Professor Ramamurti
Shankar: Stable equilibrium,
and if you disturb them,
they rock back and forth and
there are two simple examples.
The standard textbook example
is this mass on spring.
This spring has a certain
natural length,
which I'm showing you here.
It likes to sit there.
But if you pull it so the mass
comes here, you move it from
x = 0 to a new location
x;
then, there is a restoring
force F,
which is -kx,
and that force will be equal to
mass times acceleration by
Newton's law,
and so you're trying to solve
the equation d^(2)x/dt^(2) =
-k/m times x,
and we use the symbol,
ω^(2) = k/m,
or ω square root of
k/m.
And what's the solution to this
equation?
I think we did a lot of talking
and said look,
we are looking for a function
which,
when differentiated twice,
looks like the same function up
to some constants,
and we know they are
trigonometry functions.
Then, we finally found the
answer.
The most general answer looks
like A cos ωt - φ.
A is the amplitude,
φ is called the phase.
That tells you what your clock
is doing when x is a
maximum, and we can always
choose φ to be 0.
That means when the time is 0,
x will have the biggest
value A.
This is an example of simple
harmonic motion.
But it is a very generic
situation, so I'll give you a
second example.
Now, we don't have a mass but
we have a bar,
let's say, suspended by a
cable, hanging from the ceiling,
and it's happy the way it is.
But if you come and twist it by
angle θ,
give it a little twist,
then it will try to untwist
itself.
So, now we don't have a
restoring force but we have a
restoring torque.
What can be the expression for
the restoring torque?
When you don't do anything,
it doesn't do anything,
so it's a function of θ
that vanishes when θ is
0.
If θ is not 0,
it'll begin as some function of
θ, but the leading term
would be just θ.
But now, you can put a constant
here.
It'll still be proportional to
θ, and you put a minus
sign for the same reason you put
a minus sign here to tell you
it's a restoring torque.
That means if you make
θ positive,
the torque will try to twist
you the other way.
If you make θ negative
it'll try to bring you back.
So, you have to find this
κ.
If you find this κ then
you can say -κ times
θ is I times
d^(2)θ over
dt^(2).
Mathematically,
that equation's identical to
that, and θ then will
look like some constant.
You can call it A cos (ωt-
φ, and ω [squared]
now will be the ratio of this
κ at the moment of
inertia,
because mathematically that's
the role played by κ and
I.
They are like k and
m.
Now, if someone tells you this
is κ,
then you are done.
You just stick it in and
mindlessly calculate all the
formulas.
You can find θ,
you can derivatives and so on.
Sometimes, they may not tell
you what κ is.
In the case of a spring,
they would have to give you
k or they may tell you
indirectly if I pull this spring
by 9 inches,
I exert a certain force,
they're giving you F and
they're giving you x and
you can find k.
But in rotation problems,
the typical situation is the
following.
Let us take the easiest problem
of a pendulum.
The pendulum is hanging,
say a massless rod,
with some mass m at the
bottom.
So, it's very happy to stay
this way.
If you leave it like this,
it would stay this way forever.
No torque, no motion.
Now, you come along and you
disturb that by turning it by an
angle θ.
If you do, then the force is
like this, the separation vector
r from the pivot point is
this,
and r cross F
will be non-zero because the
angle between them is not 0,
and you remember that torque is
equal to rF sin θ,
and I told you for small
angles,
this can be rF θ,
and r happens to be,
in this problem,
the length of the pendulum,
so it's really -mgl θ.
So, you have to do some work to
find θ [correction:
should have said κ].
It wasn't just given to you on
a plate.
You did this piece of thinking,
and namely, you disturbed the
system from equilibrium and
found the restoring torque.
Then, you stared at the formula
and say hey, this guy must be
the κ I'm looking for
because that number times
θ is the restoring
torque.
Then the ω here will
be--this κ is a generic
κ that goes into
whatever it is for that problem.
For our problem,
κ happens to be
mgl.
That's something you should
understand.
It's not a universal number
that's known to everybody like
the spring constant of anything.
In the case of mass and this
pendulum, it depends on how long
the pendulum is,
how big the mass is,
but you can work it out and
extract κ.
Downstairs, you need the
momentum of inertia for a point
mass m at a distance
l from the point of
rotation,
which is ml^(2).
So, you can cancel the
l, you can cancel the
m, and you get ω
is root of g over
l.
Let me remind you guys that
ω is connected to what
you and I would call the
frequency by this 2π,
or what you and I would call
the time period by 2π
over T.
Okay, so this is the situation.
Let me give one other example.
This is a homework problem,
but I want to give you a hint
on how to do this.
Don't take a pendulum with all
the mass concentrated there.
Take some irregularly shaped
object, a flat planar object.
You drive a nail through it
there, hang it on the wall.
It'll come to rest in a certain
configuration and you should
think about where will the
center of mass be if I hang it
like this on the wall?
Center of mass is somewhere in
the body here.
So, I want you to think about
it.
I'll tell you in a second.
The center of mass,
they claim, will lie somewhere
on this vertical line going
through that point.
I know that by default because
if the center of mass is here,
for example,
we know all the force of
gravity can be imagined to be
acting here [pointing to
drawing].
Then if you do that separation
and do the torque,
then you will find this is able
to exert a torque on this point,
but that cannot be in
equilibrium.
So, center of mass will align
itself.
The body would swing a little
bit and settle down;
the center of mass somewhere
there.
If you now disturb the body,
I don't want to draw another
picture, but I'll probably fail.
This is the rotated body,
off from its ideal position,
then there will be a torque.
What will the torque be?
It'll be the same thing minus
mgl sin θ,
which I'm going to replace by
θ.
l is now the distance
between the pivot point and the
center of mass.
That's your torque.
So, you can read the κ.
In fact, it's just like the
pendulum.
In other words,
as far as the torque is
concerned, it's as if all the
mass were sitting here.
The difference will be in a
moment of inertia.
Moment of inertia of this is
not ml^(2),
so don't make the mistake.
All the mass is not sitting
here.
All the mass is sitting all
over the place.
So, you should know that if you
want the moment of inertia,
it is I with respect to
center of mass plus this
ml^(2).
That's the old parallel axis
theorem.
So, take that moment of inertia
and put that into this formula
here;
take this κ and put it
there, you'll get some
frequency.
That's the frequency with which
it will oscillate.
So, every problem that you will
ever get will look like one of
these two.
Either there is something
moving linearly with a
coordinate that you can call
x,
or it's rotating or twisting by
an angle you can call θ.
