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visit MIT OpenCourseWare at ocw.mit.edu. YEN-JIE LEE: OK, so
welcome back everybody. Happy to see you again. So today, we are going to
continue our exploration and understand single
harmonic oscillator. And this is actually
a list of our goals. And we would like to
learn how to translate physical situations into
mathematics so that we can actually solve the
physical problem, so we actually have
and predict what is going to happen afterward. And we also sort of
started this course by solving really
simple examples, single harmonic oscillators. And as a function
of time, you will see that, for our next
class, the next lecture, we are going to bring in
more and more objects. And of course, more objects
means more excitement also, in terms of phenomena,
but also more complication on the mathematics. So we will see how things go. And then after
that, we are going to go through infinite
number of oscillators to see what will happen. Of course, we will
produce waves. That's very exciting. Then we'll do all kinds
of different tricks to do those waves. So what we have we
learned last time? So last time we
went over example, a simple harmonic oscillator. It says you have a
rod fixed on the wall, and you can actually
go back and forth. And we also introduced a model
of the drag force, or drag torque, and that's actually
proportionate to the velocity of the motion of
that single particle. And the interesting thing
we learned last time is that we have three
completely different behaviors if we actually turn
on the drag force. The first one is on that damped. Damping is actually very small. Then we have the
solution in this form. It's oscillating. The amplitude is
decaying exponentially. As we make the drag
force larger and larger, you will pass a critical
point, which actually give you a solution, which you don't
have oscillation anymore. The cosine disappeared. Finally, if you actually put
the whole system into water or introduce something
really dramatic-- a very, very big drag force-- then you have
overdamp situation. And there you see that
the solution is actually a sum of two
exponential functions. So this is actually
the one equation which actually works for all the
damped situation we discussed up to now. And this is actually the map. Basically, if gamma
goes to zero-- gamma actually controls
the size of the drag force. Then we got no damping. Then you have a pure,
simple harmonic motion. And as we increase
the gamma, then you get see that the
behavior is changing as we increase the gamma. So you can see that we
can use a quantity, which is called Q. Q is actually
defined as a ratio of omega at zero, which is basically
the natural angular frequency of the system. And gamma is a measure of
how big the drag force is. If we make a ratio
of this to quantity, then you'll see that,
at Q equal to 0.5, it reaches a critical point,
which actually the behavior of the whole system changed. And you can see that the
oscillation completely disappeared. So that is actually what
we have learned last time. So what are we
going to do today? We have been really
doing experiments really with our hands, hands-on, right? So basically we will
prepare the system. Then we release it. Then we don't touch it again
and see how this system actually evolves as a function of time. So that's what we
have been doing. So today, what we
are going to do is to start to
drive this system. We can introduce some
kind of driving force and see how the system will
respond to this external force. So that is actually
what we are going to do. And that will bring
us to the situation of damped driven
harmonic oscillator. So let's immediately
get started. So we will use the example
which we went through last time as a starting point. So set example from
the last lecture is a rod, which is
fixed on the wall. And the lens of this
rod is over two. And I define a counter-clockwise
direction to be positive. And I measure the
position of the rod by this theta,
which is the angle between the vertical direction
and the pointing direction of the rod. And we have went
through with the math, and we got the
equation of motion without external
force, which is already shown on the blackboard. So now, as I mentioned
at the beginning, I would like to add
a driving force, or driving torque, tau drive. This is equal to d0
cosine omega d t. So I am adding a driving torque. The amplitude of the
torque is actually d0. And there's actually also a
harmonic oscillating force, or torque, and the angle
frequency of this torque is omega d. And that means our
total torque, tau of t, will be equal to tau g t-- this is actually coming from
the gravitational force-- plus tau drag, which is to
account for the drag force. So this time we are
adding in a tau drag. So I'm not going to go over
all the calculations on how did the right from the
beginning to the end. But I will just continue
from what we actually started the last time. So if I have additional
driving torque there, that means my equation of motion
will be slightly modified. This time, my equation
of motion will become theta double dot plus
gamma theta dot plus omega 0 squared theta, and that
is equal to d0 divided by I. This is actually
divided by I because, in order to get the
acceleration, I'll need to divide my torque by
a movement of inertia of this system and
cosine omega d t. This is actually the
oscillating frequency of the driving torque. And just a reminder, gamma is
defined to be equal to 3b m l squared. And the omega 0 is
actually defined to be square root of 3g over 2l. So as I mentioned
in the beginning, this is actually
giving you a sense of the size of the drag force. And the right hand
side, the omega 0 is actually the natural
angular frequency. So, of course, we can
actually simplify this by replacing this term, or
this constant, by symbol. So the symbol I'm
choosing is f0. And this is defined to be
