# 1. Simple Harmonic Motion & Problem Solving Introduction

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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.