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visit MIT OpenCourseWare at ocw.mit.edu. BOLESLAW WYSLOUCH: Good
morning, everybody. I'm Bolek Wyslouch. I'm a teacher substitute for
Professor Lee, who is now at some conference
in China, and he asked me to talk to you
about coupled oscillators. I understand that he introduced
the concept last time. You worked through
some examples. So what we are
going to do today is to basically go
through one or two examples of very straightforward
coupled oscillators, where I will introduce various kinds
of systematic calculational techniques, how
to set things up, how to prepare things
for calculations. And also, we may, depending
on how much time we have, start driving, have driven
coupled oscillators. And we will work on two,
again, simple physical systems, one that consists of two pendula
driven by forces of gravity, each of them. And then they are
connected with the spring. So each of those pendula,
each of those masses, will feel the effects of
gravity and effects of springs at the same time, and they
will talk to each other. There will be
coupling between them. So that's one physical
example which we'll consider. The other physical
example consists of two masses in the horizontal
frictionless track connected by a set of springs. So they are driven
by forces of spring. And those two systems are
very similar to each other, almost identical in
terms of calculations, and they exhibit
the same phenomena, and I will be able
to demonstrate several of the neat new things. And this particular
system is set up to introduce external
driving force, which will create a new set of phenomena. And we'll talk about it today. And what I would
like to stress today when we go through all
those calculations is, A, how do you convert
a given physical system with all the forces, et cetera,
into some sort of fixed form, fixed type of notation,
with which you can treat all possible coupled oscillators? And also we will discuss
various interesting-- even though the system
is very simple, just two masses, a spring, a little
bit of gravity on top of that, the way they behave could
be extremely complex, but it can be
understood in terms of very simple systematic
way of looking things through normal modes
and normal frequencies, so the characteristic
frequencies of the system. So let's set things up. So we'll start. This will be our workhorse. And by the way, once
we understand two, we will then generalize to
infinite number of oscillators, which is actually-- so
this model, which consists of weights hanging under
the influence of gravity plus the springs
will be then used for many applications of the
concepts later in this course. So let's try to convert
this physical system into a set of equations. So we have a mass, m, hanging
from some sort of fixed support, another mass here,
same mass for simplicity. We connect them with
a string, and we know everything about this system. We know the length of each
of those pendula, which is the same. We know masses. We know spring constant of a
spring connecting those two things. The spring is initially
at its rest position such that when the two pendula
are hanging vertically, the spring is relaxed. But if you move it
away from verticality, the spring either
compresses or stretches. And everything is in Earth's
gravitational field, g. We assume that this is an
ideal system, highly idealized. We only consider motion with
small angle approximation, only small displacement. There's no drag force assumed. The spring is ideal,
et cetera, et cetera. Of course, this thing here
is very far from being ideal, but hopefully basic
behaviors are similar. It's approximately ideal. To study the motion
of this thing, to understand how it
works, let's try to-- let's try to parameterize
it, and displace it from equilibrium, and
look at the forces, and try to calculate
equations of motion. So we will characterize
this system by two position coordinates. We will have x. We'll give this one number one. This will be number two. And we will have x subscript
1, which in general would depend on the time. This is the position of
this mass with respect to its equilibrium position. We will have x2 as
a function of t. Again, this tells us everything. And the full description
of the system is to know exactly what
happens to x1 and x2 for all possible times. And we will impose some
initial conditions. We can come back to that later. So again, so the
coordinate system is this. When we start talking about
the system in principle in the case of
somewhat larger angles, you have to worry about
vertical positions as well. So we will introduce. So there is also a coordinate y,
which we will need temporarily to set things up. So x is, as I say, x is
measured from equilibrium. Y is positioned vertically. So to calculate the
equations of motion, we have to look at the forces. So let's look at what
are the forces acting, for example, on this
mass, the mass, which is-- if it's displaced
from a vertical position. Let's say this mass, mass 1,
has moved by some distance away from thing Temporarily,
let's introduce an angle here to characterize this
displacement from vertical. And let's write down all
the forces acting on this - force diagram
acting on this mass. So there is a tension in
the string or the rod. Let's call it T1. There is a force
of spring acting in a horizontal direction. This is a vector. And there is a force of
gravity acting on this in the vertical direction. We can write down those forces. We know a lot about them. This one is minus mg y-hat. This one is equal to k x2 minus
x1 in the x-hat direction. So this is the force which, when
the spring is displaced from equilibrium, there is a
spring force, Hooke force, in the direction of -- in the usual direction. In this case, it's actually
in the opposite direction. And then there is a
tension the spring, which has to be calculated
such that we understand the acceleration of this object. So let's write
down the equations in the x-hat direction. This is m acceleration