d0 divided by I. Therefore, I arrive at my final
equation of motion-- theta double dot plus gamma
theta dot plus omega 0 squared theta, and that is
equal to f0 cosine omega d t. So I hope this looks pretty
straightforward to you. So this is our
equation of motion you can see from this slide. So we have three
terms in addition to the theta double dot. The first one is
actually related drag force, or drag torque. The second one is actually
related to so-called spring force. So that is actually be
related to the spring constant or, because of the
restoring force of the gravitational force. The third one is actually
what we just add in as a driving force. So one question which I
would like to ask you is-- so now I bring one more
complication to this system. So now I am driving this system
with a different frequency, which is omega d. The question is, what would
be the resulting oscillation frequency of this driven
harmonic oscillator? What is going to happen? Well, this system
actually follows the original damped oscillator
frequency, omega, which is actually close to omega 0. Or what this system actually
follows the driving force frequency. Finally, maybe
this system chooses to do something in between. We don't know what
is going to happen. So our job today is to solve
these equation of motion and to see what we can
learn from the mathematics. Then we can actually
check those results to see if that agrees with
the experimental result, which is through those demos, OK? So as usual, I have this
equation of motion here. So one trick, which
I have been using, is to go to complex
notation, right? Therefore, I can now re-write
this thing to be Z double dot plus gamma Z dot plus
omega 0 squared Z, and that is equal to f0
exponential i omega d t. So basically, I just go
to the complex notation. And we would like to
solve this equation. So in order to
solve this equation, I make a guess, a test function. I guess Z of t has
this functional form. This is equal to A-- some kind of amplitude-- exponential i omega
d t minus delta. Delta is actually
some kind of angle, which is actually to account
for the possible delay of the system. So if I start to try-- for example, this is a system,
which I am interested-- I start to drive this
system, it may take some time for the system to react
to your driving force. So that's actually accounted
for by this delta constant. And the amplitude is actually
what we were wondering, what would be the amplitude. Therefore, you have some
kind of a constant in front of the exponential function. And you can see that this
exponential is actually having angular frequency, omega d. And that is actually designed to
cancel this exponential i omega d t here in the drag force. So now we can as you
calculate what z dot of t would be equal to i omega
d Z. Z double dot t will be equal to minus omega
d squared Z. With those, we can now plug that back
into this equation of motion and see what we can
actually learn from there. So basically, what
I'm going to do is to insert all
those things back into the equation of motion. And that is actually
going to be like this. Basically, the first
term, the double dot, you get a minus omega
d squared out of it. The second term, gamma Z dot-- I have Z dots here. Basically, I would get
plus i omega d gamma out of the second term. That's third term, I get
omega 0 squared out of it. And that is actually
multiplied by Z. And this is equal to f0
exponential i omega d t. All right, and we also
know from this expression Z is equal to A exponential
i omega d t minus delta. That's the test function. So this is actually equal
to A exponential i omega d t minus delta. So now what I can do is-- I have some constant
in the front. Multiply it by
exponential i omega d t. And now I can actually cancel
this exponential i omega d t with the right hand side term. Very good. The whole equation is
actually exponential free. Now I don't have any
exponential function left. And exponential i delta is
actually just a constant. So now this equation is
actually independent of time. So what I getting is
like this-- basically if I multiply the both sides
by exponential i delta, then I get minus omega d squared
plus i omega d t plus omega 0 squared A. And
this is going to be equal to f exponential i delta
because I multiply both sides by exponential i delta. And this is equal to f cosine
delta plus i f sine delta. Just your last equation. Any questions so far? So look at what I
have been doing. So I have this
equation of motion. As usual, I go to
complex notation. Then I guess Z equal to
A exponential i omega d t minus delta because my
friends from the math department already solved this, and
I'm just following it. Then I can calculate all
those terms, plug in e, and basically, you will
arrive at this equation. This equation is a
complex equation. So what does that mean? That means one equation
is equal to two equations because you have the real part,
you have the imaginary part. Therefore, that's very nice
because I have two unknowns. The first one is A, a constant. And the second one is delta. Now I have two
equations I can solve what would be the functional
form for A and the delta. And let me go immediately
solve this equation. So if I take the real
part from this equation, basically what I'm going
to get is omega 0 squared-- this is real-- minus
omega d squared-- this is also real-- times A. A is actually
some real number. This is actually
equal to f cosine-- f0. Sorry, I missed a zero here. So that zero I missed. This should be f0. f0 cosine delta. And I can also
collect all the terms, which is imaginary terms. Then I get only the second
term from the left hand side is with i in front of it. Therefore, I get omega d gamma
A from the left hand side. And from the right
hand side, there's only one imaginary term. Therefore, I get-- this
is equal to f0 sine delta. So now I have two equations. I have two unknowns. Therefore, I can easily
