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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.