Okay, so what problem were we
trying to solve?
We were trying to solve the
problem of masses,
coupled to springs,
and acted upon by various
forces.
And the equation that we need
is the following [writes
equation on the board].
So, this is the most general
problem we looked at.
This is F_0.
I might have just called it
F last time.
It doesn't matter;
it's a matter of this symbols,
okay?
Alright.
Let us first divide everything
by m.
Divide everything by m,
then it becomes x double
dot, plus γx dot,
plus
ω_0^(2)x,
equals F_0
over m,
cos ωt.
I'm using the fact that
ω_0 was
defined as the root of k
over n.
We used to call it ω,
but now I'm calling it
ω_0,
and you should understand why
I'm doing that,
because there are now two
ωs in the problem.
There is, the ω of the
driving force,
so you should imagine an
oscillatory force grabbing this
mass and moving it back and
forth.
ω_0 is the
natural frequency of the mass.
If you pull the spring and let
it go, that's the frequency that
it will vibrate.
There are really the natural
frequency that is fixed by
k and m and
ω is for you to decide.
You can decide to apply to it
any force you like of any
frequency.
Professor Ramamurti
Shankar: So first,
take the easy case in which the
right-hand side is 0.
No driving force.
So, I'm saying let's consider a
simple case,
F_0,
or F,
the applied force vanishes.
That's the problem I spent a
considerable amount of time on
yesterday.
You can say,
"Well, if there's no driving
force, why does anything move?"
That's actually correct.
If you took a mass in a spring
and didn't do anything to it,
the solution of this equation
is x = 0,
and that is fine.
But it also has interesting
non-trivial solutions because
you can pull the mass and then
let it go.
After you let it go,
there are no forces on the
system except the force of
friction and the force of the
spring, and that is this
problem.
Now, how do you solve this
problem?
I told you the way you solve
this problem is you make a
guess: x of t is
some number, A times
e to the αt.
See if it'll work.
And you find it will work and
you can do this in your head.
I certainly don't want to
repeat all of that.
The reason it's going to work
is that every time you take
derivative of x,
you bring down an α,
and still keep e to the
αt.
So, when you put that guess
into this equation,
you're going to get
α^(2) plus γα,
plus ω_0
square times A,
e to the αt
equal to 0.
And if this is going to vanish,
it's not because of A
and it's not because of the
αt,
it's because this guy in
brackets is going to vanish.
So, when that's 0,
you all recognize this to be a
quadratic equation.
You go back to your days in the
sand box [earlier in the course
or your life]
when you all knew what the
answer to this is;
the answer to this is minus
γ over two,
plus or minus γ over
two square, minus
ω_0 square.
These are the roots.
Now here, you have to look
at--So anyway,
what does this mean?
This means that yes,
you can find a solution to this
equation that looks like
A to the αt;
A can be any number you
like.
No restriction is placed on
A.
But α cannot be any
number you like.
α has to be the
solution to this quadratic
equation and there are two
solutions.
So, you can call them α
plus and α minus.
That's when I branched off into
two different cases.
Case one, you can see look,
many of the quadratic
equations, you know the way it's
going to fall out is depending
on whether the stuff inside the
bracket is positive or negative.
So, let's take the easier case
when it's positive,
that means γ over two
is bigger than
ω_0n.
γ, you remember,
is the force of friction so we
want it to be not only non-zero,
but bigger than this amount
controlled by the natural
frequency.
In that case,
let's not go into details,
but you can all see you put
whatever value you have for
γ, the square root of
this is some number.
The whole thing is going to be
two real choices you get.
One will be minus γ
over 2, plus whatever is inside
the square root.
The other will be minus
γ over 2,
minus what you have in the
square root.
The only thing I want you to
notice is that both these roots
are negative.
This guy's obviously negative
because it's a minus γ
over 2 minus something else,
that's definitely negative.
You have a little [inaudible]
worried [inaudible]
here, that this is minus
γ over two and this is
plus something.
But the plus something is
smaller than the minus
something, because what's inside
the square root is γ
over 2 square.
If you took the square root of
that you'll get exactly
γ over 2.
But you're taking away from it,
this number;
therefore, this square root is
smaller than this number,
so the overall thing will still
come out negative.
So, why am I so eager that it
should come out negative?
Why am I eager that the root
should be negative?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: No,
even if I had positive,
if I took two derivatives,
it would work out.
Yes?
Student: Because if
alpha is greater than one,
then the e won't
converge to cosine.
Professor Ramamurti
Shankar: Won't converge to
what?
Student: The cosine
function?
Professor Ramamurti
Shankar: This is not a
cosine function.
This is never going to be.
A cosine function is related to
e to the iωt or
e to ix.
These are all real
exponentials,
yes?
Student: If alpha is
positive t goes up
[inaudible]
Professor Ramamurti
Shankar: Yeah,
that's our main problem.
When you write exponential
function--;Usually,
people write exponentials like
this, for population growth.
Everyone says our problems are
increasing exponentially,
they always think of
exponentials as rising.
Of course, it's also possible
that if α were a
negative number,
you could write it as A
times the absolute value of
α just to show you it's
a falling exponential.
And we have a stomach only for
falling exponentials.
You've got to think if I pull
this mass and I let it go,
I expect x to go down
from whatever value I have,
not to keep growing
indefinitely.
That doesn't make any physical
sense.
And one of the great things
about the union of mathematics
and physics is the equation
automatically works out that
way.
It gives you an answer that
makes sense.
Okay, so what you get then is a
solution that looks like this.
One number A times
e to one of the roots
plus another number,
B, times e to the
other root.
I wrote them--they're both
e to the αt,
but I write it as minus the
absolute value so you can all
see they're falling
exponentials.
You understand if a number is
negative, like negative five,
that can be written as minus
one times the absolute value of
the number.
So, it is e to the the
αt, but I want to write
it in this form,
so it's falling.
So, this is a case where if you
pull the mass and you let it go,
it will just relax eventually
coming to x = 0,
which is its normal equilibrium
location.
Now, how about the numbers
A and B?
Right now, they're completely
arbitrary.
And another thing you guys
should know by now is how we
find A and B.
In order to find A and
B, we need to be given
more information other than
simply the mass and the spring
constant.
They're all sitting here,
and the friction coefficient,
but that's not enough to tell
you what it does on a given day.
It wants to know,
well, yes, it's the mass,
there's that the spring,
but did you pull it or did you
release it in this way or that
way?
That's what controls the answer.
They're called initial
conditions.
One initial condition is what
was the value of x at
time, t = 0?
That's asking you how much you
pulled it when you let it go.
The other initial condition is,
what is the velocity?
Namely, the derivative of
x at time 0.
Most of the time,
what you will find as you pull
the mass and release it,
then it's got a non-0 x
and a 0 velocity.
But it doesn't have to be that
way because I could have pulled
it and released it at t =
0,
starting to move inwards,
and that's when you jump in and
you set your clocks to 0.
Well at that instant,
x has a value and the
velocity also has a non-0 value.
They don't all have to be 0,
but they are some two numbers
that you can pick at will.
You can see that if I put
t = 0,
I'm going to get A times
e to the 0,
which is one.
