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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: I'm Wit Busza,

professor of physics at MIT. I'm joining my colleague,

Professor Walter Lewin, to help you understand

the physics of waves and vibrations. Now you may well ask, why

spend so much effort on waves and vibrations. And the answer is very simple. If you take any system,

disturb it from equilibrium, from a stable equilibrium,

the resultant motion is waves and vibrations.

So it's a very

common phenomenon. Not only is it that very

common, understanding waves and vibrations

have very important practical applications. And furthermore, the

fact that they exist, that this phenomenon exists,

has tremendous consequences on our world. If waves and vibrations were

different or didn't exist, you wouldn't recognize

our universe. What is the role I am

playing in this course? To answer that question,

I have to remind you what is the scientific method. In essence, the scientific

method has two components. The first, you

look around and you describe what you see in the

one and only language that can be used, or we

find that can be used for the

description of nature. That this mathematics, in terms

of mathematical equations. The second aspect is

since the universe is describable in terms

of mathematical equations, we can solve those equations. And that means predict

result of situations, of experiments, which

we've never seen before. Again, this is important

for two reasons. One, practical– to be able

to predict what will happen.

But the other far

more important is that it is the way we have,

the objective way we have of checking whether our

understanding of the universe is correct or not. If the predictions do not give

the right– do not correspond to what one actually sees,

you know, your theory, your understanding is wrong. My role is related

to the second part. In other words, what I would

want to help you learn, take a given situation,

convert it into mathematics, solve it, and predict

what will happen. We call that problem solving. OK. Let me immediately start

with a concrete example. What I have here,

describing a situation which we would like to understand. Imagine you have

an ideal spring, a spring that obeys Hooke's law. As I've shown here is the

spring constant k, length, natural length l0, and your

suspend it from the ceiling. Imagine that you take

a mass, a small mass, m, and you attach

it to that spring. At some instant of time, and in

the proceeds of attaching it, you may stretch the string.

So the spring may at this

instant not be stretched. But let's assume

while you're attaching and you've stretched

the spring a little bit, you're holding it, all right. At that instant, it's

velocity is 0, stationary. You let go. The question is,

what will happen? Can you predict what will be

the motion of that particle? You know, you've

seen this often. But a priori, it's not

obvious what will happen. The spring may pull the mass up. The mass may pull

the spring more down. It may oscillate. Everything, until you've

understood what's going on, you cannot predict the outcome. So let's assume that at

some instant of time, we call that time t, it's

as shown on the right. In order to be able

to describe this, I have to tell you

where these masses are at these various times. So I will define a

coordinate system.

This is a one

dimensional situation. So I only need one coordinate. And I'll call it the y. My y will be up. All right. Now I also have to measure

things from some location. So I need to define what

I mean by y equals 0. And I will define y

equals 0, the position where if the mass is at that

location, the force of gravity pulling it down and the

spring force pulling it up cancel, so that there is no

net force on the particle. So y equals 0 is the

equilibrium position.

And then the position

of the spring when it has no mass attached

is the distance y0 from that at t equals 0, the position

I will say is y initial. That's some number. So that's a known quantity. All right. And that any other instant time,

I defined it as y equals t. That is the physical situation

I wish to understand. I want to know what

happened with that spring. So now I will translate

that into mathematics. I will now try to give you

a mathematical description of that situation. So we know that we are dealing

with forces and masses. So to describe that, we

use Newtonian mechanics. So here is now my

mathematical description. The mass is a point m, of mass

m, on which two forces act. There is the force fs due to

the spring and the force fg due to the fact that this mass

is in the gravitational field, and therefore there is

a gravitational force on this mass, fg.

OK. We call this a force diagram. Or some people call it

the free body diagram. Now this mass, because

of the force acting it, its motion will change. And it will have

an acceleration, which I will call a of t. And by the way, that of course

is the second derivative of y with respect to dt. It's a vector. It's in the y direction. And in order so I don't

have to write things over and too many things

in these equations, I will define the symbol

y with two dots on it as the second derivative

of y with respect to time. y dot is the first

derivative– in other words, the velocity, et cetera. And by the way, you may notice

I'm going very slowly here. I'm doing that intentionally. I'm going to go here in gory

detail every part, you know, because often I know

that when one goes to a lecture, or studies

in a book, et cetera, you look at some step

from one step to another.

