the string. When the spray paint
is going to spray, I’m going to pull
on that string, and if I can do that
at a constant speed, then you get horizontally a time
axis and of course vertally… eh, vertically you will get
the position of x. So I want you
to just see qualitatively what kind of a weird curve
x as a function of time is, which then will have to satisfy
that differential equation. All right. It’s always a messy experiment
because the paint is dripping, but I will try to get
the spray paint going. There we go. Okay. Now I’ll pull… All right Will you give me a hand? Yeah, could you, please? I will cut it here
and then you… Be very careful,
because it’s… it is messy. Oh, let’s put it…
let’s take it out this way. Aah… Okay, just walk back. Just walk– yeah, great. now hold up the top
so that we can see it fine. Okay. What is… what does it
remind you of? (student responds ) LEWIN:
Sinusoids– reminds me
of a cosinusoid, by the way. Sinusoid or a cosine–
same thing. All right. Let’s try to substitute
in that equation a sinusoid’s
or a cosine solution, whichever one you prefer. Makes no difference. So I’m going to substitute
in this equation– this is my trial function– that x as a function of time
is a constant, A– I will get back to that
in a minute– times cosine omega t plus phi. This A we call the amplitude. Notice the cosine function is…
the highest value is plus one and the lowest value is minus
one, so the amplitude indicates that is… the farthest
displacement from zero on this side would be plus A and
on this side would be minus A. So that’s in meters. This omega we call
the angular frequency. Don’t confuse it
with angular velocity. We call it angular frequency,
and the units are the same. The units are
in radians per second, the same as angular velocity. If I advance
this time little t– if I advance that
by 2 pi divided by omega… if I advance this time
by 2 pi divided by omega, then this angle here increases
by 2 pi radians, which is 360 degrees, and so
that’s the time that it takes for the oscillation
to repeat itself. So this is the period
of the oscillation, and that is in seconds. And you can determine…
if you want to, you can define the frequency
of the oscillations which is one over T, which we express always
in terms of hertz. And then here we have
what we call the phase angle, and I will return to that. That’s in radians. And this trial function
I’m going to substitute now into this equation. So the first thing I have to do,
I have to find what the second derivative is
of x as a function of time. Well, that’s my function. I have here first
the first derivative, x dot. That becomes minus A omega. I get an omega out because
there’s a time here, and now I have to take the derivative
of the function itself so I get the sine
of omega t plus phi. Of course I could have started
off here with a sine curve; I hope you realize that. I just picked the cosine one. x double-dot. Now I get another omega out,
so I get minus A omega squared. The derivative of the sine
is the cosine. Cosine omega t plus phi. And that is also
minus omega squared times x, because notice I have
A cosine omega t plus phi, which itself is x. So now I’m ready
to substitute this result into that differential equation. This must always hold
for any value of x, for any moment in time. And therefore, the only way that this can work
is if omega squared is k over m. So omega squared must be
k over m. And therefore, we now have
the solution to this problem. So we have omega equals
the square root of k over m and the period is 2 pi times
the square root of m over k. And what is striking,
really remarkable, that this is independent
of the amplitude, and it’s also independent of this angle phi,
this phase angle. What is this business
of this phase angle? It’s a peculiar thing
that we have there. Well, you can think
about the physics, actually. When I start this oscillation,
I have a choice of two things. I can start it off at a certain
position which I can choose. I can give it a certain
displacement from zero and simply let it go. But I can also, when I let it
go, give it a certain velocity. That’s my choice. So I have two choices: where I let it go
and what velocity I give it. And that is reflected
in my solution: namely, that ultimately
in the solution I get the result of A
and the result of phi, which doesn’t determine
the period, but it results from what we call
my initial conditions. And I want to do an example
whereby you see how A and phi immediately follow
from the initial conditions. So in this example, I release the object at
x equals zero and t equals zero. So I release it
at the equilibrium. At that moment in time, I give it a velocity, which is
minus three meters per second. My units are always
in MKS units. The spring constant k equals
ten newtons per meter, and the mass of the object is
0.1 kilograms. And now I can ask you what now
is x as a function of time, including the amplitude A,
including the phase angle phi? Well, let’s first
calculate omega. That is
the square root of k over m. That would be
