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YEN-JIE LEE: OK, so
welcome back, everybody.
Welcome back to 8.03.
Today, we are going to
continue the discussion
of the harmonic oscillators.
And also, we will add
damping force into the game
and see what will happen, OK?
So this is actually what
we have learned last time
from this slide.
We have analyzed the physics
of a harmonic oscillator, which
we actually
demonstrated last time.
And you can see the
device still there.
And Hooke's law,
actually the Hooke's law
is actually far more general
than what we saw before.
It works for all
small oscillations
around about a point of
equilibrium position, OK?
And that can be demonstrated
by multiple different kinds
of physical systems.
For example here, I have a
mass, which actually can only
move along this track here.
And if I put this mass
set free, then this thing
is actually exercising
harmonic oscillation, OK?
We can do this with
large amplitude.
We can also do it
with small amplitude.
And you see that,
huh, really, it works.
Hooke's law actually works.
And it predicts
exactly the same motion
as to what you see
on the slide, OK?
And we also have a little
bit more complicated system.
For example, this
is some kind of rod.
And you can actually fix one
point and make it oscillate.
And you see that,
huh, it also does
some kind of
harmonic oscillation.
But now, what is actually
oscillating is the amplitude.
The amplitude is actually
the angle with respect
to the downward direction.
And finally this is actually
the vertical version
of this spring mass
system, which you will be
analyzing that in your P-set.
And you see that, huh, it
actually oscillates up and down
harmonically.
So that's all very nice.
And we also have
learned one thing which
is very, very interesting.
It's that a complex exponential
is actually a pretty beautiful
way to present the solution.
And you will see it works
also when describing
the damped oscillators.
And we will see how it
works in the lecture today.
I received several questions
during my office hour
and through email or Piazza.
There were some confusions about
doing the Taylor expansion, OK?
So in lecture last time,
the equilibrium position
is at x equal to 0.
Therefore, I do Taylor
expansion around 0, OK?
But in this case, if
the equilibrium position
or the minima of the
potential is at x equal to L,
then what you need to do
is to do a Taylor expansion
around x equal to L,
just to make that really,
really clear, OK?
OK, I hope that will help
you with the P-set question.
OK, so let's get
started immediately.
So let's continue the discussion
of the equation of motion
we arrived at last time.
So we have M x double-dot and
this is equal to minus kx, OK?
That is actually the
formula from last time.
And we can actually
calculate the kinetic energy
of this spring and mass system.
And basically, this is
going to be equal to 1/2 M
times x dot squared.
OK, and we can also calculate
the potential energy
of the spring.
Potential energy, and that
is equal to 1/2 kx squared.
We also know what would
be the total energy.
The total energy would be
a sum of the kinetic energy
and of the potential.
Basically, you get this
formula, 1/2 M x dot
squared plus 1/2 kx squared.
One last time, we have solved
this equation of motion, right?
So the solution we got is x
equal to A cosine omega 0 t
plus phi.
Well, omega 0 is equal to
a square root of k over M.
Therefore, we can
actually calculate
what would be the total energy
as a function of time, right?
So if we calculate
that, we'll get
E will be equal to 1/2 M A
squared omega 0 squared sine
squared omega 0 t plus phi--
so this is actually
the first term here--
plus 1/2 kA squared cosine
squared omega 0 t plus phi, OK?
Then, we also know that
this coefficient here
is just kA squared, right?
Because omega 0 is actually
equal to the square root of k
over M. And if you replace this
omega 0 squared by k over M,
then you actually arrive
at kA squared, OK?
So that is actually very good.
So that means I can
simplify the total energy.
And what we are going to
get is 1/2 kA squared.
I can take this factor out.
And that will give me, inside
these brackets, I will get sine
squared omega 0 t plus phi
plus cosine squared omega
0 t plus phi.
And this is actually
equal to 1, right?
Just a reminder, sine squared
of theta plus cosine squared
of theta is always equal to 1.
So that gives me
this result. This
is actually 1/2 kA squared, OK?
So that is actually the
result. What does that mean?
That means, if I actually
pull this mass harder,
so that initially it
has larger amplitude,
then the total energy
is actually proportioned
to amplitude squared, OK?
So I am storing more
and more energy.
If I increase the
amplitude even more,
then I am storing the
energy in this system.
And it's proportional
to A squared.
And also, if the spring
constant is larger,
the same amplitude will
give you more energy.
So that means that you can
store more energy if you have
a larger string constant, OK?
The most surprising thing
is that actually this
is actually a constant, right?
What does that mean?
The total energy is
actually not variating
as a function of time.
You see?
So total energy is constant, OK?
So you can see from this slide
the total energy is actually
showing us the sum,
which is the green curve.
And the kinetic energy
and the potential energy
are shown as red
and blue curves.
You can see that the total
energy is actually constant.
But this system
is very dynamical.
You see?
So that energy is actually
going back and forth
between the spring and the mass
in the form of kinetic energy
and in the form of
potential energy.