I'm going to get A +
B is x(0).
That's one condition on
A and B.
I will take a derivative,
namely, find the velocity,
then put T = 0,
and set that equal to some
pre-assigned number,
you'll get a second equation
for A and B,
and we all know we can solve
those two equations to find
A and B.
That's how you fit these three
parameters to the solution
tailor-made for that particular
situation.
You should also ask another
question.
Why is it that I can pick these
numbers, x_0
and v_0,
but why not keep going?
Maybe they're the acceleration
of times 0 is some number.
Why is that not up to me to
decide?
Can anyone tell me why that's
not negotiable?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Depends upon what?
Can I pick the acceleration of
a body at will?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah,
I'm not sure I heard the exact
answer but it sounds close
enough to what I had in mind,
which is Newton will tell you
what the acceleration of a body
is, namely mass times
acceleration is the force,
and the force,
once it's given in any given
problem, which is given by where
the body is,
and what the velocity is in
this example,
you have no choice on what the
acceleration is.
But initial position and
velocity are not given by
Newton's law.
Newton's law says,
if the force is so and so,
the acceleration is so and so,
but he doesn't tell you what
should be the velocity at a
given time, what should be the
position at a given time.
That's why they are two free
numbers, that's why you have an
answer with two free numbers
that you can pick to bring it in
accord with these two numbers.
Okay, this solution to the
equation with F = 0 from
now on, I'm going to call
x_c,
where c stands for the
complimentary function.
Now, there's another thing that
could have happened here,
which is in fact another
alternative.
You can think about what I'm
saying.
The other alternative is when
this γ over 2 is smaller
than this ω_0.
Then, of course,
you have the square root of a
negative number.
So, then you go back to the
stage where you're finding the
root, and you say α
equals minus γ over two,
plus or minus the square root
of γ over two^(2) minus
ω_0^(2),
but that number is negative.
So, that can be returned as a
negative sign,
times the square root of this
number,
and the square root of minus
one will come out as an i
out front.
So, either you will get two
real roots, this is a
mathematical result,
it has nothing to do with
physics,
a quadratic equation with real
coefficients either will have
two real roots,
or it will have two roots which
are complex conjugates of each
other.
You learned the notion of
complex conjugation,
which is to exchange i
to minus i,
you can see the α plus
and α minus have the
same real part,
minus γ over 2,
and have opposite imaginary
parts.
So, in this example,
α plus is the complex
conjugate of α minus and
I've told you this relationship
works both ways;
therefore, α minus is a
complex conjugate of α
plus.
So, what's the solution going
to look like?
It's going to look like
Ae to the minus
γt over two,
e to the iω
prime t,
but I'm going to call this
fellow in brackets as ω
prime.
Then, there's the second
solution.
The real part is exactly the
same, the imaginary part is
opposite, minus iω prime
t.
ω prime is a new
frequency connected to the
natural frequency and the
frictional force.
That's why, we've got three
ωs.
We've got to keep that in mind.
ω_0 is a
natural frequency of vibration.
If I have a driving force,
I reserve the symbol ω
for the frequency of the driving
force.
Here is yet another quantity
with units of frequency which
I'm calling ω prime.
So, what does this guy look
like?
But first of all,
it's full of complex numbers
and we don't like it.
We know the oscillator should
have a real coordinate,
so you ask, I've not picked
A and B yet,
how shall I pick them so that
x is real?
Well, forget A and B for
a minute and look at this number
and this number.
Do you guys realize that
they're complex conjugates of
each other, because i has
gone to minus i.
So for example,
if A and B were
equal numbers,
this would be a real number
because number plus its
conjugate,
you killed imaginary part
because they have opposite
imaginary parts.
So, A and B real
[and equal as stated above]
is a good answer,
but it's too restrictive.
All we will demand is that if
A looks like absolute
value of A times e
to the iφ,
B should look like the absolute
value of A times e to the
minus iφ.
In other words,
B should be the complex
conjugate of A.
That means the same absolute
value, but opposite phase.
I remind you,
in the complex plane,
if a number z is sitting
there, the complex conjugate is
sitting here,
they have opposite angles,
φ.
So, if you do that,
then you get A absolute
value, e to the minus
γ over two,
times e to the
iω prime t plus
φ, plus e to the
minus iω prime t
plus φ.
Now we have made it real.
A and B,
both being fully real means
this φ is 0,
but you see φ doesn't
really have to be 0.
That is too much of a
restriction.
φ can be whatever you
like provided the φ that
A has,
the phase that A has,
is the same [correction:
should have said "minus"]
the phase that B has.
That means the angle in the
complex plane of the number
A and the number B
have to be opposite of each
other.
This, you guys should know what
it is.
I've told you many times.
This is unforgivable if you
don't know what this is.
e to iθ,
plus e to the minus
iθ over two,
is cos θ.
Since you don't have the over
two, you can write it as
2A, e to the minus
γ over two,
cosine of whatever is in here,
ω prime t plus
φ.
At this point,
I'm going to change my notation
a little bit in view of what's
to come.
Instead of calling this
φ, I'm going to call it
φ_0.
Alright, because another
φ is going to come in
and I want to separate those
two.
So, two times A is a new
number, I'm going to call that
C, and this is the bottom
line.
The answer looks like e
to the minus γt over
two, cos ω prime
t plus
φ_0.
So, we had done all this
yesterday.
I've given you the notes,
but I want you to understand
where we're going with all of
this.
If you pull a mass and let it
go, it turns out there are two
things that can happen.
I really wish I had time to
bring you these different masses
and show you.
If you have no friction,
everyone knows what's going to
happen.
You pull the mass and let it go;
it oscillates forever.
In real life,
there's always some friction,
so let's turn on a little
friction.
What'll happen is what's very
familiar and expected.
We start with an amplitude of
five, go to four,
go to three,
go to two, rock back and forth
and slowly it may wiggle around
but you won't be able to see it.
That's called a damped
oscillation.
That's really what's described
by this solution.
Here, it looks like a cosine of
something, but the number in
front of it itself is falling
with time.
And the picture for that will
look like the following:
if you just draw an
exponentially falling function
as your amplitude,
and you multiply that by an
oscillating cosine it will do
this.
This is a damped oscillation.
But another extreme case which
you may not think about is,
if you crank up the friction
more,
‘till γ over two is
bigger than
ω_0,
then what happens is if you
pull the mass and let it go,
it doesn't even go to the other
side of x = 0.
It just slowly gets to 0,
in an infinite amount of time,
you understand?
If you pull it from here,
you let it go.
No friction is this [moving
hand back and forth]
for eternity.
Small amount of friction has
slowly damped the oscillation.
Too much of friction,
you come back slowly and you
come to rest.
These are different cases and
the mathematics tells you the
different cases so that's what
we like about the union of
mathematics and physics.
Once you wrote down the basic
law of Newton,
you have no choice as to where
it takes you.
You go where it takes,
and these are the different
solutions.
Okay, so this is what I did
earlier on in the day.
But now, let's go back to the
more general problem I had,
which is a right-hand side
which, is not 0.
So, let's write the right-hand
side which is not zero as
x double dot,
plus γx dot,
plus
ω_0^(2)x,
equals F_0
over m,
cos ωt.