ten radians per second. The period T, which is
2 pi divided by omega, would be roughly 6.28 seconds. And the frequency f would be
about 0.16 hertz, just to get some numbers. 1.6 hertz– sorry. This is not my day. This is 0.628,
and this is 1.6 hertz. 2 pi divided by omega;
you can see this is ten. Six divided by ten is about 0.6. All right, so now I know that
at t equals zero, x equals zero. So I see my solution
right there. Right here I put in
t equals zero, and I know that x is zero. So I get zero equals
A times the cosine of phi. Well, A is not zero. If I release that thing
at equilibrium and I give it a velocity
of three meters per second, it’s going to oscillate. So A is not zero. So the only solution is
that cosine phi is zero, and so that leaves me
with phi is pi over two, or phi is 3 pi…
phi is 3 pi over two. That’s the only
two possibilities. Now I go to my
next initial condition, that the velocity is
minus 3. Now, here you see
the equation for the velocity. This is minus 3
at t equals zero. So minus 3 equals minus A,
and A is… we don’t know yet. Minus A, and then
we have omega squared. Omega– sorry– which is ten. t is zero. I get the sine of phi. If I pick pi over 2,
then the sine of phi is 1. And so you find immediately
that A equals plus 0.3 And so the solution now,
which includes now phi and A, is that x equals plus 0.3 times
the cosine of omega, which is 10t plus pi over 2. So you see that
the initial conditions… what the conditions are
at t equals zero, they determine my A and
they determine my phase angle. If you had chosen this as
the phase angle– 3 pi over 2– that would have been fine. You would have found
a minus sign here, and that’s exactly the same. so you would have found nothing different I want to demonstrate to you
that the period of oscillations, nonintuitive as that may be, is independent of the amplitude
that I give the object. And I want to do that here
with this air track. I have a… an object here. This object has a mass–
186 plus or minus 1 gram. Call it m1. I’m going to oscillate it and we’re going to measure
the periods. But instead of measuring
one period, I’m going to measure
ten periods, because that gives me
a smaller uncertainty, a smaller relative error
in my measurements. So I’m going to do it
as an amplitude, which is 15 centimeters. Let’s make it 20 centimeters. So I get 10T,
I get a certain number, and I get an error
which is my reaction error, which is about
a tenth of a second. That’s about the reaction error
that we all have, roughly. Then I will do it
at 40 centimeters. We get a 10T, and we get, again,
plus or minus 0.1 seconds. And we’ll see
how much they differ. They should be the same,
if this is an ideal spring, within the uncertainty
of my measurements. You see the timing there. I’m going to give it
a 20-centimeter offset, which is here, and then I will start it
when it comes back here. So I will allow it
one oscillation first. That’s easier for me to see it
stand still when I start it. There we go. One… two… three…
four… five… six… seven…
eight… nine… ten. What do we see? 15.16 15.16 seconds. By the way, you can derive the
spring constant from this now because you know the mass
and you know the time. Now I’m going to give it
a displacement, an amplitude which is twice as high. So I make it 40 centimeters. So this is ten. 40 centimeters–
a huge displacement. Now… one… two…
three… four… five… six… seven…
eight… nine… ten. 15.13 Fantastic agreement within the
uncertainty of my measurements. They’re within
3/100 of a second. Of course if you try it
many times, you won’t always get that close,
because my reaction time is really not much better
than a tenth of a second. Now I will show you something
else which is quite interesting, and that is how the behavior
of the period is on the… on the mass of the object. I have here another car
which weighs roughly the same. Uh, I’m going to add
the two together, and so we get m2 is about
372 plus or minus 1 gram. The plus or minus 1 comes in because our scale is no more
accurate than one gram. So we put them both on the scale and we find this to be
the uncertainty. So now I’m going to measure the ten periods of this object
with mass m2, so twice the mass. So that should be the square root of m2
divided by m1 times 10 times the period of m1. And so I can make a prediction because this is the square root
of two, and I know what this is. So I will take my calculator and I will take the square root
of two, and I multiply that by, uh, let’s take 15.15. And so that comes out
to be 21.42 21.42. It’s not clear that
this two is meaningful. And now comes the $64 question:
What is the uncertainty? This is a prediction. And this now becomes
a little tricky. So what I’m telling you now
may confuse you a bit. It’s not meant to be, but I really won’t hold you
responsible for it. You may now think that the uncertainty
in these measurements follows from the uncertainty in this,
which is true, which is about 0.6%, and
from the uncertainty in this, so this has about
an uncertainty of 0.6%. I got it low because I measured