And if you want to find out the
frequency of vibration,
you have to disturb it from
equilibrium,
either by pulling the mass or
by twisting the cable or
displacing this pendulum from
its ideal position here,
then finding the restoring
torque.
Yep?
Student: What can we say
[inaudible]
its restoring torque in the
twisting example?
How do we describe that?
Professor Ramamurti
Shankar: That's a very good
point.
His question is,
if I gave you a cable and I
twist the cable by some angle,
how are we going to calculate
restoring torque?
Here is the good news.
This problem is so hard we'll
give it to you.
In other words,
there is, of course,
an underlying answer.
Given a cable,
made of some material and its
torsional properties,
how much of a torque you'll get
if you twist it by some amount,
but it's not something you can
calculate from first principle,
so they'll simply have to give
it to you,
okay, in such problems.
The only time you'll have to
find the torque on your own is
in a problem like this where I
believe you know enough to
figure out the torque.
Okay, what you will find is,
if you leave it alone,
it'll go to a position where
there is no torque.
If you move it off the
position, there will be a
torque, and the torque will
always be proportional to the
angle by which you've displaced
it.
You read off the
proportionality constant and
that's your κ.
You should look at the forces
too.
It is pretty interesting.
This body, when it is hanging
in its rest position,
has two forces on it.
The nail, which is pushing up
and the weight of the body which
is pushing down,
and they cancel each other.
The nail will keep it from
falling.
The nail will not keep it from
swinging, because the force of
the nail acting as it does at
the pivot point is unable to
exert a torque,
whereas the minute you rotate
the body, mg is able to
exert a torque.
That's why if you twist it,
it'll start rattling back and
forth.
Alright.
So, what I'm going to do now is
to go over more complicated
oscillations using some of the
techniques we learned last time.
So, here is the mega formula.
I'm going to give it to you
guys.
You're allowed to tattoo it on
your face.
You can carry it with you.
I will allow you to bring it
but you cannot forget this
formula.
I think you know the formula
I'm driving at.
Ready?
e to the ix,
or θ,
I don't care.
Let me call it θ here,
is cos θ + I sin θ.
That's a great formula from
Euler.
From this formula,
if you take the complex
conjugate of both sides,
that means change every
i to minus i,
you will get e to the
minus iθ is cos θ -
i sin θ.
That also you should know.
If you got this in your
head--By the way,
this is something you're not
going to derive on the spot.
It takes a lot of work.
If you tell me what are the few
things you really have to cram,
that you don't want to carry in
your head, well,
this is one of them.
Okay, this is a formula worth
memorizing, unlike formula No.
92 of your text.
It's not worth memorizing.
This is worth memorizing.
Once you got this,
you should realize that in any
expression involving complex
numbers,
you can get another equation
where every i is changed
to minus i.
That'll give you this one.
That's called complex
conjugation.
I generally said,
if you have a complex number
z, z star is equal
to x - iy,
if z is equal to
x plus Iy.
So, in our simple example,
this fellow here is z.
This is the x and this
is the y.
Now, you should be able to
invert this formula.
To invert the formula,
you add the 2 and divide by 2.
Then, you will find cos
θ is e^(iθ) +
e^(-iθ) over 2.
If you subtract and divide by
2i, you will find sin
θ is e^(iθ) -
e^(iθ) over 2i.
In other words,
my claim is that this funny
exponential, sum of exponential,
is the family of cosine.
It'll do everything your cosine
will do.
It'll oscillate,
sine squared plus cosine
squared would be 1,
it'll obey all trigonometric
identities,
like sine 2 θ is 2
sin θ cos θ,
everything comes out of this
expression.
So, you should know,
here is the main point you
should learn.
We don't need trigonometric
functions anymore.
Once you got the exponential
function, provided you let the
exponent be complex or imaginary
you don't need trigonometric
functions.
This is one example of grand
unification.
People always say Maxwell
unified this,
and Einstein tried to unify
this.
Unification means things that
you thought were unrelated are,
in fact, related,
and there are different
manifestations of the same
thing.
When we first discovered
trigonometric functions,
we were drawing right-angle
triangles and opposite side and
adjacent side.
Then, we discovered the
exponential function which,
by the way, was computed by
bankers who were trying to
calculate compound interest
every second.
The fact that those functions
are related is a marvelous
result, but it happens only if
you got complex numbers.
So, this is another thing you
should know.
Okay, now, armed with this we
are ready to do anything we
want.
Let me just tell you that every
complex number z can be
written as its absolute value
times some phase.
In other words,
if my complex number z
is here, it is x +
iy;
that's one way to write it.
You can also write it as its
absolute value.
It'll be e to the
iφ φr,
re to the iφ.
r is the radial length
which also happens to be the
length of the complex number.
That's called the polar form of
the complex number.
x + iy is the Cartesian
form of the complex number.
I may have called them r
and θ,
r and φ.
I just don't remember.
I'm going to use these symbols
back and forth because people do
use both the symbols,
which is why I'm sometimes
calling this angle as θ
and sometimes this angle as
φ,
but the basic idea is the same.
This entity here can do it
either in this form or in this
form.
This is something you should
know.
If you don't know,
or you don't understand,
you should stop and ask.
Everything is built on this.
So, what I'm going to do now is
to go back to this rather simple
equation, d^(2)x/dt^(2) =
-ω^(2)x.
By the way, today I'm going to
call this ω as
ω_0.
It's the same guy.
I'm going to call it
ω_0 because we
will find as the hour
progresses, there are going to
be many ωs in the game.
Two more ωs are going
to come in.
So, we've got to be able to
tell them apart.
As long as there's only one
ω in town,
you just call it ω.
If there are many,
you call this
ω_0 to mean
it's the frequency of vibration
of the system left to its own
design,
and you pull it and let it go.
What's the frequency?
An angular frequency,
that's what we call
ω_0.
Now, how did we solve this
equation?
The way I tried to solve it for
you is to say,
turn it into a word problem.
I'm looking for a function,
x(t) with a property
that two derivatives of the
function look like the function,
and we rake our brains
[inaudible]
and we remembered,
hey, sines and cosines are the
property.
One derivative is no good.
Turn sign into cosine.
Two derivatives bring back the
function you started with,
which is why the answer could
be sines or cosines.
But now I'm going to solve with
a different way.
My thinking is going to be:
I know a function that repeats
itself, even when it
differentiated once.
Namely, the function has a
property;
its first derivative looks like
the function itself.
It's obvious that his 90-second
derivative will also look like
the function because taking the
derivative leaves the function
alone,
except for pulling out some
numbers.
So, why not say,
"I want an answer that looks
like this: x equals some
A to the αt."
