WEBVTT
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PROFESSOR: Welcome
to this recitation
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on damped harmonic oscillators.
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So here you're asked to
assume an unforced, overdamped
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spring-mass-dashpot that
started at x dot of 0
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equals to 0, so rest,
and to show that it never
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crosses the equilibrium
position, x equal to 0, for t
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larger than 0.
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The second part of
the problem asks
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you to show that regardless
of the initial condition,
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this overdamped
oscillator can not
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cross the equilibrium position
more than one time, or more
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than once.
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OK.
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So why don't you
pause the video,
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try to think about this
problem, and I'll be right back.
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Welcome back.
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So the system that
we're looking at
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is a spring-mass-dashpot
that would
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be written in this form-- second
order differential equation.
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Let's assume that we have
positive constant coefficients
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here.
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And basically, to
solve this, you
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would be considering the
methods that we saw before.
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And the general
solution would just
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be written in the form of
c_1 exponential lambda_1,
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the c_2 exponential lambda_2,
with c_1, c_2 just two
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constants that
would be determined
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by the initial condition.
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Lambda_1, lambda_2
would be here the roots
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of the characteristic
polynomial that you
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would have found there.
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Given that we're looking
at an overdamped, unforced
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spring-mass-dashpot,
you can actually
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show that lambda_1
and lambda_2 would
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be both real and negative.
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And here, we can just
say that basically,
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lambda_1 and lambda_2
would be less than 1.
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And we'll keep that aside for
now, and I'll use this later.
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So this is just
setting up the problem.
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So now, what was the question?
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The question was to show that
if we start this system with
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initial condition x
dot of 0 equal to 0,
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which corresponds to
lambda_1*c_1 plus lambda_2*c_2
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equals to 0, then the system
cannot cross the equilibrium
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position x equal to 0
for t larger than 0.
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So let's just start by assuming
that the system crosses
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the equilibrium position.
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So for example, let's look for
t-star such that x of t-star
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is equal to 0.
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So x of t-star, we
know its form already.
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We have the general
form of x of t-star.
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That's basically c_1 exponential
lambda_1*t-star plus c_2
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exponential lambda_2*t-star.
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And so we can massage
this equation,
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and basically end up with
minus c_2 over c_1 equal
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to exponential of lambda_1
minus lambda_2*t-star.
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So now let's just
find our t-star
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by applying the log of both
sides of this equation.
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So we get t-star equals to
the log of minus c_2 over c_1,
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and we divide by the
lambda_1 minus lambda_2.
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So here this tells us that if
t-star exists, which means,
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if the log is defined, and
this minus c_2 over c_1
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basically is positive, then we
only have one value of t-star
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possible.
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So here, we actually are
answering the second part
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of this question,
number two, which
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was telling us that regardless
of the initial condition-- so
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regardless of the
coefficient c_1,
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c_2 that we would
have-- if t-star exists,
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there is only one.
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And so that means that
the system would not
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cross this equilibrium
position more than once.
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But I'll come back on that.
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But now let's go back
to what we were asked
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to do in the first part,
where we basically now go back
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to our x dot of 0 equals to
0, which basically gave us
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that minus c_2 over c_1 is
equal to lambda_1 over lambda_2,
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and the way we defined
lambda_1 over lambda_2
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here gives us that minus
c_2 over c_1 is less than 1,
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which means that the log
is going to be negative.
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What happens in the denominator?
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Lambda_1 minus lambda_2
would be positive.
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So with this initial
condition, we
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would end up with a t-star
that would be negative.
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So basically, x is never equal
to 0 again for t larger than 0,
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given these initial conditions.
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So that finishes this
first part of the problem.
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So I'll go back on the physics
of it in a moment with a graph.
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So starting from this
initial condition,
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x can never be equal to
0, because the only t-star
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we can find would be negative.
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So for t larger
than 0, it does not
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cross the equilibrium point.
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The second part of the
problem-- so this was one--
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just comes from the fact
that, if I label this star,
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star tells us that if minus
c_2 over c_1, strictly
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larger than 0, then
only one t-star exists.
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So if we have a solution,
there is only one.
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And this is regardless
of the initial conditions
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that we would be given.
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So the system cannot cross the
equilibrium position more than
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once.
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So now let's look at what
we're doing here, graphically.
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Let's assume that we're,
for example, starting
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with initial
condition here, where
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we're stretching our spring,
but we start with 0 velocity.
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So x dot of 0 equal to 0.
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That was the system that we had.
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Then this is an overdamped
case where both basically
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exponentials are
decaying to 0, and so we
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would have a solution that
would go to 0 quickly.
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It would be damped.
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And so this would be part 1.
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Now let's look at what
would happen if we started
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with other initial conditions.
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So for example, starting
from the same point
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with a much bigger velocity.
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Then the system would go up, but
eventually, it has to go down.
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It wouldn't have this
shape, but basically, it
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would have to go
down to 0's position.
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And when it reaches,
you can show that again,
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the derivative of x
can reach 0 only once.
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And at that point, you're then
back to the initial conditions
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that you had in the first
part of the question,
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and so you can then, from here,
argue again that you cannot
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cross the zero, the
equilibrium point,
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after reaching a maximum.
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Now what if we
had a stretch that
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would be giving a
negative velocity
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to the mass, and a very
strong negative velocity?
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Then the system also wants to go
back to 0, but could overshoot.
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And the overshoot would
also generate a unique time
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at which the derivative
would be equal to 0,
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and after that point, you would
be back to the same argument
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we had before,
where the solution
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would have to go toward
0, but never crosses it.
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So we can have various
configurations.
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And here I start
with this point,
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but you could also start with
other initial conditions, where
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you could have,
as well, something
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that would be, for example,
a very strong positive,
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where again, here you
would have an overshoot,
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but then the solution would be
attracted by the x equal to 0
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solution.
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And of course you could also
start from the equilibrium.
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If you're not imposing
any initial velocity,
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you just stay there,
because this is not forced.
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But if you're
imposing a velocity,
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then you would have other
trajectories of the kind,
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for example, like this, where
again, it would go up, but then
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be attracted back
by the 0 solution.
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So that's the typical behavior
for a damped oscillator, where
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basically there's
no oscillation,
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but the solution is
attracted to rest.
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And you could have
cases of overshoot.
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Then you can show that
after the overshoot,
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velocity would
reach 0, at maximum,
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and then would be attracted
back to the zero solution
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with never crossing it.
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And that ends this recitation.