ten oscillations, you see? The uncertainty is only
one out of 150, which is low. You may think
that the uncertainty in there equals the square root
of 372 plus or minus 1 divided by 186 plus or minus 1. And now you may argue, and it’s completely reasonable
that you would argue that way– you would say,
“Well, this is roughly “a quarter of a percent error
here under the square root and this is roughly
half a percent error.” One out of 200 is about half. So you would add up
the two errors– a quarter plus half,
that’s about 0.7– and because of the square roots,
that becomes 0.35%, and that’s wrong. And the reason why
that is completely wrong– that has to do with the fact that these two errors
are coupled to each other. See, we… the 186 is included
in the 372. The best way
I can show you this– suppose I measured
m1 divided by m1, which would be 186 plus or minus 1
divided by 186 plus or minus 1. That number is one
with a hundred zeros. This number is one. You have the mass of one object; you divide it
by the same object. Whereas if you would say, “Ah,
this is a half a percent error and this is a half a percent
error,” you would say the ratio has an error of 1%,
and that’s not the case. So I will not bother you
with that. I will not hold you
responsible for that, but it turns out
that if you do it correctly and you take the error of this
into account, of about 0.6%, that the error in this ratio is
really much less than .2%. You can almost forget about it. I will allow, generously, for
a 1% error in the final answer, and so I stick to my prediction that the 10T of double the mass
is going to be like this. And now we’re going to get
the observation: 10T times m2, which is double
the mass, and that, of course, always has my uncertainty
of my reaction time. There’s nothing I can do
about that. And we will compare
these two numbers. So I will put the other mass
on top of it. Goes here. Tape them together
so that they won’t fall off. There we go. So, I hope I did that correctly. The square root of two
times 15.15. We’ll give it a… amplitude, something like 30,
maybe 35 centimeters. There we go. One… two– much slower, eh?
You see that. Three… four… five… six… seven… eight… nine–
I’m not looking– ten. 21.36. 21.36. You can round it off to 40
if you want to, and you see that
the agreement is spectacular. Within the uncertainty
of my measurements, it comes out amazingly well. You could have removed
this two, of course, because if you have
an uncertainty of .2 here, it’s a little silly to have
that little two hanging there. But you see that indeed this spring is very close
to an ideal spring. It obeys Hooke’s Law,
and it is also nearly massless. Here is the pendulum. Here is the mass, and it’s
offset at an angle, theta. The length of the pendulum is l,
the length of the string. There is gravity here, mg, and
the other force on the object, the only other force,
is the tension, T. Don’t confuse that with
period T; this is tension T. It’s in Newtons. Those are the only two forces. There is nothing else. The thing is going
to arc around like this and it’s going to oscillate. I call this the y direction
and I call this the x direction, and here x equals 0. Well, I’m going to decompose
the tension into the y and into the x direction
as we have done before. So this is going to be
the y component. This is the x component. So this y component equals
T cosine theta and the x component equals
T sine theta. And now I’m going to write down the differential equations
of motion, first in the x direction. Second… Newton’s Second Law:
ma equals… this is the only force
in the x direction. It’s a restoring force,
just like with the spring. I therefore have to give it
a minus sign. So equals minus T times
the sine of theta. T itself could easily be
a function of theta. So I have to allow for that. The sine of theta equals x if it’s here
at position x divided by l, and so I can write for this
minus T– which may be a function of
theta– times x divided by l. That is my differential equation
in the x direction, and I prefer always for this a
to write down x double-dot. Now the y direction. In the y direction, I have
m y double-dot equals… this is my plus direction, so
I have T cosine theta minus mg. This is equation one
and this is equation two. And so now we have to solve two
coupled differential equations, which is a hopeless task. It looks like a zoo,
and it is a zoo. And now we’re going to make
some approximations, and the approximations
that we will make which we will often see
in physics when something oscillates– is what we call
the small-angle approximations. Small-angle– we will not allow
theta to become too large. I’ll be quantitative,
what I mean by too large. When theta, which is in radians,
equals much, much less than one, we call that a small angle. If that’s the case, the cosine
of theta is very close to 1. You will say, “Well, blah,
blah, blah– how close to 1?” Okay, five degrees–
the cosine is 0.996. That’s close to 1. Ten degrees–
the cosine is 0.985. That’s only 1½% different
from one. So even at ten degrees,
you’re doing extremely well. So, this is