That certainly has a property
that if you take two derivatives
it's going to look like e
to the αt on the
left-hand side,
and e to the αt
on the right-hand side,
and then you can cancel them
and you've got yourself a
solution.
Do you guys remember why I
didn't follow this solution for
the oscillator?
Why was it rejected?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Ah.
That's a very good point.
So, maybe I will take e
to the minus αt.
How about that?
Will that give me a second
derivative which is negative?
You didn't fall for that, right?
Because the first time you'll
get -α but second time
you'll get +α^(2).
That's correct.
So, this function is no good.
Also, it doesn't look like what
I want.
Even without doing much
mathematical physics,
I know if you pull this spring
it's going to go back and forth,
whereas these functions are
exponentially growing;
they're exponentially falling,
they just don't do the trick.
But now, we'll find that if
you're willing to work with
complex exponentials,
this will do the job.
So, we're just going to take
this guess and put it in the
equation and see if I can get an
answer that works,
okay?
This is called an ansatz.
Ansatz are something we
use all the time.
It's a German word.
I'm not sure exactly what it
means, but we all use it to mean
a tentative guess.
That's what it is.
If you're lucky, it'll work.
If you're not lucky,
that's fine.
You move on and try another
solution.
So, we're going to say,
"Can a solution of this form be
found, with A and
α completely free?"
Maybe the equation will tell me
what A and α
should be.
So, let's take it, put it here.
By the way, I'm going to write
this equation in a form that's a
little easier.
x double dot is minus
ω_0^(2)x but
each dot is a derivative.
It's just more compact that way
rather than to write all of this
stuff.
So, let's take that guess and
put it here and see what we get.
I get, when I take two
derivatives, you can say,
"Do I want e to the
αt or e to the
-αt,"
you'll find it doesn't matter
so I'm going to take e to
the αt and put it in.
Now, the beauty of the x
function now is to take one
derivative, x dot.
I hope everybody knows the
x dot of this is,
A [α]
e to the αt,
and x double dot is
another α.
Aα^(2)e to the
αt.
That's the beauty of the
exponential.
The act of differentiation is
trivial.
You just multiply it by the
exponent, α.
See, this being the result of
taking two derivatives,
let me write it here now.
What do I get?
So, I'm going to go to x
double dot equal to minus
ω_0 squared
x.
Into that, I'm going to put in
the guess, x equals
Ae to the αt,
and what do I find?
In fact, let me write the
equation in a nicer way:
x double dot plus
ω_0 squared
x has to vanish.
I just brought everything to
one side.
Now, put e^(αt),
A to the αt as
your guess, then you find that
α^(2),
Ae^(α)t,
plus ω_0
squared Ae^(αt) has to
be 0.
So, we know what to do.
Let's group a few terms.
So, this means I get A
times α^(2) + 1,
e^(αt) has to vanish.
If you can make that happen,
you got a solution because
you've satisfied the equation.
No one's going to say,
well, you got it by guesswork.
Well, it turns out solving
differential equation is only
guesswork.
There's no other way to solve
the equation other than to make
a guess, stick it in,
fiddle with the parameters,
see if it will work.
So, we're not done yet because
we want this to vanish.
How many ways are there to kill
this answer?
You can say maybe A is 0.
A is 0;
you've got what you call a
trivial solution.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Ah,
yes.
Thank you.
Yep.
Thank you very much.
So, this one is α^(2) +
ω_0^(2).
Does everybody follow that?
Yes, from here to here it's
very clear what to do.
So, we cannot kill A.
Yep?
Student: [inaudible]
Professor Ramamurti
Shankar: Yeah,
yeah, yeah.
I'm trying to rule out certain
options I don't like.
The option I don't like to get
this to be 0 is to say that is
0.
That's not happening.
e^(α)t is not going to
vanish.
Maybe it'll vanish for some
negative infinity time,
but I want this to be true at
all times.
The equation has to be obeyed
at all times,
so this guy certainly is not
zero at all times,
so the only way to fix that is
to have α^(2) +
ω_0^(2) = 0.
That's the only solution,
which is not a trivial
solution.
Well, that's a simple equation,
right?
It's a quadratic equation that
means α^(2) =
ω_0^(2),
or α is plus or minus
iω_0.
So, when we were young and we
didn't know about complex
numbers, we would come to this
stage and we would quit and say,
"Look, exponentials won't work,
but now we are not afraid of
complex exponentials,
so we embrace the solution."
So now, I've got two solutions.
By the way, what's the value of
A?
Think about this.
If this condition is satisfied,
you realize A can be
whatever you like.
A is completely
arbitrary.
So, what this equation has told
me is the following.
Yes, there are solutions of
this form.
For A you can pick any
number you like,
in fact, real or complex,
it doesn't matter.
The equation is satisfied.
But your α is not an
arbitrary number.
It can be only one of two
numbers: +iω_0
or -iω_0.
So, we've got a problem now
that I look at one answer,
I got two.
So, I will write them both.
So, let's say that the solution
x_1 of
t, which is some number
A,
e to the
iω_0t,
then I got another solution
x_2 of
t,
which is equal to any other
number B,
e to the
-iω_0t.
Now, I think you guys can see
in your head that if you take
this and you put it in the
equation,
it works, and if you take this
and put it in the equation,
that also works.
Because when you take two
derivatives, iω whole
square would be minus ω
square,
minus iω_0
whole square would also be minus
ω not square,
so both would work.
Now, we have to ask,
"How do I decide between that
solution and that solution?"
It turns out that we don't have
to pick.
We can pick both,
and I'll tell you what I mean
by "we can pick both."
Now, this is a very,
very important property for all
of you who are going into
economics or engineering or
chemistry.
This or other disciplines,
which are mathematical,
the property I'm going to
mention of this equation is very
important, so please pay
attention.
This is called a linear
equation.
A linear equation will obey a
certain property called
superposition.
I'll have to tell you what it
is.
If I give you differential
equation, you know,
96 derivative of x,
plus 52 times the 37 derivative
of x, blah blah blah adds
up to 0;
what makes it a linear equation
is that throughout the equation,
either you find the function
x or its derivatives,
but never the squares of cubes
at higher powers of x or
the derivatives.
Okay?
The function appears to first
power, not to second power.
For example,
if this were the equation you
were trying to solve,
that is not a linear equation.
That's called a non-linear
equation.
If you have a linear equation,
there's a very,
very profound consequence and
I'm going to tell you what it
is,
and that lies at the heart of
so many things we do.
So, let me write down two
solutions.
One of them will be
x_1 double dot
+ ω_0
^(2)x_1 = 0.
