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visit MIT OpenCourseWare at ocw.mit.edu. YEN-JIE LEE: And
welcome back, everybody, to this fun class, 8.03. Let's get started. So the first thing
which we will do is to review a bit what
we have learned last time. And then we'll go
to the next level to study coupled oscillators. OK. Last time, we had learned a lot
on damped driven oscillators. So as far as the course
we've been going, actually, we only
study a single object, and then we introduce
more and more force acting on this object. We introduce damping
force, we introduce a driving force last time. And we see that
the system becomes more and more difficult to
understand because of the added component. But after the class
last time, I hope I convinced you that we can
understand driven oscillators. And there are two very important
things we learned last time. The first one is the
transient behavior, which is actually
a superposition of the homogeneous solution
and the steady state solution. OK. One very good news is that
if you are patient enough, you shake the
system continuously, and if you wait long enough,
then the homogeneous solution contribution goes away. And what is actually left over
is the steady state solution, which is actually much simpler
than what we saw beforehand. It's actually just
harmonic oscillation at driving frequency. Also, I hope that we also have
learned a very interesting phenomenon, which is resonance. When the driving frequency is
close to the natural frequency of the system, then the
system apparently likes it. Then it would respond
with larger amplitude and oscillating up and
down at driving frequency. So that, we call it resonance. This is the equation
of motion, which we have learned last time. You can see is theta double
dot plus Gamma theta dot, is a contribution
from the drag force, and omega 0 squared
theta is the contribution from the so-called spring force. And finally, that is equal to
f0 cosine omega d t, the driving force. And as we mentioned
in the beginning, if we prepare this system
and under-damp the situation, then the full solution
is a superposition of the steady state
solution, which is the left-hand side, the
red thing I'm pointing to, this steady state solution. There's no free parameter in
the steady state solution. So A, the amplitude, is
determined by omega d. Delta, which is the phase is
also determined by omega d. There's no free parameter. OK. And in order to make the
solution a full solution, we actually have to add in
this homogeneous solution back into this again. And basically, you
have B and alpha, those are the free
parameters, which can be determined by the
given initial conditions. OK. So if we go ahead and
plot some of the examples as a function of
time, so the y-axis is actually the amplitude. And the x-axis is time. And what is actually
plotted here is a combination or
the superposition of the steady state solution
and the homogeneous solution. And you can see that the
individual components are also shown in this slide. You can see the red thing
oscillating up and down harmonically, is steady
state solution contribution. And also, you have
the blue curve, which is decaying away
as a function of time. And you can see that if you
add these two curves together, you get something
rather complicated. You will get some kind of
motion like, do, do, do, do. Then the homogeneous
solution actually dies out. Then what is actually left
over is just the steady state solution, harmonic oscillation. And in this case, omega d
is actually 10 times larger than the natural frequency. And there's another example
which is also very interesting. It's that if I make the
omega d closer to omega 0-- OK, in this case it's actually
omega d equal to 2 times omega 0, then you can produce
some kind of a motion, which is like this. So you have the oscillation. And they stayed
there for a while, then goes back, and
oscillates down, and stay there then goes back. OK, as you can see on there. The homogeneous solution part
and steady state solution part work together and produce
this kind of strange behavior. OK. And that's just another example. And if you wait
long enough, again what is actually left over
is the steady state solution. OK. So what are we
going to do today? So today, we are
going to investigate what will happen if we try to
put together multiple objects and also allow them
to talk to each other. OK. So if we have two objects, and
they don't talk to each other, then they are still
like a single object. They are still like simple
harmonic motion on their own. But if you allow them
to talk to each other, this is the so-called
coupled oscillator, then interesting thing happen. So in general, coupled systems
are super, super complicated. OK. So let me give you
one example here. This is actually
two pendulums that are a coupled to each
other, they are actually connected to each
other, one pendulum, the second one is here, OK. And for example, I can actually
give it an initial velocity and see what is going to happen. You can see that the
resulting motion-- OK. Remember, we are just talking
about two pendulums that are connected to each other. The resulting motion
is super complicated. This is one of my
favorite demonstrations. You can actually stare at
this machine the whole time. And you can see