consequence number one of the small-angle
approximation. But there is
a second consequence of the small-angle
approximation. Look at the excursion
that this object made from equilibrium
in the x direction. That’s this big. Look at the excursion it makes
in the y direction. It’s this small. It’s way smaller than the
excursion in the x direction, provided that
your angle is small. I’ll give you an example. At five degrees, this excursion
is only 4% of this excursion. At ten degrees, this excursion
is only 9% of this excursion. And since the excursion
in the y direction is so much smaller
than in the x direction, we say that the acceleration
in the y direction can be approximated
to be roughly zero. There is almost no acceleration
in the y direction. With these two conclusions, which follow from
the small-angle approximation, I go back to my equation
number two, and I find that zero equals T, which
could be a function of theta. The cosine of theta is
one minus mg. So I find that T equals mg. Notice it’s no longer
even a function of theta. So I simply have,
in my small-angle approximation, that I can make T
the same as mg. It’s approximately, but I still
put an equals sign there. I substitute this back
in my equation number one. And so now I get
that m times x double-dot– and now I bring this
on the other side– plus– T is now mg– mg times x divided by l
equals zero. And now comes
the wonderful result: x double-dot plus g over l
times x equals zero. And this is
such a beautiful result that it almost makes me cry. This is a simple
harmonic oscillation. This equation looks like a carbon copy
of the one that we have there. Here we have k over m,
and there we have g over l. That’s all. Other than that,
there is no difference. So you can write down
immediately the solution to this
differential equation. x will be some amplitude times
the cosine of omega t plus phi, just as we had before, and omega will now be
the square root of g over l. And so the period
of the pendulum will be 2 pi times
the square root of l over g. Just falls into our lap,
because we did all the work. I want you to realize that
these results for a pendulum have their restrictions. Small angles,
and we discussed quantitatively how small you would like
to allow, and also the mass has to be
exclusively in here and not in the string. We call that a massless string. To give you some rough idea
of what these periods will be, substitute for l, one meter. And you take for g 9.8,
take the square root and you multiply by 2 pi, and what you find is that the
period is about two seconds. So a pendulum one meter long has
a period of about two seconds. One… two… three…
four… five… six. So to go from here to here
is about one second. If I make it four times
shorter– l four times shorter– the square root of four is two. Then the period is ch…
the period is changing. Four times shorter, the period
must be two times shorter. I make roughly 25 centimeters. I’m not trying to be
very quantitative here. Now the whole period must be
about one second. One… two… three…
four… five… six. Roughly one second. So, you see that the period is
extremely sensitive to the length of the string. I now want to compare with you the results that we have
for the spring with the results that
we have from the pendulum to give you
some further insight. We have the string,
and we have the pendulum. And I’m only going to look
at the period T, which here is 2 pi divided
by the square root of m over k, and here is 2 pi times
the square root of l over g. If I look here,
there is a mass in here. If I look here,
it’s independent of the mass. Why is there a mass in here? That is very easy to see. If I take a spring and I extend the spring
over a certain distance, then there is a certain force
that I feel. That force is independent of the mass that I put
at the end of the spring. The spring doesn’t know what the
mass is you’re going to put on. All it knows is “I am too long and I want to go back
to equilibrium.” That force is a fixed force. If I double the mass,
that fixed force will give, on double the mass,
half the acceleration. If the acceleration goes down, the period of oscillation
goes up. It’s very clear. So you can immediately see
that with the spring, the mass must enter
into the period. Now go to the pendulum. If I double the mass of my bob
at the end of a pendulum, then the vertical component
of the tension will also double. That means this restoring force, which is proportional with the
tension, will also double. So now the restoring force
doubles and the mass doubles, the acceleration remains
the same, the period remains the same. So you can simply argue that there should be no mass
in here, and there isn’t. How about this k? If k is high,
then a spring is stiff. What does that mean,
a stiff spring? It means that if I give it
a small extension, that the spring force is huge. If I have a huge spring force, the acceleration on a given mass
will be high. If I have a high acceleration,
the period will be short, and that’s exactly what you see. If k is high,