But they are doing it so well,
such that the sum is actually
a constant.
So the energy is
actually constant, OK?
So that is actually
pretty beautiful.
And it can be described very
well by these mathematics.
Any questions from here?
OK, so I would like to say
simple harmonic motion,
actually, what you
are going to get
is the energy is actually
conserved and independent
of the time.
And later, you will see
an example with damping.
And you will see that
energy conservation
is now no longer the case, OK?
So let's immediately
jump to another example,
which is actually involving
simple harmonic motion.
So let's take this rod and
nail system as an example.
If I actually slightly
move this rod,
and then I release
that, then actually
you will see simple harmonic
motion, also for this system.
So let's actually do the
calculation as another example.
So this is actually my system.
I have this rod, OK?
Now, I am assume that the
mass is actually uniformly
distributed on this rod and
is nailed on the wall, OK?
And the length of this
rod is actually l.
So that means the center
of mass is actually at l/2
with respect to the nail, OK?
And also, this whole system
is set up on Earth, right?
Therefore, there will
be gravitational force
pointing downward, OK?
So that means you have
gravitational force, Fg,
pointing downward, OK?
So this is actually
the system, which
I would like to understand.
And just a reminder, what are we
going to do afterwards in order
to turn the whole system
into a language we
know describes the nature?
What are we going to do?
Anybody?
We are going to define
the coordinate system,
so that I can
translate everything
into mathematics, right?
So that's actually what
we are always doing.
And you will see
that we are always
doing this in this class, OK?
So what is actually
the coordinate system
which I would like to use?
Since this system is going to
be rotating back and forth,
therefore, I would
like to define theta
to be that angle with respect
to the axis, which essentially
pointing downward, OK?
So the origin of this coordinate
system uses theta equal to 0.
This means that the rod is
actually pointing downward, OK?
And also, I need to define what
is actually the positive value
of the zeta, right?
So I define anti-clockwise
direction to be positive, OK?
So it is actually important
to actually first define that,
then actually to translate
everything into mathematics,
OK?
So the initial condition
is the following.
So I actually move this thing,
rotate this thing slightly.
Then, I actually release
that really carefully
without introducing any
initial velocity, OK?
Therefore, I have two
initial conditions.
OK, at t equal to 0, there
are two initial conditions.
The first one is theta 0
is equal to theta initial.
The second condition
is the same as what
we have been doing last time.
The initial velocity
or angular velocity
is actually equal to 0.
So that gives you theta
dot equal to 0, OK?
Now, we have actually defined
the coordinate system.
Now, we can actually
draw a force diagram,
so that we can actually use
our knowledge about the physics
to obtain the equation
of motion, right?
So now, the force
diagram looks like this.
So this is actually the
center of mass of this rod.
And you have a force
pointing downward,
which is due to the
gravitational force.
Fg is equal to mg.
It's pointing downward.
The magnitude is
actually equal to mg.
And also, we know the R vector.
This vector has a length, l/2.
It's pointing from the center
of mass of this rod to the nail,
OK?
And also, we know the angle
between these vectors,
pointing from the center
of mass to the nail,
and the vertical direction,
which we have already defined,
which is actually called theta.
Therefore, now, we
can actually calculate
what would be the torque.
Tau will be equal
to this R vector
cross the force, total force
acting on the center mass.
In this case, it's just Fg, OK?
So now, we can actually
write this down explicitly.
Since the whole
system is actually
rotating on a single plane,
so there's only one plane
this is sitting on.
And it's actually going back and
forth only on this plane, OK?
Therefore, actually, I
can drop all the arrows
and write down the magnitude
of the tau directly.
And this will be equal to
minus mg l/2 sine theta t, OK?
Any questions so far?
OK, so now, we have the torque.
And we can make use of
the rotational version
of Newton's Law to obtain the
equation of motion, right?
So that should be
pretty straightforward.
Tau will be equal to I, which
is the moment of inertia
of the system,
times alpha t, OK?
And just for your
information, I already
calculated the I for you.
I is equal to 1/3
ml squared, OK?
So you can actually go
back home and actually do
a check to see if I'm
telling the truth.
And if you trust me,
then that's the answer,
which is actually
1/3 ml squared,
if the mass is actually
uniformly distributed
on this rod, OK?
So that would give me minus mgl
divided by 2 sine theta t, OK?
So that is actually
coming from this side, OK?
So now, I can actually
simplify this expression.
I can now plug in the I
value into this equation.
And I will get 1/3 ml squared
theta double-dot t, which
is actually alpha, OK?
Now, I write it as
theta double-dot.
And that will be equal to minus
mgl over 2 sine theta, OK?
I can move all the constants
to the right-hand side.
Therefore, I get
theta double-dot t.
This is equal to minus mgl.
OK, actually, I can already
simplify this, right?
These actually cancel.
And the 1/l actually cancels.
So therefore, I get minus
3g over 2 sine theta t.
OK, as you know, we
actually defined omega
to replace this constant
to make our life easier.