that, huh, sometimes it does this rotation. Sometimes it doesn't do that. And it's almost like
a living creature. So we are going to
solve this system. No, probably not, he knows. [LAUGHTER] But as I mentioned before,
you can always write down the equation of motion. And you can solve
it by computer. Maybe some of the
course 6 people can actually try and write the
program to solve this thing and to simulate what
is going to happen. So let's take a look at this
complicated motion again. So you can see that the good
news is that there are only two objects. And you can see-- look at the green,
sorry, the orange dot. The orange dot is always
moving along a semi circle. But if you focus
on the yellow dot, the yellow is doing all
kinds of different things. It's very hard to predict
what is going to happen. So what I want to say is,
those are interesting examples of coupled systems. But they are actually far
more complicated than what we thought, because they
are not smooth oscillation around equilibrium position. So you can see that now
if I stop this machine and just perturb it slightly,
giving it a small angle displacement, then you
can see that the motion is much more easier to understand. You see. You even get one of these
questions in your p-set. OK, that's good news. So our job today
is to understand what is going to happen to
those coupled oscillators. Let me give you a few
examples before we start to work on a specific question. The second example I would
like to show you is a saw and you actually connect
it to two - actually a ruler, a metal ruler,
which is connected to two massive objects. Now I can actually give
it the initial velocity and see what happens. And you can see
that they do talk to each other through this
ruler, this metal ruler. Can you see? I hope you can see. It's a bit small. But it's really interesting
that you can see-- originally, I just introduced
some displacement in the left-hand side mass. And the left-had side
thing start to move or so after a while. There are two more examples
which I would like to give you, introduce to you. There are two
kinds of pendulums, which I prepared here. The first one is there are two
pendulums that are connected to each other by a spring. And if I try to
introduce displacement, I move both masses slightly and
see what is going to happen. And we see that the motion
is still complicated. Although, if you stare at
one objects, it looks more like harmonic oscillation,
but not quite. For example, this
guy is slowing down, and this is actually
moving faster. And now the right hand side
guy is actually moving faster. Motion seems to be complicated. Also, you can look at this one. Those are the two pendulums. They are connected to each
other by this rod here. And of course, you
can displace the mass from the equilibrium position. I'm not going to hit-- not hitting each other. So you can displace the
masses from each other. And you can see that they
do complicated things as a function the time. How are we going
to understand this? And I hope that by the
end of this lecture you are convinced that you can
as you solve this really easy, following a fixed procedure. In those examples, we
have two objects that are connected to each other. And therefore, they
talk to each other and produce coupled motion. Those are a couple
oscillator examples. There's another very
interesting example, which is called Wilberforce pendulum. So this is actually a pendulum. You can rotate like this. And it can also
move up and down. It's connected to a spring. The interesting thing
is that if I just start with some
rotation, you can see that it starts to also
oscillate up and down. You see? So initially I just
introduced a rotation. Now it's actually
fully rotating. And now it starts
to move up and down. And you can see that the
energy stored in the pendulum is going back and forth between
the gravitational potential, between the potential
of the spring, and also between the kinetic
energy of up and down motion and the rotation. They're actually doing all
those transitions all the time. So you can see-- so initially it's just rotating. And then it starts
to move up and down. And this one is
also very similar. But now the mass is
much more displaced. And if I try to rotate this
system without introducing a horizontal direction
displacement, it still does is up and down
motion, like a simple spring mass system. So what causes this
kind of motion? That is because when we move
this pendulum up and down, we also slightly
unwind the spring. That can generate some
kind of torque to this mass and produce rotational behavior. And you can see
that this is just involving one single object. But there's a coupling
between the rotation and the horizontal
direction motion. So that's also special
kind of coupled oscillator. So after all this,
before we get started, I would like to say that
what we are going to do is to assume all those
things are ideal, without them being forced,
without a driving force. We may introduce that
later in the class. But for simplicity, we'll
just stay with this idea case, before the mass becomes
super complicated to solve it in front of you. And also, we can see that
all those complicated motion are just illusion. Actually, the reality
is that all of those are just superposition
of harmonic motions. You will see that by
