= Solution
For passive friction $\Gamma>0$, the time dependence of each spatial normal coordinate follows its damping regime:
* If $\omega_0>\gamma$, it is underdamped: $u_k(t)=e^{-\gamma t}[A\cos(\Omega t)+B\sin(\Omega t)]$, with $\Omega=\sqrt{\omega_0^2-\gamma^2}$. The amplitude envelope has decay time $1/\gamma=2m/\Gamma$; cycle-averaged <energy> decays as $e^{-2\gamma t}$.
* If $\omega_0=\gamma$, it is critically damped: $u_k(t)=(A+Bt)e^{-\gamma t}$.
* If $0<\omega_0<\gamma$, it is overdamped: $u_k(t)=A e^{-[\gamma-\sqrt{\gamma^2-\omega_0^2}]t}+B e^{-[\gamma+\sqrt{\gamma^2-\omega_0^2}]t}$. Both terms decay without oscillation, with two different rates.
For any fixed positive damping, sufficiently small nonzero $k$ is overdamped because $\omega_0\simeq c_s|k|$. The slow branch then has $\omega\simeq-i(mc_s^2/\Gamma)k^2$, a diffusive long-wavelength relaxation. At exactly $k=0$, uniform translation has no restoring force, so $u_0(t)=A+B e^{-\Gamma t/m}$ and the constant displacement need not decay. Thus saying every amplitude simply decays as $e^{-\Gamma t/(2m)}$ is valid only for the underdamped envelope, not for all wave numbers.
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