This is called a driven
oscillator.
Driven oscillators are present
all the time.
For example,
if you have a swing,
the swing goes back and forth.
You may choose to push the
swing back and forth.
If you've got a kid in a park,
you're applying a force which
is an oscillating force.
Whereas in the absence of
external forces,
all oscillations will
eventually die down.
Do you guys understand that?
Why?
Why do all oscillations die
down in the absence of external
forces?
Student: Friction?
Professor Ramamurti
Shankar: Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah,
well in this case,
it's not air resistance,
but friction.
But what friction does is that
in the presence of friction,
you cannot prove the Law of
Conservation of Energy.
We saw that.
Whenever friction acts,
the energy total mechanical
energy goes down.
So, kinetic plus potential will
decrease with time so the
oscillations get damped.
But if you are pushing it from
outside, you can provide a
constant source of energy so you
can maintain a steady
oscillation,
and that's the problem we're
looking at.
That is this problem.
So, how did we solve this
problem?
Well, I said I don't want to
solve this equation.
I'm going to manufacture
another problem.
z double dot,
plus γ times z
dot, plus ω_0
square z,
is F_0 over
m, e to the
iωt.
So, let's be very clear.
This is a real physics problem
that occurs all the time.
This [pointing to other
equation]
is clearly not of our world,
because in our world you don't
have a force,
e to the iωt.
All forces in the real world
are real.
So, this is a complex force.
If you break down the e
to the iωt,
you all know it's cos
ωt, which I like,
plus i sin ωt,
which I don't like.
I'm solving a problem in which
on top of the force I'm really
applying, there's an imaginary
force put on top of it.
So, what is the strategy?
Why have we taken the problem
and making it worse than it is?
Because we have a strategy and
the strategy is the following.
If you have right-hand side
cos ωt,
ask yourself what you're trying
to do.
You're looking for some
function x(t),
which will obey this equation.
You can try various functions,
maybe x is a cosine of
time.
If it's a cosine of time,
two derivatives will be a
cosine.
No derivative will be a sine.
One derivative will--I'm sorry,
no derivative will be cosine,
one derivative will be a sine.
So, you cannot make it into a
cosine.
If you took a sine,
the middle guy will look like a
cosine.
You can try to balance that
against this;
the end two terms look like
cosine, so neither sin nor a
cosine works.
It turns out something in
between will work.
But something that always works
is the following.
If the right-hand side,
instead of being cos
ωt, was e to the
iωt, you can guess the
answer immediately.
Namely z itself will
behave like e to the
iωt, because then if I
take two derivatives,
that's going to look like
e to the iωt,
times some numbers,
that's going to look like
e to the iωt,
that's going to look like
e to the iωt.
So, the beauty of the
exponential is when you take
derivatives you keep getting the
exponential.
Okay, but now,
the problem is this is not the
problem.
No one told you to solve this
problem, but our strategy is the
following.
The right-hand side has got a
real force and imaginary force.
The real force will produce the
x that I want.
The imaginary force,
purely imaginary,
i times sin
ωt, will produce a
purely imaginary answer.
So, at the end of the day,
I will find the answer and keep
only the real part.
That's the plan.
That's going to be the
strategy, so that this z,
this z is going to be
the x that I want and,
will have a part hanging
around, y,
that I don't want,
but y is being driven by
the sin ωt.
And x is being driven by
the cos ωt.
But now, let's solve it by
inspection.
So, let's make a guess that
z looks like some number
z_0,
e to the iωt.
Now, it's very easy to take
derivatives of exponential.
Just pull down iω every
time you see a derivative.
So, z double dot will
become iω square,
which is minus ω
square.
z dot will become
iω, there's a γ
there, and
ω_0^(2) is
just multiplying by a number.
So, this is what I get on the
left-hand side,
and I want it to equal this on
the right-hand side.
And you're almost there because
the time dependence of the two
sides are identical,
so you can cancel it and what
you find is, yes,
I can satisfy the equation
that's given to me,
provide the number
z_0,
the number in front of it will
obey condition.
That's trivially solved,
because we all know how to get
the solution to that one.
z_0 is
obtained by dividing both sides
by everything that multiplies
z_0,
which happens to be
ω_0^(2) minus
ω square plus
iωγ.
So, the beautiful property of
the exponential function is that
you take what is a differential
equation and you turn it into an
algebraic equation.
Because the act of taking
derivatives generally is a nasty
operation.
Sines become cosines,
x cubed,
becomes three x square.
But exponential remains
exponential.
That's the property you're
trying to exploit here.
So now, we've got
z_0 to be this
one.
And z itself,
the complex function I'm
looking for is
F_0 over
m,
e to the iωt,
divided by this quantity
downstairs.
This quantity downstairs I'm
going to call I(ω).
I(ω) is a complex
number.
It's got a real part and an
imaginary part.
The real part is
ω_0^(2) -
ω^(2).
The imaginary part is
ωγ.
It's not a fixed complex
number, but every frequency of
the driving force ω,
it is a different complex
number, but if it's sitting at
one frequency of the driving
force,
it's a particular number.
Everyone should know how to
work it out.
ω_0 is the
root of k over m;
ω is given to you,
that's the real number.
γ is given to you,
that's an imaginary part.
If you actually put in numbers,
it might look like six plus
five i.
In the end, the thing I called
impedance or I,
is just a complex number.
But it varies from frequency to
frequency.
You can see that.
Okay, now we've got z of
t.
If the problem given to us was
z we're done.
You don't have to do any more
work.
Here's the answer.
But we were told to take the
real part of this answer because
x was the real part of
the answer;
y was not part of the
game.
How do you take the real part?
Your temptation may be well,
e to the iωt is
cos ωt plus i sine
ωt, so drop the i sin
ωt.
But that would not be right.
Now, why is that not right?
Anyone know why that's not the
way to get the real part yet?
Yep?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar Yes.
This number downstairs is not a
real number.
It's a complex number.
So, if you have a complex
number on the top,
divided by the complex number
in the bottom,
the real part is not simply the
real part of the top.
You got to write the whole
thing so you can clearly
separate the real and imaginary
parts, and that's what I'm going
to do next.
Every complex number we know
can be returned as an absolute
value times e to the
iφ.
Right?
Every z looks like
[inaudible]
e to the iθ,
because then it's [inaudible]
cos θ which is
x,
plus i times [inaudible]
which is iy.
So, every complex number can be
written that way and what will
be the magnitude of this guy?
Well, by the Pythagoras'
theorem, it'll be the square of
the real part,
plus the square of the
imaginary part under root.
Every complex number that you
have will have an absolute value
which is that square plus that
square under the root,
and it'll have a phase
φ, obeying the condition
tan φ,
is the imaginary part divided
by the real part.
I want you to realize that in a
given problem,
with a given mass,
given friction,
given frequency,
both these are known.
They're simply a magnitude and
the phase of I can by
found by this process.
Now, write it like that,
then you get
F_0 over
m, divided by the
absolute value of I,
e to the iφ,
and I got e to the iωt
on the top.
I'm almost there.