So I can now define omega 0
equal to square root of 3g
over 2l, OK?
And that will give
you theta double-dot
of t equal to minus omega
0 squared sin theta t.
Any questions so far?
A lot of calculations.
But they should all be
pretty straightforward.
And actually, we are done now.
We are done.
Because we have the
equation of motion.
And the rest of the job
is to solve just it.
So it is actually now the
problem of the math department.
So can anybody actually tell me
the solution of the theta of t?
Anybody?
AUDIENCE: Unfortunately,
we'd have to approximate it.
YEN-JIE LEE: That's
very unfortunate.
So now, we are facing a
very difficult situation.
We don't know how to solve
this equation in front of you.
I don't know, OK?
Of course, you can actually
solve it with a computer,
or, if you want to go fancy,
solve it with your cellphone,
if it doesn't explode.
But it's not really nice
to do this in front of you.
We don't learn too much.
OK, so what are we going to do?
So what we can do is actually
to consider a special case.
So we know that this equation
of motion is exact, OK?
So if you solve it,
it would describe
the motion of this rod.
Even with a large
angle, it works, OK?
And now, in order
to actually show
you the math in the
class, therefore
actually I will do a
small approximation.
So actually, I would
only work on the case
that when the
amplitude is very small
and see what is going to happen.
So now, I'm considering
a special case.
Up to now, everything is exact.
And now, I am now going
to a special case.
Theta t goes to 0, OK?
Then, we can actually get this.
Sine theta t is
roughly theta t, OK?
Based on the Taylor
expansion, you
can actually verify this, OK?
So in this case, if we take
theta equal to 1 degree,
then the ratio of the
sine theta and the theta
is actually equal to
99.99%, which is very good.
If I take it as 5 degrees,
then it's actually 99%.
Even at 10 degrees,
it's actually 99.5%.
Now, that shows you that sine
theta is so close to theta, OK?
We are pretty safe.
Because the difference
is smaller than 1%.
OK, so that's very nice.
After this approximation, I get
my final equation of motion.
Theta double-dot t equal to
minus omega 0 squared theta t.
Just a reminder, omega 0
is equal to square root
of 3g over 2l, OK?
We have solved this equation
last time, last lecture, right?
It's exactly the same.
OK, it happened to
be exactly the same.
Therefore, I know
the solution will
be theta of t equal to A
cosine omega 0 t plus phi.
From the initial conditions,
which I have one and two,
I am not going to go over
these calculation again.
But again, we can
actually plug in 1 and 2
to solve the unknown
A and the phi.
If you do this
exercise, you will
conclude that A is
equal to theta initial.
And phi is equal to 0, OK?
So the solution would be
theta of t equal to theta
initial cosine omega 0 t.
You can see that this actually
works for this system.
Simple harmonic
oscillation actually
described the motion of this
system as a function of time.
You can also see a few
more examples shown here.
Two of them you are going to
really work on in your P-set
and also another one
involving circuits.
If you have a capacitor
and you have an inductor,
actually the size of
the current is also
doing a simple
harmonic motion, OK?
And as we actually
discussed before,
the energy is always conserved.
And that is actually stored
in different components
of the system, OK?
So we have done this.
What is actually new today?
What we are going to do
today is let's actually
observe this phenomenon here.
So this thing is actually
going to go back and forth.
But it's actually not going
to do that forever, right?
Something is happening, which
actually slows the motion down.
I can also make use
of this system, OK?
I start from here.
And I'm not worried
that this actually
goes out of this track.
Because I know for sure
it will stop there.
Why?
Because the initial
amplitude is not going to--
the amplitude is not
going to be larger
than the initial
amplitude, right?
So I'm not worried at all, OK?
But you can see that
the amplitude is
changing as a function of time.
Apparently,
something is missing.
And that is actually a
direct force, or friction,
which is actually not
included in our calculation.
So let's actually try to make
the calculation more realistic
and see what is going to happen.
So now, I will
introduce a drag force,
which actually introduces
a torque tau drag, t,
which is equal to minus b--
b is actually some
kind of constant,
which is given to you--
theta dot t, which is actually
proportional to angular
velocity of that rod, OK?
And also of course, I keep
the original approximation.
The theta is very
small, such that I
don't have to deal with the
integration of sine theta, OK?
So solving this, theta
double-dot equal to minus
omega 0 squared sine theta
is a complicated function.
You may ask, why do I actually
introduce a drag force
proportional to the velocity?
And why do I put a
minus sign there?
That is actually because, if you
have a minus sign, that means,
when this mass or that rod
is actually going downward,
then the drag force
is really dragging it.
Because it's actually in
the opposite direction
of the velocity of the
mass or the angular
velocity of the rod, OK?
So I need a minus
sign there, OK?
Otherwise, it's not
a drag force anymore.
It's actually accelerating
the whole thing.
Secondly, why do I choose that
to be proportional to theta dot
or velocity?
There's really no
much deeper reason.
