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visit MIT OpenCourseWare at ocw.mit.edu. BOLESLAW WYSLOUCH:
Let's get started. So today hopefully
will be a busy day, with lots of interesting
insights into how things work. We talked about coupled
oscillators last time. We developed a
formalism in which we can find the most general
motion of oscillators. So let's remind ourselves what
are the coupled oscillators. Coupled oscillators, there
are many examples of them, and they have more or less
the following features. You have something
that oscillates-- for example, a pendulum. You have to have more than one,
because for coupled oscillators you have to have at least two. So let's say you
have two oscillators. So each of them
is an oscillator, which in, for example, in
the limit of small angles, small displacement
angles, undergoes a pure harmonic motion
with some frequencies. And then you couple them
through various means. So for example, two masses
connected by a spring is an example of a
coupled oscillator. We could have two masses on
a track and another track, also connected by
several springs. This is also an example
of a coupled oscillator. Each of those masses
undergoes harmonic motion, and they are connected together
such that the motion of one affects motion of the other. You can have slightly
more complicated pendula. For example, you can hang
one pendula from the other. Each of them-- again, in the
limit of small oscillations-- will undergo harmonic motion. And they are coupled together
because they are supported one on top of each other. And you can have-- we have another example
of two tuning forks sitting on some sort of boxes. Each of them was an oscillator,
with audible oscillating frequency, and by putting them
next to each other they coupled through the sound waves
transmitted through the air. So one of them felt the
oscillations in the other one. This was an example of
coupled oscillation. Two masses and the thing. You can build oscillators
out of electronics. Some capacitor and
inductor together, with a little bit of-- maybe without resistors. You have two of those. They constitute a
coupled oscillator if you put a wire between them. So there are many,
many examples. And of course, these
are all examples in which you have two
oscillating bodies, but it's very easy to have three
or more oscillating bodies. Then basically the
features of the system are the same, except the math
becomes more complicated, and we have more types of
oscillations you can have. And there's a couple
of characteristics which are the same for
all oscillating systems. And it's very
important to remember that we are learning
on one example, but it applies to very many. Number one, any motion-- I can maybe summarize it here. So if you look at the motion of
an oscillator, you can have-- let's say arbitrary oscillation. Arbitrary excitation. Excitation means I-- I kick it in some sort
of arbitrary mode. I just come in and set up
some initial condition such that things are moving. And motion in this
arbitrary assertion is actually-- looks
pretty chaotic. It looks pretty
variable, changing. It's difficult to
understand what's going on. So And it clearly
doesn't look harmonic. Non-harmonic. There is no obvious
single frequency that is driving the system. If you look at
amplitude of the objects here-- for example, two
pendula, pendulum one and two. At any given moment of
time they are oscillating, there's a characteristic
amplitude. But what we saw is
that motion changes, looks like things are flowing
from one to the other. One of them has
a high amplitude. After some time, it cools
down, the other one grows. So the amplitudes
are changing in time. So they are variable. Are variable. And also, we didn't
calculate things exactly, but you know from study of
a single oscillator that if the things are moving,
it has a certain amplitude, there's certain energy
involved-- with some potential, some kinetic-- and it's proportional to
the square of amplitude. So it's clear that energy
is moving from one pendulum to the other. This one was
oscillating like crazy. So all energy was sitting here. After some time,
this one stopped. So its energy is zero. And the other one was
oscillating like crazy. So the energy's flowing
from one to another. It's not sitting in one
place, but it's flowing. This one has lots of
energy right now, but now that one is picking up. So the energy-- you see
the energy flowing here. And this one will
eventually stop-- well, this is a pretty
crappy oscillator, but it will eventually
stop, and this one will have all the energy. And this is, again,
characteristic in every system. We can see energy flowing
around from one to the other, growing, stopping. So it's-- in general, in
the most general case, it's a complicated system. Energy is migrating
between different masses. However, every single one
of those coupled oscillating systems has a magic. There's a magic involved, namely
the existence of normal modes. Every single coupled oscillator
system has normal modes, and those modes are beautiful. Those modes are-- everything
is moving in sync. So this is normal
mode excitation. There's a very special
way, a special setting of initial conditions, that
leads to the-- that results in a pure harmonic motion. So this is a harmonic motion,
with a certain frequency omega, characteristic frequency
for this particular motion. The amplitudes remain fixed. Once you set initial
conditions, you get it moving, everything is moving,
simple harmonic motion means its amplitude is constant. So if I-- and remember,
for example, this system. It was something like this. Symmetric or
antisymmetric motion. And if not for the
friction, the amplitudes would remain constant
forever, if it will be a perfect oscillator. So amplitudes-- in fact, it's
not amplitudes themselves, but amplitude ratio. The ratio of amplitude
between the different elements in the system is constant. So in a sense, every
harmonic motion has a characteristic shape. And then by-- since everything
is constant, nothing changes, this energy stays
in the place it is. So energy is--
once you put energy to mass number one, mass
number two, mass number three, the energy sits there. The energies are
constant, as the system undergoes harmonic motion. Energy does not migrate. So this is a very nice-- and
there is another beautiful feature, that any arbitrary
excitation can be made out of some linear sum-- sum of normal modes. Linear sum, of
superposition of normal. Any arbitrary excitation with
all its complicated motion can be made into
some of normal modes. So since normal modes are
easy and simple and beautiful, the description of motion of
any coupled oscillator, the best way to approach it
is to decompose it, to find all possible
normal modes, and then decompose the initial