the period will be short. g Imagine that you have
a pendulum in outer space, that there is
no gravity, nothing. The pendulum will not swing. The period of the pendulum
will be infinitely long. Go into to the shuttle where the perceived gravity
in their frame of reference– perceived;
they’re weightless, remember– their perceived gravity is zero. You take a pendulum
in the shuttle and you put it at this angle,
you let it go, it will stay there
forever and ever and ever. The period is infinitely long. But take a spring in the shuttle
and let the spring oscillate, and it does. So you can actually measure the
mass of an object using a spring on the shuttle and let it oscillate
if you know the spring constant, and that’s the way
it’s actually done. So, you see indeed
that these things make sense when you think about it
in a rational way. We have here in 26-100
the mother of all pendulums. It is a pendulum… (object clangs ) Oops. It is a pendulum
which is 5.1 meters long, and there is a mass at the end
of it which is 15 kilograms. The length is 5.18 meters and the uncertainty is
about five centimeters. We can’t measure it any better. And the mass at the end of it, which doesn’t enter into the
period, is about 15 kilograms. The period, which is 2 pi times
the square root of l over g, if you substitute in
your length of 5.1 meters, you will find 4.57 seconds. 4.5 second… seven. Since you have a 1% error in l, you’re going to have a half
a percent error in your period, so that is about 0.02 seconds. So this is my prediction. And now I’m going
to oscillate it for you and I’m going to do it
from two different angles. I’m going to do
at a five-degree angle and I’m going to do it
at a ten-degree angle. In order to get
my relative error down, I will oscillate ten times. So I’m going to get at an angle theta maximum
of roughly five degrees. I get ten T equals something
plus or minus my reaction time, which is 0.1 of a second. And then I will do it
from ten degrees and I will do again ten T, and again my reaction time is
not much better than 0.1 second. So, let’s do that first. I will move this out of the way because if that 15-kilogram
object hits this, that is not funny. All right. Zero. I have a mark here on the floor. This is about five degrees,
and this is about ten degrees. I will first do it
from five degrees. I will let it swing
one oscillation, and when it comes to a halt
here, I will start the timer. That’s, for me, the easiest. But I count on you
when it comes to counting. You ready? You ready? You’re sure? I’m ready, too. Okay, Now, keep counting and
don’t confuse me again, now. You’re completely responsible
for the counting. So you only have to tell me
is when… when eight or nine is coming up. That’s all I want to know. Don’t even bother me with three. Don’t even bother me with four. Just let me know when I have to get in position
for the final kill. Notice there’s almost no denting
on this pendulum. The amplitude remains almost
the same, whereas with the… with the air track
you could actually see that there was already some kind
of friction… Where are we now? STUDENTS:
Nine. LEWIN:
Nine? Nine, right? STUDENTS:
Oh, my God! LEWIN:
45.70 45.70 Where is my chalk? 45.70 What was my prediction? (students responding ) LEWIN: Yeah! (applause ) LEWIN:Yeah! Yeah, exactly. You get the picture. That is pure luck, because my accuracy is no better
than a tenth of a second. Now we do from ten… ten degrees, and I want
to show you now that the effect on the angle–
you go from five to ten– is small, so small
that you cannot measure it within the accuracy
of your measurement. Yeah! Okay. Again, relax and count. Aah, nerve-wracking! Ooh! Where are we now? STUDENTS:
Seven. LEWIN:
Eight. STUDENTES: Nine. Ten. Oh! (applause ) LEWIN:
Did you expect anything else? (students laugh ) LEWIN:
45.75. 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 bob
and I swing down with the bob 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? (students respond ) LEWIN:
Some of you think it’s the same. Have you thought about it, that I’m a little bit taller
than this object and that therefore
maybe effectively the… the length of the string
has become a little less if I sit up like this? And if the length of the string
is a little less, the period would be
a little shorter. Yeah? Be prepared for that. On the other hand,
I’m also pre… well, I’m not quite
prepared for it. (students laughing ) LEWIN:
I will try to hold my body
as horizontal as I possibly can in order to be
at the same level as the bob. I will start
when I come to a halt here. There we go. (students laughing ) LEWIN:
Now! (students laughing) You count! This hurts! Aah! (students counting
in background ) (laughter continues ) LEWIN:
I want to hear you loud! STUDENTS:
Four! LEWIN (groaning ):
Halfway! LEWIN:
Thank you! STUDENTS:
Six… LEWIN (groaning ):
Seven… Eight. LEWIN (groaning ):
Nine… Ten! LEWIN:
Oh! (cheering and applause ) LEWIN:
Ten T with Walter Lewin. 45.6 plus or minus 0.1 second. Physics works, I’m telling you! I’ll see you Monday. Have a good weekend. (applause )