And you can't figure it out. The reason for it often is not

that you are not smart enough to do it, is but the

because the teacher or whoever wrote

the book, et cetera, is so familiar with

the material he will do several steps in

his head or her head, and you don't know about it. For this first

example, I will try to avoid anything of that kind.

Later on, I'll go faster. And I'll do the same

as everybody else. But the moment, as I say,

I'm going in gory detail. OK. So this is the diagram,

this free body diagram, of the situation. And I know from

Newtonian mechanics that if there are forces

acting on that mass, that mass will have

an acceleration, which will be equal to the

net force acting on it, divided by the mass, the

inertia, of that system. So that is what a will be. I further know the

force is vector, so the net force

acting on this mass is the sum of those,

the vectorial sum, of those two forces. So f is the sum of the

force due to the spring and due to gravity. Next, we also know

something about the spring. I told you at the

beginning that I'm considering an ideal spring. So for the purpose

of this problem, I'm assuming I have this

fictitious thing, a spring which essentially has

no mass, massless, which obeys exactly Hooke's law.

And here I can't help

digress and point out to you that that's a terrible misnomer. There is no Hooke's

law of nature. It is an empirical relation

which tells you the force that the spring exerts when you

stretch it a certain distance, all right. But anyway, it's

stuck historically. It's Hooke's law. So Hooke's law,

from Hooke's law, I know what will be the force,

fs, when the situation is as shown over there,

all right, at time t.

So at this time, this

extension of this spring will be, of course, y0 minus yt. OK. And so I get that the

force due to the spring will be the spring constant

times its extension at that instant of time. It is a vector. And y0 is a bigger number than

yt, this is a positive number. Therefore, the stretched

spring will pull the mass up. So this is in the y direction. This is plus. How about the

gravitational force? Well, that is, of

course, the minus mg, the force of the

gravitational field on that. And it's minus in

the y direction here, because it's

pulling this mass down. OK. Now what else do we know? We know that we could get

everything done very carefully. We know that we

defined y equals 0 to be the equilibrium position. Therefore, when y is 0, we know

that the second derivative of y is 0. It's not accelerating. So that's a condition

we must not forget.

Another thing we know that

initially, in other words, at t equals 0, the position

of that mass is y initial. Finally, I told you that

the velocity of that mass was 0 at t equals 0, stationary. So this is the beginning of our

translating all the information we gathered here

into mathematics. Let me continue now

using this information and try to reduce it to the

minimum set of equations. From a equals fm, from this,

I get that the acceleration is the total force divided by m. I can now replace

these two forces from the information

I wrote over there. And so that is equal, 1

over– this is f over m.

It's 1 of m times

the net force, which is the force due to the

spring minus the force due to the gravity, OK. So from this, I can now actually

write an algebraic equation, rather than the vector one. Notice – ha ha. I have noticed myself

even something. Here this has to be

in the direction of y. OK. This is a vector equation,

but all the parts are in the same direction,

in the y direction. Therefore, I can rewrite this

just the equation for the one component and not bother to

write the y hats throughout. So this equation I've rewritten

now just removing y hat. So this is how the mass

will be accelerating.

Unfortunately, it's

a single equation, but I have more than

one unknown in it. Because I don't know y0. And I don't know y of t. Clearly, I won't be able to

solve that equation, all right. But at that stage, I go

back to the information I told you at the beginning. We defined y equals 0

to be the place where a y double dot, the

second derivative, is 0. Therefore, I can write

that at position. When y of t is 0, this is 0. So 0 is equal to 1 over

m, into ky minus mg. And I immediately from

this get that ky0 is mg. Therefore, I have

found what y0 is. OK. Great. So there's only

one unknown here. So using this

information in here, I end up immediately

with this equation, that the second derivative

of y with respect to time, the

acceleration of the mass is equal to minus k

over m times y of t. At this stage, I will really

find this quantity, k over m.

For the time being, you can look

at it as just for convenience, less to write on the board. But later, you see this

will help us understand how to deal with

different situations. But for the time

being, you can just think of this as a

convenience, so I can write less on the board. And I end up finally

with one equation. The second derivative of

y with respect to time is equal to minus a constant,

that's k over m, right, times the value of y times t.

