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PROFESSOR: Welcome to this
recitation on pure resonance.

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So here we're given an operator
p(D) equals D square plus 4I,

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where D is the
differential operator

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and I is the identity operator.

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And you're asked to consider
the equation p(D) applied to x

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equals F_0*cos(omega*t),
where F_0 is a constant.

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So the first question is
what is the natural frequency

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of the system.

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The second one is to use the
exponential response formula

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to solve for p(D)*x
equals F_0*cos(omega*t).

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And here you need to
be careful and do it

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for both cases omega equal
to 2 and omega equals not

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equal to 2.

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And the last question is
just to sketch the graph

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for the response of this
system: p(D)*x equals cos 2t,

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with the initial conditions x
of 0 and x dot of 0 equals to 0,

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basically, rest
initial conditions.

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So why don't you pause the
video, take a few minutes,

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and work through this problem.

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Welcome back.

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So first what is the natural
frequency of this system?

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So let's just rewrite
our system here.

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This is the left-hand side.

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So basically, this just
gives us an x dot dot

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plus 4x on the left-hand side.

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So the system that we're solving
is simply x dot dot plus 4x

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equals F_0*cos(omega*t).

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So the first question asks
us for the natural frequency

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of this system.

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The natural frequency
of this system

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can be found regardless of what
you have the right-hand side,

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just by looking at the
characteristic polynomial

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of your equation.

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The characteristic polynomial
here would be s squared plus 4.

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When this characteristic
polynomial is equal to 0,

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we can solve for s and find
what are the natural frequencies

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of the system, if basically we
get complex solutions, which

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is the case here.

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Gives us s square equal minus 4.

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So s equals plus or
minus i*2, or 2i.

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So the natural frequency of our
system would be omega equals 2,

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because we only
consider frequencies

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that are positive here.

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Second part.

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Now, we're asked to
look at the full system

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with the forcing on
the right-hand side.

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And using the exponential
response formula,

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find one solution
to this system.

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So here we're talking
about a particular solution

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with the exponential
response formula.

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So what does the ERF tell us?

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The ERF, if you recall here,
the base of it for this system

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for example, is the fact
that cosine is the real part

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of the exponential i*omega*t.

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So we can rewrite this whole
equation as x dot dot plus 4x

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equals F_0
exponential i*omega*t.

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And we would get then
a particular solution,

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if I ignore any particular
value of omega at this point,

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which would have the form of
the amplitude that we have

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on the right-hand side at
0, exponential i*omega*t,

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which is basically our forcing.

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Over the characteristic
polynomial of the equation,

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so s squared plus 4, evaluated
at the frequency here that

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would appear at the forcing
in the exponential form,

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so with the i*omega*t.

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So here you can see right away
that we would have a problem,

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using this formula, if i*omega*t
was a pole or basically a zero,

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to this characteristic
polynomial.

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And so that's why
you were asked to be

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careful with the value of omega
equal to 2 or not equal to 2.

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So here let's consider
omega not equal to 2,

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so that I can actually write
down 1 over p(i*omega),

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because we know that
p(i*2) is equal to 0.

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So if omega is not equal to 2,
were out of the danger zone.

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And from this point, we can just
basically plug in our values,

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i*omega*t, and p(i*omega) would
just give us 4 minus omega

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squared.

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So here again, the omega equals
2 danger zone approaches,

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where we would be
dividing by 0 if we

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didn't take the constraint
omega not equal to 0.

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So this is the complex form
of this particular solution.

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But we're dealing with
a real-valued problem,

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so we want to take
the real part of this

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to have the solution to
the problem we were given.

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And so that would just give
us F_0, 4 minus omega squared,

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cosine omega*t.

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So now let's take the
case omega equals to 2.

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OK So what happens?

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If omega equals to 2, this
formula that you're given

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fails, and you need to
seek for the derivative

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of the characteristic
polynomial.

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And we basically have
to-- 2i equals to 2.

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So what about p prime of 2i?

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So p prime of s is simply 2s.

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So if we evaluate p
prime at 2i, we simply

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have 4i, which is
not equal to 0.

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So at this point we can use the
resonant exponential response

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formula that you saw.

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Just change my chalk.

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We're here.

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We would again, same
trick, the cosine

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is just the real part
of the exponential.

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So we can use this formula.

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And we have now to introduce a
t, F_0 exponential i*omega*t,

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because we're solving here for
the complex value equation.

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And now we can
divide by the p prime

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evaluated at 2i, which is 4i.

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And so basically, I can
end up with a minus i

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at the numerator.

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So to take now the
real value solution,

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we need again to take
the real part of z_p.

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So here now we have an i,
so we need to be careful.

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We're going to have
solution in sine.

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So let me just write
down what know.

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t*F_0 over 4.

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This with the Euler formula
would be cosine plus i sine.

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The i sine would be
multiplying this i,

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the two minus would cancel out.

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And so we would end up with
sine omega*t, t*F_0 over 4.

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And this would then
give us the solution.

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And here, note that
I actually chose

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the value omega equals to 2, so
we can even be more explicit.

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For this case, we actually
have omega equals to 2t.

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So the last part
of the problem was

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to sketch the solution for the
initial conditions x of 0 or x

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dot of 0 equal to 0, so the
rest initial conditions.

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So here are two ways to proceed.

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The long way would be
to seek the solution

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to the homogeneous equation
without the right-hand side,

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the forcing cosine, introduced
two constants of integration,

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and then seek these
constants of integration

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on the general solution.

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And you would find that these
two constants of integration

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would be 0 with these
initial conditions.

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The other fast way to test
your particular solution

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and verify that it actually does
satisfy the initial conditions

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that you were given,
and so you can

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then right away the solution
as being simply sine 2t.

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Here you can see that at 0,
we would have basically a 0.

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And then if you do
a differentiation,

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you just need to
be careful here,

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because you have a
product function,

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and you end up also with a 0.

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So this actually is
our general solution

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for this particular initial
condition, And to sketch this,

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we can grow-- so here if I
just pick F_0 equal to 1,

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I'm just going to do
t/4 for the envelopes.

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At t equals 0, we start with 0.

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And we know that we're going to
have the first extrema at pi/4

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and the first zero at pi/2.

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And so basically, we end up
with something like that.

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So basically, it's sine
of circular frequency 2,

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and with an envelope
prescribed by t/4.

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Or if we had another value
of F_0, it would be F_0 t/4.

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So the oscillation
is ongoing as t

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goes to infinity
with an envelope that

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diverges to infinity.

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So this is basically
a solution that

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would not be convergent to 0.

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So this ends this recitation.

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And before I finish, I
just want to point out

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that the fact that it
diverges is due to the fact

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that we are forcing
this system very

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close to its natural frequency.

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And so this is a
typical phenomenon

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that you can associate
with a resonance,

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because we're basically
forcing a system

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close to its natural frequency.

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So it's having this huge
amplification in the response.

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And that's what these
increasing envelopes mean.

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So this ends this recitation.

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And the key here
was to realize how

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to use your exponential
response formula, how to move on

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to use the resonance exponential
response formula by testing

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for the first order derivative.

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And if that test
failed, you would

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be going to higher orders.

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And then, given an
initial condition,

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how to basically
sketch the function

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and have a physical
understanding of what

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the resonance means.

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That ends this recitation.