I'm almost there because
e to the iφ in
the bottom can be written as
e to the minus iφ
on the top.
So then, this whole thing
becomes F_0 over m
divided by the absolute value of
i, e to the iωt,
minus φ.
Now, it's easy to take the real
part because this is cos
ωt minus φ plus
i + sin ωt minus
φ.
These are all real numbers.
So, they'll just multiply the
whole thing.
But the real to imaginary
separation now comes from just
taking the cosine part of this
exponential.
And that is our final answer.
That's the answer we've been
looking for, which is x(t) is
equal to F_0
over m,
divided absolute value of
I, cos ωt -
φ.
Why did we go through the whole
business of complex numbers?
We went there because the
answer is not a cosine and it's
not a sine.
Cos ωt - φ,
if you know your trigonometry,
is cos ωt,
cos φ - sin ωt sin
φ.
So, cos ωt has got some
sin ωt and that's why
it's hard to guess that answer.
But if you go through complex
numbers, it comes out very
easily.
So, the next thing is to
analyze the properties of the
answer you got.
You can write this whole thing
in front as some overall
amplitude x_0 cos
ωt - φ and you ask,
"What does this function look
like?"
Well, this function is obvious
to all of you,
it's oscillating.
Notice that even though it is
oscillating like a cosine,
it's not in step with the
applied force.
The applied force is cos
ωt.
Its maximum occurs at
0,2π,
4π, et cetera.
This guy's maximum occurs
somewhat later.
The first maximum occurs when
ωt = φ.
So, it's said to be lagging the
function.
You can imagine two functions
oscillating.
One is behind the other one.
It reaches the maximum slightly
later.
So, the first property is in a
driven oscillator,
the x will not be in
step with the F.
It will be lagging.
Or, it could be leading depends
on what φ is.
Then, what about the amplitude
of oscillation?
That's the thing that's very
interesting.
How big is the oscillation?
It looks like
F_0 over
m divided by the absolute
value of I.
Let me write the absolute value
of I for you.
So, the amplitude of
oscillation depends on the
frequency.
You've got to draw the graph.
That's the function of
frequency.
What does the one over this
look like?
Pardon me?
Student: [inaudible]
Professor Ramamurti
Shankar: Yes,
but (A minus B)^(2) = (B minus
A)^(2);
that's why I've not been paying
attention.
Okay.
Yeah, so I'm pretty careless on
what I put in,
but this is one occasion where
I sort of know it's okay.
I'm going to square this,
I don't have to remember who is
ω_0 and who's
ω.
I have to be very careful where
I don't square things like here.
But if I'm squaring it,
it doesn't matter.
Okay, so go back here and plot
something divided by this
denominator and ask,
"What is it at various
frequencies?"
When ω is 0,
that goes away,
that goes away,
you get
ω_0^(2),
and ω_0^(2)
is k over m,
and I claim what you will find
initially is this will become
F_0 over
k.
That'll be the value of this
whole amplitude at 0 frequency.
I hope that makes a lot of
sense to you.
Zero frequency means the force
of not even oscillating.
It's a constant force.
What x are you going to
get?
You're going to get the force
divided by the force constant.
That's the x to which
you'll pull the spring.
That's why if you put all the
ωs and masses back into
their proper place,
you'll find there's
F_0 over
k.
Then what happens is as
ω increases,
this denominator,
one of these guys,
a denominator is starting to
vanish.
If γ were a very,
very tiny number,
when ω_0 =
ω,
you will be dividing by a very
small number and the answer will
have a huge rise.
Then, you can see when
ω goes to infinity,
very large ω,
you're diving by ω
square, it'll fall off like
this.
Now, one of the homework
problems is to find out exactly
where it peaks.
But if γ is very small,
you don't have to do much work
to know it peaks at ω =
ω_0,
but otherwise it's near that.
That's the phenomenon that's
called resonance.
Resonance is the following.
You are applying a force,
F_0 cos ωt.
You go on varying the frequency
and in response to your force,
the system oscillates.
The amplitude of oscillation is
the largest when the applied
frequency is very near the
natural frequency.
If you take the limit of a very
small γ,
this maximum is very close to
ω_0.
That's called resonance.
If γ is really tiny,
you can get graphs where it's
incredibly peaked.
So, that's when systems start
vibrating a lot.
That's why when your car goes
over certain bumps,
they're periodic,
they're hitting your car at a
certain frequency,
and if that frequency is
anywhere near the natural
frequency or vibration of your
suspension system in the car,
the car begins to resonate.
There's this famous bridge in
Tacoma that was getting hit by
these winds, they were coming at
some frequency and by bad luck,
they coincided with the natural
frequency of the bridge.
So, there is this famous movie
made for high school,
in which the bridge oscillates
more and more and more and
twists and turns and collapses.
That's another example of
resonance.
People always say,
"Yeah I can resonate with what
the person is saying."
What does that mean?
A person is saying whatever the
hell he wants.
Some people think it's
interesting and some people
think it's boring.
It's the same thing here.
The force is acting at a
certain frequency.
If your natural frequency is
close to that,
your response is high.
If it's low, you don't care.
Same thing.
That's mechanical systems
selecting a frequency.
In fact, this is how radios
work.
You have a radio,
if there's only one radio
station, you build an electrical
circuit in which the currents
like to go back and forth and
oscillate at a certain
frequency,
a radio station should send
signals at that frequency,
your response will be very
high.
But what if there's more than
one radio station,
which is a fact of life?
What do they do?
What they do is each one sends
a signal at a different
frequency.
You take your radio and there
are dials inside.
Well, I forgot,
do you kids--do you know what a
dial is?
You're not just pushing some
other button.
There is a time when you had to
turn a dial.
When you turn a dial,
you're changing the properties
of the electrical circuit and
changing its natural frequency.
You move the natural frequency
until it lines up with the
frequency of the radio station,
with one particular station.
Then, the response from that
station is very high and that's
the sound you hear.
Other stations are also
present, but you're not near
their maximum,
they are sitting,
some right here and some right
here.
So, you want the station
frequency to come under this
resonance peak.
So, if you've got a bad radio,
it will pick up sound from many
stations because it's not very
refined.
If the peak is broad,
you will get sound from many
stations.
You want to build a system
which is very finely sensitive
to the frequency.
That requires γ to be
very small.
If γ is very,
very small, you can see if
γ is going to 0,
the denominator is one divided
by 0, it's going to be infinite.
So, if γ is very tiny,
one over very tiny,
is very big,
and you can get enormous spikes
in this part and that's what
people need.
Okay, the last thing you need
to do this problem set.
By the way, I know this problem
set is very nasty.
Towards the end,
it got tough,
you got email,
I agree with you that this
whole course is somewhat more
difficult than it has always
been.
Okay?
I'm aware of that fact,
and I will bear this fact in
mind when you go into your final
because for some reason I wanted
to use problems not in our book
but for technical reasons so I
got them from here or there or
made them up.
They generally tend to be more
challenging and more time
consuming.
So, if you thought this was a
tough course,
well, it's supposed to be a
tough course but maybe not this
hard or not this time consuming.
So, I will remember that in the
end when I make the final exam
or look at your grade,
I will have to compare you not
simply to your predecessors,
but I will have to cut you some
slack and then compare you to
previous generations.