of object number 1 in x direction is equal to
minus T1 sine theta 1 plus k x2 minus x1. And in the y-hat
direction, we have m y1 direction is equal to
T cosine theta 1 minus mg. At the small angle for theta
1 much, much smaller than one, we can assume, that cos theta
1 is approximately equal to 1 and sine theta 1
is equal to angle. We do the usual thing. So basically, in
this approximation, and also by looking
at the system, it's clear that the
system does not move, and the vertical
direction can be ignored. Yes? AUDIENCE: How do you know
which way [INAUDIBLE]?? BOLESLAW WYSLOUCH: Excuse me? AUDIENCE: The [INAUDIBLE]. How do you know which
way it [INAUDIBLE]?? BOLESLAW WYSLOUCH:
How do I know? AUDIENCE: Yeah. [INAUDIBLE] BOLESLAW WYSLOUCH:
The spring force is-- well, you have to
look at the mass 1. You are just looking at mass 1. So the spring is
connected to mass 1. And the force of
the spring on mass 1 is k times however the spring
is squashed or stretched, all right? So it knows about the existence
of mass 2, but only in a sense that you have to know
the position of mass 2. So we just assume
that x2 is something, and we just look
where the spring is. So that's why--
the force of spring depends on the difference
of position x1 minus x2. So this is written here. And in fact, interestingly,
the position of the mass 1 itself is a negative sign here. So if you move mass
1, the spring force is in the right
direction, minus kx. All right? So there is no
motion x1, so we can conclude from here the T cosine1
is approximately equal to 1. So T is simply equal to mg. So the tension in the spring
can be assumed to be mg. We don't have to worry about it. And then we just plug in-- also the angle can be
converted into position by realizing that the
distance times the angle is equal to displacement,
the usual geometry. The net result is that
by simplifying things, I can write down
equations for acceleration in the horizontal
direction for mass 1 is equal to minus mg x1
over l plus k x2 minus x1. OK? So this is an equation
of motion for mass 1 in our coupled system. And I could say
most of the terms have to do with a
motion of mass 1 itself. Mass 1 is its own pendulum. And mass 1 is feeling the
effect of the spring force. But because the
force of the spring depends on the difference
between positions, there is this coupling-- so the motion of mass 1
knows of where mass 2 is. And motion of mass 2 influences
the motion of mass 1. That's how the
coupling shows up. So for most of those
problems, what you do is you simply focus on
the mass in question. You take all the forces,
you calculate them, and then this
coupling will somehow appear in the equations. So we can repeat exactly
the same calculation focusing on mass 2. And then the equation which you
will get will be very similar. Let me just slightly
rewrite this equation here to kind of combine
all the terms which depend on the position
of mass 1 with terms that depend on mass 2. So where m x-acceleration
is equal to minus k plus mg over l times
x1 plus k times x2. So this is the coupling term. This is what makes
those pendula coupled. All right? And then I can write almost
exactly the same equation of mass 2 with the proper
replacement of masses. So let me write this down
in the following way-- kx1 minus k plus mg over l times x2. So the motion of
mass x1 depends on x1 itself multiplied by
something with a spring term and gravitational
term and depends on the position of mass 2
only through the spring. Mass 2 also is mostly driven
by its own gravitational force of itself plus the spring
depends on the position of x2. But there is this
coupling term that depends on position of mass 1. So both of them feel the
neighbor on the other side, right? So if I keep this one
steady of x2 equals 0, then basically the
forces here is just the spring plus the gravity. If I move this one and keep
this one at 0, the force on this spring spring and gravity. But if this one is displaced,
and I move that guy, the forces on this one
are affected by the fact that number 2 changed. OK? Again, I was able to
determine those coupling terms by simply looking at
mass 1 itself, mass 2 itself. All right, so this is the
set of two coupled equations. I have accelerations
here for x1, x2, and I have positions here. It's like an oscillator
of position acceleration with a constant term except that
things here are a little mixed. And the trick in this
whole mathematics, and calculations,
and the way we do things is how do you solve
those coupled equations? OK? So what I would like to do is-- and there is multiple
ways of doing that. So let me do everything. Let's write down everything
in the matrix form, because it turns out
that linear matrices are very useful for that. We will use them very, very-- in a very simple way. So let's introduce to
them and show vector, which consists of x1 and x2. So basically, all the
position x1 and x2 are here. So we will be monitoring
the change of this x2 as a function of time. We will introduce
a force matrix k, which is equal to k plus mg over
l minus k here, minus k here, k plus mg over l there. This is a two by two matrix. And then we need a third
matrix, mass matrix, which simply says that masses
are mass of first object is m and the other one
is also m, right? So these are three
matrices that basically contains exactly the same
information as out there. I probably need another matrix. I need an inverse matrix
for mass, which basically is 1 over m, 1 over m, 0 and 0. This is a inverted matrix. OK, and it turns out that after
I introduced these matrices, this set of equations
can be written simply as X, the second derivative
of the vector capital X, is equal to minus
m to the minus 1, this matrix, multiplying
matrix k and then multiplying vectors x again. All right? So this is exactly the
same as this, just written a different way. So it's only the
question of notation. So it turns out
it's very convenient to use matrix calculation
to do things faster. So instead of repeating writing,
all the x1s, x2, et cetera, instead I just stick them into
