If you have a mass-spring simple harmonic oscillator, with a mass of $m$, spring coefficient $k$ and damping coefficient $c$, the critical damping condition as shown here and here is defined by:
$$ω = \sqrt{\frac{k}{m}}$$
$$c = \sqrt{4mk}$$
Let's say you have a hypothetical such critically damped oscillator of a known $m$, $k$ and $c$ which is released with zero velocity from an initial displacement of $x$.
It will eventually settle out to a final displacement of $0$. Can you calculate the maximum velocity that will be achieved along the way? If so how?
I could "solve" this by running it through a stepwise simulation with a given sampling rate, but I am wondering if there is an actual solution mathematically for this problem so it is not so inefficient and I can be more precise.
Is there an equation that describes the displacement curve of a critically damped oscillator released from zero velocity at a certain starting point? If so then I presume I would just have to find the derivative of that and get the maximum absolute value.
Any thoughts? Thanks.
Best Answer
the ODE is:
$$\ddot x+\frac cm\,\dot x+\omega^2\,x=0$$
with $$c=2\sqrt{m\,k}\quad,\omega=\sqrt{\frac km}$$
you obtain the solution $~(x(0)=x_0~,\dot x (0)=0)~$
$$x(t)=\frac{x_0}{m}\left[{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}} \left( m+\sqrt {m}\sqrt { k}t \right)\right] \quad\Rightarrow\\ v(t)=-\frac{x_0}{m}\left[k{{\rm e}^{-{\frac {\sqrt {k}t}{\sqrt {m}}}}}t\right]$$
hence $~|v(t)|~$ is maximum at $~t_m=\sqrt{\frac mk}$
you can obtain the time where the velocity is maximum by solving this equation
$$\frac{d}{dt} v(t)=0$$
for t, $~\Rightarrow~t_m=t$ and
$~|v_m|=\frac{\omega\,x_0}{e}$