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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?enter image description here

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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.

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  • $\begingroup$ Nice explanation, but it doesn't completely address the question, i.e., what if the damping is non-linear? $\endgroup$
    – Bernhard
    Apr 1 '14 at 7:21
  • $\begingroup$ Hmm. So do you mean you want damping to be an arbitrary $g(v)$ instead of $-cv$? $\endgroup$
    – Ruslan
    Apr 1 '14 at 8:28
  • $\begingroup$ Not me, but I can imagine that it makes a difference. $\endgroup$
    – Bernhard
    Apr 1 '14 at 8:53
  • $\begingroup$ @Bernhard Ah, confused you with the OP. Can't say analytically yet, but if you take (with $k=1$) $g(v)=-2v+2v^3$, then it indeed goes to rest quicker (checked with NDSolve in Mathematica). Adding even more odd powers makes the speed even higher. $\endgroup$
    – Ruslan
    Apr 1 '14 at 8:56
  • $\begingroup$ What if damping is not viscosity but friction? Will it be faster or slower? $\endgroup$ Apr 4 '14 at 7:25

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