solve A and delta. So I call this
equation number one. I call this equation number two. So now I can-- sounding in
quadrature the two equations-- in quadrature. And the left-hand
side will give you A squared omega 0
squared minus omega d squared squared plus omega
d squared gamma squared. That is actually coming
from the second equation. That gives you the
left hand side. It's a square of the sum the
first and second equation. And the right hand side will
become f0 square cosine delta cosine squared delta
plus sine squared delta. And this is equal to 1. So that's actually the
trick to get rid of delta. Then I can get what will be
the resulting A. A is actually a function of omega d. Omega d is given to you. It's actually
determined by you-- how fast do you want to
oscillate this system. And this is equal to f0
divided by square root of this whole thing. So this will give
you omega 0 squared minus omega d squared
squared plus omega d squared gamma squared. Then we can also calculate
what would be the delta. The trick is to take a ratio
between equation number two and the equation number one-- 2 divided by 1. Basically, you will
get tangent delta. This is sine divided by cosine. f0 actually cancel. This is equal to what? Equal to the ratio
of these two terms. After you take the
ratio, A drops out. Basically, what you
get is gamma omega d divided by omega 0 squared
minus omega d squared. So we have solved A and the
delta through this exercise. So what does that mean? Originally, I assume my solution
to be A exponential i omega d t minus delta. Therefore, I would like to go
back to the real world, which is actually theta. So basically, if I
take the real part, I would get theta of t, which
is actually the real part of Z. And that will give you A omega d
cosine omega d t minus delta is also a function of omega d. So we have done this exercise. And you can see that the
first thing which we see here is that there's no free
parameter from this solution. A is decided by omega d. And delta is also
decided by omega d. There's a lot of
math, but actually we have overcome those and
that we have a solution. But it is actually clear to you
that this cannot be the full story. Because you have a second-order
differential equation, you need to have two free
parameters in the solution. What is actually missing? Anybody can tell
me what is missing. AUDIENCE: The
homogeneous solution. YEN-JIE LEE: Very good. The homogeneous
solution is missing. So that's actually
why we actually have no free parameter here. Once is the omega
d is determined, once the f0 is given, then you
have the functional form which decides what is actually theta. So what in actually
the full solution? A full solution should
be, as you said, a combination of
homogeneous solution and the particular solution
which we actually got here. So if I prepare the system to be
in a situation of, for example, underdamped situation. Then what I'm going to do
is actually pretty simple. What I am going to do is to just
copy the underdamped solution from last lecture
and combine that with my particular solution,
which I obtained here. So that actually to
see what actually the full solution looks like. I have A omega d
cosine omega d t minus delta is a function of omega d. This is actually so-called
steady-state solution. And, of course,
as you mentioned, I need to also add the
homogeneous solution and basically the no
-- this actually -- according to what
I wrote there, I have a functional form of
exponential A exponential minus gamma over 2t equals
sine omega t plus alpha. So I changed A to B because I
already have the A there just to avoid confusion. Then basically, you get
B exponential minus gamma over 2t cosine
omega t plus alpha. Basically, they are two free
parameters, B and alpha. Those two free parameters
can be determined by initial conditions. So, for example, initially
I actually release the rod at some fixed angle
of theta initial. And also the initial
velocity is 0. Then I can actually
practice solution A using those initial conditions
to solve B and alpha. Any questions so far? Yes. AUDIENCE: Are we assuming
it's underdamped? YEN-JIE LEE: Yeah, I'm
assuming it's underdamped, the situation. So that's the assumption. So it depends on the size
of gamma and the omega 0. Then you have actually four
different kinds of solution. If gamma is equal to 0, then
what you are going to plug in is the solution from no damping
as a your homogeneous solution. And if you prepare
this system underwater, damping is colossal,
it's huge, then you actually plug in
the overdamped solution to be your homogeneous
part of the solution. Any other questions? Very good question. So now maybe you
got confused a bit. I have now omega d. I have also omega. And there's another one
we just called omega 0. What are those? So omega 0 is the
natural angular frequency without given the drag force. If you remove
everything just like without considering any
drag force, et cetera, and that is actually the
natural frequency of the system. And what is omega? Omega, according
to the function, omega is defined to be omega
0 squared minus gamma squared over 4 square root of that. That is actually the oscillation
frequency, which we actually discussed last
lecture, after you add drag force into it again. Finally, omega d is how fast
you actually drive this system. So that is actually
the definition of these three omegas. So you can see that, if
I prepare my solution to be underdamped situation,
then basically you will see that this is actually
so-called a steady-state solution because A
omega d is a constant. So it's going to
be there forever. And the second term is actually
B exponential minus gamma over 2t. It's decaying as a
function of time. So if you are patient
enough, you wait, then this will be gone. So that is actually how
we actually understand this mathematical result.
