WEBVTT
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GILBERT STRONG: This is the
second video on second order
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differential equations,
constant coefficients,
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but now we have a
right hand side.
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And the first one was free
harmonic motion with a zero,
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but now I'm making this motion,
I'm pushing this motion,
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but at a frequency omega.
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This is my forcing term.
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So I think I'm having a
forcing frequency, omega,
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and remember that for this
one, for the no solution,
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there was a natural
frequency omega n.
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It's very important
are those close,
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are those well separated?
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That governs whether the
bridge that you're walking over
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oscillates too much
and eventually falls.
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Or in the extreme
case, are they equal?
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If omega n is equal to omega
that's called resonance.
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Let me put that word in.
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Resonance.
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When omega equals omega n.
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And we're not going
to deal with today,
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but you should know that
always the formula has an omega
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minus omega n dividing by that.
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So if that is 0, if
omega equals omega n
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our formula has to change.
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Today, this won't happen.
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No.
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So what's the formula?
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What is yp?
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I'm looking for a
particular solution.
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That's a nice function and
also important in practice.
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So I would like to hope that
the particular solution could
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be some multiple of
that cosine omega t.
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And in this problem
that's possible.
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Because if I have a cosine, I've
got a cosine on the right hand
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side, and if that cosine comes
here, it's on the left side,
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and the second derivative of
the cosine is, again, a cosine,
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I'm going to have a match
of cosine omega t terms.
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And then I'll just choose
the right number capital Y.
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I won't be able to
do that when there's
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a first derivative in there,
because the first derivative
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of cosine will bring in signs.
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I'll have a mixture
of cosines and sines
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and then I better
allow for that mixture.
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But here I don't have to.
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There's the forcing function.
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Response, this is
the forced response.
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I'd like to get used
to that word, response,
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for the solution.
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Here's the input, the
response is the output.
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So let me just plug
that into the equation
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and find capital Y.
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So here I have m, second
derivative is going to be a Y,
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and second derivative
will bring out a minus
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omega squared times the cosine.
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And here I have kY is Y times
the cosine equal the cosine.
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I could have a constant
there, but the whole thing
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would be no more interesting,
no more difficult than with a 1.
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So what do I do?
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The nice thing is here
I have all cosines,
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so I'm just going to have
minus omega squared m and a k.
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So it's k minus m omega squared.
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Can I write it that way?
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Times Y. I'm going to
cancel the cosines.
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That's just a 1.
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On the side is a 1.
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I've canceled the
cosine, so I've kept kY.
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I've kept the 1, and I've
kept a minus omega squared mY.
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So that tells me Y right away.
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It's just like plugging in
an exponential and canceling
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exponentials all the way along.
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Here, I'm canceling cosines
all the way because every term
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was a cosine.
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So I know Y. So I
know the answer.
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So the final answer
is Y(t) is Yn.
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Well, let me put Y
particular first plus Yn.
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So I've just found Y particular.
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Y particular is this
capital Y cosine omega t.
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So it's cosine omega t times
Y and Y is 1 over this.
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Here's goes Y. Down below I
have k minus m omega squared.
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Right?
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That's what we just found,
that particular solution.
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The capital Y, the
multiplying constant,
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was 1 over that constant.
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And now comes the C1
cosine of omega nt
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and the C2 sine of omega nt.
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Remember, omega n is
different from omega.
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Actually, this is
pretty nice here.
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I could write that
another way so you
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would see the important here.
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So remember, what
is omega n squared?
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Can I just remember that
omega n squared is k over m.
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Right?
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Yup.
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k is the same as-- I'm going
to put that m up here--
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k is the same as
m omega n squared.
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k is the same as
m omega n squared
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and here I'm subtracting
m omega squared.
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You'll see the whole point of
resonance or near resonance
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when the bridge is getting
forced buy a frequency close
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to its resonant frequency.
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This difference, omega
squared, the difference
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between the two frequencies
squared is in the denominator
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and will be small and
then the effect is large.
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And if we get those too close,
the effect is too large.
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So we'll see this cosine omega
t over this is, I would call,
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the frequency response
is this factor.
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1 over m omega n squared
minus omega squared.
