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