This is the equation of

motion for this mass. It tells me in mathematical

form how the motion of that mass changes with time. I can now actually

predict what will happen in this

particular situation. Because I know what was

the motion of it at time 0. I know that at time 0, the

position was y initial. And the velocity was 0. These three lines are completely

equivalent from the point of view of understanding

the motion of the mass to our original description. This is a physical

description of the situation. This is a mathematical

description of the same situation. So we've achieved step one. We've translated a

physical situation into a mathematical one. Let me now try

from this, I should be able to predict

what this mass will do. OK. I'm now switching into

the world the mathematics. As I just am

repeating here, I've gone away from a

physical description to a mathematical description. This is pure mathematics.

I have an equation, a

mathematical equation, for y of t. It's a second order

differential equation. I had the boundary conditions,

or initial conditions. I can solve that

using mathematics. OK. Let's do that. So I'm now doing

pure mathematics. I don't want to teach you math. That's the role of the

math department, all right. So how do I solve that equation? And let me tell

you how I solve it. I am make use of the

so-called uniqueness theorem. I know, or the

mathematicians have told me, that if I find a solution to

that equation which satisfies– if I find a solution which

satisfies that equation, and if it has the right

number of arbitrary constant, then I have found the one

and only general equation, which is a solution of that.

Let me be concrete. y of t equals to a cosine omega

t plus phi, where A and phi are arbitrary, are arbitrary. They are some arbitrary numbers. But a number is there. This can be 7 and this

can be 21 degrees, or whatever, but any number. This equation satisfies

my differential equation. If you don't believe me, try it. Differentiate this

twice, all right, for any value of A and phi and

you'll satisfy that equation. So this is a solution which

satisfy that equation. It has the right number

of arbitrary constants, that two arbitrary

constants in here. And therefore, this

is the only solution in the universe

of that equation. OK. Now, so being a

physicist, I don't care how I got the solution. Once I had the

solution, if I know it's the only one

that exists, I'm home.

Now you can say well, I

suppose I didn't guess it. Well, there are many ways. You go on the web and find it. You go in the book and find it. You ask your friends

what it is, all right. That's mathematics. And once you've found the

solution, we can go on. All right, so this is the

solution of that equation. Next, if that's

y, what is y dot? What is the rate of change of t? That's going to equal minus

omega 0 A sine omega 0 t plus phi, OK. Can I predict what will happen? All right.

I still need, in order to

be able to predict what will happen, I need

to find out what are the values of

A and phi which satisfy the other

information right here. See, I told you that we

reduce that physical situation to a differential equation,

the equation of motion for this mass, including the

information about where it was at some instant of time, how

it was moving, et cetera. So I need to make sure that

this equation satisfies these boundary conditions. In other words, it its

thees boundary conditions which will determine

what are the A's and phi for the particular problem

that I had there, OK.

And so what I do is– let's,

for example, takes here. Because I see 0. Y dot t is 0, all

right, at t equals 0. Well, when t equals 0, this

is minus omega 0 A sine phi. OK. Therefore, I immediately

conclude that phi is 0. OK. Next, I know– so now

that I know that phi is 0, I can go back to this equation. This is now 0. And we know that y at t

equals 0 is y initial. But that t equals

0 cosine of 0 is 1. Therefore, A is y initial. And so I get finitely y of t is

equal to y initial, all right, times cosine omega 0. Let me now replace

it with the 1– well, let me leave it as

omega 0 t pluse 0.

That, and I can rewrite

this, putting all the numbers that I have, y initial cosine. And I'll now even

replace omega 0 by k over n, square root

of k over n, times t. Notice there are no

unknown quantities in here. This tells me two things. At any instant of

time, I can calculate where this mass will be. It's given by this equation. Secondly, I can describe

the kind of motion it does. What is this equation? As a function of

time, this corresponds to an oscillating position y. So this mass, when I

let go, will oscillate. What will be the period? How long will it take before it

comes back to where it started? Well, the period t

will be how much time do I have to add to this

t, so that the angle here changes by 2 pi? Well, that's obviously

2 pi root of m over k.