the end of this class. So that is really amazing. OK. Let's immediately get started. So let's take a
look at this system together and see
if we can actually figure out the motion
of this system together. So I have a system with
three little masses. So there are three little
masses in this system. They are connected to
each other by spring. Those springs are highly
idealized, the springs. And they have spring constant k. And the natural length's l0. And they are placed on Earth. And I carefully
design the lab so that there's no friction
between the desk and all those little masses. So once you get started
and look at this system, you can imagine
that there can be all kinds of different
complicated motions. You can actually, for
example, just move this mass and put the other two on hold. And they can
oscillate like crazy. They can do very
similar kind of motion. There are many,
many possibilities. But if you stare at
this system long enough, you will be able to identify
special kinds of motion which are easier to understand. So what I would like
to introduce to you is a special kind
of motion which you can identify from the
symmetry of this system. That is your
so-called normal mode. So what is a normal mode,
a special kind of motion we are trying to identify? That is actually
the kind of motion which every part
of the system is oscillating at the same
frequency and the same phase. So that is your
so-called normal mode, and is a special kind
of motion, which I would like you to identify with me. And we would later realize that
those special kinds of motions, which are easier to
understand, actually helps us to understand the
general motion of the system. You will realize that the most
general motion of the system is just a superposition of all
the identified normal modes. And then we are
done, because we have a general solution already. So that's very good news. That tells us that
we can understand the system systematically,
and step by step. And then we can write the
general motion of the system as a superposition of
all the normal modes. So let's get started. So can you guess what are
the possible normal modes of this system? So that means each
part of the system is oscillating at the same
frequency and the same phase. Can anybody and any one of
you guess what can happen, each part of this
is an oscillating at the same frequency? Yeah? AUDIENCE: If the two masses
on that side are displaced the same amount
and then they're -- YEN-JIE LEE: Very good. So he was saying that now
I displace the right hand side two masses all together
by a fixed amount, and also the left hand side, right, by
a fixed amount and then let go. So that's what
you're saying, right? OK. So the first mode we have
identified is like this. So you have left hand side
mass displace by delta x. And the right hand
side two masses are also displaced by delta x. So basically you hold
this three little masses and stretch it by the same-- introduce the same amplitude to
all those three little masses, and let go. So that is actually
one possible mode. And if we do this,
then basically what you are going to see
is that this is actually roughly equal to this system. They're connected to each
other by two springs. And the right hand side part
of the system, both masses are oscillating at the same
amplitude and the same phase. They look like as if they are
just single mass with mass equal to 2m. And if you introduce a
displacement of delta x, then what is going to
happen is that if I take a look at the mass, left
hand side mass, and the force acting on this
mass, the force will be equal to minus
2k times 2 delta x, because that's the
amount of stretch you introduce to the spring. And that will give
you minus 4k delta x. And we have already solved
this kind of problem in the first lecture. So therefore you can
immediately identify omega, in this case,
omega a squared will be equal to 4k divided by 2m. This is actually the
effective spring constant, and this is actually the mass. So that is actually the
frequency of mode A. Can you identify a second kind
of motion which does that? So in this case, what
is going to happen is that the three masses will-- OK, one, two, and three. The three masses will
oscillate as a function of time like this with angular frequency
of square root of 4k over 2m. What is actually a
second possible motion? Yes? AUDIENCE: All masses being
stretched [INAUDIBLE] YEN-JIE LEE: Compressed. AUDIENCE: Compressed the same-- YEN-JIE LEE: Very good. I'm very lucky that I'm in front
of such a smart crowd today. And we have
successfully identified the second mode, mode B.
So what is going to happen is that the left hand
side mass is not moving. And you compress the
upper one slightly and you stretch the lower
one, the lower little mass to the opposite direction. The displacement is delta
x, and the displacement of the second mass is delta x. So what is going to happen? What is going to happen
is that the left hand side mass will not move at all
because the force, the spring force, acting on this
mass is going to cancel. And apparently, these
two little masses are going to be doing
harmonic motion. Since this left hand
side mass is not moving, it's as if this is a wall and
this were a single spring, k, that's connected
to a little mass. And it got displaced by delta x. So what will happen is that this