one or two element objects. I use matrices to
multiply things, and if I want to
know x1 and x2, I can always go, OK,
the top component of vector x, lower component of
vector x gives me the solution. Simple. Right? So let's try to use this
terminology to find solutions. So the question is how
do we find solutions to coupled oscillations. What is the most efficient way
of finding the most general motion of a coupled system? Anybody knows? What's the first thing? Yes? AUDIENCE: [INAUDIBLE]. BOLESLAW WYSLOUCH:
Introduce what? AUDIENCE: [INAUDIBLE]
using complex notation. BOLESLAW WYSLOUCH: Coupled? AUDIENCE: Complex. BOLESLAW WYSLOUCH:
Complex oscillation. Yes, that's right. So all right, let's do it. But hold on. But what form of oscillation? OK, all kinds of complex numbers
can write, but any particular-- AUDIENCE: [INAUDIBLE] BOLESLAW WYSLOUCH:
That's something. That's the physics
answer, all right? Complex notation is a
mathematical answer, how to solve a
mathematical equation. But the physics answer is to
find fixed frequency modes us such that the system,
the complete system, oscillates at one frequency. Everybody moves together. This is so-called normal mode. It turns out that
every of the system, depending on number
of dimensions, will have a certain number of
frequencies, normal modes, that would-- the whole system oscillates
at the same frequency, both x1 and x2, undergoing
motion of the same frequency. We don't know what
the frequency is. We don't know it's
amplitude, et cetera. But it is the same. OK? So this means that I can
write that the whole vector x, both x1 and x2, are
undergoing the same oscillatory motion. So I propose that-- so of course, we use the
usual trick that anytime we have a solution
in complex variables, we can always get back to real
things by taking a real part. So I understand you've
done this before. So let's introduce
variable z, just kind of a two-element vector, which
has a complex term, a fixed frequency, plus a
phase, a rhythm complex, multiplying vector A,
a fixed vector A. OK? And vector A is simply has two
components, A1, A2, or maybe I should write it differently. So vector A contains
information about some sort of initial conditions
for position x 1 2. Anyway, these are
two constant numbers. And also, we will, because
we have this phase here, because we keep phase
in this expression, we can assume and require
that is a real number. So A is real. It's a slightly different
way of doing things, but we can assume
this for now, right? So the solution which
is written here-- it's some two numbers,
oscillatory term, with both x1 and x2 oscillating
with the same frequency, and this is our
postulated solution. So we plug it into the
equation, and we adjust things until it fits. So let's plug this into
our matrix calculation. And what you see here is that-- so what do we have? So this is the term,
which is second time the derivative vector
X. And because vector-- or vector Z really. So I have to do-- so I plug this here. So Z double dot is simply equal
minus omega squared times Z. Right? Like this. So this is a simple thing. When I plug this in
here, my equation becomes an equation for
A. So I have minus omega squared z-hat, which maybe
I just write it immediately in terms of a complex term
by times the vector A. So I have e to i omega t plus
y times A is equal to minus M to minus 1 K times e to the i
omega t plus phi times vector A. OK? And this term is
a proportionality constant at any
given moment of time. So it goes through the
matrix multiplication. So you can just delete this. You can divide both sides. You have signs here. And then I have
an equation which is a linear matrix
equation, which is M minus 1 K times vector A. And I can rewrite it
a little bit again. So I can rewrite in this
minus 1 K minus omega squared times unity matrix times
vector A is equal to 0. So this is the
equation which we need to solve to obtain the solutions
to at least one normal mode, and we expect that there will
be two normal modes, because we have two masses. So now, this is-- so this is some matrix,
two by two matrix, which we can know very
easily how to write. Multiplying a
vector gives you 0. It turns out that for this
to work, there are two-- there is a criterion,
which has to be satisfied, namely the determinant
of the two by two matrix has to be equal to 0,
because if you take the determinant on both sides,
you have to have 0 on this side to be able to obtain
0 on the other side. So mathematically, the way
to find out the oscillating frequency is you take a
determinant of m minus 1 K minus i omega squared
must be equal to 0. So let's try to see how
to calculate things. So let's write down this matrix
explicitly using this and that. So let's write this down. So I take a big
object like this. And so in this
element here, I have to multiply this
matrix times that. If I multiply this
matrix, I simply divide all those effectively
multiplication of m minus 1 times this matrix
divides all the elements here by m. That's all there is to it. I just divide everything by m. So the first M minus 1 K
is k over m plus g over l. This is minus k over m minus k
over m k over m plus g over l. So this is multiplication. This is this term here. And then I have to do minus
unity matrix times omega squared. All this will do is it will
subtract omega squared here. I should write this. OK? So this is in this one. Maybe it would be more clear
if I move it over here. All right, so this
is the matrix that contains all the information
about our system, the mass, the gravitational
acceleration, the length, the spring strength, et cetera. And we assumed they oscillate
with a fixed frequency. So I have to find the
determinant of this matrix equal to 0. So how do I get that? And by the way,
you have a matrix, and you want to make sure
that its determinant is 0. It turns out the
only variable which we have to change
parameters of this matrix-- you know, the spring constant
and the mass this affects is given. The system has been built.