Second one obeys
x_2 double dot
+ ω_0^(2)x
_2 = 0.
Don't even have to look at
these two solutions.
Take a general problem where
you have a linear equation that
are two solutions.
Add the two equations.
On the left-hand side,
I get a second derivative of
x_1 plus a
second derivative of
x_2,
and remember that is,
in fact, the second derivative.
Now, I go back to the old
notation of x_1 +
x_2.
This has to do with the fact
the derivative of a sum is the
sum of the derivatives,
and the sum of the derivative
is the derivative of the sum.
Yep?
Student: So,
because you have these two
linear solutions,
does that mean you can generate
an infinite number of
[inaudible]
Professor Ramamurti
Shankar: Yes.
I'm coming to that in a minute.
I'm first doing a more modest
goal of showing that their sum
is also a solution.
In the second term,
I get ω_0
squared times x_1 +
x_2 = 0.
You stare at this equation.
Look, this plus this implies
what I've written down simply by
adding.
But say in words what you found
out.
What you found out,
is that if x_1
satisfies the equation,
x_2 satisfies
the equation x_1 +
x_2,
let's call it x plus,
is x_1 +
x_2 also
satisfies the equation,
and the proof is right in front
of you.
And the key was the derivative
of a sum is the sum of the
derivatives.
Now, I'm going to generalize it
more and say,
suppose I multiplied all of the
first equation by A,
and all of the second equation
by B.
Well, that's certainly still
true, right?
You take something equal to 0,
multiply both sides by A,
it's still going to be 0.
Now, imagine adding this to
this [adding the two equations
together].
Add this to this,
and what do you get?
You will find
d^(2)/dt^(2) of
Ax_1 +
Bx_2 +
ω_0^(2),
Ax_1 +
Bx_2 is 0.
So, here is the punch line.
If you give me two solutions to
this equation,
I can manufacture another one
by taking the first one times
any number I like,
plus the second one times any
other number I like.
So, the story is a lot more
complicated than it looked.
It looked like there are two
solutions;
actually, there's infinite
number of solutions because you
can pick A and B
any way you like.
So, linear equations typically
have infinite number of
solutions and you build them up
by taking a few building blocks.
They're like unit vectors
I and J.
You can take any multiple of
this unit, vector any multiple
of that unit vector,
and this combination will also
solve the equation.
Just so you don't think this is
going to happen all the time,
let me remind you.
You don't have to write this
down but you guys follow what
I'm saying.
Suppose this quantity here was
not x_1 but the
square of x_1
and here.
Okay.
You don't have to write this
because this is not a linear
equation.
We're not interested in this.
But notice this is non-linear,
and you guys should be able to
pick it up.
It's non-linear because I've
got the square of the unknown.
Now, can you see if I add these
two equations,
forget A and B,
let them both be 1.
I get x_1
squared plus
x_2 squared,
but that does not equal to
x_1 +
x_2,
the whole thing square.
So, even though the second
derivative of the sum is the sum
of the second derivative,
the sum of the squares is not
the square of the sum.
So, for a non-linear equation
you cannot add solutions,
but for linear equations you
can combine solutions with
arbitrary coefficients.
That's the lesson you have
learned today.
So, harmonic oscillating
equation is a linear equation,
so feel free to combine them
and get the following solution:
x(t) is equal to Ae^(iω0) ,
plus B e^(-iω0t).
A and B are whatever you like,
but ω_0 is
the original root of k/m.
But if I gave you this
solution, will you be happy to
take it as a solution for the
mass spring system,
and if not, what is it you
don't like?
Yes?
Anybody have a view?
I mean, would you take that as
a good solution?
Yes?
Student: Wouldn't you
need to figure out how to choose
the right A and B?
Professor Ramamurti
Shankar: In order to achieve
what?
Student: Because your
system starts at some amplitude
[inaudible]
Professor Ramamurti
Shankar: Well,
he's raising a point.
Let me repeat his point.
A and B in
general--they are arbitrary.
For this mass and this spring
you can pick any A and
B you like.
But on a given day,
when you pull it by 9
centimeters and release it,
the answer has to be chosen so
that our t = 0,
x becomes 9,
and the velocity,
let's say, was 0 when you
released it,
so when you take the derivative
of this, the velocity should
vanish at t = 0.
Or maybe it will have some
other velocity.
But you can fit initial
coordinate, initial velocity,
the two numbers,
by picking these two numbers.
You understand that.
But that's not enough.
I've got another problem.
Yes?
Student: It has
imaginary numbers in it?
Professor Ramamurti
Shankar: Yes.
That should bother you.
The answer is manifestly not
real, okay, and we know x
is the real function.
That is not a mathematical
requirement of the equation,
but a physical requirement that
when you pull a mass by 9
centimeters and you release it,
it's going to oscillate with
some real x and not a
complex x,
so you've got to fix that.
So, to say that x is
real really means the following.
You remember that in a
complex--world of complex
numbers, a complex number x +
iy has a complex conjugate
x - iy and the property
of real numbers,
is that when you take the
complex conjugate,
nothing happens because
i goes to minus i.
If the number was purely real,
it satisfies the condition
z = z star.
So, real numbers are their own
complex conjugates.
In general, a complex number is
not its own conjugate,
but if you wanted me to draw
you a picture,
that's where the z is,
and that's where the z
star is, but fellows on the
x axis have the
properties,
z is the same as
z star.
z star is a reflection
of the x axis of
z, and therefore,
if the number is real,
it's its own reflection.
So, I'm going to demand that
this solution,
in addition to satisfying the
basic equation,
also is real.
To do that, I'm going to find
the complex conjugate.
It's denoted by the x
star of t.
I want to conjugate everything
in sight.
The complex conjugate of
A I'm going to call
A star.
Complex conjugate of e^(iωt)
is e^(-iω0t).
Why?
Because the i goes to
minus i,
ω_0 and
t are real numbers.
Nothing happens to them.
And B becomes B
star and this becomes e
to the plus
iω_0t.
And I demand that these two
fellows are equal.
If they are both equal,
then I take this exponential
and demand that those be the
same.
That means I demand that
A be the same as B
star, because if A is the
same as B star,
this will go into this,
and then I will demand that
B is the same as A
star.
If these conditions are
satisfied, x star will
become x.
So, the trick is to take the
function, take its complex
conjugate, equate it to itself,
and look at the consequence.
Now, I want to tell you that if
A = B star,
then you don't have to worry
about this as an extra
condition,
because I'm your complex
conjugate, you are my complex
conjugate, but the way it works
is if you took A and
changed every i to minus
i,
right, or if you took B,
whatever its complex form was,
change i to minus
i you get A.