mass will experience a spring force, which is F equal
to minus k delta x. Therefore, we can
immediately identify omega b squared will
be equal to k over m. So you can see that we have
identified two kinds of modes, which every part of the
system is oscillating at the same frequency
and the same phase. Everybody agree? No not everybody agree. Look at this guy this
guy is not moving. How could this be? This is not the normal mode. Isn't it? OK. I hope that will
wake you up a bit. I can be very tricky here. I can say that this mass
is also oscillating, but with what amplitude? AUDIENCE: Zero. YEN-JIE LEE: Zero amplitude. Right So the conclusion is
that, aha, everybody is actually oscillating at the
same frequency, but these guy with
zero amplitude. AUDIENCE: Are they oscillating
at the same phase as well? YEN-JIE LEE: Yeah. Oh very good question. Another objection I receive. So life is hard for me today. Hey. This guy is oscillating
out of phase. These two guys are out of phase. But I can argue that the
amplitude of the first mass is actually has a minus sign
compared to the second mass. Then they are again in phase. So very good. I like those questions. And I hope I have convinced you
that everybody is oscillating, although you cannot see it,
because the amplitude is small, is zero. And they are all oscillating
at the same phase. Yes. AUDIENCE: How come
there's only one mass? YEN-JIE LEE: Oh,
the right hand side? AUDIENCE: Yeah. YEN-JIE LEE: Oh, yeah. That is because the left
hand side mass, the 2m one, is actually not moving. Because they are
two spring forces, one is actually
pushing the mass, the other one's
pulling the mass. And they cancel perfectly. Therefore, it's as if
those two guys are not-- they don't find each other. And then it's like,
they are just tools mass connected to a wall along. And then you can now identify
what is the frequency. OK. Very good. So we make the made a lot of the
progress from the discussion. And now I would like
to ask you for help. What is the third oscillation? Yes? AUDIENCE: There's no
third normal mode. YEN-JIE LEE: There's
no third normal mode. AUDIENCE: There's
no third normal mode because there are
restricted to one dimension. I can not imagine another
mode that would not displace the central mass. YEN-JIE LEE: Very good. That's very good. On the other hand,
you can also say, I also take the
center mass motion as one of the normal mode. I think that's also
fair to do that. Very good observation. You can see that the whole
can move simultaneously. I can also argue that
they are oscillating at the same frequency
and the same phase, because they are
all moving together. So these are the 2m
connected to mass one. All of them are moving
in the same direction. So now I can
calculate the force. What is the force? F is 0. Therefore, omega c is 0. So you can the small
limit of omega. So of course, I can pretend
that those mass are connected to a really, really small spring
to the wall with is a spring constant k'. And I have k' goes to zero. And they are actually
going to oscillate with omega c goes to zero. So in this case,
the amplitude is going to increase forever,
because you have A sin omega c t. And this roughly A omega c t. And this is just vt. So what I want to argue
is that this is actually also oscillation, but with
angular frequency zero. And the general motion can
be in written as vt times c, for example, some constant. Any questions? So what I'm going to
do next may amaze you. Very good. So we have identified three
different kinds of modes. We have mode A, which is with
omega a squared equal to omega a squared. Where is omega a squared. There. It's 4k over 2m. And also, the
motion is like this. x1 equal t A cosine
omega a t plus phi a. x2 is equal to minus A, because
they have different sine. So if the motion is in the
left hand side direction, then the two masses
are oscillating in the opposite direction. So therefore, I get a minus
sign in front of A. Cosine omega a t plus phi a. x3 will be also equal to minus
A cosine omega a t plus phi a. Of course, I need to define
what this x1, x2, x3. That's why most of you
got super confused. So the x1, what
I mean is that is that the displacement of
the mass 2m, I call it x1. The displacement of the upper
mass, the upper little mass, I call it x2. And finally, the displacement
of the third mass, I call it x3. Therefore, you can
see that mode A, you have this kind of motion. The amplitude of the first
mass is A. Therefore, if I define that to be A,
then the second and third one, or the amplitude will
be defined as minus A. And you can see that all
of them are oscillating at fixed angular
frequency, omega a, omega a, omega a; and also
fixed phase, phi a, phi a, phi a. Of course, we can
also write down what we get for
mode B. For mode B, the left hand side mass
is not moving, stay put. And the other two
masses are oscillating at the frequency of omega b. And amplitude, they
differ by a minus sign. OK. Omega b squared is
equal to k over m from that logical argument. And then we get x1 equal
to 0 times cosine omega b t plus phi b. x2, I get B cosine
omega b t plus phi b. x3, I get minus B cosine
omega b t plus phi b. Any questions here? Finally, mode C.
All the mass, x1 is equal to x2 is equal to
x3, is equal to C plus vt. So you can see that we have
identified three modes, mode A, mode B, and the mode C.