I choose this form
because I can actually
solve it in front of you, OK?
The reality is actually
between proportional
to theta dot and theta dot
squared, for example, OK?
This is actually a model
which I introduced here,
which I can actually
solve it in front of you.
On the other hand, you'll
see that it's actually not
bad at all.
It actually works and
describes the system,
which will actually work to
perform the demo here, OK?
And once we have introduced
this, the equation of motion
will be modified.
So let's come back to
the equation of motion.
So you have to
theta double-dot t
originally would be equal to
tau total t divided by I, OK?
And now, this will become tau
t plus tau drag t divided by I.
So there's an
additional time here.
OK, if I simplify
this whole equation,
then I get minus mgl
over 2 sine theta.
And this is actually
roughly theta minus
b theta dot divided by
1/3 of ml squared, OK?
So you can see that I still
make this approximation sine
theta roughly equal to theta.
Then, I can actually write this
equation in the small angle
case.
OK, I get minus
3g over 2l theta t
minus 3b over ml
squared theta dot t.
OK, and now, as usual, I define
omega 0 squared equal to 3g
over 2l.
And I can also define
gamma is equal to 3b over
ml squared, just to make
my life easier, right?
Finally, we will arrive at this
expression, theta double-dot
plus gamma theta dot plus
omega 0 squared theta.
And that is equal to 0.
So what you can see from here
is that we have actually derived
the equation of motion, OK?
We have derived the
equation of motion.
And actually, part of the
work is actually really
just solving this
equation of motion.
And you don't really
have to solve it.
Because you already get
the result from 18.03
actually, if you remember.
And we are going to
discuss the result.
But before that, before I really
try to solve this equation,
I would like to take a vote, OK?
So here, I have two
different systems.
They have equal amounts of mass.
They are attached to a spring.
If you do the same equation
of motion derivation,
you will actually get exactly
the same equation of motion
in that format, OK?
So the form of the
equation of motion
will be the same between this
system and that system, OK?
I would like to ask you a
question about the oscillation
frequency.
So you can see that one of
them is actually a better mass.
It's like a point-like particle.
And the other one is
wearing a hat, OK?
What is going to happen is
that this Mexican hat is
going to be trying to
push the air away, right?
Then, you may think,
OK, this Mexican thing
is not really very important.
Therefore, the
oscillation frequency
may be the same, right?
How many of you think the
oscillation frequency,
if I actually tried to
perturb these two systems,
would be the same?
Raise your hands.
1, 2, 3, 4, 5, 6, 7, 8--
OK, we have 11.
So the omega, the
predicted omega,
will be equal to omega 0--
11 of you.
How many of you will think
that, because of this hat,
this pushing this air away,
it's a lot of work to be done.
Therefore, this is going to
slow down the oscillation.
How many of you think
that is going to happen?
1, 2, 3--
OK, 17.
It may happen to
you that you think
this idea of wearing a
hat is really fashionable.
Therefore, it got
really exciting
and it oscillates faster.
Can that happen?
How many of you actually
think that is going to happen?
OK, one-- you think so?
Two.
Very good, we have 2.
What do you think?
Where are the rest?
Only 30 of you actually think
that is going to happen.
OK, all the rest
think of the class
think that this one
is going to-- pew!
Disappear to the moon, OK?
So that is actually the opinion.
And we have completed the poll.
And what we are going
to do is that we
are going to solve
this system and see
what is going to happen.
And we will do that experiment
in front of you, OK?
All right, so that's very nice.
So now, we have this
question of motion.
And now, I will pretend
that I'm from the math
department for a bit and help
guide you through the solution.
So now, I can use this trick.
I can actually say
theta is actually
the real part of the z,
which is a complex function.
And as we learned
before, z of t,
and I assume that to be
exponential I alpha t.
So alpha is actually
some kind of constant,
which I don't really know
what is the constant yet.
OK, I can now actually
write the equation of motion
in the form of z.
Then basically, what I get is
z double-dot t plus gamma z dot
t plus omega 0 squared z of t.
And this is equal to 0, OK?
So remember, exponential
function cannot be killed
by differentiation, right?
Therefore, it's
really convenient.
You can see from here.
Now, I can plug in
this expression--
which I did this and guessed
to this equation of motion.
Then what I am going to
get is minus alpha squared.
Because you take I
alpha I alpha out
of this exponential
function, right?
Because you do double
differentiation.
So you get minus alpha
squared plus i gamma alpha--
because this is only
differentiated one time--
plus omega 0 squared.
And all those
things are actually
multiplying this exponential
function, exponential i alpha t
equal to 0, OK?
So we will write
this expression.
That is very nice.
And we also know that,
this expression is
going to be valid all the time.
No matter what t you put in,
it should be valid, right?
Because this is the
equation of motion.
And we hope that this solution
will survive this test.
So I can easily conclude
that this one is actually not
equal to 0.
It can be some value, not 0.
So what is actually equal to 0?
This first term is
actually equal to 0, OK?