So, this use of complex numbers
is something from which I will
not, however,
back down.
It turns out that yes,
you can do all of this without
using complex numbers.
Why?
Because in the real world,
everything has got real
numbers.
So, you can in fact,
by an orgy of trigonometric
identities solve it with sines
and cosines.
But eventually,
I'll give you a problem where
you just cannot do it in a
meaningful way.
The power of complex numbers is
they turn differential equations
into algebraic equations.
All you have to do is find the
roots of a quadratic equation,
whereas what you started out
was to find a solution of a
differential equation.
That advantage you will lose if
you stop using complex numbers,
and when you do quantum
mechanics in the second term,
the complex numbers are in the
equations of motion.
The analog of F = ma,
there's an i sitting
there.
So, you cannot escape the
i.
It's there, so I'm saying,
depending on how much exposure
you've had, you have to do more
or less work,
but you just got to get used to
complex numbers.
You got to get used to the
absolute value,
the phase angle,
how to manipulate.
If you already know it, good.
If you don't know it,
you have to do more work than
other people but I'm not going
to let off on it.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: No.
Anything you need to know with
differential equations is what
I've done for you in the class
today.
That's it.
That's the only part you know,
and with complex numbers,
you turned it into ordinary
equations.
So, you have to go over these
steps and work them out.
That's the part I cannot do for
you.
That's something you have to do.
Alright, what is finally
missing is that x of
t;
if I wrote for you what I wrote
above, which is
x_0,
I don't want to write the
x_0 again.
It's F_0 over
mi;
maybe I'll write the whole
thing.
F_0 over
m, divided by absolute
value i times cosine
ωt minus φ.
It turns out this is not the
full answer to the question of
what happens when I push a mass.
Let's understand why.
Suppose I go this mass and
spring system.
I grab at a t = 0 and
start shaking it.
Well, at t = 0,
x was 0 because I
grabbed it at t = 0.
But a t = 0,
if I go to this solution,
it's not giving me 0.
It's saying it's
F_0 over mi
cos φ.
So, what happened to the
freedom I had to assign the
velocity and position
arbitrarily?
Well, it turns out that I can
still do that.
That's because this solution
has got a part that's actually
missing, and the part that's
missing is this complimentary
solution,
Ae to the minus
γt over two,
cosine of ω prime
t minus
φ_0.
You can always add the solution
to this solution.
Why is that?
Because if x obeyed a
certain equation,
if I put that in,
I'm saying adding this part
will not change the fact that it
obeys the equation.
Let's ask ourselves why.
Go back to the top line there.
x double dot +
γx dot +
ω_0^(2)x,
will be equal to
F_0 over m
cos ωt,
if you put this in.
Now, what if you keep adding
this to the solution?
Will you mess it up?
Imagine adding this to the
left-hand side,
but I claim on the left-hand
side there'll be no extra terms
because this guy double dot plus
γ,
times this guy dot,
plus ω_0^(2)
times this guy is 0.
This is the function which you
will say is annihilated by the
operations on the left-hand
side.
It's like saying the following.
If you want a function whose
derivative is equal to sin
x, then I'm going to say oh,
I know the function is minus
cos x,
and you all know you can add to
that a constant.
Why do you add the constant?
Why doesn't it screw up
everything?
Because a derivative of a
constant is 0.
Well, here this is not a
constant and yet,
the combination of derivatives
appear in the equation act on
this to give you 0.
That's why you can add it
without screwing up what you've
done so far.
So, the way you do these
problems is you first find this
function that depends on the
applied force,
and you add to this solution
you get for free,
even in the absence of an
applied force.
Yes?
Student: What does the
x_C stand for
again?
Professor Ramamurti
Shankar: It's called a
complimentary function and this
is called a particular solution.
This particular solution
depends on the force and this is
the function.
I don't know why it's called
complimentary.
Maybe it compliments the
answer, or it's a freebie,
you don't pay for it with any
force, but it's called
complimentary.
But you've got to add the two
to get the final answer.
So, you will still have the
numbers A and φ
_0 hanging around,
and how do you select them?
You select them if on top of
it, somebody tells you that this
driven oscillator had a certain
coordinate and certain velocity
at t = 0.
So, you evaluate the whole
expression of t = 0 by
putting t = 0 everywhere,
you'll get one equation
involving A and φ
_0 and you'll take the
derivative for the velocity,
set that equal to some given
number.
You get a second equation
involving A and φ
_0.
You'll juggle the two equations
and solve for the two numbers,
A and φ
_0.
Okay, so that is the whole
story, okay?
That's what I tried to do
yesterday, but this is not the
first time it's happened to me
that I tried to go fast;
in the end I lost time because
I was not happy with the way it
was explained,
and I usually come back and say
sorry and then do it all over
again, okay?
But there's only so much I can
do.
The rest is up to you guys.
You have to do the work and you
have to do more problems.
Okay, so now,
we are moving off to another
topic.
The other topic--Somebody was
sighing with relief.
It's not really different.
Okay, these are waves.
But I think it's going to be--I
think this probably is the
hardest point of what we're
doing in this term,
at least mathematically.
Everything else we do,
heat, hot water,
cold water, that kind of stuff
you've done in high school,
that's all in your future,
it's going to be a lot easier.
Okay.
Waves.
So, everyone has a good feeling
for waves.
You go to a lake;
you know if you're a hit man
for the mafia,
you're dropping things in with
some weight.
They go down,
what's the first thing they do?
When bodies are falling into
the water, it's carving out some
body of water and diving in so
the water next to it dives in to
fill that hole.
But then the water,
even next to it,
tries to fill in the hole left
behind by these guys.
For you, it looks like a bunch
of ripples traveling outwards
from the center.
So, this is the classic example
of a wave.
If you've got one example,
you can latch onto it every
time I say wave.
So, what's the wave in this
problem is that if I draw a plot
of the height of the water,
versus the place where it
dropped, the rock--maybe I
should do a top view here.
This is the top view,
after a while you've got these
ripples emanating from this
point.
Or, if you kept your eye next
to the water you will find these
ups and downs going outwards.
So, the wave is some
displacement of a medium.
That's what people say.
That's not a perfect definition
because we know electromagnetic
waves have no medium.
They travel in a vacuum.
But I think for this purpose in
this course, you should imagine
waves as what happens when you
excite a medium.
So, here's the way to think
about the wave.
You all understand the harmonic
oscillator pretty well,
right?
Mass and spring system in
equilibrium will sit there
forever.
If you give it a kick,
it'll start vibrating around
its equilibrium position.
There, you have to keep track
of only one variable x,
which is the location of the
mass.
The wave is an oscillation of a
medium that means that every
point in space there is
something that's ready to
oscillate.
So, you start out a system with
infinite number of degrees of
freedom because the height of
the water at each point is an
independent variable.
It can jump up and down with
time.
And that's what is oscillating.
So, if you don't do anything to
the water, it'll stay at the
height, but if you fiddle with
it or drop something,
you hit it with something,
it'll start vibrating.
But except it's a vibration not
out of one variable but of a
variable in space and time.
x was a function of time
and the name we use for the
wave, if something is
oscillating such as the water
measured from some point,
x.
