Damped Simple Harmonic Motion — Question 6

Question 6

Very strong damping can make the final relaxation slower

For $m\ddot x+b\dot x+kx=0$ with $b^2>4mk$, the two decay rates are $\lambda_{\pm}=[b\pm\sqrt{b^2-4mk}]/(2m)$, and the modes decay as $e^{-\lambda_{\pm}t}$.

1. Derive the slow and fast time scales when $b^2\gg4mk$.

2. For $m=1\,\mathrm{kg}$ and $k=4\,\mathrm{N/m}$, compare the slow rate at $b=10$ and $20\,\mathrm{kg/s}$. State the initial-condition qualification when discussing a late-time slow mode.

Question 6 - Solution

Very strong damping can make the final relaxation slower

(a) Derivation and calculation

$$\sqrt{b^2-4mk}=b\sqrt{1-4mk/b^2}\simeq b-2mk/b.$$$$\lambda_-=\frac{b-\sqrt{b^2-4mk}}{2m}\simeq\frac kb,\quad\lambda_+\simeq\frac bm.$$$$\boxed{\tau_{\mathrm{slow}}\simeq b/k},\qquad\boxed{\tau_{\mathrm{fast}}\simeq m/b}.$$

(b) Result and interpretation

For $m=1,k=4$,$$b=10:\quad\lambda_-=(10-\sqrt{84})/2=.417424,\quad\tau_-=2.39564\,\mathrm s;$$$$b=20:\quad\lambda_-=(20-\sqrt{384})/2=.202041,\quad\tau_-=4.94949\,\mathrm s.$$$$x=C_-e^{-\lambda_-t}+C_+e^{-\lambda_+t}.$$The slow rate controls late times only when $C_-\ne0$; a specially chosen initial state can eliminate that mode.

Physical check

The strong-damping estimates $b/k=2.50$ and $5.00\,\mathrm s$ approach the exact slow time scales.