condition to correspond to this linear sum
of normal modes. Once you know the normal
modes, you add them up, and then you can predict
exactly the motion. And this is what we've done. So we have a-- we have introduced a
mathematic mechanism in which we put all the
information about forces and masses in the system in
some sort of matrix form. In our example, it was
a two by two matrix, but if we have three
masses or four masses, the dimensionality of the
matrix will have to grow. But the equation
will remain the same. So this equation of motion,
we rework it a little bit. Since we are looking
for normal modes, we know that normal modes occur
with this one single frequency. So we postulate an
oscillation with a frequency. We plug it in. We obtain a simple
algebraic equation. Doesn't have any
time dependence, doesn't have any exponents. It's a simple algebraic
equation, basically a set of linear equations, which
we can solve and find the eigenvalue, or the
characteristic frequency for normal modes. And you can show that the number
of those frequencies in general is equal to the number of
masses involved in the system. And you solve it,
and then once you know the characteristic
frequencies, then you can find shape, you
can find the eigenvectors. What is the ratio of amplitudes
which corresponds to the mode. And in case of our two pendula,
there are two of such things. One is where both
amplitudes are equal, and this corresponds
to oscillation in which two pendola are
moving parallel to each other, with a spring being-- not paying any roll. So this is one mode. And then amplitude is-- as I said, any given moment is
the same, so the ratio is 1. And then you have a motion
in which the two pendula are going against each other. So any given moment
of time, they're in their negative position,
so the ratio is minus 1. The motion of one of
them can be obtained by looking at where
the first one is and multiplying by minus 1. So these are the two modes,
and any arbitrary-- any complicated, nasty excitation
with things moving around is a linear sum of
the oscillation. So we know that. We've worked it out. We used this example. And by the way, today, we'll
be using two examples-- one which is the same thing
with two pendula and the spring, and the other one
with two masses, or maybe later three masses. And the exact values of
coefficients in matrix k are different in
two different cases. But in all types of other
motion, the shape of motion, the behavior of the
system is identical. So the solutions to the
two cases are identical. The difference is basically
numerical in how the spring constants and masses come in. So we can in fact
treat those two systems completely the same. So I'll be jumping
from one to another, but we don't have to worry. But let's now look
on the system. So what we are trying
to do today is, we are trying to
apply external force so we'll have a driven
coupled oscillator. And I assume that
you know everything about driven oscillators. So the idea was that you
come with an external . In 8.03, we assumed
that this external force is harmonic force. So there's a
characteristic frequency which is given by external-- let's say by me. It has nothing to do with normal
frequencies of the system. It's an external frequency,
omega d, which I apply. Driven frequency. And then I look at how
the system responds. And I look for steady
state oscillations-- the ones where everything
oscillates with the same driven frequency-- trying to look for solutions. And as you know from a
single oscillator, what we were calculating is what is
the the response of the system? What is the amplitude? And the certain
frequencies that-- you wiggle it and the
system doesn't do anything, but if you apply a certain
resonant frequency, then the response is very large. The system starts moving
like crazy, et cetera. And the same type of
thing will happen here, except that we have
multiple frequencies. So there will be a
possibility of a resonance for several frequencies. All right? So let me quickly set this up. Just-- yeah. Doesn't matter. So there were some-- let's just start
working on the example. So just a reminder,
this is our system. A pendula of some
length L. There are two identical
masses, M. There is a spring of constant k. They are all-- and
for simplicity, we assume that we are all in
Earth's gravitational field. So we don't have to
worry about traveling to Jupiter or the moon. And-- except that
the difference will be that we apply an external
force to one of those masses. How, it doesn't matter, but
there is an external force F-- F with subscript d,
which is equal to some-- it has some amplitude
F0 cosine omega d times t, along
the x direction. And this is applied to mass one. OK. And there is a little
bit of just a warning. We will be assuming that there
is no damping in the system. For the single oscillator,
there was always a little bit of damping. So between you and me, remember
there's always a little damping. So in case we need damping-- it will come in
and will help us, but if we try to use
damping in calculations, calculations become horrendous. So for the purpose
of calculations, we will ignore damping. It'll get some. But if things go bad with the
results, like dividing by 0, then we will bring in damping
and say no no, it's not so bad. Damping helps you. We are not dividing by 0. OK? So let's write those
equations of motions. Equations of motion. So we have-- so the forces
and accelerations on mass one is the same as before. There was a spring. There is mg over l. That's the pendulum by itself. Depending on position x1. There is the influence of
a spring, which depends on where spring number two is. And, plus, there is this new
driven term, F0 cosine omega d times t, where omega
d is fixed, arbitrary, externally given. So both F0 and omega d are
decided by somebody outside of the system. Now, the second mass
M X2 dot dot, is-- actually has feels position
of x1, through the spring. And there is this-- its own
pendulum effect plus a string, depending on position x2. Interestingly, there
is no force here, because the force is
applied to mass one. So mass two a priori doesn't
know anything about the force. But of course it will
know through the coupling. Yes? Questions? Anybody have questions so far? So it's the same as
before, with the addition of this external force. Again, this is writing all
coordinates one by one. We immediately switch
to matrix form. We write it MX double
dot, where X is the same as we defined before, minus KX. I think I will stop writing
these kind of thick lines. But for now, let me-- F cosine omega d times t. So this is now a matrix
equation for the vector XD. And let's remind ourselves
what those matrices are. Matrix M is M 0
0 M. This is just mass of the individual systems. We use M minus 1, which