OK. So I've achieved

what I wanted to do. I've taken a physical situation. And I have predicted if I

let go what will happen. This is the motion

it will experience. This is the period. I can predict the

time, et cetera. At this stage, let's

stop for a second and consider what we've done. Because it's the

essence of– this is a good example of the essence

of the scientific method. We have taken a

physical situation. We've described it in

terms of mathematics. Then we made an act

of faith that if I take the mathematical

equations and I solve them, that the resultant

answer will actually correspond to what

nature will do. If you stop to think

about that, it's amazing. Nobody understands that fact. Why that's true. Why it happened. In other words,

nobody understands why nature can be described

in terms of mathematics.

OK. But it is that fact which makes

the scientific method possible. Finally on this note, let me

give a quotation from Einstein which beautifully summarizes

what I've just said. And that is the following, "The

most incomprehensible thing about the universe is that

it is comprehensible." The fact that we can follow

this procedure is amazing. OK. Let me at this stage go and

take another example, all right. So let's take another example. Consider the

following situation. I take something like

a ruler, a uniform rod. And I put a nail through

it, some kind of a pivot. There is some pivot. I pivot the ruler on it. And it's hanging like this. OK. Let's assume the mass

is m of the ruler. The length is l. It's a uniform ruler,

a rod of some kind. And at t equals 0, I

give it an impulse. I give it a little impulse,

so we are now at t equals 0. We give it an impulse. At that instant, the ruler

is still hanging vertically. Let me, just so that when

you look on the board, you may be confused

in which plane I am.

This is the vertical plane. So this is up. So I give it an impulse. So at that instant

of time, this ruler will have an angular velocity

which I will call theta dot. This is at time equals 0. And it has some

number as a result, depends how big an

impulse I gave it. And so that you remember

what I'm talking about, I like to give this,

instead of using a symbol, I'll call this angular

velocity at t equals 0. So this is some number, so

many radians per second. That's at t equals 0. And I'm now going to

follow this method again. I want to know what will

be the motion of this. What's going to

happen to this ruler. Is it going to start spinning

around this, like this forever? What will happen? So I will try to translate

this problem into mathematics.

Because of the

mechanical constraint, at some instant of time,

the ruler may be doing this. Let's call this the time t. This is time t. And time t is like this. And I've got to define

some coordinate system. So I'll take this angle

from the vertical. And I call that theta at time t. That's why I call

this theta dot. This is the rate

of change of that. So at some instance of time,

it will be at this position, all right. At that instant of

time, it'll have a velocity in this direction. And we'll have an acceleration

in that direction. So for example, the

acceleration will be theta double dot at time t. And just so that at this stage,

I will still to remind you that's alpha, alpha time t. Because often alpha is

used as the acceleration. So at the moment,

I just want you so you can easy for you to

see what I'm talking about. So at some instance of time,

that is the physical situation. I would like to now convert

this into mathematics.

Follow the same

procedure as before. I need to write the

equation of motion for this. And I need to write down

the initial conditions. So how do I do with that? So now I start off by

the free body diagram. Here is the pivot. That's the route. This angle is theta t. There will be a force acting. We're now dealing with

rigid body motion. So today we did Newtonian

mechanics for masses and forces through a single

point mass and forces. Now we are doing a Newtonian

dynamics for rigid body motion. You know that if a rigid body

is in the gravitational field, the gravity acts force fg.

We can analyze it, as

if there was a force fg g acting through the center

of mass here of the body. So this length now is l over 2. So there will be

a force fg acting. And as a result, there will

be torques about this point. Now let me say the following. We are dealing here with motion,

rotations, in a single plane. And so we are dealing

about rotations, about an axis

through this point p. We're not dealing with

three dimensional rotations, but simple situation

where all the motion is about a single axis, which is

perpendicular to this point, p.

There will be a

torque about p because of the gravitational force. And as a result, there's going

to be the acceleration, which as we've said over there,

is theta double dot of t. Now, we know that torques gives

rise to angular acceleration. Let me define that we will take

clockwise motion, clockwise motion, clockwise

rotations to be positive. So any rotation, this angle,

for example, I am sorry. I meant anti-clockwise. Anti-clockwise is positive. Look at this. If this rotates like

that to this angle, this I take to be a

positive number, it's an anti-clockwise rotation. Similarly, if this acceleration

is a positive number, it's accelerating

in this direction.