  1. I always prefered Lagrangian Mechanics when dealing with pendulums, especially pendulums on pendulums.

  2. Watched the whole video just to know the concept behind banana on his shirt😂.. And x°° is amazing… I used to write d^2 x / dt^2😍👌

  3. What's that banana meaning?😋, any specialty of banana at your shirt of your thinking. Or just fun

  4. Dear sir ,
    Please sir help me to understand explaination of reflection of a longitudinal wave by a rarer medium. I searched a lot on internet but found nothing sir…

    Thanking you

  5. Sir could you please explain the part at 25:15 ( that number is one with a hundred zeros… Part ) about the errors..
    Thank you

  6. Great pendulums experience!!! Thank you so much, professor!!! I will never forget that T will just relate to the length!

  7. I am preparing for my medical entrance exam (NEET-UG) from home, without joining coaching. And sir, you are my biggest saviour.

  8. Hello Walter Lewin! Big time fan for a few years here. For as long as i have watched your lectures i have always wonder what brand of chalk do you use Professor. I really enjoy practicing Physics on chalkbords like you did in these classic lectures. I can't find quality looking chalks like the ones you use:)

  9. There we see the science in action…. The vocabulary shortfalls to explain the ecstatic enthusiasm to see your demonstrations… 😇


  11. Why does the gyroscope rotate 15 degrees per hour everywhere on earth but a foucault pendulum only does so at the pole? Speed of the gyroscope spinning vs long period of the pendulum?

  12. I’m a high schooler taking physics and calculus. I am so happy that I’ve understood the derivation of the harmonic motion formula! By far the best explanation I’ve ever seen!👍🏻 thank you for sharing this priceless knowledge for free. 😊😊😊

  13. its funny to me how during this time you dont see kids wip out their phones to post on their stories…..these were the good days

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