Therefore, I can now
solve this equation;
minus alpha squared plus
i gamma alpha plus omega 0
squared equal to 0.
I can solve it, OK?
If I do that, then
I would get alpha
is equal to i gamma plus/minus
square root of 4 omega 0
squared minus gamma
squared divided by 2.
This is actually the
second order polynomial.
And that is actually equal to 0.
Therefore, you can
actually solve it easily.
And this is actually
the solution.
And I can write it down in
a slightly different form.
i gamma over 2 plus/minus
square root of omega 0
squared minus gamma
squared over 4, OK?
Any questions so far?
Am I going too fast?
Everything's OK?
OK, So you can see that alpha
is equal to this expression.
And I would like to
consider a situation
where omega 0 is much,
much larger than gamma, OK?
Just a reminder of
what is gamma, OK?
Maybe you've got
already a bit confused.
What is gamma?
Gamma is related to the strength
of the direct force, right?
It is actually 3b
over ml squared, OK?
b is actually determining the
size of the direct force, OK?
So I would like to
consider a situation.
The first situation is if
omega 0 squared is larger
than gamma squared over 4.
So in that case, the
drag force is small.
It is not huge.
It's small, OK?
If that is the case, this
is actually real, right?
Because omega 0
squared is larger
than gamma squared over 4.
Therefore, this is real, OK?
So now, I can actually
define omega squared,
define that as omega 0 squared
minus gamma squared over 4, OK?
And this will become i gamma
over 2 plus/minus omega, OK?
So that means I would
have two solutions coming
from this exercise.
Z plus of t is equal to
exponential minus gamma over 2
t exponential i omega t, OK?
And the second solution, if
I take one of the plus sign
and one of the minus
sign solutions,
then the second solution
would be exponential
minus i gamma over 2 t
exponential minus i omega t,
OK?
Any questions so far?
OK, so we would like to
go back to theta, right?
So what would be the theta?
So that means I would have a
theta 1 of t, which is actually
taking the real part.
So it's theta plus maybe,
taking the real part of z plus.
And that will give you
exponential minus gamma
over 2 t cosine omega t, OK?
I'm just plugging in the
solution to this equation, OK?
Theta minus t would be
equal to exponential,
and this gamma over
2 t sine omega t, OK?
Finally, the full
solution of theta of t
would be a linear combination
of these two solution, right?
Therefore, you will get theta
of t equal to exponential
minus gamma over 2 t a
(is some kind of constant)
times cosine omega t
plus b sine omega t.
And of course, from the
last time, as you will know,
this can also be written as A
cosine omega t plus phi, OK?
Any questions so far?
OK, very good.
So we have actually already
solved this equation.
And of course, we can
actually plug this back
into this equation of motion.
And you will see
that it really works.
And I'm not going
to do that now.
But you can actually
go back home and check.
And if you believe me, it works.
And also at the same time, it
got two undetermined constants,
since this is a second
order differential equation.
Therefore, huh, this
thing actually works.
It has two arbitrary constants.
Therefore, that is
actually the one and only
one solution in
the universe which
satisfies the equation of motion
or satisfies that differential
question, OK?
So this thing actually
has dramatic consequences.
The first thing
which we learn is
that, as a function of time,
what is going to happen?
The amplitude is now becoming
exponential minus gamma
over 2 t times A. This is
actually the amplitude.
The amplitude is
decreasing exponentially.
So that is actually
the first prediction
coming from this exercise, OK?
The second prediction is
that this thing is still
oscillating.
Because you've got the cosine
omega t plus phi there,
you see?
So the damping
motion is going to be
like going up and down, up
and down, and get tired.
Therefore, the amplitude
becomes smaller, and smaller,
and smaller.
But it's never 0, right?
It's never 0, OK?
It's actually going to be
oscillating down, down, down,
so small I couldn't see it.
But it's still oscillating, OK?
Finally, we actually
have also the answer
to the original
question we posed, OK?
So now, you can see that the
oscillation frequency is omega,
OK?
Originally, before we
introduced the drag force,
omega 0, which is the
oscillation frequency,
is actually an
angular frequency.
It's actually the square
root of 3g over 2l.
And you can see that the
new omega, the oscillation
frequency with drag force,
is the square root of this,
omega 0 squared minus
gamma squared over 4.
So what this
actually tells us is
that this is going
to be smaller,
because of the drag force, OK?
So that's a prediction.
Let's do the experiment and
see what is going to happen.
So let's take a look
at these two systems.
They have the identical mass,
which our technical instructor
actually carefully prepared.
They have the same mass,
even though one actually
looks a bit funny.
The other one looks normal, OK?
Now, what I'm going to do
is to really try and see
which one is actually
oscillating faster, OK?
So let's see.
I release them at the same time.
And you can see
that originally they
seem to be oscillating
at the same frequency.
But you can see very clearly
that the one with the hat
is actually
oscillating slower, OK?
So you can see
that, OK, 17 of you
actually got the correct answer.