And now, of course, you can actually take
a look at this. This is actually just assuming
some kind of initial condition and plug in the solution and
plot it as a function of time. And you can see that this
function looks really weird, looks a bit surprising. What does that mean? It looks really strange. But at some point,
this superposition of these two functions-- because one of the functions
actually dies out, disappears-- then you will see that,
if you wait long enough, then you actually only see
a very simple structure, which is oscillation
frequency of omega d. And that means a large t. In the beginning, the
system will not like it. You drive it, and the
system don't like it. Like if I go and shake
you, in the beginning, you would not like it-- maybe. And if I shake you long enough,
and you say, come on, OK, fine. I accept that. So that is actually what is
going to happen to the system. So now I would like to go
through a short demonstration, which is actually the air
cart, which you seen before. There's a mass and there
are two springs in the front and the back of this cart. And, of course, as usual,
I would turn on the air so that I make the
friction smaller, but there's still some
residual friction. And you will see that
this mass is actually oscillating back and forth. And the amplitude can
become smaller and smaller as a function of time. Now in the right hand
side, I have a motor, which actually can drive this-- I can actually
shorten or increase the length of the
right hand side string. Then I actually
introduce a driving force by the right hand side motor. If I turn it down, this is
what is going to happen. So we can see now this motor is
actually going back and forth. And it has a slightly
higher frequency compared to the natural frequency. So the frequency of
the motor is higher. And you can see that this
cart is actually oscillating. But you can see that sometimes
it pulls and sometimes it moves faster. So you can see that
it's actually moving. And it stops a little
bit because they are all superposition of two
different kinds of oscillating functions come into play. You can see that now. It got slowed down,
and it can become faster and slower and faster. But eventually, if
you wait long enough, what is going to happen? What is going to happen? If we wait long enough-- AUDIENCE: [INAUDIBLE]. YEN-JIE LEE: Exactly. So basically, if you wait
long enough, as you said, you will actually just oscillate
at the frequency of the driving force. You can see that this motion
looks really bizarre, right? Sometimes it stops,
and sometimes it actually continues to move. And are you surprised? Probably you are not
surprised anymore because we know math is the
language to describe nature. And indeed it predicts
this kind of behavior. That's really pretty cool. In order to help you to learn
a bit how to actually translate a physical situation into
mathematics, what I am going to do is to introduce
you another example so that actually we can
actually solve it together. So now I would like
to drive a pendulum. So I prepare a pendulum
at time equal to 0. This is a string attached to
a ball with mass equal to m. And the length of the
string is equal to l. And the angle between
the vertical direction and the direction of
the string is theta. And, of course, I can actually
give you initial condition X initial, which is
actually measured with respect to the vertical
direction, and time equal to t. This is actually the
original vertical direction, the same as this dashed line. And I can actually move the top
of the string back and forth to some position. And, of course, this string
is connected to the ball. And this system is actually
driven from the top by the engine's
hand, so the engine is actually shaking this
system from the top. And I do it really nicely. So basically, I define
that the displacement, d, as a function of time, is
equal to delta sine omega d t. So that is actually
what I'm going to do. OK, and I would like
to see what is going to happen to this pendulum. So, as usual, the first step
towards solving this problem is to define a
coordinate system. So what is actually
the coordinate system I'm going to use? So now I define
pointing upward to be y. I define the
horizontal direction pointing to the right hand
side of the board to be x. So that's not good enough. I still need the origin, right? So now I also define my origin
to be the original position of the ball which is
actually completely addressed before I do the experiment. So that this is actually
the equilibrium position of this system actually. Then I define here to be 0, 0. So once I have that
defined, I can now express the position of
this mass of this ball to be xt and yt. And see what we
are going to get. Of course, as
usual, we are going to analyze the force
actually acting on this ball. So therefore, as usual, we
will draw a force diagram. So basically, you have
the little mass here, and you have actually two forces
acting on this little mass, or little ball. This is Fg pointing downward. It's a gravitational force. And now this is actually
equal to minus mg y. And there's also
a string tension, T. Since we have this definition
of theta here, basically I have a T which is
actually pointing to the upper left
direction of the board. Oh, don't forget--
actually there's a third force, which