But if you do it one more time,
you come back to where you are.
In other words,
for any complex number,
if you star it and you star it
one more time,
you come back to where you are.
Therefore, if A = B
star, take the complex conjugate
of the left-hand side,
you'll find A star is
equal to B star star
which is B.
So, you don't need these two
conditions.
You do need A = B
star.
Okay, so you have to put that
extra condition now.
So, in the real world,
for people who want a real
answer, you want to write it as
Ae^(iω0t) + A
star e^(-iωt).
In other words,
B is not an independent
number.
B has to be the complex
conjugate of A.
Then, all of you can see at a
glance that this is now real,
because whatever this animal
is,
this is the complex conjugate
of that, which got all the
is turned into minus
i.
When you add them,
all the is will cancel;
answer will be real.
But A is not necessarily
real.
The complex number A,
which is some complex number,
has a length and it can have a
phase,
because every complex number
can be written in this form.
So, if you remember that,
then you find x looks like
absolute value of
Ae^(iω0t) p+ φ
plus absolute value of
Ae^(-iωt) + φ.
Well, maybe that's too fast for
you, so let me repeat what I
did.
In the place of A,
write the absolute value times
e^(iφ).
e^(iφ) will combine it
e^(iωt) and form this
exponential.
Second term will be just a
conjugate of the whole thing.
You don't even have to think.
Now, what is this function I
have?
I want you to think about it.
Do you recognize this creature
here as something familiar?
Yes?
Do you have any idea?
Yeah?
Yep?
Student: [inaudible]
Professor Ramamurti
Shankar: Without the
A.
Student: [inaudible]
Professor Ramamurti
Shankar: You remember this
great identity.
Where is it?
Here?
This one.
e^(iθ) + e^(-iθ) is 2
times cos θ.
So, this becomes 2 times
absolute value of A cos ωt +
φ, and if you want you can
call 2 A as another
number C cos ω_0t
+ φ.
Student: [inaudible]
Professor Ramamurti
Shankar: Oh,
but for any complex number.
You can pull certainly this
number out of both.
You pull it out and you got
e to the i
something plus either the minus
i something--
Student: [inaudible]
Professor Ramamurti
Shankar: So,
this is a long and difficult
way to get back to the old
answer.
You got this by filling it out
and guessing and arguing
physically.
But I want to show you that
with an exponential function,
you would have come to this
answer anyway.
So, your point of view is,
you know, we don't need this.
We've got enough problems in
life;
we're doing well with cosines,
thank you.
Why do you bring this
exponential on us?
Well, now I'm going to give you
a problem where you cannot talk
your way out of this by just
turning it into a word problem.
The word problem I meant,
we asked a word question,
find me a function where the
second derivative looks like
itself,
and you can either start at
exponentials and differentiate
twice, or sines and cosines.
So, you don't need exponentials.
But look at the following
problem.
This is a problem of a mass
m, force constant
k, moving on a surface
with friction.
The minute you got friction,
you have an extra force.
So you find m,
x double dot equal to
minus kx,
which would be all the force
due to the spring.
Friction also exerts a force
which has got some coefficient
b, but it's multiplied by
the velocity.
We know that if you're moving
to the right,
the force of the left you are
moving to the left the force is
to the right,
frictional force is velocity
dependent.
So, the equation you want to
solve, when you've got friction,
is really mx double dot
plus bx dot plus
kx equal to 0.
I'm going to divide everything
by m.
Take the m,
put it here,
and put it here.
Then, I'm going to rewrite for
us the equation we want to solve
with friction.
x double dot,
plus γx dot,
plus ω_0
squared x = 0,
where γ is just b over
x.
So, this is the equation you
want to solve.
Can you solve this by your
usual word problem?
It's going to be difficult,
because you want a function
which, when I take two
derivatives,
it looks like the function plus
some amount of its own
derivative.
If you take cos ωt,
it won't do it.
If you take sin ωt,
it won't do it.
So, we can still solve it
because even this equation,
you solve by the same mindless
approach,
which is to say,
let x look like
A, Ae to the
αt,
and it'll work.
Let's see how it'll work.
It has to work.
You can see why because when I
take two derivatives of
x, I'll get
α^(2).
Let me pull the A out of
everything plus αγ plus
ω_0^(2),
e to the αt = 0.
So, what we learn is yes,
there are solutions of this
form to this equation.
Once again, A can be
whatever number you like,
because if A vanishes
you've killed the whole
solution.
e to the αt is
not going to vanish at every
instant in time,
so the only way is for this to
vanish.
That means α that you
put into this guess is not any
old number, but the solution to
this quadratic equation,
α^(2) + αγ +
ω_0^(2) = 0.
So, we want α^(2) +
αγ not square is
ω_0^(2) = 0.
So, α must be the root
of this equation,
and we all know a quadratic
equation will have two roots,
and we will get two solutions,
and we can add them with any
coefficient we like.
That'll also be a solution.
So, let me write it down.
So, in other words,
I'm going to explicitly solve
this quadratic equation.
So, go back to the old days and
remember that the solution of
that equation will be α
= -γ,
plus or minus square root of
γ^(2) -
4ω_0 square
over 2.
I would like to rewrite this as
minus γ over 2,
plus or minus square root of
γ over 2 square,
minus ω_0
square.
And let's call the two roots,
one with a plus sign and one
with a minus sign,
as α plus and minus.
This is a shorthand for this
whole combination.
You understand that if I give
you the mass and the spring
constant, and I give you the
coefficient of friction,
the number b,
α plus and α
minus are uniquely known,
so you'll get two precise
numbers coming out of this whole
game, and your answer then to
this problem will be x of
t is any number A
times e to the minus
α plus t with any
number B times e
to the minus α minus
t,
where α plus or minus
are these two numbers.
I want you to notice both
numbers are falling
exponentials.
Very important,
because now I'm talking about a
mass where I pull the spring,
there's a lot of friction,
and I let it go,
I don't want an exponentially
growing answer.
That makes no sense.
The x should eventually
vanish and that will,
in fact, happen.
Let's make sure you understand
that.
Look at these two roots.
Minus γ over 2,
plus and minus,
well, that is this solution
minus γ over 2 square,
minus ω_0
square.
I'm going to assume in this
calculation that γ over
2 is bigger than
ω_0.
Let's do that first.
Then, you can see that that's a
negative number,
that's a negative number,
the whole thing is negative,
so α minus is
definitely, let me see how I
wrote it.
Oh, I'm sorry,
I made a mistake here.
