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GILBERT STRANG: OK.

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Now, I'm going to have
differential equations, systems

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of equations, so there'll
be matrices and vectors,

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using symmetric matrix.

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They'll be second order.

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So second order, second
derivative, that y

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is the vector.

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And S is the symmetric matrix.

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And that's the first
time we've been

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prepared for the most
fundamental equation

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of physics, of mechanics,
oscillating springs--

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so many applications--
rotating torques.

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It's very important
in applications.

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The finite element, giant
finite element codes,

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are solving equations
like that all the time.

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And we don't have a
damping term here, so

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it-- or a forcing term,
so it's the null solutions

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that I'm going to look for
to match initial conditions.

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I don't have a forcing term.

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OK.

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So the real central equation
always looks like that.

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This is-- Newton's law, is
what this is-- mass times

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acceleration.

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So M will be a matrix,
often a diagonal matrix,

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telling me the masses.

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Remember, I have
n equations here,

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so I have n masses,
as you'll see.

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And I have, let's say,
a bunch of springs

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connecting those masses.

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And then there's a matrix
K in multiplying y itself,

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and that's always called
the stiffness matrix.

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So, actually, in
applications, the first job

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is to take the problem
and create these matrices.

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I'll give you an example, but
let's suppose we've got them,

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and how do we solve them?

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We look for, as we always
do, solutions where time

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is separate from the vector x.

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I substitute that
into the equation,

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So I get M, second derivative
will bring down the i omega,

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twice.

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E to the i omega t x, right?

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Plus, this term, K times e to
the i omega t x, should be 0.

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So I'm just substituting the
expected form for the solution.

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In that form, that
exponential factor can cancel.

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And, I see, I have an
eigenvalue problem.

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Let me just look at
that eigenvalue problem.

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I'm going to put that
on the opposite side.

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But what i squared is
giving me minus one.

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I'm just going to
be left with Kx.

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Well, let me put this
on the other side,

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because it's got a minus.

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And then when I put it
over there, will be a plus.

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M omega squared x.

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That's an eigenvalue problem.

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Here is the eigenvector.

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There is the eigenvalue.

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Oh, but we have two matrices.

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That's something a little new.

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Not new to MATLAB however.

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The MATLAB command to
find these eigenvalues,

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let me call those
eigenvalues lambda,

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so lambda will now
be omega squared,

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because two derivatives
brought down omega twice.

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But we have our two matrices,
so the MATLAB command

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would be i of K and M. If you
define the matrices, K and M,

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and you call that command, it
will produce the eigenvalues

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and the eigenvectors x
for this, you could say,

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generalized eigenvalue problem,
two matrix eigenvalue problem.

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It's got a K, as usual,
and then it has an M.

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But many, many times, M will
be a multiple of the identity

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and present no problem.

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OK, so that-- this is
the eigenvalue problem

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that we reached.

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And that's the command
that would solve it.

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OK.

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That's the first step,
is to look for solutions

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of that special form.

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Now let's do a little count.

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How many solutions
are we expecting?

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How many initial
conditions do we have?

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So we initially, we
give y at 0 of course,

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the initial condition,
the position.

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But we also give, in a
second order equation,

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we also give the initial
velocity, y prime of 0.

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And those are vectors,
because those tell us

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the initial condition
of n masses.

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And so I have n numbers from
y of 0 and n more numbers.

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Two n, all together,
initial conditions.

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I'm going to need
two n solutions.

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I'm going to need
two n solutions

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to match two n
initial conditions

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and solve the--
solve the equation.

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OK, so what do they look like?

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All right.

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Here, I've gone ahead to
put an application up.

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So, again, I'm taking
the masses to be equal.

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Here are the masses, M, M, and
M. And so I end up with a 3

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by 3 matrix.

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I have three unknowns,
n is 3 for this problem.

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And I've got 4 springs.

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These are springs-- maybe I make
them look a little springier.

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OK.

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And they're connected at
the top to a fixed support

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and at the bottom
to a fixed support

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and they're connected
to each other.

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Do you see what will happen?

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As I, maybe I, the
initial condition,

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I drag all those masses down?

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As my initial
condition, I let go?

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Then they will go up
and down, up and down,

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just the way springs always do.

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And they will solve,
their position

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will be the solution to
my differential equation.

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M y double prime
plus Ky equals 0.

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So I'm not, I'm just
starting the motion,

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and then backing off.

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So I'm not forcing--
it's not forced motion.

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It's pure oscillation,
pure oscillation.

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Right.

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But coupled, several
oscillators are

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a couple, that's what's new.

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We know all about this equation
when y is just a scalar, just

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one equation.

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We know that, and that led us
to the square root of K over M.

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That was just a 1 by
1 eigenvalue problem,

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and now we will have a 3
by 3 eigenvalue problem.

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The mass matrix is simple.

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Here's what the stiffness
matrix would look like,

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if all those springs
were the same.

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Just, I wanted to see
what kind of a matrix

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shows up in the problem.

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And time to write
down solutions.

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OK, so what are we
remembering from solutions?

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We're remembering that
solutions look like this.

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I have, but I'll have three
possible eigenvectors,

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because I have 3 by 3 matrices.

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So that will give
me three solutions.

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But I want six because I have
six initial conditions all

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together.

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Let me write those
six solutions down.

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Y, the solution is, sub constant
times the cosine of omega t,

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times the first eigenvector.

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And another constant
times the sine of omega

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t, times the second
eigenvector--

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times the first
eigenvector, sorry.

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The first eigenvector, I
have three eigenvectors,

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and for each of them, I get
two solutions, one a cosine

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and one a sine.