Since we are dealing with

rotations about a single axis, we don't have to go to

the vector formulation. We can consider it

just the magnitude. And we know that

the acceleration is equal to the torque divided

by the moment of inertia. Or you may have seen

it the other way. Torque Equals I alpha. I prefer it this way. For me, it's more logical. The angular acceleration is

a consequence of the torque. So I write it like that. So this is the dynamic

equation, which tells you how the motion of this

mass changes with time. All right. So alpha is theta

double dot of t. OK. What is the torque at

that instant of time? Well, you know general torque

is r cross f, all right.

That's true in three dimension. So it will apply here. So the torque is going to be

this force times this distance. OK. So it's going to

be– let's write it. The force is mg right,

times l over 2 sine theta, theta of time t. OK. That's the torque about this

axis p on this rod, all right. And it's divided by I, where

I is the moment of inertia of this rod about

an axis through p perpendicular to the board. OK. Now we need to

calculate the moment, in order to continue further,

we need to calculate I. Since we know this

mass of the rod. And we know it's a uniform rod. And we know it's length

l, we can calculate it. You know how to do it. If you don't, you can look it

up in the book on mechanics, all right.

Or just look up the

moments of inertia. And you will find that

the moment of inertia, you will find that the moment

of inertia I for a rod like that is 1/3 the mass times

the length squared. OK. So now I have to continue. But I've run out of board space. So I'm going to erase

the board at the far end. And we'll continue from there. So I erased the board. And then so that you don't have

to look backwards and forwards, I've started rewriting it and

I realized that I actually missed the negative sign.

So I'm going to correct it here. So that's why it's

completely written out. So let me just remind you. The situation we

have is this rod, which at time t, we

define this angle to be theta t, the

rotation of the rod. It has an acceleration,

theta double dot t. And we are considering rotations

about an axis perpendicular to the board through

this point here. OK. We know, that was the

last thing we did, that the acceleration is

given by the torque divided by the moment of inertia.

All right. The torque is mg, l over

2 sine theta, I derived it for you before, divided by I. But what I neglected

to put a negative sign. And that you could do

in your head, right. Consider we've taken all

the rotations to be positive if they're anti-clockwise. So this angle is a

positive rotation. This would be, this direction

would be a positive rotation. But the torque if

you look at this, there is a force

acting down on this. So about this point,

it's trying to rotate this in the clockwise direction. And so it's minus. And I didn't– it would have

naturally come out if I did the full vector calculation,

the torque is R times F.

It would have come out, the

sign would have come out. So that's where this

minus sign comes in. OK, so this is where we got

on the board over there. And now let's continue. We can replace I from here. And we get that

theta double dot of t is equal to minus, all right,

3 halves, 3 l, 3 halves, not l. It's divided by 2. l is at the bottom. Sorry. 3 halves g over l times

sine theta, theta of t. OK. I'm sorry. Sine theta of t. OK, as before, to simplify it,

I will write omega, I'll define. Let's define omega 0 squared to

be equal to 3 halves g over l. With this definition,

we get that theta double dot t is equal to minus omega

0 squared sine theta of t. OK. So this is our equation of

motion for this problem. That's the equation of motion. And these are the

boundary conditions.

So these three equations

are a translation of this problem in the

language of mathematics. If we now want to

predict what will happen to this rod at

some other time, we have to solve these equations. And admit now, I have a problem. If you remember, when we

did it for the spring, the equation of motion was one

where I guessed the answer. I don't know what the

answer is of this. If you go into

books, you will find that this is not one of

the differential equations which you can

analytically solve. It's, in fact, a second

order differential equations with transcendental

functions in it. So this is not something

we know the answer to. So the only thing if I want to

now predict what will happen, I have to numerically

solve this. And then I can– I have

enough information. I can numerically solve this

equation with these boundary conditions and predict

what will happen.

That's not very instructive

for the purpose of course at the moment. So let me do something else. OK, let me modify the problem. Rather than take the

problem we took, let me say, how about if I took

this rod and gave it only a very tiny impulse. So this angle is small. Let me make the angles

sufficiently small, such that sine theta of t

is always approximately equal to theta of t. Depends how well you

want to approximate this. But typically, if you use

your calculator or computer, up to about 10 degrees, that

approximation is pretty good. So I will now change my problem. And I said OK, let's see whether

we can predict analytically the motion of the rod where

I give the impulse, which is sufficiently small, that

this angle is always small. Under those conditions, note

that my equation of motion becomes theta double dot of t

is equal to minus omega squared times theta at t.