I'm going to use the symbol
ψ of x [and t]
to denote the height of the
water from its normal position
at the point x at time
t.
So, x is not a dynamical
variable here.
x labels the point.
Ψ is the dynamical
variable.
Ψ is the thing jumping
up and down.
Okay.
Now, there are two kinds of
waves;
there's the longitudinal wave
and the transverse wave.
And we define them as follows.
The wave travels out with some
velocity.
I'm going to use the symbol
v for the velocity of the
wave.
That's something for which all
of you would have intuitive
understanding so no need to mess
it up right now.
You go to the lake,
you drop a rock,
you see when the waves get to
the shore, you take the distance
divided by time,
that's the velocity.
In sound, you know,
you light up a firecracker,
after a while,
you hear it,
you can see how long it took
and how far you are and you can
find the velocity.
That's the velocity of sound in
air.
And all these velocities are
all variable.
Unlike the velocity of light
that doesn't seem to depend on
anything.
All other velocities depend on
various conditions.
The velocity of sound depends
on the temperature,
for example.
Or you can take the following
thing, you can take an iron rod,
which you know is made up of a
lot of atoms and you take a
hammer and you hit it here.
You basically compress these
atoms and they compress the
atoms next to them and the
shockwaves travels through this
rod;
that's also called sound.
That travels at a speed which
is much faster than the speed of
sound and air.
But it's technically also
called sound.
If you were a creature living
inside the iron bar,
you would think of it as sound.
Now, here's another way you can
hear it.
For example,
here's a standard example.
Here's the railway track,
and there is some kind of a
mountain here,
and you are over here and the
train is coming here.
It turns out,
if the train lets out a sound,
the sound will travel around
the bend and you will hear it.
But actually,
the pounding of the train on
the tracks will set up
shockwaves in the track that you
can hear by putting your ear to
the track.
And you will hear the train
before you hear it by,
depending on the sound coming
through the air,
because the waves going through
the solids travel much faster.
Okay, so I would recommend that
if you're going to do this,
you do this somewhere here
because you won't have enough
time to move if you do it here
[laughter].
So, most of the things--please
don't do this at home,
this should be done by a
trained person but I certainly
did it as a kid,
listened for trains before they
come around the bend and you can
actually hear the track.
Or in water,
the explosions in water,
you can hear them faster if
you're under the water than if
you're outside the water because
the waves travel faster in water
than in air.
Thunder and lightning is the
most famous example.
Something happens in the sky,
the lights get to you first,
and the sound gets to you;
they're traveling at different
velocities.
Okay.
So now, let's talk about
longitudinal transverse waves.
When I talk to you,
I think you sort of know what
happens.
The air in this room is at some
pressure, and that pressure is
constant, and the pressure is
pushing your eardrum from the
outside and going through your
Eustachian canal pushing the
drum from the inside,
and the two pressures balance
and you don't feel anything.
If I talk, the vocal chords
move back and forth and they
increase and decrease the
pressure as I talk.
If I push out,
I increase the pressure,
if I go backwards,
I decompress the air and reduce
the pressure.
Those shockwaves propagate to
your ear.
They hit your eardrum,
but the eardrum is getting the
pressure difference from the
outside, but the inside pressure
is fixed.
So, what happens is when I
decompress, the eardrum comes
out and goes in.
It goes back and forth and the
eardrum moves.
Then, behind the eardrum are
the little bones,
and then the bones go and there
is a tube and the doctors take
over from this point.
So, some loose hair is shaking
in this fluid;
then, it picks up some nervous
things and goes to your brain.
Then, you figure out and
actually know what the person's
saying.
You don't know why he's talking
in this particular fashion,
but you can at least understand
and translate.
That's a long chain of
transducers of energy from sound
to the stuff in your ear into
your brain.
I draw the line at the eardrum.
I don't want to go on the other
side of the eardrum because it's
very shaky turf.
But this part I know very well.
It's a sound wave and it's a
longitudinal wave for the
following reason.
If you look at the pressure
waves, the molecules of
air--they're going back and
forth,
follow my hands,
and the signal from me to you
is also going from me to you,
so the motion is in line with
the velocity.
The motion of the medium is in
the same direction as the motion
of the wave.
A transverse wave is what you
find in water if you are--or on
a string.
Take a string,
clamp down that end and you
give it a little flick like
that.
This bump after a while will
end up here.
So, the rope is moving,
transfers but the velocity if
the signal is to the right.
So, do not confuse the motion
of the medium with the motion of
the signal, okay?
The rope, at any given point,
is either at rest or going up
and down, but the signal is
going from left to right.
So, the medium doesn't have to
actually move in the direction
of the wave.
For example,
if a bunch of people line up
and each one tells the person
next to the next person some
secret and the secret is
eventually transmitted all the
way to the end,
what has traveled is a secret.
The actual person in the first
place has not gone over
physically to the end.
So, what happens is as you're
sitting there not doing anything
and suddenly you hear something,
you pass it on to the next
person, then you've gone back to
doing nothing.
Then, the signal travels.
That's what will happen in the
rope here.
This section of the rope is not
doing anything.
When this pulse hits it,
it'll jump up and down for a
while.
Then it'll quiet down and
somebody to the right will start
moving.
Okay, so there are the examples
of longitudinal and transverse
waves.
So, we have the notion of
velocity.
Now, I want to consider one
concrete example of a wave,
so you people have a good
feeling of how to study waves in
physics.
So, there are many, many waves.
Water waves,
electromagnetic waves,
elastic waves.
I'm going to do one.
I think to do one thing in some
depth will give you a feeling
for how we analyze waves.
That's going to be waves on a
string.
So, you have to imagine that I
have a string that's been
clamped at two ends,
and somebody plucks the string,
say, into that shape and lets
it go, we know the string is
going to vibrate back and forth.
Now, let's measure the
coordinate x,
which is the point on the
string, and ψ is the
displacement of the string at
that point.
You guys got that this is
really the x,
this is not my string.
That is my string.
Each point is labeled by an
x value that it will have
for the string to relax to the
normal position.
So, we are asking what is the
displacement of the string at
the point x,
at time, t.
So, ψ is our new
dynamical variable,
namely, the variable for which
we like to write the equations
governing its motion.
And we want to ask how does
ψ vary with time?
What will it do?
If I start it out,
what will it do?
Do you understand the parallel
to the mass and spring system?
The mass and spring system,
you got a mass and you got a
spring, you pull the mass out to
some new location,
you let it go and you want to
know what it does.
The answer was cos ωt.
Now, at every point x,
I have some string,
and I displace all those
degrees of freedom and I let
them go,
and they're going to do
something and I want to know how
they're all varying.
So previously,
we had only one coordinate
which I called x,
which varied with time.
Now with every point,
x, I have a coordinate
ψ that varies with time.
And our job is to find the
equation satisfied by ψ.
So, let us take a string and
the string has to be under some
tension.
That means you hang some
weights on the end or you
tighten it with some screws.
There's a tension so that the
string is dying to break apart
if you cut it.
So, T is the tension on
the string.
Without the tension none of
this would work.
So, here is the string.