is 1 over M, 1 over M, and diagonal 0 and 0. So this carries
information about masses, inertia of the system. Matrix K contains information
about all the springs in the system, and
some pendula effects. So we have a k plus
mg over l, minus k, minus k, k plus mg over l. And now there is
this new thing, which is this vector F. Vector F
is equal to F0 0 cosine omega d times t. So this is in a vector
form, this external force, which is applied only
to mass number one. OK? So these are the elements
which are plugged in. So now the question is, what
do you want to do with this? So we have the
equation of motion. And so what do we do with this? So there are two steps
that we have to do. Number one, we have
to remind ourselves what are the normal modes
of the system, in case-- because we will need-- the information about normal
modes will come in as-- into solutions for
a driven motion. So let's remind
ourselves what this was. Well, this was a solution. I'll just rewrite
it very quickly such that we have it for the record. It should fit here. Now let's try. So there were two solutions. There was omega 1 squared,
which was equal to g over l. And the corresponding normal
mode was a symmetric one. It was 1, 1. OK. So this was one
type of solution, where the two masses
were moving together. There was a second frequency
which was equal to g over l. The square of it was
equal plus 2k over m. And this was the
characteristic normal frequency for the second type
of oscillation, which you can write it 1, minus 1. And the criterion for when we
were looking for solutions, we would find them by
calculating the determinant of this two by two matrix. It was the determinant of m
minus 1 k minus omega squared times unit matrix
was equal to 0. So this was the
equation that had to be satisfied for frequencies
corresponding to normal modes with zero external force. Interestingly, if you
do the calculations, it turns out you can-- algebraically, you can write-- after you know the
solution itself, you can write it in
a very compact way. So this determinant can be
written in the following way-- omega squared minus
omega 1 squared, times omega squared
minus omega 2 squared. And this is-- the
condition was zero. And you see explicitly that this
is a fourth order in frequency equation, fourth order
frequency, which is 0 for omega 1 and for omega 2. In a very explicit way. So this is a nice,
compact form of writing this particular
eigenvalue equation. And again, as a reminder,
the motion of the system-- the most general motion of the
system with no external force was a superposition of
those two oscillations, which we can write as
some sort of amplitude-- 1, 1 cosine omega
1 t plus phi 1, plus beta 1, minus 1 cosine
omega 2 t plus phi 2. So this is oscillations of
two different frequencies. This is the shape
of oscillations, the relative amplitude
of one versus the other. And then there's the overall
amplitude alpha and beta, which has to be determined. And then there are
arbitrary phases. So there are in
fact four numbers, which can be determined from
four initial conditions. So typically two positions
for the two masses, and two initial
velocities for two masses. So everything matches. So this a so-called
homogeneous equation. Homogeneous solution. What about driven solution? Driven solution, as we remember
from a single oscillator, results in a motion in which
all the elements in the system are oscillating at
the same frequency, and that's the driven frequency. It's a fact. I come in, I apply
100 Hertz frequency, and everybody oscillates
on the 100 frequency. That's the solution for a
driven oscillating system. And we saw it for a
one-dimensional oscillator, and we will see it here as well. There's one frequency, omega d. So we will be now looking for
a solution which corresponds to the oscillation of the system
with this external frequency, which a priori is
not the same as one of the normal frequencies. So the complete motion of the
system consists of two parts. One is this homogeneous
self-oscillating motion with two characteristic
frequencies. And there will be a
second type of motion, which is a driven one. So how do we go
about solving that? So equations of motions of
course will be the same. The solution, the way that we
solve it will be very similar. So lets try-- start working. Maybe we can work on
those blackboards here. So what is going on? So we know that if we apply
external frequency omega d, everybody in the system, all
the elements will be oscillating with the same frequency. So we can then
introduce a variable Z, which will be defined
B e to the i omega d t. This will be the
oscillating term. And this will be the amplitude
of oscillation, which we'll try to make real for simplicity. And then we plug this
into the equation of motion, which is listed
up there on the screen. So the equation of motion
is Z dot dot plus M minus 1 K times Z is equal to now M
minus 1 force e to i omega d t. You see our external
force is F cosine omega d t, with a vector 1, 0. But of course, in the complex
notation, this is exponent. So this is the challenge,
what we would like to have. And we assume that all the
elements in the system-- position,
acceleration-- oscillate at the same frequency omega d. If you do that,
then the equations become somewhat simpler, because
the oscillating term drops out. So when you plug this type
of solution into here, what you get is minus
omega d squared-- that's from second
differentiation with respect to time-- plus M minus 1 K, multiplying
vector B e to i omega d t. This must be equal to M
minus one F e to i omega d t. This is vector B,
this is vector F. And there is this
oscillating term. But both sides oscillate
at the same frequency. That's what we assume. So we can simply
divide by this, and we are left with an
equation that equates what's going on
in the oscillating system with the external force. So now, let's see
here what is known and what is unknown
in this equation. M minus 1 K carries information
about the construction built of the system of accelerators. Strength of springs, masses,
gravitational field, et cetera. So this is fixed. This is given. Omega d is the external driving
frequency, and it's also given. It's a number. I said this is externally given. I just set it at some computer. Say 100 Hertz, and it's
driven at 100 Hertz. So we know that. We know exactly
what this number is. External force, we
know what it is. We defined it. It's F0. We know what its magnitude-- so everything is known except
for vector B. And vector B are the amplitudes of oscillation-- remember, everything
oscillates at omega d-- of mass one and mass two. So in general, if I
apply external force, this guy will oscillate
with some amplitude. That guy with some amplitude,