I think these are called just
plus and minus α,
α has two roots and I
have to write them as they are.
Okay.
I'm sorry.
So please remove this,
I put unwanted sign there.
Because they are the two
answers for α.
So, I put e to the
α plus t.
So again, I should change my
answer here.
I'm really sorry about this one.
Now, I'm going to write it
below so you have a second
chance.
So, here is what it looks like.
Ae to the minus
γ over 2,
plus square root of γ
over 2 square,
minus ω_0
square t,
plus Be to the minus
γ over 2 minus γ
over 2 square minus
ω_0 square
t.
Student: You forgot the
minus.
Professor Ramamurti
Shankar: Pardon me?
Student:
You forgot the minus in.
Professor Ramamurti
Shankar: I think I did mean
this.
Or did I get it wrong?
These are the two values of
α.
So, they both have the first
number to be minus γ
over 2, and the square root
comes for the plus sign for one
of them and minus sign for the
other one.
I probably have A and
B mixed up.
Let me check that.
No, I think even that's okay.
Yes?
I'm willing to be corrected if
I got A and B
mixed up.
I know these are correct.
I want to make sure this is
what I call B.
B has a coefficient
α minus,
which would be this one.
Student: [inaudible]
Professor Ramamurti
Shankar: Oh,
yes.
Yes.
So, what should I do,
call this A?
Call this B?
How is that now?
Let's see if that works out.
This should be Be to the
α minus t.
B over the minus
γ over 2 minus that.
Yes.
Student: [inaudible]
Professor Ramamurti
Shankar: Pardon me?
Student: [inaudible]
Professor Ramamurti
Shankar: Okay.
The point of writing it this
way is to show you that both
these powers,
this is clearly a positive
number so it's e to the
minus a negative number.
I just want to show you that
this object--that's all I wanted
to emphasize.
Inside the square brackets is
also positive.
Okay, this is γ over 2.
You can see this is a number
smaller than γ over 2
square.
So, if you take the square root
of that number,
it'll be smaller than γ
over 2, so this cannot overturn
the sign of this.
It'll still be a positive over
all signs to be negative;
so, if you draw the picture
here, it's a sum of two
exponentials,
it'll just die down after a
while.
And how do I find A and
B?
To find A and B,
what I have to do is to take
extra data, one of them may be
initial position,
x of 0 is given to me.
If x of 0 is given to me
as some number,
I take the solution,
I put t = 0 and have it
match this.
If I put t = 0,
well all this exponential will
vanish and I just get A +
B, because e to the 0
is 1.
So, I already have one
condition on A and
B.
Their sum must be the initial
position.
How about their initial
velocity?
Go back to this x and
take dx/dt.
What do you get?
dx/dt is α plus
A, e to the
α plus t,
plus α minus B,
e to the α minus
t.
The derivative of my answer is
this.
Evaluate it at t = 0.
At t = 0,
you put t = 0 here,
you find x dot of 0,
equal to α plus
A, plus α minus
B.
So, here are the two equations
you need to find A and
B.
Solve these two simultaneous
equations.
In other words,
I will tell you the initial
position and I will tell you the
initial velocity.
You will take then this number,
known, and that number known.
This is a linear simultaneous
equation for A and
B.
You will fiddle around and
solve for A.
Yes?
Student: Are A
and B still complex
numbers?
Professor Ramamurti
Shankar: Ah.
Now, we have to ask the
following question.
In this problem,
if everything is real--See
previously, what happened is
when I took the complex
conjugate of this function,
the conjugate turned into the
other function,
and the conjugate of that
function turned back into this
one.
That's why A and
B were related,
by complex conjugation.
Here, if I take the complex
conjugate of x,
e to the α plus
t, if it's real,
it remains itself.
So, A just goes into
A star,
B goes into B
star, and you want the answer to
be unchanged.
That requires A equal to
A star and B equal
to B star.
Thanks for pointing out that
here.
We do want A equal to
A star and B equal
to B star.
So, what I'm telling you is go
take the solution,
conjugate it and equate it to
itself,
and remember that now this
exponential remains real to
begin with so it doesn't go into
anything.
It remains itself,
and I compare this exponential
before and after its coefficient
that went from A to
A star,
they have to be equal so
A is A star and
B is B star.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Well,
look at these functions.
There are no complex numbers
here.
That's why everything is real.
So, what problems have I solved
now?
First, I put γ = 0.
No friction.
And I just re-derived the
harmonic oscillator because when
I got the cos (ω_0
t - φ)--Then I took a
problem with friction.
Then, I got a solution that was
just exponentially falling
because even though it looks
like α plus,
that's a negative number,
that's a negative number,
in the end both are falling.
It says you pull the mass and
let it go, it'll relax to its
initial position,
it'll stop.
But all of you must know that
there's got to be something in
between.
In the one case,
I have a mass that oscillates
forever, the other case I have a
mass when I pull it and let it
go,
it goes back to equilibrium and
reaches 0 at infinite time.
But we all know that the real
situation you run into all the
time when you pull something and
let it go;
it vibrates and it vibrates
less and less and less and less
and then eventually comes to
rest.
Where is that solution?
It's got to come out of this
thing.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: That's correct.
So, you've got to go back to
the roots I found.
For α,
it's minus γ over 2,
plus or minus square root of
γ over 2 square,
minus ω_0
square.
I've taken γ over 2 to
be bigger than
ω_0.
So, what I did was I turned on
friction but I didn't turn on a
small amount of friction.
I turned on a rather hefty
amount of friction so that the
friction term was bigger than
the ω_0 term.
But if you imagine the other
limit where you have no friction
and you turn on a tiny amount of
friction,
then the tiny γ over 2
will be smaller than this
ω_0.
Of course, then you've got the
square root of a negative
number, right?
So, you should really write it.
So, let's take γ over 2 less
than ω_0,
then you will write this as
minus γ over 2,
plus or minus i times
ω_0 square,
minus γ over 2 square.
In other words,
write this stuff inside as
minus of this number now,
and take the square root of
minus 1 and write it as
i.
So now, what do the solutions
look like?
They look like x equal
to Ae to the minus
γ over 2.
Let me write the exponentials
if you like.
Ae to the minus
γ over 2,
t plus iω prime
t, but I'm going to call
this combination as ω
prime.
That's why we have
ω_0 and
ω prime.
Plus Be to the minus
γ over 2 minus
iω prime t.
Now e to the minus
γ over γt over 2
is common to both the factors.
You can pull it out.
Imagine pulling this factor out
common to the whole expression.
Then, you just got A to
the iω prime,
plus B to the minus
iω prime.