It's hanging over there. I cannot change anything. So the only parameter here,
which I can change, or adjust, or find is omega square. So I will try all possible
matrices of this type until I find one or two that
have a determinant equal to 0. But if I find them,
this would correspond to the normal frequencies. OK? So how do I calculate
the determinant of a two by two matrix? I do this by this minus
this by that, right? So that of this matrix
is equal to k over m minus g plus g over
l minus omega squared. The two identical terms
so I can put the square and then minus this minus
k squared over m squared must be equal to 0. Right? So this is the equation
which we need to solve. We need to find which
parameter omega sets this to 0. And then this is a pretty
straightforward calculation, except if I don't have-- I'll just use this one. OK, so let's rewrite
this a little bit. So this is basically equivalent
to the following equation g over l plus k over
m minus omega squared must be equal either to
plus or minus k over m. Right? I took a square
root of both sides. If you take a square
root, you have to worry about plus
and minus signs, right? So there are two solutions
which corresponds to plus here. The other one corresponds
to minus here. So solution number
1, which corresponds to plus sign right
here, it basically says that omega squared
is equal to g over l. Right? So there is one solution,
one oscillation, that does not depend
on the spring constant, because the spring
constant cancels. And there's a second
solution which corresponds to minus, where
omega squared is equal to g over l plus 2k over m. Right? Because there are two
possible solutions. And this is what we have. So we have a-- so what this says is that if I
set my frequency to g over l, if I set the system to
oscillate to this frequency, then it will be-- I will be able to set things up
such that it oscillates forever at this frequency, one
fixed frequency forever. And this is interesting. This is a frequency. It does not depend on the
strength of the spring. How is it possible? Somehow spring is
irrelevant for this motion. And it turns out that there
is a very simple oscillation, easy to see, if
basically that this is a frequency of
a single pendulum. So basically, you
got both pendula going together,
each of them happily oscillating by themselves. And the spring is completely
irrelevant for this motion. If I cut it off, the
motion will not change. It just happens that two
identical pendula are going at their own natural frequency. So the force of
spring is irrelevant. Nothing happens. This is a normal mode. And it can go forever at
this particular frequency. OK? The other option is
usually symmetrically. I move them away
from each other. And this is the motion where,
again, it's not exactly ideal small angle oscillation,
but let me try again, I guess with less. So this is the situation
where the spring really comes in at full force. It's being stretched
maximally, because they go away from each other. So very quickly, the
spring is stretched. And they go together so it's
stretch from both sides. And the whole system oscillates
at the same frequency, and because of this
additional force of spring, the frequency is actually
higher, it's larger. It oscillates faster. All right, so that's
the first step in understanding the system. We now know that there are
two oscillations and two normal frequencies. And the next step to
finish our understanding of the system in a mathematical
way, to describe it fully, I have to know what is
the shape of oscillations. I simply showed you here
so you know what to expect. But I have to be able to dig
it out from the equations. And the way to dig it out
is to find vector A. See, our real equation of
motion is up here. This is an equation of motion. This is, I have to now
find the vector A, which when you plug it in, it works-- it satisfies this equation. So I already know what are
the two possible omegas-- they can do it, but still
I have to find vector A. So I have to solve two
separate independent problems. One is finding vector
A for this situation and then find the vector
A for that situation and see if it works. So I had to plug in the whole. I had to plug it into
the whole equation. And you can show
that if you set-- if you set omega squared to g
over l, and you plug it into-- if you plug it into
this equation, what you get is a matrix
equation which looks like this-- k
over m minus k over m minus k over m k over m. And this is because-- [I try to-- so if you plug omega
squared here equal to g over l, then this cancels out,
and this cancels out. So you plug it in
here, and you get this very simple, very simple
matrix that has k over m terms. So the question is
what sort of thing can you put here to get 0. What kind of vector you can
plug into those two places such that the matrix times
vector will end up with 0? One example is that basically
amplitude is the same. Both of them move together. So you plug 1 here and 1 here. Right? So this is a good solution. And every other solution
is a linear multiplication of this one for this
frequency, right? There is k over m times 1
minus k over m gives you 0. So this is a good solution for-- so this is solution number 1. What about this thing here? If I plug this omega
squared into this matrix, it's g over l plus 2k over m. If I plug it in here, then this
matrix is way more complicated. It will actually
look very similar, but with important differences. So this one will
look minus k over m minus k over m minus k
over m minus k over m. OK? And then again, for this second
possible normal frequency, I have to find the vector
A, which corresponds to that frequency motion. And it turns out that they are
the same, but the sign changes. So one possible solution
is 1 and minus 1. If I plug in 1 minus 1, then
this matrix times the vector gives you automatically 0. So this is the second
possible normal mode. All right? So this is a systematic
way to solve equations. You plug in all
the information you know about the system
into a two by two matrix. And then you calculate
the normal mode. And then you calculate a
shape of a normal mode. Is that clear? Any questions at this time? Right? So in principle, we know,
now, at the end of the day, I still want to know how much
1 moves, how much 2 moves. So we have to put
it all together. We have identified the frequency
and the kind of, in the matrix notation, shape of the node. But of course,
the final solution is a linear superposition
of all possible normal modes with described
position of mass 1, position of mass 2, et cetera. So let's do a little bit of-- so maybe graphically I can
write down that this is the-- this is the oscillation
that corresponds to this type of mode, to those