Because sine theta t is always

approximately equal to theta t if I take the

angle small enough. And eureka, I can

solve that one. Because that's exactly the

same equation we solved before. OK, so we get the

solution to that equation, is theta of t is some constant

cosine omega 0 t plus phi. As before, A and phi are

some arbitrary constants.

And clearly, if it worked over

there for that same equation, it works here. The only difference here is we

have theta of t instead y of t. That's just different

symbols, but the solution is exactly the same. So we know that's the

solution of this equation. We know the boundary conditions. Therefore, we can

predict what will happen. Let's continue and do that. So from here, you get theta

dot of t is equal minus omega 0 A sine omega 0 t plus phase. OK. And we have to put in

the boundary conditions. OK.

Now at t equals 0, OK, we get

that this is at t equals 0. So at t equals 0,

we get theta of t 0 is equal to A cosine phi. OK. Therefore, phi is pi over 2. That's a possible value of phi. Now that gives me that

if I is pi over 2 here, we get to that theta dot of t,

which is equal to the angular, angular velocity at t

equal to 0, all right, will be equal to minus omega 0

A sine omega ) t plus pi over 2. OK, from which I

can get that A is angular velocity over t plus 0. And so my final solution

is that theta of t is equal to angular

velocity at t equal 0 divided omega 0

sine, sine omega O t.

OK. And I want to make sure I'm not

making a sign mistake again. I'm not. All right. And so, and omega 0

we know, and so the in terms of knowing

quantities, the answer is angular velocity

of t equals 0 over. And omega 0, we have

found defined to be that, so the square root

of 3 g over 2 l times sine square root

3g over 2l times t. Now this is theta t. So we have completely

solved the problem. And we have

predicted the motion. So as before,

following this process of taking the

physical situation, describing it in

terms of mathematics, solving the

mathematical equations, including all the

information we have about the problem,

the boundary– initial conditions or

boundary conditions, we can predict what will

happen to this angle as a function of time, and

also the kind of motion this is an accelerating motion.

I can also predict, as before,

that the period of this will be 2 pi, 2 pi square

root of 2l over 3g, et cetera. OK. Now, one of the things you'll

notice, that in some ways, it seems I'm repeating myself. We took completely

different situations, and yet the result, the

equations of motion, and the results, have

good very similar form. Now this is part of the beauty

of the scientific method.

Because it turns

out that very many different physical

situations can be described by the same

mathematical equations. So once you've solved

the problem for one physical situation, you have

automatically have solved it for an almost infinite number

of other situations which are described by this

same mathematics. Finally, let me do just more

as a question of practice, one more problem of this

kind that apparently seems to be

completely different. I'll take a problem from

electricity and magnetism. Let me consider the

following situation. So now we're going to

a different problem. The physical situation is

suppose I have two plates, two metal plates, and I

connect them with a wire. Schematically, it

consists of a capacitor C connected to an inductor. This is a schematic

representation of two parallel plates

connected, short circuited by a wire. I will assume for simplicity

here that these wires have no resistance,

superconducting, all right. Any loop like has

an inductant L. And the capacity

between these is C. So this is an L C circuit. And I'm going to assume that

at time equal 0, so this is now time equals 0, I have a

charge here, minus Q0 plus Q0 here, OK.

And let's assume that at time

there's even a current flowing, so I is 0 here. So this is a system which is

disturbed from equilibrium. And what will happen

is a function of time. I will do the same and

almost boring you to tears, I'm going to, you'll

see I'm essentially doing the same problem again. I will now consider this circuit

at some arbitrary time t, derive the equation of motion

for the charges in the current, therefore translate

this physical situation, or describe this

physical situation in terms of

mathematics, deriving mathematical

equations, solve them, and predict what will happen.

So I just follow what I

just did a second ago. So at some instant of

time, that same circuit L will have some current

I of t to the charge minus Q of t plus

Q of t, all right. This at time t. So from this, I can derive

the equation of motion. Let me remind you

about Faraday's law. You know that if you have

current coil in the loop, it produces magnetic

flux in that loop. The changing flux gives rise

to an EMF around that loop. To be specific, Faraday's

law I can write. If I take this circuit of

the wire, the integral of E dot dl around a closed loop,

that is equal to minus du phi. 5. Now watch out. The Greek alphabet has a

limited number of letters so you'll find one constant

is reusing the letters. But at the moment

not to confuse you, I'm going to put

here magnetic flux, total magnetic, total magnetic.