And I'm going to analyze the
fate of a tiny segment of length
dx.
An essential parameter here is
called--which is μ,
the mass per unit length.
That means you put the string
on a weighing machine,
you find the mass and you
divide by the length.
For example,
if you've got a ten meter
string and it weighs one-tenth
of a kilogram,
then the mass per unit length
is one over thousand
[correction: should have said
"hundred"]
kilograms per meter.
And T is the tension.
The textbook may use F
for tension because T is
the time period also,
so it depends on what symbol
they use in the book.
I'm not completely sure.
I'm going to call it T
because it's been called
T in the past.
Now, I pull this thing and I
want to know what the whole
string will do.
Who is going to tell me now?
I mean, what authority will
decide the behavior of this
string?
What do we have to appeal to?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Pardon me?
Student: [inaudible]
Professor Ramamurti
Shankar: Very good.
So, he said,
find the force that will return
it to the original position and
what was the it that you
mentioned?
Student: [inaudible]
Professor Ramamurti
Shankar: Pardon me?
Right, but we are going to
apply--yes, any other answers
before I go?
Yes.
Student: Can't you model
the motion of the string by
saying that [inaudible]
Professor Ramamurti
Shankar: Yeah,
it will look like that.
That is correct.
But all I want you to
understand is,
there are no new laws that I'm
going to invoke.
I'm not going to say,
well, we studied masses and
springs, today it's time to
study strings and I'm going to
tell you a new law of motion.
There's only one law of motion.
That's F = ma.
The whole purpose of my
exercise now is to show you that
really does control,
that's why it's a super law.
Everything mechanical follows
from that law and I want to show
you F = ma is all I have
and I've got to use it properly
and that's going to tell me what
the strings will do.
So, what I do is--The string
is, of course,
a long and extended and
complicated object.
I isolate a portion of it,
of length dx,
and I'm going to apply F =
ma to that guy.
All I need to know if the
F, the total force on it,
and I'm going to equate it to
mass times acceleration.
Okay now, this force on the
little segment is the tension
pulling this way,
the same tension pulling that
way.
The tension on the string
doesn't change from point to
point.
But the angle at which it is
pulling is not necessarily the
same.
Here, there is some angle
θ + Δθ,
and here, there is an angle
θ.
This force is at an angle,
say, θ below the
horizontal, this is pulling
about θ + Δθ.
So, the string is curving;
therefore, the tangent to the
string at this end and that end
are not quite the same and
that's the key.
Because if they were exactly
the same, the same tension on
the two ends,
if you take any object like
this and apply exactly the same
force,
they will cancel because
they're opposite forces,
and F is 0,
ma will be 0,
and this thing won't move.
The reason it moves is because
even though the magnitude of the
force is the same,
the angle at which it's acting
is not the same.
So, I'm going to find out the
vertical component of the force
here and take the difference
from there [pointing at
picture].
So, the force here will be
T times sin +
Δθ.
θ is the angle with the
horizontal.
And the force on the other side
will be T times sin
θ.
θ is the angle made by
the string from the horizontal,
and that angle is simply
changing over this tiny length.
Next approximation--that's the
force.
That's going to be mass times
acceleration.
Mass of this little segment is
the mass per unit length,
times the length of the
segment.
Now, what is the acceleration
in the language of calculus?
What is the acceleration here?
Student: [inaudible]
Professor Ramamurti
Shankar: Pardon me?
Student: [inaudible]
Professor Ramamurti
Shankar: Ah,
he said d^(2)x over
dt^(2),
but the d by
dt^(2) is right.
It's not d^(2) of
x of ψ because
you've got to understand--I
think--Look,
this was not a fair trick,
but you have to be aware that
what is oscillating up and down
is the coordinate side.
x is not a real
dynamical coordinate.
x is a label on the
string.
What's jumping up and down is
ψ, so the acceleration
is the second derivative,
and I write partial derivative
because ψ can vary with
x, or it can vary with
time and I want to take the
second time derivative.
Now, come to the left-hand side
and you assume the angles
involved are very small.
If the angles involved are very
small, I remind you guys of the
falling identity,
sin θ begins as
θ plus dot,
dot, dot.
You remember the series I wrote
for you.
Cos θ is one minus
θ^(2) over two,
and we have decided that we are
going to treat θ so
small that θ^(2),
we're not interested in.
And tan θ,
which is sin θ over
cos θ,
is now θ over one,
which is θ.
So, small angle approximation
is sin θ,
approximately equal to
θ,
cos θ approximately
equal to one,
tan θ may be
approximately replaced by
θ.
So, first nice break.
We don't have to work with
sine, we just want T of
θ + Δθ minus that,
which is T of θ.
Δθ is a change in the
angle from the left hand of the
string bit to the right end of
the string bit.
Now, I got a dx here,
so what I'm going to do is
divide by a dx and
multiply by a dx.
We are almost there in deriving
the equation.
Look, I've been calling it
dx and Δx so
pardon that.
Just call it dx.
μ, dx,
d^(2)ψ over
dt^(2).
Now, what is θ?
I'm going to say θ is
the same as tan θ in the
small angle approximation.
tan θ is dψ/dx.
You know from the calculus,
if you move horizontally by
dx and you move
vertically by dψ,
dψdx is a tan of the
angle.
That's one of the first things
you learned in calculus.
So, this is really d by
dx of dψdx,
that becomes d^(2)ψ
over dx^(2),
equal to μd^(2)ψ over
dt^(2).
Now, some of you may find the
derivation hard.
Then forget about the details.
You're not responsible for that.
But some of you may say look,
I don't ever want you to tell
me something which you cannot
prove unless it's an axiom or a
law of nature.
Well, this is not a new law of
nature, so I have to tell you
where I got this.
I got this by applying F =
ma to a tiny portion of the
string.
And you can ask yourself,
when you pull a string up,
why does it come down?
It comes down because there's a
tension on the string and the
tension here and the tension
here have vertical components
that don't quite match.
You see the string is
flattening out like this;
the vertical downward component
of this is bigger than the
vertical upward component of
this force.
If it is resolved into the
vertical or horizontal part the
downward part is bigger than the
upward part in this picture.
Therefore, there's a net force
bringing you down.
That one is bringing you down
is because the derivative here
and here are not the same,
because the derivative is the
slope.
So, the whole thing depends on
the rate of change of the rate
of change, and that's why you
get this.
But the mass of the tiny
segment also depends on
dx, so dx cancels
in the end.
You drop it and you get this
equation.
So, what I want you to think of
is dθ/dx is really d by
dx of dψ/dx.
Well, I should actually use
partial derivatives,
because ψ depends on
x and t,
so if you're real careful with
that, you get d^(2)ψ
over dt^(2).
So now, I'm going to write down
the final wave equation.
From this point on,
you have to get engaged in the
problem because even though you
may or may not follow the
derivation,
you should know what is called
a wave equation.
This is the wave equation:
d^(2)ψ over
dx^(2) is equal to 1 over
v^(2),
d^(2)ψ over
dt^(2),
where v^(2) is equal to
t over μ,
or v is square root of
T over µ.
I just invented a symbol
v.
This is very important.