two masses move together. And this is oscillation
that corresponds to the mode where masses move in
opposite directions. At any moment of time,
in this normal mode, at any moment of time,
wherever mass 1 is, mass 2 is minus the distance
away from its own equilibrium. So if this one is plus
1 centimeter here, the other one is
minus 1 centimeter. This one is minus 5. This one is plus 5 and so on. Whereas in this mode, both
of them move together. All right? So let's try to go back to the-- you can get rid of this one. Let's try to go back, and
now with this knowledge, let's write down x1 and x2 for
positions of the two masses. So x1-- so basically,
the x will have to be-- I used z there. So x will be real of vector z. So I take my complex numbers
and take a real of them. So from an exponent, I
will end up with a cosine appropriately and so on. And then I will use
the [INAUDIBLE].. So this is real part
of e to the i omega plus phi where omega is one
of the two possibilities. Omega t plus phi times vector
A, which we've identified here, and times some additional-- these are those vectors A
this is one possible amplitude of notation. But in general, it
can be anything. You can multiply. You can have small oscillations,
large oscillations. So there is some
overall amplitude. But the shape always
has to be simple. They either go together,
or they go opposite. So to make it more
general, I have to give some multiplicative
factor there. So if I do everything, I
end up with x, the mode 1 will in general have some
sort of overall constant C, cosine omega 1 t plus phi
1 times the vector 1, 1. This will be for x1. This will be for x2. And the mode number 2 will
be C2 cosine omega 2 times t plus phi 2 times 1 minus 1. All right? So let's see what
things are adjustable and what things are fixed. So the omega 1 and
omega 2 are fixed given by the construction of
the two coupled oscillators. This shape, 1 and
1, and 1 minus 1 is fixed, because these are the
shape of normal modes, which corresponds to
those frequencies. So we have only four
constants-- overall amplitude c1 for normal mode 1. Overall amplitude c2 for
normal mode 2 and then the relative phase of
those two normal modes. And the superposition
of x1 plus x2 gives you the most general
combination of possible motion. So if I write this
down now in terms of position of number
1 and number 2, so I have a position of
x1 as a function of time. In general, it will
look like this. It will be some sort of
constant alpha, cosine omega 1 t plus phi plus constant
beta cosine omega 2 times t plus phi 2 plus phi 1. So mass number 1, this
is position of mass 1, will in general be a
superposition of the two possible oscillations. The position of mass 2
will be very similar, but there will be a very
important difference between the alpha
cosine omega 1 t plus phi 1 minus beta
cosine omega 2t plus phi 2. This is very important
to understand exactly how this equation came about. You see, this is the influence
of the symmetric mode, where the two
things are together. So they are multiplied
by alpha, some sort of arbitrary constant, but
with exactly the same sign. And this is the part
which corresponds to a second mode, which is
with different frequencies. And there is an opposite
sign between this amplitude and that amplitude. So you have only
four coefficients-- alpha, beta, phi 1, and phi
2, which are determined, which need initial conditions. So any arbitrary mode-- this
is the most general motion of the two coupled
oscillator systems. And to describe it in
specifically-- defined for a specific
configuration, you will have to determine the
values of alphas and phis. OK? So what I want to do is I want
to write down a specific motion for the following situation. So I keep position of x1 at 0. It's not moving, so
the velocity is 0. I displaced this one by
a small positive amount. So the position of number 2 at
t equals 0 is different than 0-- some displacement
x0 or something. And its velocity is 0. And then I let it go. Again, this is not the ideal
decoupled oscillator, right? OK, and then you see
the things start moving. Let me try to show it again,
because it's not exactly here, so this one will be going on. So let's say this
one is running, and then I let this one go. And what you see here is
that this one is moving, and then that starts to move. This one stops. That starts moving. It starts being
complicated, right? It's kind of complicated motion. But whatever this
motion is, we know that it's simply those cosines
which are kind of adding up to give you this impression of
rather a complicated motion, right? So again, I let this one out. I let it go. This might be 0. So this one slows down. This starts going. And this one then slows down. The other one starts going. They kind of talk to each other. And it's this
combination of cosines. All right, so let's try
to write to simplify this for a specific case of
specific initial conditions. So I said x1 equals 0, to
equal 0 x1 velocity at 0 is equal to 0. So those ones are not moving. X2 at 0 is equal to some sort
of x0 and x2 velocity at 0 is equal to 0. So this one is displaced. They are all stationary. This one is at position 0. If I plug this in, it turns
out without lots of details that what you will get
to is that alpha will be equal to x0 divided by 2. Beta will be equal to
minus x0 divided by 2. And phi 1 will be equal
to phi 2 equal to 0. You can check. If you plug it into
those equations, if you plug t equals 0, phi
is equal to 0, et cetera, you will see that it works. So you can write down the
specific case of x1 of t to be x0 over 2 cosine omega1
t minus cosine omega2 t. It's because beta
has a negative sign. And x2 of t will be equal
to x0 over 2 cosine omega1 t plus cosine omega2 t. OK? So each of those objects
effectively feels the effects of
omega 1 and omega 2, but in a slightly different way. That's why their relative
motions are different. So what I will do now is I
will show you an animation. Hopefully, it works. And we will have time-- since on the computer, you
can make things perfect. Let's do it. So I'll have-- running
a MatLab simulation. Let's see how it goes. Large. So what is going on
here is the following. I took some initial conditions. I'm not sure if it's
exactly the same. This was for the course
that I taught some time ago. What you see here
is the following-- you have the green is the
normal mode, number 1. The magenta is
normal mode number 2. And blue and the red
are the actual pendula. All right? And the motion of
blue and red is simply a linear sum of the two. And what you see here is-- and then I plot the
position of the blue and red in color, the function of time. So you see this-- the fact that let's