a priori different. And this will be
B1, this will be B2. And we don't know
that at this stage. So this equation will
allow us to find it. And it is possible because-- this is actually a very
straightforward equation. It contains-- actually, to
be very precise, I have to-- this is a number,
this is a matrix. So I have to put a
unit matrix right here. So it's omega d
times unit matrix plus this matrix that carries
information about the system. And so we can write
this down again in some sort of more open
way, for our specific case. So this will be k
over m plus g over l, minus omega d squared, minus
k over m, minus k over m, k over m, plus g over l,
minus omega d squared. So this is this matrix here. This matrix is applied to
vector B, which is our unknown. Let's call it B1 and B2. These are the amplitudes
of oscillations of individual elements
in our system. And this is equal to m-- the inverted mass matrix
times vector F, which-- without its
oscillating part, which is simply F0 over m and 0. All right. So this is the task
in question, and we have to find out those
two values depending on these parameters and the
strength of force, et cetera. So this is actually
not a big deal. It's a two by two equation, two
equations with two unknowns. We solve it, and we are done. However, we want to
learn a little bit about slightly more general
ways of calculating things. So let's call this one matrix
E, with some funny double vector sign. Let's call this one vector B,
and let's call this one vector D, because we will use this-- use it later. And what we are
trying to do is, we are trying to use the
so-called Cramer's rule to find those coefficients B1 and B2. And for some historical
reasons, 8.03 really likes Cramer's rule. I like MATLAB or Mathematica. I just plug things in, and it
crunches out and calculates. But it turns out
that for two by two, you can always do it quickly. Even for three by three, if
you just sit down and do it, you can actually work it out. It's not scary. By five by five-- but even four by four, I'm sure
you are mighty students who can just do it in the exam. I have never seen an 8.03
exam with four masses, unless they're
general questions. But three-- well... All right. So do we go about
finding this B1 and B2? Because, again, this is a
simple two by two question. So maybe just to again
bring it even closer to what we are used
to, let me just quickly write this down as a set
of two by two equations. So there is a coefficient here,
k over m plus g over l minus omega d squared, which
is-- this is a number, times B1 minus k over m times
B2 is equal to F0 over m minus k over m B1 plus k over m plus g
over l minus omega d squared is equal to 0-- times B2 is equal to 0. So you see two equations
with two unknowns. Couple of coefficients,
all fixed. You can eliminate variables. You can calculate B2
from here, plug it into-- you can work it
out if you want to. However, there is,
again, a better way. It's Cramer's rule or method. Should have known if
it's method or rule. Rule. Right. And so the way you do
it is the following. So you look at those questions--
you calculate all kinds of determinants, and by taking
the set of two equations and plugging into-- replacing columns in the matrix. So B1, what you do is you take
the original matrix, which is here, and you replace the
first column of the matrix with vector B. So you-- no wait, with-- sorry, with
vector D. Take this matrix, and you plug in this. So what you do is-- so it turns out-- so B1 can be
explicitly calculated, but taking the determinant
of the first column replaced, F0 over M0, and keeping the
second column, which is minus k over m. m and then k over m plus g
over l minus omega d squared. So this is-- you
calculate the determinant of this thing, where-- original
matrix with the first column replaced. And you divide it
by the determinant of the original matrix. Let's call it E. So you
calculate this determinant again for the frequency omega d. So this can be written very
nicely, in a very compact way. This determinant is easy. It's just this times that. So have 0 over m multiplying
k over n plus g over l minus on I got the squared
remember this is a given number divided by n Here
comes this nice compact form for the determinant, which is
omega d squared minus omega 1 squared, times omega d
squared minus omega 2 squared, where omega 1 and omega 2 were
the normal mode frequencies. Yes? AUDIENCE: Where are you getting
the minus k in the [INAUDIBLE]?? BOLESLAW WYSLOUCH: This one? AUDIENCE: Yeah. [INAUDIBLE] BOLESLAW WYSLOUCH: This one? This is the second column. See? I'm taking-- so this is the
first column, second column. I take the first
column, I replace it with driven equation--
with a solution. I plug it here. So I have F0 for M0. And I keep the second column. All right? That's for the
first coefficient. For the second coefficient
what you do is, you put a driving term here
and you keep the first column. All right? So this is actually an
explicit solution for B1. This is magnitude
of oscillations of the first element. And you can do the
same thing for B2. And I'm not trying
to prove anything, I'm not trying to
derive anything. I'm just using it. And I'll show you a nice
slide with this to summarize. So B2 is the determinant of-- I keep the first column. It's k over m plus g
over l, minus omega d squared, minus k over m. That's the first column. And I'm plugging in F0
over M here, and 0 here. So this is-- and divided by
omega d squared minus omega 1 squared times omega d squared
minus omega 2 squared. That's the determinant
of the original matrix. And this one is also very
simple It's this time this is 0. I have minus that. So I simply have F0 k
over m squared divided by omega d squared minus
omega 1 squared, omega d squared minus omega 2 squared. All right. So we have those
things, and also what? Do you see anything
happening here? Yeah, there are some numbers,
but what do they mean? What does it mean? Yes, we can calculate it. You can trust me. These are the-- I'm not sure that
you can trust it, but most likely these
are good results. And so we know the oscillation
of the first mass, oscillation of the second mass as they are
driven by the external force. Now, one of the
interesting things to do is to try to see
what's going on. One of the-- when we
talked about normal modes, the ratio of amplitudes
carried information. Remember, we had those
two different modes. Either amplitudes were the same,
or they were opposite sign. So let's ask ourselves, what
is the ratio of B1 and B2? So let's just divide
one by the other. So let's do B1 over B2. Let's see if we learn
anything from this. If you divide B1 over B2,
this bottom cancels out, and I have k over m plus
g over l minus omega d squared over k over m. And-- yeah. So now comes the