And there are three angular frequencies which we identified
for all of those normal modes, omega a, omega b, and omega c. And you can see that
we also identified how many free parameters. One free parameter, two,
three, four, five, and six. If you careful, you
write down the equation of motion of this system,
you will have three coupled differential equations. And those are second order
differential equations. If you have three variables,
three second order differential equations. If you manage it magically,
with the help from a computer or from math department
people, how many free parameter would you expect in
you a general solution? Can anybody tell
me well how many? I have three second order
differential equations. Yes? AUDIENCE: 6? YEN-JIE LEE: 6. So look at what we have
done we identified already 1, 2, 3, three normal modes. By there are 1, 2, 3, 4,
5 6, six free parameters. That tells me I am done. I'm done. Because what is the
general solution? The general solution is just a
superposition of mode A mode B and mode C. You have six
free meters to be determined by six initial conditions,
which I would like-- I have to tell you what are
those initial conditions. So isn't this amazing to you? I didn't even solve the
differential equation, and I already get the solution. And you can see another
lesson we learned from here is that,
oh no, you can imagine that the
motion of the system can be super complicated. This whole thing can do
this, all the crazy things are all displaced,
and the center of mass can move, as you said. But the result is actually
very easy to understand. It's just three kinds of motion,
the displacement, and two kinds of simple harmonic motion. We add them together. And then you get the general
description of that system. So everything is so nice. We understand the
motion of that system. But in general,
life is very hard. For example, now I do
something crazy here. I change this to 3. What are the normal modes? Can anybody tell me? It becomes very, very
difficult, because there's no general symmetry of
that kind of system. So we are in trouble. One of the modes
maybe still there, which is actually mode
B. But the other modes are harder to actually guess. So you can see that that already
brings you a lot of trouble. And you can see that I can now
couple not just two objects, I can couple three objects,
four objects, five objects, 10 objects. Maybe I will put that in your
p set and see what happens. And you can see
that this becomes very difficult to manage. So what I'm going to do in
the rest of this lecture is to introduce
you a method which you can follow in general
to solve the question and get the normal mode
frequencies and normal modes. So we will take a
four minute break. And we come back at 12:20. So if you have any
questions, let me know. What we are going to do
in the following exercise is to try to understand a
general strategy to solve the normal mode frequencies
and the normal mode amplitudes, so that you can apply this
technique to all kinds of different systems. So what I am going to do today
is to take these three mass system, and of course as
usual, I try to define what is this coordinate system? The coordinate system
I'm going to use is x1 and x2 an x3 describing
the displacement of the mass from the equilibrium position. And the equilibrium
means that there's no stretch on the spring. The string is unstretched. It's at their own
natural length, l0. So once I define that, I can
do a force diagram analysis. So that starts from
the left hand side mass with mass equal to 2m. I can write down the equation
of motion, 2m x1 double dot. And this is going to be equal
to k x2 minus x1 plus k x3 minus x1. So there are two spring
forces acting on this mass, the left hand side mass. The first one is
the upper spring. The second one is coming
from the lower one. And you can see
that both of them are proportional to
spring constant k, and also proportional to
the relative displacement. And you can see that the two
relative displacement, which is the amount of
stretch to the spring, is actually x2 minus
x1, and the x3 minus x1. Am I going too fast? OK. Everybody's following. And you can actually
check the sign. So you may not be sure. Maybe this is your x1 minus x2. But you can check that,
because if you increase x1, what is going to happen? This term will
become more negative. More negative in this
coordinate system is pointing to the
left hand side. So that makes sense. Because if I move this
mass to the right side, then I am compressing
the springs. Therefore, they are
pushing it back. Therefore, this is actually
the correct sign, x2 minus x1. The same thing also applies
to the second spring force. So that's a way I double
check if I make a mistake. Now, this is actually the
first equations of motion. And I can now also
work on a second mass. Now I focus on a
mass number two. The displacement is x2. Therefore the left hand
side of Newton's Law is m x2 double dot. And that is equal
to the spring force. The spring force, there's
only one spring force acting on the mass. Therefore, what I am going
to get is k x1 minus x2. Everybody's following? You can actually check
the sign carefully, also. And finally, I have the
third mass, very similar to mass number two. I can write down the
equation of motion, which should k x1 minus x3. So that is my coupled second
order differential equations. There are three equations. And all of them are
second order equations. So this looks a bit messy. So what I'm going
to do Is no magic. I'm just collecting all
the terms belonging to x1, and put them together, all
the terms belonging to x2, putting all together, and
just rearranging things. So no magic. So I copied this