say red is now stopped, and the blue is at maximum. And now, the red is picking up. And now the blue
stopped, and the red is going full swing, et cetera. And this is exactly what-- this is the computer simulation
that shows you that one of them is going up, the other
one down, et cetera. And this is for the
certain combination of initial conditions. I could go change initial
conditions in my program and have a different behavior. But whatever happens,
I would be able to-- it will always be a
combination of the two motions. Now, is there a way to disable
one of the normal modes? How would you disable
one of the normal modes? Is there a quick
way to set things up such that the second normal
mode, whichever you choose, doesn't show up in
their equations at all? AUDIENCE: You said [INAUDIBLE]. BOLESLAW WYSLOUCH: Hmm? AUDIENCE: [INAUDIBLE] BOLESLAW WYSLOUCH:
Yeah, so what you do, is you just change the
initial conditions. So you set it up
at T equal to 0. I have initial conditions
that basically favor or demand that only in this
general equation either alpha or
beta is equal to 0. So for example, one possibility
is I move both of them at the same distance, and I
just let them go like this such that the spring is
irrelevant, right? How would I do it in my program? I don't know. I can, for example-- I can, for example, set one
of the initial conditions to-- this is still running. The old one is still running. So this is the moment. So what I did is I just
changed the initial condition. And you see, this is the type
of motion where one of the modes has stopped, just
you switched it off, and the other one is going
on, and then, of course, the total motion
is equal to that. And both of them happily go
with a constant amplitude. There is no shifting of
energy from one to another. So you can have all kinds of
motions by simply adjusting initial conditions. And those motions can be
done a very different way. So do you know-- so this is how we can have
different shape of motion, depending on the
initial condition. Is there another
way for me to change the way this system behaves? Let's say I take-- I have exactly
this system, and I want to change, for example,
the frequency of oscillations. How will I do it? It could be a very
expensive proposition, yes? AUDIENCE: Drive it? BOLESLAW WYSLOUCH: Yes, but
I don't want to drive it yet. I just want to have
it free oscillation. Yes? AUDIENCE: [INAUDIBLE] BOLESLAW WYSLOUCH:
Yeah, I could come and scratch it
away a little bit. And yes, the equations
depends on the mass. But I don't want to touch. I want to just have this thing. I don't want to make any
physical modification to the system. However, I can move it
into different places, any place you can think
of where I could really modify the solution. Yeah? AUDIENCE: To the moon. BOLESLAW WYSLOUCH:
To the moon, exactly. I could put it with me some
spaceship, and go to a place where the gravity
is different, right? Why not? So what would happen? So if gravity changes, then
basically what will happen is both this term and
that term will change. The spring will
remained the same. The mass will remain the same. So the relative magnitude
of omega 1 and omega 2 will change. OK? So let's say, in fact, do
I have it in this one here? Yes. So let's say I do again. So this is what I
had before, right? So this is the one here which
is operating here on earth, and I let it go. I displaced it by
a certain distance. Let's say 1 millimeter,
and that's how it's gone. So now, let's take it
to, for example, Jupiter. So what do you think will
happen when we go to Jupiter. Jupiter, g, is much larger. OK? So what would happen to those? So the frequency
would be larger. Things will be faster, right? That's the higher frequency. But also the difference
between two frequencies will be smaller. And what happens when the
difference in frequency is smaller? You saw that there's the
fact that the energy was moving from one to the other. The thing would take-- so one of them was oscillating,
the other one is stationary, then the other one would
pick up, et cetera. Do you think this
transfer of energy will be faster or slower? Two omegas closer to each other. Any guesses? AUDIENCE: Smaller. BOLESLAW WYSLOUCH:
Take kind of longer. Let's see what happens, right? So we go on the rocket,
and nowadays, you don't have to go to the rocket. Just remove one comment. And I went from about 10
meters per square second to 25 meters per square second,
and this is what is happening. Look at this. So first of all, this
identical system-- everything at the same time. It's the same. And so you see that
oscillations are much faster. So a number of amplitude
changes per second is larger. But it takes much
longer for the energy. So the red one is now stopping. It's now slowly coming up. So because the two frequencies
are closer to each other, they stay-- it takes longer for them to
shift from one to the other. OK? So we are done at Jupiter. Let's now go to the Moon,
which has much lower gravitational acceleration. Let's see what happens. Again by logical
argument-- if something-- so the smaller
gravitation accelerations means that the
frequency is now lower. So the pendula will move slower. However, the difference
between frequency will be larger, because
the spring is still the same strength. So it turns out that even
though everything is slower, but the energy transfer
will actually be faster. So let's try to see what
happens on the Moon. It's OK. It's a little bit not completely
clear what's going on, but you see, actually the motion
is kind of a little strange. Look at the red one. The red one is stopping. Then it's going halfway out. It looks kind of
messy, doesn't it? And so it doesn't show
up here very well, because the parameters have
changed so much that I have-- I have those fixed
pictures which are-- just a second. I'll show you. So this is the picture on the-- some sort of stationary
picture on the Earth. I saw one of them
up, the other one-- you see them shift
from one to the other. And you can see kind of the
frequency of how the energy shifts from one to the other. And also you can see the
frequency going up and down for the same exact conditions. This is now, just a
moment, this is a Jupiter. So Jupiter, you see that
the frequency itself it's much higher. And the energy transfer between
the two things takes longer. And on the Moon however,
the oscillations actually look really weird. This is an example
of one of them. It's kind of, you know, the two
frequencies are so far away, and it's really not even
a nice oscillatory motion. It's some sort of-- it's much less
obvious that this is a superposition of two cosines,
because they kind of are exactly out of phase. So the motion is