interesting question. This omega d can be anything. So let's say omega d is-- so we
can analyze it different ways. So for example,
when omega d is-- you can look at small, large,
and so I can compare it. But one of the
interesting places to look is, what happens when omega
is very close to one of the-- to the characteristic
frequencies? Because, remember, when we
analyzed a single driven oscillator, the
real cool stuff was happening when you are near
the resonant frequency. Things, you know, the bridges
broke down, et cetera. So let's see if we can do
something similar here. Now we have two choices. We have omega 1, omega 2. So let's see what happens
if I plug in omega 1. Omega d being very,
very close to omega 1. Let's say equal to omega 1. Omega 1 is-- omega 1
squared was g over l. So if I plug omega 1 here, I
have k over m plus g over l. So I have k over m plus
g over l, minus g over l, divide by k over m,
which is equal to what? Those two terms cancels. k over m, it's plus 1. That's interesting. So if I drive at a frequency
which corresponds to omega 1-- and omega 1 was the
oscillation where both masses were going together. So the characteristic
normal mode had the ratio of two
masses equal to one. And here I'm getting the system
to drive at this type of mode. Again, I have-- the
driven amplitudes are the ratio is equal to one. So what happens if I drive
at omega d close to omega 2? Omega 2 squared was equal
to g over l plus 2k over m. If I plug it in here, I get
that the ratio is minus 1. Again, the ratio is strikingly
similar to the ratio of the normal mode corresponding
to frequency omega 2. So it's like I'm inducing
those oscillations. So what does this all mean? There's, by the way,
a little catch here for all of your mathematicians. What happens to equations if I
set omega d equal to minus 1-- to omega 1, for example? I just plugged it here,
and nobody screamed. But there was something
fishy about what I did. Yes? AUDIENCE: --coefficient
[INAUDIBLE] BOLESLAW WYSLOUCH: If you took-- AUDIENCE: Oh, sorry. [INAUDIBLE] BOLESLAW WYSLOUCH: Exactly. So the ratio of the two was one,
but both of them were infinite. So infinite divided by
infinite equals what? I mean, this happens. So what's going on? Why can I do it? One-- we should
not really scream. Damping. Exactly. This is where the
damping comes in. So the amplitude is enormous,
but it's not infinite, because there's always
a little damping. The system will
not go to infinity. So in real life, there's
a little term here that makes sure things
don't blow up completely. There's a little damping here. Yes? AUDIENCE: Does it
at all matter-- also the fact that
those equations are inexact in the
first place, because we had made theta smaller-- BOLESLAW WYSLOUCH: No. That's not-- no. This doesn't actually matter. It's the absence of damping that
makes things look nonphysical. AUDIENCE: But as the frequency--
as the amplitude increases, when we're in resonance,
eventually those equations wouldn't hold any
longer, and perhaps-- BOLESLAW WYSLOUCH:
Yeah, that's right. But you could-- that's true. That's true. But you can come up with, for
example, an electronic system which has a huge range of-- enormous range of possibilities. And then-- or of amplitudes. Many, many-- so the damping is
much more important in that. So in reality, there is some
damping here and so forth. All right. So why don't we do,
now, the following. So let's try to see
how this all works out. First of all, such
that we can get started, I will make
a sketch for you. I'll calculate these formulas-- just a second-- and display
you as a function of frequency, such that we can
analyze what's going on. So where is it-- OK. It's still slow. All right. So this is what those-- OK, so let's say-- I don't know which is which,
but let's say B1 is the red one, B2 is the blue
one, or vice versa. It doesn't matter. These are the numbers which
I plug in for some values for some system. So we see that-- and this is as a
function of frequency. So first of all, you see
a characteristic frequency around one, characteristic
frequency around three on my plot. And in the region in the
vicinity of frequency number one, you see that
both the blue and red, the individual amplitudes
are basically close together. So the ratio is close to one. If you look at this plot,
you should believe me that it's plausible
that if you are very close to the
frequency, basically the red and blue
will move together. If you go around the
second frequency, you see that red goes up,
blue goes down, or vice versa on the other side. So the ratio is minus 1. So this plot actually
carries in formation. And in fact, what
you see also is that there is some sort
of resonant behavior. So the amplitudes
are enormous if you are close to any of those
characteristic frequencies, but they're much smaller
if you're further out. There is some motion, but
not as pronounced as when you're at the right
driving frequencies. All right. So let's try to see it. Why not? So let me go to another system-- a system which
consists of two masses, has the same type of behavior,
slightly different parameters. There is no g here, but
everything looks the same. It's just much easier to show. And I can remove
most of damping. And you'll see there
are again two modes, one which is like this-- that's number one,
that slow motion. They move together. And the other one, which
is like this, where the amplitudes are minus 1. This is the
frequency number two. So now let's try to drive it. How do I drive it? I have some sort of engine here
which is applying frequency. So let's start with some
sort of slow motion. So you see they are
moving a little bit. Very small, minimally. Just a tiny motion. But they're kind of together,
more or less, right? Slowly, but together. And this is what-- this is this area here. I don't know if you see that. This is this area. I'm driving at a
very slow frequency. I'm somewhere here. The two masses kind of go
together, but very slowly. So let me now crank
up the frequency and try to be in the
region of oscillation. So you see? All I did is I
changed frequency. The effect is enormous. I'm somewhere here now. You see? Enormous resonance. And very soon, I
will hit the limit. The system will break. OK, so we are somewhere here. I'm driving it. Interestingly, this really
looks like a harmonic motion of first type. There is no other things. OK, so now let's swing
by and get to this area. So all I'm doing is, I quickly
change frequency to- this one. So now what you
see is that there were some random
initial conditions, so we have a homogeneous
equation going, but the driven is coming in. All I did is I
changed frequency. And suddenly the system knows
that it has to go like that. Isn't that cool? So this is the region here. And all I'm doing is I'm