What you have to know is that
ψ, the displacement of
the string at the point labeled
x horizontally from the
left-hand,
at the time t,
varies with time according to
this equation.
This is a very famous equation
and it turns out you take any
elastic medium and you disturb
it a little bit from
equilibrium,
always seems to obey equations
of this type.
So, our job from now until some
time is to analyze the equation
and see what it means.
What are the consequences of
this wave equation?
And even though I called it
v, as units of velocity,
I have no right to argue that
it really is the velocity of the
wave.
We'll have to find out.
It's got dimensions of velocity;
you can check by putting the
units in.
Okay but now,
let me write down a solution to
this equation;
then, we'll understand what can
happen.
So, this is like saying I've
got some equation,
second derivative of x
is equal to some number times
x,
and we said the answer is
cos ωt and sin
ωt.
But now, we've got second
derivatives with the respect to
space, second derivatives with
respect to time and I know that
this guy is going to oscillate
up and down.
But it's going to oscillate,
if you imagine a string,
take a string or a water wave
that is traveling back and
forth;
the ripples are traveling.
If you took a snapshot of the
wave at a given time,
it would be going up and down
in space.
Or if you stood at one point in
the water and have the wave go
past you, the water will go up
and down in time at a given
space.
You understand?
If the wave is coming towards
the shore, you stand in one
place, you don't have to move.
The wave will go past you and
where you are,
the water will go up and down
with time.
Or, if you took a picture of it
at one instant,
the wave will go up and down in
space at a given time.
So, waves oscillate in space
and time and you have to guess
the answer.
Now, if I had more time,
I can tell you guys how to
deduce the answer.
It's not very hard,
but I just don't have that kind
of time.
So, I'm going to just write
down a solution.
In fact, there are many,
many solutions to the equation.
It's very general,
so I'm going to write a
particular solution.
The solution I'm going to write
down looks like this.
It is some number A,
which is the amplitude and it
looks like cosine,
some number called k
times x minus ωt.
Unfortunately,
this k is not a force
constant.
It is a new symbol.
And ω is another symbol
whose meaning we will deduce
shortly.
At least you can see that I
just cannot write a wave like
x - t or something,
because cosine should have an
argument which has no
dimensions.
So, if I have an x,
I should multiply it by some
number that's got units of
inverse x.
And if I have a time,
I should multiply it by
something that looks like a
frequency.
From dimensional analysis,
I look for a solution of this
form to these equations.
Once again, whether you solve
it--By the way,
this called a partial
differential equation because it
involves partial derivatives.
All I'm going to do--I'm going
to be content with showing you
that this solution,
when I stick it into this
equation actually works.
Let us first verify that it
works.
Then, we'll come next Monday
and analyze in detail what does
this really do?
We'll come to that in a moment.
Let's verify that it works.
So, let's take dψ/dx.
What happens when you take
dψ/dx,
when you take a partial
derivative with respect to
x it says forget about
time;
treat x as the only
variable, then you'll get -a
sin kx - ωt,
times the derivative of what's
inside the argument which
happens to be k.
Now, if you differentiate the
second time,
d^(2)ψ/dx^(2),
you do the whole thing again;
then, you'll get -Ak^(2) sin
kx - ωt.
Students: [inaudible]
Professor Ramamurti
Shankar: I'm
sorry--;yes--thank you
[correction: professor changes
Ak^(2) sin kx - ωt to
Ak^(2) cos kx - ωt].
Now, if you take
d^(2)ψ/dt^(2),
you will get -ω^(2) A cos
kx - ωt.
If you take one over
v^(2) times that,
you'll get this.
What does the wave equation say?
It says, the second derivative
of an x should be one
over v^(2) times the
second derivative in time.
And for my solution,
that's a second derivative in
x;
that's a second derivative in
time;
I want them to match.
Well, they're almost there.
A matches A,
cosine matches the cosine,
but ω and k are
not independent numbers.
ω over v should
be equal to k.
In other words,
this wave equation for waves on
a string will have solutions
that look like this,
where A can be any
number you like,
k and ω,
between the two of them,
there is one degree of freedom.
You can pick k at will,
and pick ω so that
ω will equal the
kv,
or pick ω freely,
and choose k so that
k is ω over
v.
But there is a relation between
the two of them,
which you can write in any
number of ways.
So, let us put that knowledge
back into the solution.
Let us say ω is in fact
kv, so that we write it
explicitly in terms of the
k, which is completely
free.
So, ψ now looks like
A cos,
k times x
minus--;now, here is where you
guys got to do a little mental
algebra here.
If I pulled the k out,
can you see that t will
be multiplied by ω over
k?
But ω over k is
v.
So, this is my solution.
A cos [k(x – vt)].
And that is a very pretty
result and I just--I will
conclude by telling you what
that is telling you.
It says that ψ of
xt, which depends on
x and depends on
t,
depends on x and
t through this
combination x - vt.
In other words,
if x - vt has a certain
value, ψ has a certain
value.
If I change x and if I
change t,
keeping x - vt constant,
the function doesn't change,
right, because it's a function
only of this combination.
I will argue that that really
means it's a wave that's
traveling at a velocity v.
Let me show you why.
Let's take a function,
e to the -x^(2).
You must have seen this
function in statistics.
It looks like this.
It likes to have x = 0.
Take the function that x
is replaced by x - vt.
Where does this function have
its maximum?
It has a maximum when this guy
on top vanishes,
that is to say,
when x = vt.
So, if you took the function,
e to the (x -
vt)^(2) with the minus sign,
at t = 0 it'll be here,
but other time it'll be here,
but at every instant,
the bump will be moving to the
right, you can see with the
velocity v,
because the peak is at a
location vt.
So, in a wave,
the whole profile moves at a
velocity and that,
for this symbol we use here is
in fact the velocity of the
wave.
Maybe I'll give you one extreme
example.
At t = 0,
it just looks like cos
kx, and we all know what
cos kx looks like.
It's got a maximum of x
= 0.
If you wait a little longer,
where is this maximum a little
later?
If you increase x a
little bit, if you increase time
a little bit,
you got to increase x a
little bit so that these two
still cancel each other.
Originally, x was at 0,
and t was at 0;
you were at a maximum.
A little later,
after some time Δt,
if you go to the point
Δx, so that these two
numbers cancel,
we are still riding the
maximum.
But that means that Δx
over Δt is v.
That means, if you move to the
right, after ψ
Δt, you will ride the
crest.
That means the crest is moving
at the velocity v.
So, I think to summarize what
I've shown you so far,
is that when you take a string
and you have a tension on it,
you pull it and let it go,
it vibrates according to this
equation;
that is the first statement.
This is a solution.
I'm not telling you it's the
most general one or the only one
or anything.
I want us to at least have one
concrete solution.
And I've not told you yet how
to visualize the function,
but I have told you whatever
the function is,
it is drifting to the right at
a speed that we call v.
Because any function of x -
vt has a property that if
[initially]
it did something x = 0,
it does the same thing at x
= vt.
That means the whole shape is
bodily transported to the right.
So, this really signifies a
wave traveling to the right.
If you wanted a wave traveling
to the left, you should put a
plus here;
it'll go backwards.