So phi is the

total magnetic flux linking this circuit,

all right, dt. So Faraday's law tells us

that the integral of ED all around this loop

will be equal two minus the rate of

change magnetic flux. This is the dynamic

equation which tells you how this behaves. It is the analogous to

Newton's law f equals ma in the case of our

mass, or 2 torque is I alpha in the case of

rotations, et cetera.

This is the non

dynamic equations. So let me calculate this. And now I'm going through

the– around this. And you find since this wire

I'm assuming is superconducting, there can be no electric

field inside it. So the contribution

to this line integral is 0 when I go through the wire. So the only place where this

line integral is non-zero is between the

plates, all right. And that is simply the

potential difference between those,

which is q over c. OK. Q at time t over c, where

c is the capacitance is. That is the integral of

EDL around that loop. And that's going to be equal

to minus D dI at time t dt. All right. Because the magnetic

flux, this is by definition of the

inductance or first inductance is that the total flux

linking the circuit when the current flowing in it

is I of I, the total flux is L times I.

Here, I have the

rate of change of that flux, so it's equal to this. OK. Now we know by

charge conservation, that the current I of t. What is the current? It's the charge is

flowing per second will be equal to the number of

charges per second that arrive at this plate here or the part

from there is equal to dQ dt. OK. Or in other words, Q dot. OK, let me continue. So from these two equations,

right, the dIdt therefore, this is the second thing,

so I end up from there that Q double dot,

second derivative of t, is equal to minus right

1 over LC times Q of t. Eureka. We have once again

the same equation. This I can define as

before, omega O squared. If I define that as

one over LC, all right, then what I have here is Q of

t double dot is equal to minus omega ) squared Q of t.

Again, we have come

to the same equation. This is the same

equation of motion as we came in the

other two situations. So the answer will be the same. The variables will

be different here. It'll be the charge that

will be changing with time, while there in the one

case was the angle. In the other case was

the position of the mass. OK, and the solution

to this problem, I can now write

immediately, is Q of t is A cosine omega 0 t plus fe. Note that this is the fe is

nothing to do with that phi. OK. And Q dot of t, which

by the way is I of t, is equal to minus omega o

A sine omega o t plus phi. OK. Now as before, what

actually happens depends on the

initial conditions. And we look at that

picture on the top board. We know that initially Q is Q 0. And we know that

initially Q dot t is I0. To save time, I'll

just immediately write. You can do that in your head. And if you the

write that out, you find that if you've used

those two conditions, you find that time phi this

time is equal to minus IO over Q0 omega 0 and A is equal

to Q0 over cosine phi.

I saved time without just

solving algebraic equations. Take these two equations. Consider t equals 0, the

values of those quantities, and just solve for the two

unknowns and you get this. And so once again, we have

predicted what will happen. And what I would like

to just at this stage emphasize that although

we have taken three different physical

situations, in each case, we took the system, displaced

it from equilibrium, let go and we wanted to see

what will happen. In all the cases,

it turned out that the mathematical description,

the mathematical equations are identical in form.

And so they gave,

not surprisingly, the same kind of motion. This motion that we

see in all those cases, we called simple

harmonic motion. It has the characteristic that

if you displace the system from equilibrium, it oscillates

with harmonic motion, meaning it oscillates

as sine or a cosine of different phases, et cetera. If you tell me any

one of these systems where it was at any

instant of time, I can predict it

forever in the future. Now finally, the

last few minutes. Some of you may have noticed

that in each of these– or I told you at

the beginning that I can take a physical

situation and describe it in terms of mathematics, and

thereby predict the future. But in each case, I in

some ways almost cheated. I said let's consider

an ideal spring. We'll assume it has no

mass, that it exactly obeys Hooke's law. Or when it came to

that rod, I assumed that it's only displaced

by small amounts, so that's sine theta equals–

I can approximate with theta.