bringing the amplitude up, because this is close to zero. And then I'm keeping
the ratios close to the characteristic modes. So I think-- to be honest,
this is one of the coolest-- all I'm doing, just
changing frequency. And the system just
responds and starts going with a resonance
of one particular mode. So imagine a system
that has 1,000 masses, and you come in with
1,000 frequencies. You tune one frequency, and
suddenly everything starts oscillating in one go. And imagine you have
multiple buildings, each with different frequency,
and there's an earthquake. And the frequency is
of a certain type, and one building collapses, and
all the other ones are happily standing. Why? Because the earthquake just
happened to hit the frequency that corresponded to one
of the normal frequencies of that particular building. And it's an extremely
powerful trick. It fishes out normal modes
through this driving thing. And we are able to
calculate it explicitly. So now what I will do is,
I will modify the system and I will make it into
a three mass thing, which will have a somewhat
more complicated set of normal modes. And then I will show
you that I can in fact go with three
different frequencies, and pull out those
even complicated modes. So this will be it. So this is a three mass system. Now before, since we didn't
calculate it, what I will do is, I'll go to the web and I
will pull out a nice example. Let me go to my bookmarks. Normal modes. So this is a nice
applet from Colorado. And you can-- I suppose preso ENG will
send you links, et cetera. You can simulate-- you
can do everything with it. So it has two masses. It has different amplitudes,
different normal modes. And you can see nothing happens. So I have to give it
some initial condition. Sorry, I have to
change polarization. Where is polarization? Here. I give it some
initial condition. So this is basically
what you just saw. I'm just demonstrating to you
that this applet looks the same as our track. So this is you can
see normal modes. It's a combination
of normal modes. There's one which is first
frequency, second frequency. This is first normal mode. This is second normal mode. You can very quickly
see what happens. So this is what
we just looked at. This is what we calculated,
more or less, and so on. Now I want to show you
three masses where things are somewhat more complicated. In general, three normal modes. For the three mass elements, the
first normal mode is like that. All the three masses
move together. And slightly different--
the ratio of amplitudes is slightly different. The second mode of operation
is actually quite interesting. The central mass is
stationary, and those two are going forth and
back, like this. And then I have
a third frequency where the middle one is
going double the distance, and the two other
ones are going up. So this is the
third normal mode. All right. So this is the system which
we now have standing here. Let's quickly see if
it works in reality. So this is the first-- so this is the first mode. This is the second one. All right. And the third one will be-- Sometimes I do five of them,
and then it's really difficult. OK. But-- so we have a computer
model, we have a real model. Let's now do the calculation
of the frequencies, the ratios, such that we can
see what happens. So I'm coming here, I'm
changing mass to three. I'm running my-- the
terminal calculating thingy. OK. It's very slow. It's busy, busy, busy. Imagine-- OK. Spectacularly slow. Where is it? I hope it's not--
oh, here it is. OK, so this is
what's coming out. So this is the same
calculation as we did, except for three masses. So what do we have here? Where's my pointer? So we have, again, three
characteristic frequencies, we have three masses, and
the same type of behavior. See, if you are far
away from resonance, if you have very low frequency,
everybody goes together. I haven't shown
you this one here, which is also interesting. I'll show you in a second. And then-- so
presumably if you are close to the first frequency,
you see all three of them go together. And this is the first mode. So I should see, if I
set the proper frequency, the thing should respond
in mode number one. This is the one where two of
them go opposite to each other, and the red one is stationary. It doesn't move. And then you have those
things where they're kind of more complicated. It's difficult to
read them from here. And I can do it for
more masses, et cetera. So generally it's calculable. It can be calculated and can
be actually demonstrated. So let's try it. So-- 32. So there's this magic
frequency number one. I'm setting frequency
by turning a knob. That's omega d. I'm a supervisor
of this operation. It stops because of other
reasons, but it will continue. Then I go to 56. By the way, remember
that every-- this is the particular solution. This is a steady state
distillation with omega d. But we also have all those
homogeneous solutions, which have to die down with damping. Remember, it's a combination
of homogeneous plus particular. So the motion is actually
a little bit distorted because we have this homogeneous
stuff hanging around. But hopefully, if I can start it
with little homogeneous stuff, it will be better. So you see? Pretty cool. Almost there. It's almost in assembly. Then it kind of stops. You see? I get two of those
going forth and back, more or less, and
this one going. I could probably
tune the frequency a little bit higher or lower. I'm not exactly at the
right place, but I'm close. And now let's go to the
last one, which is 68 according to my helpers here. You see? This one goes opposite
phase, and those two more or less together. Then they keep going. See now, those two move a
little bit forth and back, but they are in phase. They move together. The ratio is 1. And this one-- the
ratio is minus 2. Right? Make sense? That's the beauty. You drive it at some frequency,
and those normal modes pop out. It's actually very, very cool. And as I said, you encounter
those type of behaviors very often. Sometimes you drive a car and
something starts vibrating, it's just because the
car driving on the road creates a frequency, provides
a driving frequency which corresponds to oscillation
frequency or some piece of-- old car. Usually it happens in old cars. So I think that's
the message we can-- and we have all the machinery
to be able to do it. We can set up any
matrix at K, which has information about all the
forces acting on anything, and we can set matrix
M with the masses. We can put it all together,
we can find normal modes, and then we can use
Cramer's equation to take care of the
arbitrary external forces. And what comes out, just as a-- for summary, for
future reference, the oscillation
of the system is-- this is conveniently written. This is vector X.
