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:

Ten. STUDENT:

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:

Seven. STUDENTS:

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 ):

Oh… STUDENTS:

Five! YOUNG MAN:

Halfway! LEWIN:

Thank you! STUDENTS:

Six… LEWIN (groaning ):

Oh… STUDENTS:

Seven… Eight. LEWIN (groaning ):

Aah… STUDENTS:

Nine… Ten! LEWIN:

Ah! STUDENTS:

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 )

What a great teacher.

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

WAW GOOOOOOOOOOOOOOOOOOOOOOOOOD TEATCHER TINKYOU

Wow … very nice e

I guess he's hungry for bananas😜

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😍👌

Awesome

now I realize that

physics works

Physics works

Physics works!!!

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

He is an eccentric. Wears something on his right pocket every time. Today it is a banana.

Its so enjoyfull your explanations its not a class but is an intersting movie to me. thanks alot

If only i knew about these incredble lectures back at my school time

You made me love physics a lot

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

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

Thank you

Walter white for chemistry

Walter Lewis for physics

There's a Watermelon on his Shirt 😂

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

Sir your vedios are addictive..I can't study the school stuff caz these vedios make more sense

Happy teachers day sir. Love and respect from India. 🙏🙏🙏 You are the greatest teacher of all time.

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

Money don't matter no more!

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:)

44:05 – Professor Lewin fiercely attacking the start button, complete with a karate shout. 🙂

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

46:22 HE SAID NOW , AND I START GETTING GOOSEBUMPS TILL THE END WHEN HE SAYS "PHYSICS WORKS". TRIBUTE TO THIS GREAT TEACHER WHO IS SERVING FOR THIS WORLD ALTHOUGH HE IS GETTING SO MUCH PAIN . #Bornforgreatness

Without walter lewin I don't know where my physics would be right now.

Why is there a length in the formula for time period of pendulum?

Sir you like our school super sir

Yesmy eyes also just speedified when he drew dotted lines!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?

Love from India 👍👍👍👍❤️❤️❤️❤️❤️❤️❤️

sir which class do you read in this lecture?

i like sir

we study this in class 11 in India 😅

This is 12th class high school levels physics….

like seriously if my university professor be like him, every1 could have seen me..but now u cant see me 👌

Great prof! You are take so conceptual lecture. I am physics student I so like your lectures

I wish I could understand everything that he said. Whatever, I loved him and his passion of get teach.

Love you sir…you are the best.

Вот это учитель физики! Круто

I just simply loved it

Nice effort sir….I like the way you teach ……..dream to be in your classroom…

Really he is teaching physics,, just wooow and I am fallen in love with physics,,,

the profesor is in simple harmonic motion

We study these in 11 & 12, intermediate class in Bangladesh.

i cant believe there are empty seats in his lectures

9:50 wow

Best theory and practical class in one time.

AMZING❤

He is one of the best professor in world….

Brilliant teacher…. theory as well as experiment at same time… awesome

I wish my professors were like him…my professors just derive they dont explain concepts😔☹️

Physics works 😍😎

شكرا لك

LAW

Anything For PEACE

In PEACE

Students who are attending live classes are so lucky ….

Wish having a teacher like him ❤️

No one:

Me at 4 AM: 46:21

😍😍😍😍 I wish I were there 🙂

What if feynman take this lecture ?Would he hurt himself ? XD

46:51 that girl in the right lol

11:15

46:19

Se tivéssemos essa qualidade de aula no Brasil …,

What master class.

You are the only one English man whom English I can easily understan

Definitely better than my professor

Masterpiece 💙💙💙💙💙

👏👏👏

Wow, sitting in one of your lectures would be an honor.

If i only have a professor like him

Sir you are great thanks sir

6:55 the face you make when you don’t understand a bit. LOL

I would love to be his student.

Respect from india sir I love your way of teaching

You are insane. I wish Indian Education was like this.

Oh no dude!!! In India we know all this stuff in 12th grade 😂😂😂

brilliant master! you are wonderful. I wish the comment section knew the brilliance of this lecture.

I have one doubt…can I ask?sir I hope you'll clarify…frm India ❤

At 8:24 you sir mentioned that this equation is the most important of all of physics .why is that.

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. 😊😊😊

3:55how accurate

Great proffessor

Great teacher

You are the best teacher i have ever seen.

Love from India

Banana for beauty? or is the audience in which he is giving a lecture empty?

45:28

Physics Works!!!

6:37

just out of curiosity, why are you wearing a banana?

Where does he teach???

This is where the banana duct taped to a wall got its start at being art

2019?

put playback speed rly fast so it feels like ur super lost in a complicated class

Lol i love how when i watch this theres a whole trend of people with bananas taped to the wall/their clothes and stuff

Thanks sir for such a wonderful lecture.what is x double dotts which you replace by acceleration 'a'

This lecture is $150,000 worth in 2019, thanks to the banana 🙂

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

HATS OFF TO HIM !!!!