In the case of the

electrical circuit, it maybe not so

obvious what I assumed, but I certainly made assumptions

about that the wire is perfectly conducting, et cetera. And I didn't discuss

in detail what happens in between the

plates or the capacitor where the fields are, et cetera. One is doing approximations. In reality, if you look

at any physical situation in the world, it's always

incredibly complicated. It's never that you have an

idealized situation like this. So to what extent does

what we have just done correspond– is

it useful at all.

And the way I'll answer

it is by another example. This is the last thing I'll do

on the top of a simple harmonic motion, last problem. Suppose I'm looking

out of the window. And I see there is a tree and a

branch and a bird lands on it. Do I understand

what will happen? It's clearly

extremely complicated. The mechanics of the

branch is complicated.

There is air friction. Nothing is simple about it. And yet you and I can

predict what will happen. You know what will happen. As the bird lands, it'll

start oscillating and finally come to rest, very much

like harmonic motion. I claim I can use the word,

I understand what's going on. And the reason why I claim

that is the following. That to understand

something, all I would like to understand

the general features of what's going on. I don't need to know what

every atom in the branch is going on in the

process of trying to understand what

the bird is doing. If I want to understand

what atoms are doing, that's a different story. And so one of the important

abilities we have to develop is to be able to, when

you see some situation, model it in terms of the

most important aspects of the situation. And let me be concrete. In this case, I can say look,

I can model this approximately as the branch I'll

treat as a spring, of some spring constant k. The bird I'm going

to treat as a mass m.

And I'm going to

consider this situation to be modeled by a mass being

placed on a spring and let go. Now is that going

to be exactly this? No. But from the point of

view of understanding the general features

of this, it will be a reasonable approximation. Now how can I check, this

is the scientific method, that this is a good

approximation is the following.

Make a prediction. Suppose when I see

the bird landing, it makes five oscillations, five

oscillations in ten seconds. OK. I can predict approximately

once the oscillations have died out how much the bird

has compressed, distorted, this branch. In other words, from

the moment it landed, what distance will the

branch, its position, be lowered when

it comes to rest. So let's model it. I'll model it first

of all, as I did here, as a mass and a spring

problem using– this is now the same problem we did at

the beginning that is still on the board here.

I can calculate that for

this idealized situation, the period will be equal

to 2 pi square root m over k, for this

idealized model. In reality, there's friction. So this oscillation

will be dumped out. And you would have learned

from Professor Walter Lewin that if you have a

damped oscillator, the frequency of oscillations

does not depend significantly on how much damping it is,

provided it is weak damping.

So I would make the

assumption that the period will be given by that. I also know that at the time

when the motion has been damped out and the bird has come

to rest, at that instant, there's no net

force on the bird. And so the force

of gravity on it will be equal to the restoring

force due to the spring. So mg is equal to kl. From this, I get that

m over k is l over g. All right. But we know what the period is. We said five oscillations

in 10 seconds. So the period is two seconds. So two seconds will

equal 2 pi divided by but this, which is l over

g, the square root of l over g. Square this and calculate

the one unknown l and you get one meter.

So my prediction

is that this bird will, after it's settled down,

roughly be lower by one meter. It's certainly not going

to be one millimeter. It's not going to

be one centimeter. It's not going be 10 meters. And if you go and

you measure it, you find this is

approximately correct. The fact that I can

predict it is for me the same as saying I

understand what's going on.

I realize it's not exact. But with the approximations

that I've made, I get an answer which

is consistent with what is observed. Today I have tried to tell you

what my role in this course is, what I'm trying

to help you learn. I intentionally

went very slowly. I used the word gory detail. I tried, in particularly

in the first problem, not to miss any steps. And what we covered

today is the phenomenon of simple harmonic motion. It occurs whenever you

have any system which is displaced from equilibrium

where the restoring force is proportional to

the displacement. And it illustrated

that you could have very, very different

physical situations which, when translated

into mathematics, give essentially

the same problem. So it's a beautiful example

of the scientific method where we utilize this same–

well, once we've learned it for one system, we can apply

the results to another system.

So as I said, today I did

simple harmonic motion. Next time, we would be

considering problems to do with simple

harmonic motion, but which includes

friction, damping. We'll then go on to talk about

harmonic oscillators which are driven. So we have driven

harmonic motion. And gradually in

the course, we'll go to more and more decrease

of freedom, waves, et cetera.