is actually the F drag. F drag is actually
equal to minus bx dot in the x direction. Now I would like to write
down the expression for also the string tension,
T. The T is actually equal to minus T sine
theta in the x direction because the T is pointing
to upper left direction. So therefore, the
position to x direction will be minus sine
theta and plus T cosine theta in the y direction
because the tension is actually pointing upper left. As usual, this is actually
pretty complicated to solve. I have this cosine. I have this sine there, right? So what I'm going to do is
to assume that this angle theta is very small, as usual. So I will take small
angle approximation. Then basically,
you have sine theta is roughly to be equal to
theta, and that is actually equal to what? Equal to-- here. Basically, you can
actually calculate what will be the theta. The sine theta,
or theta, would be equal to x minus d divided by l. And of course, taking a
small angle approximation will bring cosine theta to be 1. Then after this
approximation, my T will become minus
T x minus d divided by l in the x direction
plus T in the y direction because sine theta is replaced
by this approximated value because sine theta is
actually replaced by 1. Any questions? Yes. AUDIENCE: Is that a constant
or is that the change in sine? YEN-JIE LEE: It's a constant. Yeah, I was going too fast. So this is actually a
constant of my amplitude. AUDIENCE: And the drag force
is only in the x direction? YEN-JIE LEE: Yes, it's
only in the x direction. So I'm only trying to actually
move this point back and forth horizontally. So now I have all the
components T, Fg, f drag. And, of course, you can see
that I already ignored the drag force in the y direction
from that formula because I am only
considering the system to be moving in the x direction. Therefore, I can now collect all
the terms in the x direction. Basically, you will
have m x double dot. This is equal to minus b
x dot, which is actually coming from the drag force,
minus T x minus d divided by l. This is actually coming from
this term in the x direction. Let's look at those
in the y direction. And y double dot would
be equal to minus mg plus T. The minus mg is from
the gravitational force. And this T is coming from
the y component of the string tension. And of course, since we are
taking a very small angle approximation, there will
be no vertical motion. Yes. AUDIENCE: Why did we use
the small approximation for sine theta when
we're going to use x minus d over l, which
represents psi instead of just theta? YEN-JIE LEE: Yeah,
so also in this case, they happen to be
exactly the same. And why I care is
actually the cosine theta. Otherwise, I would have
to deal with cosine theta. And also, this y
double dot would not be equal to 0, which is what
I'm going to assume here. Good question. So the question was, why do I
need to take an approximation? Because I want to get
rid of cosine theta. So now from this
y direction, I can solve T will be equal to mg
because I assume that there's no y direction motion. And I can conclude that-- originally, I don't know what
is actually the string tension. It's denoted by T. Now,
from this second equation, I can conclude that T
will be equal to mg, which is the gravitational force. Then, once I have that, I
can go back to x direction. Basically, I get m x double dot. This is equal to minus b x
dot minus mg over l x minus d. Everything is working very well. And I just have to really write
down the d function explicitly. What is d? d is just a reminder,
delta sine omega d t. So I will plug that
into that equation. And also I will
bring all the terms related to x to
the left hand side just to match my convention. All right, so now I will
be able to get the result, m x double dot plus b
x dot plus mg over l x. And that is actually
equal to mg over l d. And this is equal to mg
over l delta sine omega dt. So basically, I
collect all the terms, put it to the left hand side. And I write down T
explicitly, which is this. Then I can divide
everything by m. Then I get m x double dot plus
b over m x dot plus g over l x. And that would be equal to g
over l delta sine omega dt. Now, of course, as usual, I
will define this to be gamma, define this to be
omega 0 squared. And I would define
this to be f0, which is equal to omega 0 squared delta. It happened to be
equal like that. And then this actually
becomes x double dot plus gamma x
dot plus omega 0 squared x equal to
f0 sine omega dt. Am I going too fast? OK, everybody is following. So we see that ha! This equation-- I know that. I know this equation, right? Because we have just solved
that a few minutes ago. Therefore, I know immediately
what will be the solution. The solution is here already. I have A omega d and the tangent
delta, the function of force there. Therefore, I can now write
down what will be the A. So A is actually just
equal to f0 divided by square root of
omega 0 squared minus omega d squared
squared plus omega d squared gamma squared. So now the question is-- what does the result
actually mean. I have this function. I have that function,
tangent delta. It's solved. It's actually the amplitude
of the steady-state solution and also the phase difference
between the drag force phase and the steady-state
oscillation phase. So that's actually
the amount of lag and the size of the amplitude. But through this equation it is