thing, left hand side. 2m x1 double dot. And the dot will be equal
to minus 2k x1 plus, I collect all the times
related to x2, plus k x2, there's only one term
here, then plus k x3. I'm just trying to
organize my question. So you can see that I
collect all the terms related to x1 to here. Minus k, minus k,
I get minus 2k. And the plus k for
x2, plus k for the x3. And I can also do that
for m x2 double dot. That will be equal to k x1
minus k x2 plus zero x3, just for completeness. I can also do the same thing
for the third mass, m x3 double dot. This is equal to k x1 plus 0 x2. There's no dependence on
x2, because x1 and x2-- x3 and x2 are not talking
to each either directly. Finally, I have the third,
which is minus k x3. Now our job is to solve
those coupled equations. Of course, you have
the freedom, if you know how to solve
it yourself, you can already go
ahead and solve it. But what I am
going to do here is to introduce technique,
which can be useful for you and make it easier to follow. It's a fixed procedure. So what I can do
is the following. I can write everything
in them form of a matrix. How many of you heard the
matrix for the first time? 1, 2, 3, 4. OK. Only four. But if you are not being
familiar with matrix, let me know, and I can help you. Let the TA know. And also, there's a section
in the textbooks, which I posted on announcement,
which is actually very helpful to understand matrices. But sorry to these
four students, we are going to use that. And maybe, you already learn
how it works from here. So one trick which we
will use in this class is to convert everything
into matrix format. What I am going to do
is to write everything in terms of M,
capital X, capital M, capital X double dot equal
to minus capital K capital X. Capital M, capital
X, and capital K, those are all matrices. Because I write this
thing I really carefully, therefore we can already
immediately identify what would be M, capital
M and capital X and a K. So I can write down immediately
will be equal to 2m, 0, 0, 0, m, 0, 0, 0, m. Because there's only one in each
line, you only have one term. X1 double dot, x2 double
dot, x3 double dot. And also, you can
write down what will be the X. This
is actually a vector. X will be equal to x1, x2, x3. Finally, you have the K? How do I read off K? Careful, there's
a minus sign here, because I would like to make
this matrix equation as if it's describing a simple
harmonic motion of a one dimensional system. So it looks the same, but they
are different because those are matrices. But therefore, I
have in my convention I have this minus sign there. Therefore, when you read off
the K, you have to be careful. So what is K? K is equal to 2k. You have the minus 2k
here in front of x1. But because I have a minus
sign there, therefore this one is actually taken out. So we have 2k there. Then you have minus k,
minus k, minus k, k, 0. Minus k, plus k, 0. And finally, you can
also finish the last row. You get minus k, 0, k. K becomes minus k. Minus k becomes k. So we have read off all
those matrices successfully. So you may ask,
what do they mean? Do they get the meaning? M, K, X, what those? M, capital M matrix, tells
you the mass distribution inside the system. So that's the meaning
of this matrix. X is actually
vector, which tells the position of individual
components in the system. Finally, what is K? K is telling you how each
component in the system talks to the other components. So K is telling you
the communication inside the system. So now we understand a
bit what is going on. And as usual, I will go
to the complex notation. So I have xj, the small
xj are the position of the mass, x1, x2, and x3. xj will be real
part of small zj. Small xj equal to
real part of zj. Therefore, I can now write
everything in terms of matrices again. So now I can write
the solution to be Z, the capitol z is a matrix,
exponential i omega t plus phi. This is the guess
the solution I have. A1, A2, and A3. Those are the
amplitudes, amplitude A of the first mass, amplitude
of the second mass, amplitude of the third mass,
in their normal mode. And all of those are oscillating
at the same frequency, omega, and the same phase, phi. Does that tell you something
which we learned before? Oh, that's the definition
of the normal mode. I'm using the definition
of the normal mode. Every part of the
system is oscillating at the same frequency
in the same phase. And we use that to
construct my solution. The complex version is
exponential i omega t plus phi, oscillating at the
same frequency, oscillating at the same phase. And those are the amplitude,
which I will solve later. OK, any questions? I hope I'm not going too fast. If everybody can
follow, now I can go ahead and solve the
equation in the matrix format. So now I go to the
complex notation. So the equation M X double
dot equal to minus KX becomes M Z double
dot equal to minus KZ. And also, I can immediately
get the Z double dot will be equal to minus
omega squared Z, because each time I
do a differentiation, I get i omega out of the
exponential function. And I cannot kill that
exponential function, so it's still there. Therefore, I get
minus omega squared in front of Z. I hope
that doesn't surprise you. So that's very nice
and very good news. That means I can replace this
Z double prime with minus omega squared Z. Then I get
minus M omega squared Z. And this is equal to minus KZ. And I can cancel the minus sign. That becomes
something like this. So, I can now cancel the
exponential i omega t plus phi, because I have Z in
the left hand side. And exponential i omega t
plus phi is just a number. So therefore, I can cancel it. So what is going to
happen if I do that? Basically, what
I'm going get is I get M omega squared