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And the frequencies,
there would be omega 1,

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and you are remembering
that lambda,

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lambda is omega squared.

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So if I write
omega 1 there, it's

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the square root of lambda 1.

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That's the square
root of lambda 1.

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So I've got two
solutions so far,

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coming from the first
eigenvector at its eigenvalue,

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and with the cosine and a sine.

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And then, I'll also have
for the A2 and a B2,

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and an A3 and a B3.

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So I just, briefly, A2, B2 with
omega, using omega 2 and x2.

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And then they'll be
an A3 and a B3 using

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omega 3 and the eigenvector x3.

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That's pretty stupid looking.

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This is what I
meant to represent.

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I don't want to rewrite all that
with 2's, and then rewrite it

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again with 3's.

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That's what the
solution looks like.

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OK.

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How do we-- when we match
the initial condition,

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what happens?

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I said t equals 0, right?

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When I said t equals
0, the sines disappear.

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So it's the A's
that match y of 0.

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So A's match y of 0.

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And when I look at the initial
velocity, the derivative,

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the derivative of the cosine
is the sine, and it's 0.

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But the derivative of the
sine is the cosine and it's 1,

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so I'll see that the
B's match y prime of 0.

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I'm trying to get the
total count to be six.

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So there are three,
three initial positions,

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and three A's.

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There are three
initial velocities,

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cause there are three masses,
and there are three B's.

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I get a perfect
match, a total of six.

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Six constants, six
numbers to match.

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It all works.

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OK.

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Now, do I want to
try an example?

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Sure.

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Let me end with a
particular example.

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I better go to 2 by
2 for an example.

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So can I do the
same problem 2 by 2?

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So my problem is going to
be y double prime plus S--

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there's a K, there's
a division by m,

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and there's a 2, 2 minus
1, minus 1 y equals 0.

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That's my equation.

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Now, I'm speaking
now, about the problem

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with a spring, a first mass
m, a spring, the second mass

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m, and a spring.

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So two masses, two equal
masses, three equal springs.

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That's my equation.

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So what's a solution?

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How do I solve it?

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I need the eigenvalues and
the eigenvectors of my matrix.

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So here's my matrix S. That's
my matrix S. It's symmetric,

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it's got physical
constants there.

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The stiffness of
the springs divided

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by the masses, k over m, we're
expecting that same k over m

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that always shows up.

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And we need the
eigenvectors there.

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And what are they?

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The eigenvalues of that
are 1 times k over m,

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And 3 times k over m.

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Cause the trace is 4.

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2 plus 2 is 4.

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The determinant is 3.

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4 minus 1 is 3.

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Those are the two eigenvalues.

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And these are the omega
squareds, remember.

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OK.

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That's how I start
the system from rest.

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So physically I pull
down these masses,

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or maybe I push that one
up and pull that one down,

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whatever I like.

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I hold them for a
moment and I let go.

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I don't give them
an initial velocity.

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They start from rest,
so the B's will be 0.

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So y prime of 0, y prime of
0 is going to be 0 from rest.

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And that will give me B's are 0.

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So my solution will
be, my A cosine of,

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so the eigenvalues cosine,
I take the square root of k

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over m t times the
first eigenvector--

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one of the eigenvectors.

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The eigenvectors of
this, probably are 1, 1.

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And then the other
eigenvector, and that

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00:14:44,320 --> 00:14:48,840
will have a cosine of
the square root of omega,

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I have to take the
square root of that,

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so that's the square
root of 3, k over m t,

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times its eigenvector,
which I think is 1 minus 1.

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It's going to be
perpendicular to that one.

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I've solved the problem.

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The A1 and the A2 are determined
by the initial condition.

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Now do you see what's
happening in the motion?

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That's the last thing,
last point for this video.

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This motion with a 1, 1
eigenvector, the two masses

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are in sync.

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They're growing
together, up and down.

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That's one eigenvector
of the problem.

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And it has a certain
frequency that they

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go up and down-- a square root
of k over m, our old friend.

251
00:15:36,710 --> 00:15:41,080
But also, with two masses,
they can go against each other,

252
00:15:41,080 --> 00:15:43,350
like this motion.

253
00:15:43,350 --> 00:15:46,210
That's coming from
this eigenvector,

254
00:15:46,210 --> 00:15:49,220
and it happens at
a higher frequency.

255
00:15:49,220 --> 00:15:52,840
So those are going--
the final solution

256
00:15:52,840 --> 00:15:56,570
is a combination of the
masses moving together

257
00:15:56,570 --> 00:16:00,980
at a little slower oscillation,
and the masses moving

258
00:16:00,980 --> 00:16:04,200
opposite each other at
a faster oscillation.

259
00:16:04,200 --> 00:16:07,990
Some combination of those
two is the solution.

260
00:16:07,990 --> 00:16:11,280
And then if we had
three masses, there

261
00:16:11,280 --> 00:16:13,860
would be three oscillations.

262
00:16:13,860 --> 00:16:16,840
One where all three
are going together,

263
00:16:16,840 --> 00:16:22,870
one where the outside
ones are opposite,

264
00:16:22,870 --> 00:16:26,930
and one where all three
are, I see opposite signs.

265
00:16:29,830 --> 00:16:32,230
It's a beautiful subject.

266
00:16:32,230 --> 00:16:37,280
Highly developed, and highly
important applications.

267
00:16:37,280 --> 00:16:41,130
But that's the nicest
solution you could hope for.

268
00:16:41,130 --> 00:16:44,290
OK that's a second
order system, solved.

269
00:16:44,290 --> 00:16:45,840
Good.