very difficult to understand. So what I'm going to do
is to take some limit so that actually we can help you
to understand what is going on. So suppose I assume
that omega d goes to 0. So what does that mean? This is the engine's hand
and is moving really slowly and see what is going to happen. If I do this, then you
will find that A omega d-- since omega goes to 0,
this is gone, this is gone. Therefore, you will see
that omega A will be equal to omega 0 divided by-- I'm sorry-- of f0 divided
by omega 0 squared. And that is actually
equal to g delta over l divded by g over l. And that will give you delta. So what does that mean? This means that, if I drive
this thing really slowly, then the amplitude
of the mass will be equal to how much I
actually move, which is delta. OK. Do you get it? In addition to that,
tangent delta-- since I am taking the
limit omega d goes to 0. Therefore, tangent delta
will be equal to 0, and that means delta
will be equal 0. Any questions? So that means there will
be no phase difference. The system has enough time
to keep up with my speed. The second limit, obviously,
omega d goes to infinity. What does that mean? That means I'm going to
hold this as a string and shake it like
crazy really fast and see what will happen, OK? So in that case,
you will get A omega d, and that one goes to
0, because omega d goes to infinity. This one goes to 0. And also, tangent delta
will go to infinity. Therefore, delta will go to pi. So that means they
will be out of phase. Any questions so far
in these two limits? OK, so what I'm
going to do now is to take a small toy, which
I made for my son, who is one-year-old now. Because I would like him to
learn wavelength vibration before he goes to
quantum, right? Hey? So I made this toy. And he looked at it. So you can see now,
I can demonstrate what is going to happen when
omega is approaching to 0. OK? I am already doing it. Can you see it? No? It's a very exciting experiment. Can you see that? You see that this is the
origin vertical direction. If I do it really,
really, really slowly, you can see that the
amplitude of the ball is actually exactly the same as
the displacement I introduced. So that's kind of obvious. So now, let's see what
is going to happen if I drive this system like crazy. OK, not going up and down. Eeeee-- that's the
maximum speed I can do. Maybe you can do it faster. But you can see that
nothing happened. So amplitude is close
to 0, because what you have been doing is-- disappear, you sort
cancelling each other. And it's actually not
going to contribute to the motion of this ball. So now, you can see that I can
also test, what is actually the natural frequency? And what I am going
to do is to oscillate at around the natural frequency
to see what is going to happen. Let's see what is
going to happen. You can see that the delta
is really small, right? Can you see the delta. It's really small-- very small-- very small. But you can see that the
amplitude, the A, is huge. What does that tell us? What does that tell us? Yeah? AUDIENCE: Well, we're
experiencing resonance. YEN-JIE LEE: Yes, we are
experiencing resonance. And also, that also
tells you that the system is under-damped very much. The Q value is very big. So if I calculate
the amplitude, A-- now, I can calculate
the amplitude, A, at natural frequency. What I'm going to get is-- now, I can actually plug in
omega d equal to omega 0. So if I plug in omega d-- omega d equal to omega 0-- then what is going to happen? So this term is working. So you have A is equal to
f-0 divided by omega 0 gamma. Omega-d is now equal to omega 0. And that is going to give you-- so f-0 is actually
omega 0 square delta divided by omega 0 gamma. And the one omega
0 actually cancels. Then, basically, you
will get Q times delta. What is Q? Just a reminder, it's
actually the ratio of omega 0 and the gamma. When the Q is very large,
what does that mean? That means it's so close
to an idealized situation that direct force is very small. You can see that in the example
which I have been doing. So you can see that, ah,
it is really the case. So you can see that if my delta
is something like 1 centimeter, but the amplitude is actually
at the order of 1 meter, maybe. What does that mean? That means that Q is
actually, roughly, 100. So you can actually even get
a Q out of this experiment. Any questions so far? OK, that's very good. So now, we can go
ahead and take a look at the structure of
the A and the delta. As we demonstrated
before, we make sense of those three different
kinds of situations-- omega d goes to 0, omega
d goes to infinity. And of course, I would
like to know the force structure of A and delta. Therefore, what I'm going
to do is to plug A omega d as a function of omega d. So what I'm going to get is
this will be equal to delta when omega d goes to 0. We just demonstrated that. And this will increase
to a large value and drop down to 0, when
omega d goes to infinity. And you can see that
this is around omega 0. And you are going to get a huge
amplitude at around omega 0. But not quite. The maxima is actually
slightly smaller than omega 0. You can actually calculate
that as part of the homework. So that makes sense. Now, I can also plug the delta,
which is the phase difference-- and you can see that
this phase difference will be, originally, 0, when