In general, this homogeneous solution this plus
the particular solution, which is plus vector B, which
is very important. Vector B depends on
the driving frequency. Those amplitudes of a particular
solution during emotions are dependent on
driving frequency. Cosine omega d times t. So in the most
general situation, we have some homogeneous
solution here, and there is this
driven solution which we observed in action,
with proper amplitudes. So in fact, what
you've seen is the sum of both, because this depends
on the initial conditions. Now, in reality, as with
a single oscillator, this homogeneous equation,
there's always a little damping, which we ignore it. And the damping
comes in, and it only affects the
homogeneous solution. So this part will
eventually die down, whereas a driven
solution is always there. There's external force that
is driving the system forever and ever. So this part, this steady
state or particular solution will remain forever,
because there's an external source of energy
which will always provide it. So these guys will die down. And of course, because of
damping the exact value of coefficients B will
be slightly modified, because as you know from the
from a one oscillator example, the presence of damping
actually slightly modifies the frequency. Whereas here, we--
for simplicity-- if we introduce damping
here, those calculations are really amazing. So we don't want to do it. All right. Any questions about it? Yes. AUDIENCE: If we were doing
Cramer's rule with a three by three matrix, would we
only replace the column that corresponds to the B that
we're trying to find, and then keep the other two? BOLESLAW WYSLOUCH: Yes. So it's always-- you'll
be doing always that. In fact, I should have
some slides from Yen-Jie on Cramer's rule. Let's see. OK. So this is some
reminder of last time. So this is Cramer's--
there's Mr. Cramer. So this is an example of what-- this is the two by
two, three by three. OK? That's what you do. Question? AUDIENCE: So it makes sense that
the Cramer's rule [INAUDIBLE],, but what does that mean
for physical system? BOLESLAW WYSLOUCH:
Well, basically-- so the Cramer's rule
is Cramer's rule. The question is
what do you plug in? And what you plug in
depends on the omega d. So it is true that if you
insist on plugging in omega d exactly equal to one of
the normal frequencies, then things blow
up mathematically. In reality, there is-- this is the situation
of resonance. So as I discussed this
before, in reality there is a little
bit of damping. So those equations
have to be modified. There will be some small
additional terms here that will prevent this from
being exactly equal to 0. So this will be a
very large number. The amplitude will be enormous. If I would have a
little bit more time, I'll fiddle with
frequency, I could actually break the system, because those
masses would be just swinging forth and back like crazy. So you basically go out
of limit of the system. So physically, there's always
a little bit of damping. You do not divide by zero. On the other hand, it's so
close that, for simplicity and for most of the-- to get
a feeling of what's going on, it's OK to ignore it. Just have to make sure
you don't divide by 0. So you can do this Cramer's
rule with arbitrary omega d. Make sure you don't
divide by 0, you solve it, and then you can
interpret what's going on. Again, Cramer's rule has
nothing to do with physics. It's just a way to solve
those matrix equations. As I say, you can do
it anyway you want. Two by two, you can do it
by elimination of variables. Five by five I do by
running a MATLAB program. Anything you want. But for some historical reasons,
8.03 always does Cramer's rule. All right? And, yeah, it's useful,
especially for three by three. All right? OK. So I have to start
a new chapter. I'm much slower than
the engine, by the way. I don't know if you noticed. And that is the-- there's a very
interesting trick that you can do which is of an absolutely
fundamental nature in physics, which has to do with symmetry. You see, many things
are symmetric. There's a circular symmetry. There's a left and
right symmetry. Example, two little
smiley faces are mirror images of each other. There is some-- this
thing is symmetric along this vertical axis. This one is symmetric around
rotations by 30 degrees. That house seems to be
symmetric along this way. This is part of our
experiment in Switzerland, also kind of symmetric
in the picture. The rotational symmetry, there's
reflection symmetry, et cetera. It turns out, if you have
a system that is symmetric, then the normal modes
are also symmetric. And there's a way to dig out
normal modes just by looking at symmetry of the system. So let me explain
exactly what this means. So let's take our system here-- OK, so we have one
mass, the other mass. There is a spring here. This one is x1, this one is x2. If I take a reflection
of a system-- let's say this mass is
displaced by some distance. Some x2. This one's some x1. If I do the
following transform-- I replace x1 with minus
x2, and x2 with minus x1, this is mirror symmetry. I basically flip
this thing around. In other words, what I do here
is I look at the system here-- hello-- and I go to
the other system. Hello. Right? I did a mirror transform. I looked at it from
this side, that side. Now, when I look
at it I see the one on the left, one on the right. I call this one x1, this one x2. It's oscillating. You are looking at it, this
is your x1, this is your x2. When I move this one, is it-- it's your negative x1. For me this is positive x2. This one is positive x2 for you. It's negative x1 for me. Do we see a different system? Does it have different
oscillations? Does it have a
different frequency? No. It's identical. They're completely identical. So the physics of
those two pendula doesn't depend on if
he's working on it or if I'm working on. That's the whole thing. And this is how you
write it mathematically. And if you have a
solution which-- x1 of t, which consists of
some sort of x1 of t, x2 of t. Let's say we find it. Now it's over there. We know alphas,
betas and everything. Because of the symmetry, I
know that for sure the equation which looks like this-- x1-- no, it's not x1. It's x of t. That's the vector x of t. I have another one with a tilde,
which is identical functions, everything is dependent, except
that this one is minus x2 of t minus x1 of t. And I know for sure that if
this is the correct solution, this is also a correct solution. Why? Because he did x,
and I did x tilde. But the system is the same. Completely identical. And you don't have to know
anything about masses, lengths, springs, anything like that. Just the symmetry. All right. How do you write
it in matrix form? You introduce a symmetry
matrix, S, which is 0, minus 1, minus 1, 0. And then x tilde of t is
simply equal S, x of t. And we can check that. That's simple you just
multiply the vector by 0, minus 1, minus 1, 0, and
you get the same thing. Turns out-- and if
this is symmetry, if this is a solution,
this is also a solution. So we can make solutions
by multiplying by matrix S. So what does it mean? So let's look at
our motion equation. The original motion
equation was-- equation of motion was minus
1 k matrix times x of t. This is what we use
to find solutions. Usual thing, normal
modes, et cetera. Let's multiply both sides by
matrix S. I can take any matrix and multiply by both sides. So I get here S X
double dot of t. And of course, S
is a fixed matrix, so it survives differentiation. And this is equal to minus
S M minus 1 k x of t. Just multiply both sides by S. However, if MS is equal to
SM, and KS is equal to SK-- in general matrices, the
multiplication of matrices matters. But it turns out that if
the system is symmetric, if you multiply mass M
by S, you just replace-- it will just change
position of two masses. So nothing changes. Also, if the forces are the
same, then multiplying mass S, you flip things. Nothing changes. And mathematically, it
means that the order of multiplication
does not matter. It means that they
are commuting. And of course, M minus 1
S is equal to S M minus 1. If this is the case, then I
can plug it into equations and see what happens. So I can take this equation,
and I can take this S here and I can just move it around. I can flip it with M1 position,
because the order doesn't matter. So I can bring it here. And I can flip it with K,
because the order doesn't matter. I can bring it here. So after using those
features, I get that S X dot dot is equal
to minus M minus 1 K S X, which means that X dot dot-- remember, this was-- I'm using this expression. I'm just-- S times a
variable x gives me X tilde. X tilde dot dot is equal to
minus M minus 1 k X tilde. X tilde. Which basically proves--
this is a proof-- that x tilde is a solution. So if a system is symmetric,
it means that it commutes-- that mass and K matrices
commute, and you can-- and this means that
this holds true. If I have one solution,
the symmetric solution is also there. All right? Let's say x-- yes? AUDIENCE: So in the
center equation, you introduced negative S.