A equal to K A. I'm trying to go extremely
slowly, because this is the first time we
go through matrices. So now you have this expression. Left hand is a matrix, M, times
some constant, omega squared. I can actually get omega
squared in front of it, because this is
actually just a number. A is just a vector, which is
A1, A2, A3, also a matrix. K is actually how the individual
components talks to the others. So that's there, times A. Now I would like
to move everything to the right hand side,
all the matrices in front A to the right hand side. Then I multiply both
sides by M minus 1. So I multiply M minus 1
to the whole equation. M minus 1, what is M minus 1? The definition is
that the inverse of M is called M minus 1. M minus 1 times M is equal to
I, which is actually 1, 1, 1. Therefore, if I do this thing,
then I would get omega squared. M minus 1 times M
becomes I, unit matrix. And this is equal to M
minus 1 K A. And be careful, I multiply M minus
1, the inverse of M, from the left hand side. That matters. So now I can move
everything to the same side. I moved the left hand side
term to the right hand side. Therefore, I get M minus
1 K minus omega squared I. Those are all matrices. Times A, this is equal to 0. Any questions? So a lot of manipulation. But if you think about it,
and you are following me, you'll see that all
those steps are exactly identical to what we have been
doing for a single harmonic oscillator. Looks pretty familiar to you. But the difference is that now
we are dealing with matrices. AUDIENCE: What is A? YEN-JIE LEE: Oh, A. A
is actually this guy. I define this to be
A. And that means Z will be exponential i omega
t plus phi times A. I didn't actually write it explicitly. But that's what I mean. Any more questions? Yes? AUDIENCE: [INAUDIBLE]
is for [INAUDIBLE]?? YEN-JIE LEE: Can
you repeat that? AUDIENCE: So this whole
process, this is mode A, right? YEN-JIE LEE: Yeah. So this whole process is
for, not really the for mode A. So that A may be confusing. But in general, if
I have a solution, and I assume that the amplitude
can be described by a matrix. So it's in general. And you'll see that
we can actually derive the angular frequency
of mode A, mode B, and mode C afterward. I hope that answers
your question. So you see that for in
general, what I have been doing is that now, all
those things are equivalent to the original
equation of motion. What I am doing is
purely cosmetic. You see, make it beautiful. So all those things, this
thing is exactly the equivalent to that thing, up there. Up to M X double dot
equal to minus K x. Cosmetics. Beautiful. Looks-- I like it. All right. Then what I have been
doing is that now I introduce using a
definition of normal mode. I guess the solution will
have this functional form. Z equals to exponential i
omega t plus phi, everybody oscillating at the same
frequency, the same phase. Frequency omega, phase phi. And everybody can have
different amplitude. You can see from this
example, normal modes, they can have
different amplitude. The amplitude is what? I don't know yet. But we will figure it out. Then that's my assumption. The definition of normal mode. And I plug in to the
equation of motion. Then this is what we
are doing to simplify the equation of motion. There's no magic here. If I plug in the definition on
normal mode to that equation, this is actually going to bring
you to this equation, matrix equation. So if you have learned matrices
before, you have something, some matrix, times Z.
This is equal to zero. A is not zero. I hope. If it's zero, then the
whole system is not moving. Then it's not fun. So if A is not zero, then
this thing should be-- this thing times A should
make this equation equal to 0. So what is actually
the required condition? I get stuck, and of course again, my
friend from math department comes to save me. That means if this
thing has a solution, this equation has
a solution, that means that determinant of
M minus 1 K minus omega squared I has to be equal to 0. So that is the condition
for this equation to satisfy this
to be equal to 0. And just to make sure that I
don't know what is the angular frequency omega yet. I don't know what
is the phi yet. We can actually solve the
angular frequency, omega. So now, turn everything around. And basically now,
using this normal mode definition, and some
mathematical manipulation, the condition we need for this
equation to satisfy equal to 0, is determinant M minus 1
K minus omega squared I. I can write down M
minus 1 K minus omega squared I explicitly, just
to help you with mathematics. M minus 1 K is equal to 1 over
2m, 0, 0, 0, 1/m, 0, 0, 0, 1/m. It's just the inverse
matrix of the M matrix. Therefore, now I can write
down the explicit expression of M minus 1 K
minus omega squared I. This will be equal to k
over m minus omega squared, minus k over 2m, minus k
over 2, minus k over m, k over m minus omega squared, 0. I will write down all
the elements first. Then I will explain to you how
I arrived at the expression. OK. So this is M minus 1 k. The definition of
M minus 1 is that. And the definition of
K is in the upper right corner of the black board. Therefore, if you
multiply M minus 1 K, basically, the first
column will get-- wait a second. Did I make a mistake? No. OK. So basically, what you
arrive at is k/m, k/m, k/m. And also, the minus k
over 2m for the rest part of the matrix. And the minus omega
squared I will give you the diagonal component. Yes? AUDIENCE: Why do you have
to take the determinant and set it equal to
0 instead of just setting that equal to zero? AUDIENCE: This is a matrix. So these are the matrix. So a matrix times A
will be equal to zero. The general condition