And the most important
thing is that you
can see that this
simple mass actually
describes and predicts
what is going to happen
in my little experiment.
So that is actually really cool.
And I think it's time
to take a little break.
And then, we will come back
and look at other solutions.
And of course, you
are welcome to come
to the front to play with
those demonstrations.
So there are two
small issues which
were raised during the break.
So the first one is that, if you
actually calculate the torque
from this equation--
so I made a mistake.
The R vector should be
actually pointing from the nail
to the center of mass, OK?
So I think that's
a trivial mistake.
So if you do this,
then you can actually
calculate the tau
equal to R cross F.
Then, you actually get
this minus sign, OK?
So if I make a mistake in
pointing towards the nail,
then you will get
no minus sign, then
that didn't really work, OK?
So very good, I'm very
happy that you are actually
paying very much attention
to capture those.
The second issue is that--
so now, I'm saying
that, OK, now I
have the solution in
the complex format.
So I have a Z plus and
I have a Z minus, OK?
And then I would like to go
to the real world, right?
Because the imaginary
thing is actually
hidden in some kind of motion
in the actual dimension,
et cetera, I would like
to go back to reality, OK?
And what I said in the class
is that I take the real part
of one of the solutions.
And I can also take
a real part of i
times one of the solutions.
But of course, you
can also do this
by doing a linear combination
of the solutions, right?
As we actually
discussed last time,
the linear combination
of the solutions
is also a solution to the
same equation of motion,
since this one is
actually linear.
Therefore, what I
actually do is actually
to sum the two solutions, Z plus
and Z minus and divide it by 2.
Or actually, I can actually do
a minus i/2 times Z plus minus Z
minus, OK?
And then I can also extract
this sign term here, OK?
So that should be the correct
explanation of the two
solutions in the real axis, OK?
Any questions so far?
Thank you very much
for capturing those.
Ok, so now, you can
see that we have
been discussing the equation of
motion of this functional form.
And the one thing which is
really, really interesting
is that the solution, when we
take a small drag force limit,
actually we arrive at a
beautiful solution that
looks like this, A
exponential minus gamma over 2
t cosine omega t plus phi.
That actually predicts
the oscillation, OK?
At the same time, it also
says that the amplitude
is actually going to drop
exponentially, but never 0, OK?
Finally, we also know that
this solution actually tells us
that, if we have a spring mass
system oscillating up and down,
if we have a rod like what
we actually solve in a class,
this object is going to pass
through 0, the equilibrium
position, an infinite
number of times, right?
Because the cosine
is always there.
Therefore, although
the amplitude
will become very
small, but it's still
oscillating forever until
the end of the universe, OK?
All right, so that's actually
what we have learned.
And also, one thing which
we learned last time
is that simple harmonic
motion, like this one, which
we were just showing
here, or this one,
which is actually a mass
oscillating back and forth
on the track, is actually just a
projection of a circular motion
in a complex plane, OK?
And what we are really
seeing here in front of you
is actually a projection
to the real axis, OK?
So that's actually a
really remarkable result
and a beautiful picture.
And of course, we can actually
also plug in the solution
with damping.
So what is actually the
picture in this language,
in this exact same language?
If we actually follow
the locus, then basically
what you are going to see
is that this thing actually
spirals.
And the amplitude is actually
getting smaller and smaller
and is sucked into this
black hole in the 0, 0, OK?
So you can see that
now the picture
looks as if there is
something really rotating
in the complex plane.
And it's actually approaching 0.
Because the
amplitude is actually
getting smaller and smaller.
But this whole thing
is still rotating, OK?
OK, that's really nice.
All right, so now, this is
actually a special case.
When we actually assume that
gamma is actually pretty small.
So you have very
small drag force, OK?
So let's actually check
what would happen.
If I now start to
increase the drag force,
make this hat larger,
larger, and larger,
introducing more and more
drag, what is going to happen?
OK, so now, I consider
the second situation,
omega 0 squared equal
to gamma squared over 4.
OK, so when the
gamma is very small,
what we see is that this is
actually underdamped, right?
So the damping is really small.
But if I increase the
gamma to a critical value,
now omega 0 squared happens
to be equal to gamma
squared over 4, OK?
I call this a critically
damped oscillator, OK?
So what does that mean?
That means omega is
equal to 0, you see?
This is our definition
of omega, right?
If omega 0 squared is equal
to gamma squared of over 4,
then omega is equal to 0.
That is actually
the critical moment
the system stops
oscillating, OK?
So it is not
oscillating anymore.
So now, I can actually
start from the solution
I obtained from 1, OK?
Then, I can actually
now make use
of these two solutions, the
theta plus and the theta minus.
Theta plus t would be equal
to exponential minus gamma
over 2 t cosine omega t.
When omega goes to
0, what is going
to happen is that this is
actually becoming, which value?
Anybody know?
If omega is 0, what
is going to happen?
1, yeah.
OK, 1, right?
