Finding the damping force for a critically damped oscillator

[issacnewton]

Homework Statement

A critically damped simple harmonic oscillator starts from an amplitude of 5.0 cm and comes to rest at equilibrium 3.5 s later. The SHO is made of a 0.58 kg mass hanging from a spring with spring constant 150 N/m. Assuming the friction force is in the vertical direction, how big is the friction force?

Homework Equations

Equations for the critically damped oscillator. We have $b = 2\sqrt{mk}$ for critically damped oscillator and general equation for critically damped oscillator is
$$ x(t) = e^{-bt/2m}\left(c_1 + c_2 t\right) $$
where $c_1, c_2$ are parameters which depend on initial conditions

The Attempt at a Solution

Now I reasoned that $x=0$ when $t = 3.5 s$, so we get the equation
$$0 = e^{-3.5b/2m} \left(c_1 + 3.5c_2 \right) $$
which leads us to conclude that $c_1 = -3.5 c_2$. Now, we also know that at $t=0$, $x$ is equal to the amplitude of 5 cm. This gives us the equation
$$5cm = c_1$$
So, using these equations, we can solve for $c_1 = 0.05m$ and $c_2 = -0.014286\;m/s$. Using $b = 2\sqrt{mk}$, I get $b = 18.655$. So I get all the parameters for the general solution. Now the friction force or the damping force here has the form of $F = -b v$. And I have no idea how to get that from the given data.

 

[scottdave]

I haven't checked your calculations, but the gist of it looks correct. Can you find v(t) by taking the derivative of x(t)? Then you just need to find its maximum value, since they want to know "how big" the force is.

 

[issacnewton]

so do I equate the first derivative of $v(t)$ to zero and solve for $t$ ?