kind of complete. Anyway, so this is-- actually, so the lesson
is that the exact shape, the exact motion, we
know that can always be decomposed into simple motions. If you put them
together, things may get really interesting
and complicated, depending on what sort of
frequencies we are running and what sort of-- what sort of initial
conditions we have. All right? Yes? Any questions? Yes? AUDIENCE: It's talking about
the center mass of the system or just one of the two --? BOLESLAW WYSLOUCH:
This one, I think, this one is just one of them. Actually, the one--
on the difference-- it normally doesn't matter. What matters this
is the frequency and how these move to the other. OK? Let's just forget about it. Just keep it. So let me now talk
about this thing, which is called beat phenomenon,
because when you look at the motion of one
of those objects, or the difference between
them or whatever, there's something kind of interesting
which can be extracted for those equations. Let's look at these
equations here. Let's look at mass 1. This is mass 1 and mass 2. So I can rewrite those solutions
a little bit different. And so what I want
to do is I want to-- you see, this is a
difference of two cosines. This is a sum of two cosines. There are lots of neat
trigonometrical identities which we can use. So we just-- we do
zero physics here. We just rewrite the
trigonometrical formulas. So I do exactly this,
but I rewrite it. I use, for example, some of--
you have cosine alpha plus cosine beta is equal to-- two cosine-- is equal to two
cosine alpha plus beta divided by 2 multiplied by cosine
alpha minus beta divided by 2. Right? That's the
trigonometric identity. Right? So let's just use this to write
this down and what you get is x1-- x1 of t is equal to
minus x0 sine of omega 1 plus omega 2 divided by 2 times
sine omega 1 minus omega 2 divided by 2 times t. And x2 t is equal to x0,
some amplitude cosine omega 1 plus omega 2 divided by 2
cosine omega 1 minus omega 2 divided by t. So again, we did
zero physics here. We just rewrote the simple
trigonometric equations. But what you see is
something interesting here. So there is-- we have
those two frequencies which are playing a role. And for example, at Jupiter,
those two frequencies are actually very
close to each other, because everything is
dominated by the gravity, and we have a very weak spring. So the omega 1 and
omega 2 actually are very close to each other. So this thing, this
term here, kind of goes omega 1 plus
omega 2 divided by 2 is like omega, right? 100 plus 105 divided
by 2 is about 100. Whereas this one here
carries information about the difference
of frequencies-- 100, 102, the difference
is 2, which is very small. So how would this look like? So if you make a plot
under some conditions, you can, let's say,
so the two frequencies are close to each other. So if omega 1 is
close to omega 2-- for example, omega 1 is
0.9 times omega 2, right? This is roughly what we have
on Earth in case of our system here. Then omega 1 plus
omega 2 divided 2 would be about 0.95 omega
1, omega 2, I think, which is approximately
equal to omega 2 or omega 1 and omega 1
minus omega 2 divided by 2 will be about minus
0.05 times omega 2-- much, much smaller than that. So we have-- so this term here-- it basically oscillates
at the frequency of omega, of the frequency of the
individual pendulum. And the other term is
much, much smaller. How does this look? Well, it turns out that if
you make a sketch of this, if you do signs, for
example, it looks like this. OK? So there are in fact two-- when you look at this picture,
you can see two frequencies. One which is clear the
oscillation of the-- high-frequency oscillation
of things moving up and down. But there's also this kind
of overarching frequency of much smaller
frequency, and this is what corresponds to a
difference of two things. So in a sense, if you
look at this formula here, you have oscillation,
which is happening very quickly with a typical
oscillation of the system. But this is like a
modulation of the amplitude. So the amplitude of
the signal is changing. And this is what you see here. This is exactly the
picture out there. So the system oscillates. So one of those pendula,
either of them, is moving fast. But it's going faster. It's amplitude is larger,
and after some time, it slows down to 0. It goes higher and
slows down to 0. And you've seen this. We can do it again
here that both of them oscillate at roughly
the same frequency, but their individual
amplitudes are changing. And this transmission of-- you know, one of them moving
full blast, the other one moving full blast. There's this kind of
frequency of energy moving from one to the other,
which is something called beat. This a beat system,
beat phenomenon somehow that energy is moving from
one place to another one. And we can have
some demonstration of how this happens. So we see this here. We see it on the pendula. We saw it on the
computer simulation. But now what we
are going to do is we're going to try
to hear it, right? So this is a demonstration
which maybe it works, maybe not. So let me-- it will work, OK? So let me explain what we have. So we have two speakers. And they basically go on very,
very similar frequencies, all right? So they both work at
similar frequencies. And so when I switched
on, you should hear-- hear the sound. [HUM SOUND] OK? So this is the frequency. I believe it's just one of
them is working, and you know, this is just one
pendulum that is going on that given frequency, right? Then I will switch a
second loudspeaker. [HUM SOUND] Can you hear this kind of-- wiggle? We'll change the
frequency a little. This is another frequency
of the original sound. And it's kind of the loudness of
the sound overall is changing. All right? This is faster. This is kind of extra,
extra sound which you hear is the difference of
mainly the frequency is not stable here, so I'll change it. Right? So this is, again, this is a
single one, perfectly constant frequency, no
change in amplitude, no change in loudness. Put them together, right? That's what they do. So if you have two, and I
can adjust the frequency, and the frequency is close,
then this frequency of changing is very slow. So you can actually hear it. Let me switch it off. So this is the effect of beats. I can maybe show you another
simulation of this works. Let's See. This one is oops, just a second. Let's see what it is. OK, so this is just
a single frequency. OK, again, I plot some pendulum. Then I can plot-- sorry, no this one is this. I can-- this one. OK, we'll just plot it here. Maybe we can see. So there's a red one,
and there's a blue one. And I plot two
plots independently on top of each other. So they have an amplitude of 1. And clearly, you see that they