So that will give me exponential
minus gamma over 2 t.
Theta minus t--
OK, I can do the same trick
and see what will happen.
So I take theta minus
t, which is actually
obtained from the
exercise number one
when we discussed the
underdamped system.
Then, you actually get
exponential minus gamma
over 2 t sine omega t.
When omega goes to 0, actually
then I get 0 this time, OK?
So that doesn't
really work, right?
Because if I have a
solution which is 0, then
it's not describing
anything, right?
I can always add
0 to the solution.
But that doesn't help you.
OK, so instead of taking
the limit of this function,
actually we choose
to actually do
theta minus t divided by omega.
And then, we actually make
this omega approaching 0.
Then basically, I get
exponential minus gamma
over 2 t sine omega t
divided by omega, OK?
If I have this omega
approaching to 0,
then this is actually roughly
just exponential minus gamma
over 2 t omega t over omega.
And this is actually giving
you t times exponential minus
gamma over 2 t.
Any questions so far?
Yes.
AUDIENCE: Completely
unrelated, but is
that a negative sign in front
of the theta minus negative 1/2?
YEN-JIE LEE: This one?
AUDIENCE: Yeah.
YEN-JIE LEE: Yeah.
So actually, OK, yeah.
AUDIENCE: In front of the
1/2 is that a negative sign?
YEN-JIE LEE: Yes, this
is a negative sign.
OK, any other questions?
OK, so you can see that now
I arrive at two solutions.
One is actually proportional
to exponential minus gamma
over 2 t.
The other one is actually
proportional to t
times exponential minus
gamma over 2 t, OK?
So you can see that the cosine
or sine term disappeared,
right?
So that means you are
never oscillating, OK?
So this is actually what
we see in this slide,
this so-called
critically damped, OK?
When actually,
omega 0 squared is
equal to gamma squared over 4.
And you can see that
what is going to happen
is that this mass or this
rod is going to pass 0
only one time at most, OK?
And it could actually
never passed 0,
if you actually set up the
initial condition correctly,
OK?
So one thing which I can do is
I really shoot this mass really,
really, very forcefully,
so that I have
a very large initial velocity.
And what it actually
is going to do,
like the right-hand
side diagram,
is that, oh, you
overshoot the 0 a bit.
Then, it goes back
almost exponentially, OK?
So at most, you can
only pass through 0 one
time, if you do this kind
of initial condition, OK?
So that is actually
pretty interesting.
And there are practical
applications of this solution,
actually.
So for example, we
have the door closed.
So it's also here, right?
The door closed, you would
like to have the door go back
to the original closed
mold, the position
of equilibrium position
actually really fast, OK?
So what you can do is really
design this door close
so that it actually matches
with the critical dampness
situation, of your condition, so
that actually you would go back
to 0 really quick, OK?
Any questions?
OK, so now, what we
could do is that, instead
of having a very small drag
force, or we'll a slightly
larger drag force,
so that actually
reach the critically damped
situation, what we could do
is that we put the whole
system into water, right?
Then, the drag force
will be very big, OK?
And we would like to see
what is going to happen, OK?
So in this case is
the third situation.
The third situation is that
omega 0 squared is actually
smaller than gamma
squared over 4.
So you have huge drag force, OK?
So that would give
you a situation
which is called
overdamped oscillator.
Now, I have, again,
alpha is equal to i gamma
over 2 plus/minus
square root of omega 0
squared minus gamma
squared over 4, right?
I'm just copying from here, OK?
And that will be equal to i--
I can take out the i, OK?--
gamma over 2
plus/minus square root
of gamma squared over 4
minus omega 0 squared.
Now, I can actually
define gamma plus/minus
equal to gamma over 2
plus/minus square root of gamma
squared over 4 minus
omega 0 squared, OK?
Then basically, the solution--
actually now, I already
have the solution.
So basically, the two solutions
would be looking like this.
Theta of t would be
equal to A plus some kind
of constant exponential
minus gamma plus t plus A
minus exponential minus
gamma minus t, OK?
Because this is actually
becoming already--
OK, so alpha is actually
i times gamma plus/minus.
Therefore, if you put
it back into this,
then basically what
you are getting
is exponential minus gamma
plus t or exponential minus
gamma minus t, OK?
So that's already
a real function.
And the linear combination
of these two solutions
is our final, full solution
to the equation of motion.
OK, again, what we
are going to see
is that actually the
drag force is huge.
I just throw the whole
system into water.
And the water is really trying
to stop the oscillation, really
very much.
Therefore, you can
see that, huh, again,
I don't have any
oscillation, OK?
If I am very, very
strong, I really
start the initial velocity
or initial angular
velocity really high, I actually
give a huge amount of energy
into the system,
then, at most again, I
can actually have the system
to pass through the equilibrium
position only one time.
Then, this whole system
will slowly recover,
because exponential
function we see here.
The amplitude is going to be
decaying exponentially, OK?
Any questions?
So let's actually do a quick
demonstration here, OK?