That's a very familiar problem.
If you want that to be real,
you want A to be equal
to the complex conjugate of
B.
Then, you repeat everything I
did before, so I don't want to
do that one more time.
You will get some number C,
times e to the minus
γ over 2t,
times cos ω prime
t, plus,
you can add a φ,
if you like.
There.
So, what does this look like?
What does this graph look like?
It looks like cos ωt
but it's got this thing in the
front.
If γ vanishes,
then forget the exponential
completely.
This is our oscillating mass.
If γ is not 0,
imagine it's one part in
10,000, then for the first one
second this number will hardly
change from 1.
Meanwhile, this would have
oscillated some number of times.
But if you wait long enough
this exponential will start
coming into play and the way to
think about it is,
it's an oscillation whose
amplitude itself falls with
time.
And if you draw a picture of
that, I think you will not be
surprised that if you draw the
picture it will look like this.
It's called a damped
oscillation.
So, that's the method by which
you describe the three cases.
One is no friction and one is
small friction or friction
obeying this condition,
when you have oscillatory
motion whose amplitude is damped
with time.
Then you have the other case
where the gammas has crossed
some threshold when
ω_0 is smaller
than γ over 2,
then you get two falling
exponentials.
So, what do you imagine the
mass is doing?
There's no friction if you pull
it and let it go;
it vibrates forever.
That's not very realistic.
If you turn a tiny amount of
friction, it will do this,
which is very ubiquitous.
We see it all the time.
You pull a spring and you let
it go.
After a while, it comes to rest.
This tells you it never quite
comes to rest because as long as
t is finite,
this number is going to be
non-0, but it'll be very small.
At some point you just cannot
see it.
Third option is overdamped.
Overdamped is when γ is
bigger than
ω_0 over 2,
then the answer has nothing
that oscillates.
Nothing oscillates,
everything falls exponentially
with time.
That is, you should imagine you
pull something,
let it go, it comes slowly and
stops.
When you buy your shocks in
your car, the shock absorbers,
they're supposed to damp the
vibration of the car.
It's got a spring suspending
the tires, but it's immersed in
some viscous medium.
So that is, vibrations are
damped.
So, when you hit a bump,
if your shocks are dead,
you vibrate a lot of times and
slowly settle down.
That's when your shocks are in
this regime.
When you bought them,
they were in this regime.
When you bought them they were
doing this.
Once you hit a bump and you
overshoot from your position,
you come down to 0 and you stay
there.
That's the ideal situation for
damping.
All right.
So, I'm going to the last topic
in oscillation but is there
anything that you need
clarification?
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Ah,
yes.
So, when γ over 2
equals ω_0,
this thing vanishes,
and you seem to have only one
solution, e to the minus
γt over 2.
You don't have the plus or
minus, right.
And we sort of know in every
problem there must be two
solutions, because we should be
able to pick the initial
position and velocity at will.
If a solution has only one free
parameter, you cannot pick two
numbers at will.
Now, that's a piece of
mathematics I don't want to do
now, but what you can show is in
that case, the second solution
looks like this.
It's a new function,
not an exponential but t
times an exponential will solve
the equation.
Those who want to see that this
is true can put that into the
equation and check.
In other words,
pick your γ carefully
so that the square root
vanishes.
Only a γ is left,
but that γ is not
independent of ω.
That γ is equal to
2ω_0.
For that problem you can check
that that's a solution,
so is that.
There are nice ways to motivate
that but I think I don't have
time to do that.
Yes?
Student: [inaudible]
Professor Ramamurti
Shankar: Why was A
equal to A star here?
You mean here?
Okay, take this x.
Demand that when you conjugate
it nothing should happen,
okay.
In this particular example,
when α plus an
α minus real numbers,
e to the α plus
t, remains e to
the α plus t when
you take conjugate.
So therefore,
A has to remain A
star, and B must also
remain invariant when you take
the conjugation.
In other words,
if this was an imaginary
exponential like in the other
problem,
when you take the conjugate,
this exponential becomes this
guy and that exponential becomes
that guy.
Then, by matching the solution,
you will be able to show this
must be the conjugate of that;
that must be the conjugate of
this.
But if no imaginary
exponentials appear,
each term must separately be
real.
Now, this is something you will
have to think about.
I'm not saying my repeating it
makes it any clearer,
so what I want you to do is
take the x,
take the complex conjugate of
x, equate the two sides.
There's a rule among functions
that if you got e to the
plus αt,
its coefficient should match on
both sides and e to the
minus αt should have
matching coefficients.
It's like saying when two
vectors are equal,
the coefficient of I
should match and the coefficient
of J should match.
There's a similar theorem that
if you take a sum of two
independent functions,
equate it to the sum of [same]
two independent functions,
the coefficient must separately
match.
If you impose that you will
find what I told you.
Okay, now for the really
interesting problem.
The really interesting problem
is this.
I got the mass,
I got the spring,
I got the friction,
but I'm going to apply an extra
force, F_0 cos
ωt.
This is called a "driven
oscillator."
So far, the oscillator was not
driven.
In other words,
no one is pushing and pulling
it.
Of course, you pulled it in the
beginning and you released it,
but once you released it,
no one's touching the
oscillator.
The only forces on it are due
to internal frictional forces or
spring force.
But now, I want to imagine a
case where I am actively driving
the oscillator by my hand,
exerting cos ωt force.
So ω here is the
frequency of the driving force,
okay.
That's why there are so many
ωs in this problem.
The ω prime that you
saw here is not going to appear
too many times.
It's a matter of convenience to
call this ω prime.
But this ω will appear
all the time.
When I write an ω with
no subscript,
it's the frequency of the
driving force,
and the equation we want to
solve is x double dot,
plus γ plus dot,
plus ω_0
square x equal to
F over m,
because I divided everything by
m, cos ωt.
So, we've got to solve this
problem.
Now, this is really difficult
because you cannot guess the
answer to this by a word
problem.
Now, you can do the following.
If the right-hand side had been
e to the iωt
instead of cos ωt,
you would be fine because then
you can pick an x that
looks like e to the
iωt,
and when you take two
derivatives that look like
e to the iωt,
one derivative would look like
e to the iωt;
x itself would look like
e to the iωt,
you can cancel it on both
sides.
But what you have is cos
ωt.
So, here is what people do.
It's a very clever trick.
People say, let me manufacture
a second problem.
Nobody gave me this problem.
Okay, this is a problem you
gave me.
I make up a new problem,
the answer to which is called
u.
But y is the answer to
the following problem.