have a different frequency. So the red one is going
with some frequency. The blue one is going
with some other frequency. Sometimes they agree. Sometimes they do
not agree, right? And the places where they meet-- they are on top of each other. This is where when you
add them up together, this is where they
will be large. In the places where
they're out of phase, they will cancel each other. So if you take two
of those together, same amplitude, just
slightly different frequency, and you simply make a linear-- superposition of the two, you
will get exactly the beating effect. So I just took two
of those pictures before I added them together
and got exactly that. You have a maximum,
minima, et cetera. And you see this overall beat
frequency, and the carrier, it's called carrier frequency. And this is something that,
again, happens very often. There's another
demonstration here. I have two tuning
forks, and they are very similar frequency. So first, I will show you
that they are coupled. They are coupled
because I gave this guy some initial condition. It's going. Then I stop it. But there's still sound,
because the second one picked up some energy, and it took off. Of course, you don't see them. So basically, what I'm saying is
that I [TONE] give this energy. This one is
completely stationary. Now energy is slowly
moving to the other one. I stop this guy, and
this guy is still going. So the energy is
being transferred by this air oscillating here. The coupling goes through the
air to the sound here, right? And they have very
similar frequency. So they are nicely coupled. But what we can also do-- we can [TONE]. Right? So they're both going. Do you hear the beats? [TONE] Not really. In fact, if they would have
exactly identical frequency, right? If they will be
perfectly the same, then the difference
would be 0, and there will be no beats at all. The period of beats
will be infinitely long, so it will take forever
for us to hear anything. So what we can do-- we
can break one of them. We can add some sort of weight. Some are here. There's some magic place
where it works best. So what I would do is
I will break this one. I will modify its frequency. That's another way to modify. I don't have to go to
Jupiter to modify it, because this one is just
a little mass here, right? [TONE] Ah, cool. AUDIENCE: Is that [INAUDIBLE]? BOLESLAW WYSLOUCH: Really,
this is actually a huge effect. [TONE] You can clearly see that they
are going up and down, up and down, because the frequency
is slightly different. So now, this thing is probably-- I know it's a period, a
fraction of a second, right? Yes? AUDIENCE: Should both of
those sine and cosines have Ts in their arguments? BOLESLAW WYSLOUCH:
Of course always. They are both time
dependent, yeah. This is the fast
thing, and this is this time-dependent
modulation, yeah. All right, so
where are my notes? So this is the-- this is how the-- so we were able to
set up the system, put in some of the
matrix equation, kind of solved it, found
two frequencies, et cetera. There is one more-- one additional
trick, which you can do to describe the motion
of a coupled pendula. And that is, in a sense,
force mathematically, force the normal modes
from sort of early on, to instead of, so far, when
we talked about pendula, we describe their motion in
terms of motion of number 1, motion of number 2. It turns out we can
rewrite the equation into some sort of
new variables, where, so-called normal coordinates,
where you'll simultaneously describe both of them
and then kind of mix them together to
have a new formula, just rewrite the equation
in terms of new variables. So you do change of variables. So instead of keeping track
of x1 and x2 independently, you define something
which I called u1, which is simply x1
plus x2, and I define u2, which is x1 minus x2. So instead of talking about
x1 and x2 independently, I have a sum of
them and difference. Why not? Right? Two variables. I can always go back and
get x1 and x2 if I want to. So if one tells me that u1
is 1 centimeter and u2 2 centimeters, I can always
go and get x1 and x2 if I want to, right? So I can do it. And it turns out that if
I plot those variables in, in other words, I take
the original equations, which I conveniently erased and
make a sum or difference, it turns out that this
coupling kind of separates. So I will end up having
two separate equations for this one. So in general, the
equation of motion would be-- would look
like, so let's say I can write down m x1 plus
x2 is equal to minus m g over l times x1 plus x2. OK, this is when I
add two equations. And the other equation
when I subtract them-- minus x2 is equal to minus mg
over l plus 2k x1 minus x2. I think that's
what is coming out. So if I add and subtract the two
original equations of motion, which I don't know if
I have them somewhere, and you can look
back, then you end up having those crossed
terms drop out. And you have one, which has only
this coefficient, the other one which has that coefficient. And this immediately--
and it looks-- if I now write it in terms
of normal coordinates, then I have that m u1 double
dot is equal to simply minus mg over l, u1, and m u2 double
dot is equal to minus mg over l plus 2k times u2. And if you look at
those two equations, it turns out that
they are not coupled. Each of them is a question
of a one-dimensional harmonic oscillator. The first part one
only depends on u1. The second one
only depends on u2. And you can see the oscillating
frequency with your own eyes. So no, the determinants needed
no matrices, no nothing. We just added and subtracted
the two equations, and things magically separated. All right? So sometimes, especially in case
of very simple and symmetric systems, if you
introduce new variables, you can simplify your
life tremendously, and these are called normal
variables, normal coordinates. And it turns out that
you can always do that. So you can always have
a linear combination of parameters for arbitrary
size coupled oscillators system where you combine
different coordinates, and you basically force
the system to behave in a way in which it induces
the single oscillation, single frequency. So this is, again, a
very powerful trick, but usually for most
cases, you can do that only after you have solved
it, after you've found out normal modes, et cetera. So after you know your normal
mode, then you can say, ha, ha, I can I can introduce
normal variables and make things simpler. But at the end of the day
for complicated systems that work is the same. But for simple systems like
this one where there is a good symmetry, you can do it. Anyway, so I think we
are done for today. And on Tuesday, we'll continue
with forced oscillators. All right? Thank you.