I didn't really get that. BOLESLAW WYSLOUCH: So this
negative is simply the-- Hooke's law. This is this minus sign here. AUDIENCE: Yeah, but where
did the S come from in the-- BOLESLAW WYSLOUCH: Oh, I
multiplied both sides by S. AUDIENCE: Oh, OK. BOLESLAW WYSLOUCH:
I just brought the S and I put it here. S, X dot dot, and S after-- minus commutes with S, so
I kind of shifted my minus. But then I waited before
I hit the matrices, because I wanted to discuss. OK? So now comes the
interesting question. Let's say X is a normal mode. Right? We have normal modes. Let's say X is a normal mode. It oscillates with
a certain frequency. So I have X of t. Let's say it's equal to-- let's say it's a
normal mode number one. Cosine omega 1 t. And we know that X tilde
is also a solution. So what happens to mode number
one when I apply matrix S? So X tilde-- so matrix
is a constant number. It's just a couple
of numbers I just reshuffle things, et cetera. Try So if I have X, which is
oscillating with frequency omega 1, if I multiply
by some numbers and reshuffle things
around, it will also be oscillating in number one. So it will be also
the same normal mode. So if I take matrix S, I
apply it to the normal mode, I will get the same
normal mode, with maybe a different coefficient. Linear coefficient. Plus, minus, maybe some
factor, something like that. So if this is the solution,
it means automatically that X tilde is proportional
to A1 cosine omega 1 t. And the same is
true for omega 2. So the only way
this is possible, since cosine is the
same in both cases-- matrix S to normal
solution gives me normal solution with some sign. So the only way this can
work, matrix S actually works on vectors, on A1. This is just an
oscillating factor. So we know for sure that
S A1 must be proportional. to A1. Similarly, S times A2
is proportional to A2. So let's try to see with
our own eyes if this works. So let's say S is
0, minus 1, minus 1, 0, times 1, 1 is equal to what? 0 minus 1, I get minus 1 here. This one, I get
minus 1 here, which is equal to minus 1 times
1, 1, which is vector A. So vector 1, 1, which is our
first mode of oscillation, is when you apply the
matrix S, you get a minus 1 the same thing. And similarly, if you do the
same thing with matrix S-- so you see, the simple
symmetric matrix consisting of 0s
and minus 1s has something to do with
our solutions, which is kind of amazing. So if I have 0, minus 1, minus
1, 0, I multiply by 1, minus 1, I get 1 here, I
get minus 1 here. Just a moment. Something is not right. Something's not right. No, it should be-- AUDIENCE: [INAUDIBLE] BOLESLAW WYSLOUCH: Hmm? AUDIENCE: [INAUDIBLE] AUDIENCE: It's 1, minus 1. BOLESLAW WYSLOUCH: 1, minus 1. Yes. I don't know how
to multiply here. I should be fine. OK. That's right. This is-- sorry, this
is 1, because it's minus 1 times minus 1, and
this is-- yeah, that's right. Which is 1 times 1, minus 1. So this is something that-- I get the same vector
multiplied by plus 1. So this is, of course-- these are eigenvectors
and eigenvalues. So the matrix S has
two eigenvectors, one with eigenvalue of plus
one, the other one plus 2. So we have an equation SA
is equal to beta times A, and beta is-- OK. So this is something-- so it turns out-- and I don't
think I have time to prove it, but it turns out
you can prove it-- if I would have
another three minutes-- you can prove it that the
eigenvalues of matrix S-- eigenvectors, sorry,
eigenvectors of matrix S are the same as eigenvectors
of the full motion matrix. So in other words, our
motion matrix M minus 1 K-- this is the matrix. Then we have a matrix
S. And normal modes are, you have a normal frequency
and they have a shape. You have a normal vector,
the ratio of amplitudes. And turns out that eigenvectors
here, so the A's are the same. And again, I don't
have time to show it, but you can show that
this is the case. So if you have a
symmetry in the system, then you can simply find
eigenvectors of the thing to obtain the normal modes. So if I look at my two pendula
here, the symmetry is this way, so I have to have one which
is fully symmetric, like this, and I have another one
which is antisymmetric. Plus 1, minus 1,
plus 1, minus 1. Similarly, here I have-- let's say if I have
two masses, there is one motion
which is like this, and one motion which
is like that, because of the mirror symmetry. And you can show that if you
have some other symmetries, like on a circle
et cetera, that you have a similar type of fact. So you can build up on
this symmetry argument. And finding eigenvectors of a
matrix 0, minus 1, minus 1, 0 is infinitely simpler
than finding matrix with G's and K's and
everything, right? All right. So thank you very much, and
I hope this was educational.