the omega d is very small. And this is actually
the omega 0. I hope you can see it. And this will be
increasing rapidly here and approaching to pi. So that means, when you
are shaking this system like crazy-- very high frequency-- then
the system cannot keep up with the speed. The amplitude will
be very small. And also, the amplitude will
be out of phase completely. So let's actually do us
another demonstration, using this little device here. This is actually
what you see before, the ball with a Mexican hat. And you can see that there is a
spring attached to this system. And on the top, what
I am going to do is to use this motor to drive
this system up and down, as a direct force. So now, what am I going to
do is to come from a very low-frequency oscillation. So you can see that the natural
frequency is sort of like this. And you can see that, now, I
am driving this system really slowly. You can see, this is
actually going up and down really slowly. And you see that-- huh-- the amplitude is
actually pretty small. There's no excitement
for the moment. All right, so what
I'm going to do is, now, I increase the speed of the
motor and see what will happen. So you can see,
now, it's actually driving it with higher
and higher frequency. You see that-- huh--
something is happening. You can see the amplitude is
getting larger and larger. I'm still increasing
the frequency-- increasing, increasing-- until
something-- something happened! Right? Did you see that? It starts to
oscillate up and down. Because right now,
you can see that-- look at the top. The frequency of
the motor is now really close to the natural
frequency of this system. So a resonance
behavior will happen. And what you are going
to get is that, OK, omega d around omega 0. Then you are going to
get large time amplitude. So now, what I am going
to do is to continue to increase the
driving frequency to a very large value. OK, now it's actually
doing the "mmmmm"-- doing it really fast. You can see on the top-- very fast. OK, I even get it even faster. You see that-- huh--
indeed, this system is now oscillating at
a larger frequency. It's trying to keep up
with the driving force. But you can see that the
amplitude is actually much smaller than what
had happened before. So before the class, you
may actually think that, OK, drive it really fast. Maybe we'll increase
the amplitude. But in reality, actually,
it will give you a very small amplitude. Another thing, which
is interesting to know, is that you can see that, when
the driving force is actually at the maximum. And actually the
position of this mass is actually at the minimum. So they are actually
out of phase. I hope you can see it. It's like this. OK, so what you can see is that,
when I understand the system and I try to drive it with
the natural frequency, what is going to happen
is that I'm exciting this system to a
state of resonance. So basically, you'll get
some resonance behavior. So I have shown
you that this works for driven mechanical
oscillator. It also works for the
spring-mass system. And there are many other
things which also work, which is around you. For example, if you happen
to be my office hour, you would notice that the
air-condition in my office is actually creating
a resonance behavior. You'll see low frequency sound-- "mm mm mm"--
something like that. And that is because
the pipe actually happens to have the frequency
match with the resonance frequen-- OK, the airflow actually
happened to excite the pipe, so that it's actually
oscillating up and down at that frequency. So what I did was I tried
to turn it down to low and see what happened. But unfortunately, it actually
excited another resonance. I see, now, not a
low-frequency sound, but a very high-frequency sound. I will post a video,
actually online. So my life is hard, right? But I'm a physicist. Is So I choose to
use the median. Then I actually stay
between the two resonances. Then I don't hear the additional
sound, which bothers me. Another example is that, when
I was in Taiwan as a undergrad, I was living outside
in a apartment. And with my flat-mate, we owned
a very old washing machine. So in the middle of the
night, the washing machine would started to walk
around, like my flat-mate. And we are not scared. That is because the oscillation
frequency-- actually, the rotation-- happened to match
with the frequency of the washing machine. Therefore, when we started
to wash our clothes, it start to walk
around in the room. So as a physicist,
what we have decided is to make the spin slightly
slower, or even faster. Then, actually,
you can see that, when you do that, then you get
rid of the resonance behavior. So it's not walking
around any more. We can control it. Another thing which
is interesting is that the resonance
behavior is not only in the physical objects,
which we actually deal with these days also. But either you learn
quantum mechanics and upon the field
theory, you will find that there are resonance
also in a mass wave function. So basically, you can see
that these are examples of the Z boson resonance peak. So if you scatter a
electron and positron then, basically, you'll see that the
cross-section have a resonance peak at around 90 GeV. And that is actually another
very interesting example of a resonance in
particle physics. Finally, the last example,