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That's the key
multiplier for when
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the forcing term is a pure
frequency, that frequency
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gets exploded.
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And now, of course, what are
capital C1 and capital C2?
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We find those from
the initial condition.
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At t equals 0, we
put in t equals 0,
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and that tells us
what C1 has to be.
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And we put in t equals 0 again
to match the velocity Y prime
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at 0, and that tells us C2.
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Are you OK with that?
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Just look at the beauty
of that solution.
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This is null part.
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This is the forced part,
the particular part,
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the cosine divided
by that constant.
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There's one more
equation, one more
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forcing term I'd like often
and always and now to discuss.
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And that is a delta
function, an impulse.
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So I'm going to add
one more example.
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my double prime plus ky
equal the delta function.
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Delta function.
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It's called an impulse.
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So I'd like to solve
that equation also.
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When the forcing term just
happens at one second,
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at the initial second.
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At t equals 0, the
delta function,
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I'm hitting the spring.
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So the spring is sitting or
the pendulum is sitting there.
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Actually, let's set it at rest.
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Here's my pendulum.
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I'll try to draw a pendulum.
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I don't know.
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That's not much of a pendulum.
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But it's good enough.
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This equation says what
happens if I hit it
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with a point source?
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At t equals 0, I hit it but
I give it a finite velocity.
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It doesn't move in
that instant second.
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This is where delta
functions come in
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so let me give you the
result of what happens
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and then we'll see them again.
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So what am I doing?
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I want to solve this equation
when the forcing function is
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a delta function.
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So I'm going to call y
the impulse response.
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It's the solution that comes
when the forcing function is
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an impulse.
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So y is the impulse response.
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In fact, it's so
important, I'm going
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to give it its own letter. g.
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Now, can I turn that y into a g?
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So that g is g of t is
the impulse response.
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If I can solve that equation.
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You might say, not so easy.
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With a delta function, it's
not even a genuine function.
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It's a little bit crazy.
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It all happens in one second.
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I'm sorry, in one instant.
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Not over one second,
but one moment.
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But I can solve it.
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I can solve it for this reason.
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I can think of it
as an impulse here
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or I have an option,
another way which clearly I
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can think of it as solving it
with no force mg double prime.
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Same problem, same
solution is 0.
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But I start from rest.
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Nothing's happening.
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y of 0 is 0.
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And it starts from an initial
velocity, y prime of 0.
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The impulse starts it
out like a golf ball.
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Just go.
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And there's a 1 over m there.
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I'll discuss that another time.
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What I want to see now is that
I have either this somewhat
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mysterious equation or this
totally normal equation,
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even a no equation starting
from y of 0 equals 0.
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But with an initial
velocity that the impulse
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gave to the system.
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And I should be calling this g.
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This is the g.
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We'll see impulse
responses again,
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but let's see it this time
by solving this equation.
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So I plan to solve that
equation and actually we
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solved it last time.
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You remember the
solution to this one?
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When it starts from
0, there's no cosine.
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But when the initial velocity
is 1 over m, there is a sign.
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So I'm going to just
write down the g of t,
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which is just sine of omega nt.
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And why is it the
natural frequency?
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Because I'm solving the no.
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I'm looking for a no solution.
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But the previous video on
no solutions gets me this.
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Only I have to divide by, get 1
over m as the initial velocity.
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You'll see that that will
solve the no equation.
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This is what happens to the
pendulum or the golf ball.
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Well, pendulum much better.
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Actually, golf ball
is poor example.
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Sorry about that.
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Golf balls don't
swing back and forth.
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They tend to go.
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I'm looking at pendulums,
springs going up and down.
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So the spring starts out, has
an initial velocity of 1 over m
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and then after that
nothing happens.
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So that is the impulse response.
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The response to an impulse.
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And why do I like that?
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First of all, its beautiful.
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Simple answer.
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Secondly, every
forcing function,
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and the output
comes from this one.
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We'll see that point.
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So we've introduced forcing
functions, cos omega
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t, where the particular solution
was a multiple of cos omega t.
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And now, we've introduced
a forcing function delta,
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the delta function where the
response is a sine function.
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Thank you.