The driving force is sin
ωt, but this is what you
give me to solve.
This is my artifact.
I introduced a new problem you
did not give me,
but I want to look at this
problem.
Now, here is a trick.
You multiply this equation by
any number;
it's still going to satisfy the
equation.
Let me multiply by i.
Put an i here,
put an i here,
put an i here,
put an i here.
Multiply both sides by i
and add them.
Then I have got x plus
iy, double dot,
plus γ times x
dot,
plus iy dot is
ω_0 square
times x,
plus iy.
Let's introduce a number
z, which is x plus
iy.
It varies with time.
Then this equation,
by adding the two equations,
look like z double dot,
plus γz dot,
plus ω_0
square z is F over
me to the Iωt.
So, I have manufactured a
problem in which the thing
that's vibrating is not a real
number.
The force driving it is also
not a real number.
It's a cos ωt plus I
sin ωt.
But if I can solve this problem
by some trick,
at the end what do I have to
do?
I have to take the real part of
the answer, because the answer
will look like a real part and
an imaginary part;
I'll have to dump the imaginary
part.
That'll be the answer to the
question I posed.
The imaginary part of it will
be the answer to the fictitious
question I posed.
Now, why am I doing this?
The reason is the following.
If the driving force is
e to the iωt,
I can make the following
ansatz,
or a guess.
z is some constant times
e to the iωt.
I will show you now solutions
of this form do exist,
because let me take that
assumed form and put it into
this equation.
Then, what do I get?
I get minus ω square,
Z_0e to the
iωt because two
derivatives of e to the
iωt give me iω
times iω.
Then, one derivative gives me
iω times γ,
times z_0 e to
the iωt,
and no derivatives,
just leaves it alone,
e to the iωt
equals F_0 over
m,
e to the iωt.
So, let me rewrite this as
follows.
Let me rewrite this as minus
ω square,
plus iωγ,
plus ω_0
square,
times z_0e to
the iωt,
is F over m,
e to the iωt.
This is what I want to be true.
Well, e to the
iωt, whatever it is,
can be canceled on two sides
because it is not 0.
Anything that's not 0 you can
always cancel,
and here is the interesting
result you learn.
For this equation to be valid,
the z_0 that
you pick here in your guess
satisfies the condition,
z_0 equals
F over m divided
by ω_0 square
minus ω square,
plus iωγ.
Now, parts of it may be easy,
parts of it may be difficult.
The easy part is to take the
guess I made and stick it into
the equation,
cancel exponentials and get the
answer.
What you should understand is,
x was what I was looking
for.
I brought in a partner
y, and I solved for
z, which is x plus
iy,
and z was assumed to
take this form with the
amplitude, which itself could be
complex,
times e to the
iωt.
If z_0 looks
like this, then z,
which is z_0e
to the iωt,
looks like F over
m, e to the
iωt divided by this
number in the denominator,
I'm going to call I for
impedance.
I is called impedance
and there's the following
complex number,
ω_0 square,
minus ω square,
plus iωγ.
Okay, we're almost done now.
The z looks like
F_0 over
m, e to the
Iωt divided by this
complex number I.
Imagine this complex number
I.
What does it look like?
It's a got a real part and an
imaginary part.
Imaginary part is iωγ,
real part is
ω_0 square
minus ω square.
This is the complex number
I.
So, we are told the answer to
our problem is to find this
number z,
then take the real part.
Does everybody agree that every
complex number I can be written
as an absolute value times
e to the iφ.
φ, which is going to be
here.
And I'm going to write it that
way because now things become a
lot simpler if you write it that
way.
F_0 over
m, divided by absolute
value of i,
e to the iφ,
e to the iωt.
Now, this e to the
iφ, I can delete it here
and put it upstairs as minus
iφ.
So, let me write it in that
form.
Then it's very easy.
F_0 over
m, divided by the
magnitude of this I,
times e to the
Iωt minus φ,
where φ is this angle.
Well, now that I got
z, I know how to find
x.
x is the real part of
z.
Now, when I look at real part,
all of these are real numbers
so I will keep them as they are.
F_0 over
m divided by i,
and the real part of this
function, I hope you know by
now, is cos ωt minus
φ.
And that's the answer.
The answer to the problem that
was originally given to us is
the following.
You know the magnitude of the
applied force,
the amplitude of the applied
force.
You know the mass of the
particle.
You need to find the absolute
value of I and φ.
For that, you construct this
complex number whose real part
is this, whose imaginary part is
iωγ.
Then, the absolute value of
I is just by Pythagoras'
theorem, ω_0
square minus ω square
square,
plus ω square γ
square.
And the phase φ obeys
the condition tan
φ is equal to
ωγ divided by
ω_0 square,
minus ω square square
is ω square γ
square.
I'm sorry.
It's just the imaginary part
over the real part.
So, this is the answer to the
problem that was given to us.
But there's one subtle point
you should notice,
which is the following.
Where are the three parameters
in this problem?
Everything is determined in
this problem.
φ, absolute I,
all these are known,
but we know every equation
should have two free numbers.
You know what they are?
Anybody know?
Okay, let me tell you what the
answer to that question is.
I'm saying, I can add to this
the solution I got earlier on
without the driving force,
because in this problem,
when I had a right-hand side
with a driving force,
let me add a 0 to that.
That's harmless,
and by superposition principle,
if that force will produce that
displacement,
0 will produce what?
Well, we have seen all morning
that even when the right-hand
side is 0, they are the
solutions I got for you.
e to the γ over
2 cos ω prime t
etc., so you can always add to
this,
what you call a complementary
function, which is the solution
of the equation with no driving
force,
which is what we were studying
earlier in the class.
But usually,
people don't bother with this
because they all have in them
e to the minus γt
over 2.
So, if you wait long enough,
this will die out,
and this is the only thing that
will remain.
So at earlier times,
this is not the full answer.
You should add to this the
answer when there is no driving
force and together they form the
full answer,
and the numbers A and
B that you have there
will be chosen to match some
initial conditions,
like initial position,
initial velocity.
Okay now, time is up.
I'm going to stop,
so I have not been able to
finish some parts of this,
so I'm trying to see what I can
do.
I had asked for you to give
this problem set to me on--what
is this today?
On Wednesday, right.
So, you don't have enough time.
I mean, I haven't taught you a
few other things you need about
five minutes more of work.
I don't want to keep you back.
So, here is what I will do.
I will post on the website some
notes on just the missing part
here, so you can read it and you
can do one or two problems you
may not be able to do without
that.
Then, I will come on Wednesday,
I will teach that to you again,
but you will be able to hand in
your problem set.
