= Solution
Substituting $x=Ae^{pt}$ into the <damped harmonic oscillator> equation gives
$$
p^2+\gamma p+\omega_0^2=0,\qquad p_\pm=-\frac\gamma2\pm\sqrt{\frac{\gamma^2}{4}-\omega_0^2}.
$$
For nonnegative <viscous damping>, the cases are
$$
\boxed{\begin{aligned}
\gamma>2\omega_0 &: \quad x=C_+e^{p_+t}+C_-e^{p_-t} &&\text{overdamping},\\
\gamma=2\omega_0 &: \quad x=(C_1+C_2t)e^{-\omega_0t} &&\text{critical damping},\\
0\leq\gamma<2\omega_0 &: \quad x=e^{-\gamma t/2}[C_1\cos\omega_dt+C_2\sin\omega_dt] &&\text{underdamping},
\end{aligned}}\qquad \omega_d=\sqrt{\omega_0^2-\gamma^2/4}.
$$
The PDF's request to express these conditions in terms of $\gamma/m$ is dimensionally inconsistent with its definition $\gamma=b/m$: $\gamma$ already has units $\mathrm{s^{-1}}$. With the unnormalised equation $m\ddot x+b\dot x+kx=0$, the equivalent comparisons are $b/m\gtrless2\omega_0$, or $b^2\gtrless4mk$. Also $\omega_0$ is the undamped natural <angular frequency>, not in general the driven displacement peak.
Back to article page
