Damped oscillations

A second-order model

A mass attached to a spring and viscous damper obeys

$m\ddot{x}+c\dot{x}+kx=0.$

Dividing by $m$ gives $\ddot{x}+2\zeta\omega_0\dot{x}+\omega_0^2 x=0$, where $\omega_0=\sqrt{k/m}$ and $\zeta=c/(2\sqrt{km})$. Take $x(0)=1$ and $\dot{x}(0)=0$.

Deriving the solution

Substitute $x(t)=e^{rt}$ to obtain

$r^2+2\zeta\omega_0r+\omega_0^2=0,\qquad r=-\zeta\omega_0\pm\omega_0\sqrt{\zeta^2-1}.$

For $0\leq\zeta<1$, let $\omega_d=\omega_0\sqrt{1-\zeta^2}$. The initial conditions give

$x(t)=e^{-\zeta\omega_0t}\left[\cos(\omega_dt)+\frac{\zeta}{\sqrt{1-\zeta^2}}\sin(\omega_dt)\right].$

Increasing damping changes both the decay envelope and the frequency.

Critical and overdamped motion

At $\zeta=1$, the root repeats and

$x(t)=(1+\omega_0t)e^{-\omega_0t}.$

For $\zeta>1$, both roots are real and negative. Increasing damping beyond the critical value slows the long-term return to equilibrium.