for that to be satisfied is more general. It's actually the determinant
of this matrix equal to zero. Because this is actually
multiplied by some back to A. So I think there are
mathematical manipulation. Basically, you would
just collect the terms. And then calculate
M minus 1 K first. And the minus omega squared I
will give you all the diagonal and terms have a minus
omega square there. And that is actually the matrix. And of course, I can
calculate the determinant. So if I calculate
the determinant, then basically I get this
times that times that. So what you get is k over m
minus omega squared times k over m minus omega squared times
k over m minus omega squared. So these are all diagonal terms. And the minus 1 over 2 k
squared over m squared, k squared over m squared. sorry. Minus omega squared. So that's this off diagonal
term, this times this times that. OK. It will give you
the second term. And the third one,
which survived because of those zeros, many,
many terms are equal to 0. And then the third term, which
is nonzero, is again minus 1 over 2k squared over m squared,
k over m minus omega squared. And this is actually equal
to 0, because the determinant of this matrix is equal to zero. Everybody following? A little bit of a mess. Because I have been doing
something very challenging. I'm solving a 3 by 3 matrix
problem in front of you right. So the math can get
a bit complicated. But next time, I think we are
going to go to a second order one, 2 by 2 matrix. And I think that will
be slightly easier. But the general
approach is the same. So basically, you calculate M
minus 1 K minus omega squared. Then you get what is inside, all
the content inside this matrix. Then you would calculate
the determinant. And basically, you can
solve this equation. Now I can define omega0
squared to be k/m. And I can actually make this
expression much simpler. Then basically,
what you are getting is omega0 squared
minus omega squared to the third minus 1/2 omega0
to the fourth, omega0 squared minus omega squared. Minus 1/2 omega0 to the
fourth, omega0 to the square, minus omega squared. And this is equal to 0. And you can factor out
the common components. Then basically,
what you are going to get is, you can
write this thing to be omega0 squared minus
omega squared, omega squared. Because all of them
have omega squared. And omega squared
minus 2 omega0 squared. And that's equal to 0. So I am skipping a lot of steps
from this one to that one. But in general, you can solve
this third order equation. And I can first combine
all those terms together. And then I factor out
the common components. Then basically, what you
are going to arrive at is something like this. A lot of math here. But we are close to the end. So you can see now what are the
possible solutions for omega. That is the omega,
unknown angular frequency we are trying to figure out. You can see that there are
three possible omegas that can make this equation equal to 0. The first one is
omega equal to omega0. The second one is
square root of omega 0, coming from this
expression, that omega squared minus 2
omega zero squared. If omega equal to
square root 2 omega0, this will be equal to zero. And that will give you the
whole expression equal to 0. Then finally, I take this term. And then you will get zero. Omega squared, if
omega is equal to 0, then the whole expression is 0. I have defined omega0 squared
to be equal to k over m. Therefore, I can conclude
that omega squared is equal to k over
m, 2k over m, and 0. Look at what we have done,
a lot of mathematics. But in the end, after you
solve the eigenvalue problem, or the determinant
equal to zero problem, you arrive at that
there are only three possible
values of omega which can make the
determinant of M minus 1 K minus omega
squared I equal to 0. What are the three? k/m, 2k/m, and 0. If you look at this
value, then we'll say, this is essentially what we
actually argued before, right? Omega A squared is equal
to 4k over 2m is 2k over m. Wow. We got it. The second one is,
we think about really keep a straight
question in my head and understand this system. The second identified
normal more is having omega squared be equal to k/m. I got this also here magically,
after all those magics. And finally, the third one,
the math also knows physics. It also predicted that this is
one mode which have oscillation frequency of 0. Isn't that amazing to you? But that also gives
us a sense of safety. Because I can now add
10 pendulums, or 10 coupled system to
your homework, and you will be able to solve it. So very good. This example seems
to be complicated. But the what I want to say,
I have one minute left, is that what we
have been doing is to write the equation of
motion based on force diagram. Then I convert that
to matrix format, and X double dot
equal to minus KX. Then I follow the
whole procedure, solve the eigenvalue problem. Then I will be able to figure
out what are the possible omega values which can satisfy
this eigenvalue problem or this determinant. M minus 1 K minus omega
squared I equal to 0 problem. And after solving
all those, you will be able to solve the
corresponding so-called normal mode frequencies. You can solve it. And of course, you can plug
those normal mode frequencies back in, then you
will be able to derive the relative amplitude,
A1, A2, and A3. So what we have
we learned today? We have learned how
to predict the motion of coupled oscillators. That's really cool. And then next time,
we are going to learn a special kind of motion in
coupled oscillators, which is the big phenomena. And also, what will
happen if I start to drive the coupled oscillators? So I will be here if
you have any questions about the lecture. Thank you very much.
