= Solution
The vertical acceleration in the <thin disk> is of order $\Omega^2H$. Balancing it against the characteristic <pressure> force gives
$$
\frac{P}{\rho H}\sim\Omega^2H,\qquad c_s\sim\Omega H,
$$
up to factors depending on the <adiabatic exponent> and the vertical profile. A displacement of size $\delta z$ has a restoring acceleration of order $\Omega^2\delta z$, so $\delta\ddot z\sim-\Omega^2\delta z$. Equivalently the <sound-crossing time> is $H/c_s$. Both estimates give the <vertical dynamical timescale of a disk>:
$$
\boxed{t_{\rm dyn}\sim\frac H{c_s}\sim\Omega^{-1}.}
$$
This is a mechanical response time, not an unconditional damping time. An inviscid stable layer can exhibit a <vertical breathing mode of an astrophysical disk> without settling; its adiabatic frequency satisfies $\omega_H^2=(\gamma+1)\Omega^2$. Moreover the profile in part (b) need not be convectively stable: its <specific entropy> satisfies
$$
s_{\rm gas}=c_v\log(P/\rho^\gamma)+\text{constant}
=c_v[1-\beta(\gamma-1)]\log T+\text{constant}.
$$
For $z>0$, $T$ decreases upward, so <convective stability> requires $\beta>1/(\gamma-1)$, with equality neutrally stratified. Thus interpreting the timescale as re-establishment of a stable equilibrium presupposes suitable stability and damping; the stated equations alone do not guarantee either.
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