So here, this is actually the
original little ball here,
a metal one, which
actually you can
see that this is really going
to go back and forth really
nicely.
And you can see that,
because of the friction,
actually the
amplitude is becoming
smaller and smaller, OK?
So that actually matches
with this situation, right?
So it's actually an
underdamped situation.
This ball, in an
idealized situation,
is going to go through 0
infinite number of times, OK?
So now, what I am
going to do is now
I change this ball to something
which is different, OK?
This is actually
made of magnets, OK?
And let's see what
is going to happen.
So now, you can see that,
because this is actually
made of magnets, therefore, the
drag force will be colossal,
will be very, very big.
And let's see what will happen.
You see that the
drag force is huge.
Therefore, you see I
put it here so that it
has big initial velocity.
It only passes
through 0 once, right?
Of course, it now is actually
approaching the zero really,
really slowly, exponentially.
But it is not 0, OK?
So it only passes through the
0 if you believe the math,
only once, OK?
Just to show that
this is a real deal--
OK, now, whoa, right?
Oh, I'm not trying to
destroy the classroom, OK?
So you can actually
play with this
after we finish
your lecture, OK?
I would like to
ask you a question.
After we learned this
from this lecture,
there are three situations,
underdamped, critically damped,
and overdamped, OK?
I would like to ask
you two questions.
The first one is through
this demonstration, OK?
So, now I have a system
which is nicely constructed.
I hope you can see it, OK?
You can see it.
And this system is made
of a torsional spring.
And also, there's
a pad here, which
is connected to the spring, OK?
If I actually
perturb this thing,
it's going to be oscillating
back and forth before I turn
on the power, so that the
lower part is actually
you have a magnet, OK?
It's not turned on yet, OK?
And this magnet is
going to provide
a drag force to actually change
the behavior of the system, OK?
So you can see that, before
I turn on the magnetic field,
the whole system is actually
oscillating back and forth
really nicely.
As we predicted, small
amplitude vibration
is harmonic oscillation, OK?
So that's very nice.
So now, what am I going to
do is to turn on the power
and see what is going to happen.
After I turn on the power,
there's an electric field, OK?
And this is actually
going to be--
OK, so the magnetic field
is actually turned down.
Therefore, it is actually
acting like a drag force
to this system, OK?
So let's actually see
what is going to happen.
Now, I release this.
The behavior of the
system looks like this.
It first oscillates,
and then it stops.
So the question is, is this a
critically damped, underdamped,
or overdamped system?
Anybody knows?
Yeah?
AUDIENCE: Underdamped.
YEN-JIE LEE: Yes,
this is underdamped.
How do I see that?
That is because, when
I do this experiment,
you would pass through
0s multiple times.
Therefore, there are
oscillations coming into play.
Therefore, I can conclude
that the drag force is not
large enough.
So that is actually an
underdamped situation, OK?
And the next time, we are
going to drag this system.
I have a second
question for you.
So now, your friends
know that you took 8.03.
Therefore, they will
wonder if you can actually
design a car suspension system,
to see if you can actually
make this design for them.
When you design this
car, which condition
will you consider
to set up the car?
Will you set it up as
underdamped, critically damped,
or overdamped?
How many of you actually think
it should be underdamped?
No, nobody?
How many of you actually
think it should be overdamped?
1, 2, 3, 4, OK.
How many of you actually think
it should be critically damped?
OK, the majority of
you think that should
be the correct design.
So if you have the car designed
as an underdamped situation,
then, when you
drive the car, you
are going to have
very funny style.
You are going to have this.
This is the style.
So the car is going to be
oscillating all the time, OK?
Because it's going to be there.
And it's really damping
really slowly, OK?
If you design it
to be overdamped,
it would become
very bumpy, right?
So let's take a limit
of infinitely large drag
force constant, OK?
Then, it's like, when you
hit some bump, you go woo!
Wow!
It doesn't really help you
to reduce the amplitude, OK?
So the correct
answer is you would
give the advice that you would
do it critically damped, OK?
So before we end
the section today,
I would like to pose
a question to you.
The thing which we have learned
from simple harmonic motion
is that the energy is conserved
in a simple harmonic motion,
OK?
I have the Fs, the spring force,
proportional to minus k times
x.
And the energy is conserved, OK?
But if I add a drag force in
the form or minus b times v,
energy is not conserved, right?
So you can see that it
was actually oscillating.
Now, it's not
oscillating, right?
This thing has stopped
oscillating, OK?
Why is that the
case mathematically?
OK, we know what is
happening physically
in this physical system.
Because OK, this Mexican hat
is trying to push the air away.
So what is going to happen
is that it's transferring
the energy from this system to
the molecules of the air, OK?
So it's accelerating the air.
So the energy goes away.
But why the mathematical
form looks so similar
and it does different things?
And think about it.
And I'm not going to talk
about the answer today.
And thank you very much.
And we will continue
next time to see
what we can learn if I start
to drive the oscillator.
Bye-bye.