which I am going to go through is an example involving a glass. We have prepared a very
high-quality glass here. Maybe you have seen
this glass before. They are pretty nice. And I usually use it to enjoy
my red wines, which you cannot, enjoy, probably now. So you can see that
this is the glass. And if I put a little bit of
water on my hand and I rub it-- [VIBRATING TONE] --carefully, I can
actually excite one of the resonance frequencies. So you can see that we
have all everything working on a single particle. And that will give you
one resonance frequency. If I work on two
particles, which you will see that
in the next lecture, I would get two
resonance frequencies. And this glass is made of
infinite number of particles. Therefore, I will
have infinite number of resonance frequencies. When I'm rubbing
it, I'm actually giving input of all kinds
of different frequencies. But the glass will
be smart enough so that it will pick up the
one it likes the most, which is the resonance frequency. So you can see that the sound
is actually, roughly, 683 Hertz. And you can actually
measure it with your phone. So on the TV commercial,
you may have seen that there's a lady singing. And she's singing so loudly
such that the glass-- "bragh!"-- breaks. Can we get a volunteer today
to sing in front of us? Oh-- singing. Can you sing it-- high frequencies? "Ahhhh." [LAUGHTER & APPLAUSE] Very good try. But it didn't work. OK, I guess it's
really difficult to perform that in front of
so many people unprepared. But fortunately, we are MIT. So we have designed a
device, which actually can help us to achieve this. So this device actually
contain a amplifier here. And I can now control the
frequency of the sound through this scope. And this amplifier will
actually amplify the signal and produce a sound wave and try
to actually isolate this glass. So we are going to
do this experiment. So we will need to change
the loud setting a bit. Because the sound is going
to be, probably, too loud. Just for safety-- some
of you may not survive. [LAUGHTER] So I'm handing out these. OK, who is closer? OK, maybe you. AUDIENCE: [INAUDIBLE] YEN-JIE LEE: Oh. Oh, sorry. [LAUGHTER] I'm so sorry. What? I don't need that. OK. So just for safety,
I will put this on. And what I am going to do
is also put these glass on. OK, maybe I'll do this first. AUDIENCE: [INAUDIBLE] YEN-JIE LEE: OK, so what
I am going to do now is to start
producing sound wave. [LOUD TONE] So through the
camera, you should be able to see what is
actually shown on the screen. So you can see that, if this
glass is actually moving, the wood inside
would also vibrate. So you can see that, clearly,
we don't have resonance yet. So what I am going to do is
to increase the frequency and see what happens. So now, it's actually 643. It's actually still below
the resonance frequency. Now, I have measured
the frequency. It should be 684. So now, it's actually 653-- 663 Hertz-- 673 Hertz. Can you see the movement? You cannot see the movement yet. 683-- you see? You see, now-- AUDIENCE: Yes. YEN-JIE LEE: --the
frequency of the sound is actually matching with one
of the natural frequencies of the glass. Apparently, the glass likes it. And now, you can see that
it is still vibrating. And the next step, which
we are going to do, is to try to increase
the amplitude, increase the volume of the
sound, and see what happens. Maybe you want to cover
your ears, just for safety. [LOUD TONE] OK, then the glass may
break, if we are lucky. Let us see what is
going to happen. [INCREASINGLY LOUDER TONE] Oh! [APPLAUSE] TECH SUPPORT: Good job. YEN-JIE LEE: Very good. TECH SUPPORT: Perfect. That's the quickest
one we've had. YEN-JIE LEE: Yeah,
thank you very much. Thank you, glass. [LAUGHTER] So you can see the
power of resonance. So if I tune down the
frequency slightly more, then you will be where? You'll be here. Then you will not have enough
amplitude to break the glass. And also, as we
discussed before, the quality of the glass should
be really, really high, such that the resulting amplitude
will be very large. Then you can actually break
it with a external sound wave. And if we go above the
resonance frequency, then you would not
also move a bit. Because if you go to
very large omega d, then amplitude will
be pretty small. OK, let me try to switch
back to my presentation. I think we did. Sure. So this is actually what
we have learned today. We have learned the behavior
of a damped driven oscillator. We have learned the
transient behavior. So what is actually
transient behavior is a mixture of steady
state solution, which was coming from
the driving force, and the homogeneous solution. If you wait long enough, this
will decay and disappear. Is And we have learned resonance. So an IOC circuit, which
you actually solved that in your P-set, in pendulum,
which I just show you, which helped my son to learn
wavelength vibrations-- and air condition,
washing machine, glass-- particle physics. We can see damped os-- driven oscillator or
resonance almost everywhere. So I hope that you
enjoyed the lecture today. And what we are
going to do next time is to put multiple objects
together so that you see the interaction between one
particle to the other particle and see how we can actually make
sense of this kind of system. Thank you very much. And I will be here if you have
additional questions related to the lecture.
