Faster than critical damping for harmonic oscillator? The image below shows damping for spring oscillator with Hooke law F=-kx and damped with F=-cv where:
k is spring constant
x is oscillator position
c is damping coefficient
v is velocity of oscillator
Critical damping happens when $\zeta = { c \over 2 \sqrt{m k} } = 1$ where m is mass of oscillator
The question is, is there another method for damping function to get to rested position faster than critical damping of F=-cv?
 A: Let's take $k=1$ for simplicity. Then our function $f(t)$ as a solution to equation
$$\ddot f=-f-c \dot f,$$
$$f(0)=1,$$
$$\dot f(0)=0$$
will look like:
$$f(t)=e^{-\frac{ct}2}\left(\cos\left(\frac12\sqrt{4-c^2}t\right)+\frac{c\sin\left(\frac12\sqrt{4-c^2}t\right)}{\sqrt{4-c^2}}\right).$$
Critically damped version of it is when we take limit $c\to2$:
$$f_c(t)=e^{-t}(t+1).$$
Let's now compare $f(t)$ with $f_c(t)$ at infinity. For $c>2$ we have
$$\lim_{t\to\infty}\frac{f(t)}{f_c(t)}=\infty,$$
i.e. overdamping makes getting to rest position infinitely slower in the long term*.
Now for $0<c<2$ we have:
$$0<\sqrt{4-c^2}<2;$$
and the term
$$\frac{c\sin\left(\frac12\sqrt{4-c^2}t\right)}{\sqrt{4-c^2}}$$
will oscillate with some amplitude, which goes to infinity as $c\to2$.
Comparing now the remaining factor of $e^{-\frac{ct}2}$ with $f_c(t)$ gives
$$\lim_{t\to\infty}e^{t-\frac{ct}2}(t+1)=\lim_{t\to\infty}e^{\left(1-\frac{c}2\right)t}(t+1)=\infty,$$
as $1-\frac{c}2>0$. Thus, underdamping also makes transition process slower in longer term.
So, bottomline: no, it's not possible to speedup getting to resting position better than critical damping.

* I found this limit in Mathematica. I'm pretty sure it's correct as I've also looked at plot of function of time and $c$, but feel free to ask to elaborate if needed.
