= Solution
Both sides of the dispersion relation are positive on an interval beginning at zero phase, so a zeroth branch exists for every $k>0$. Higher branches first appear when $c=c_s$, for which the right side vanishes and $mh=n\pi$. Thus
$$
k_c^{(n)}=\frac{n\pi\bar c_s}{h\sqrt{c_s^2-\bar c_s^2}},
$$
and
$$
\boxed{\omega_c^{(n)}=c_sk_c^{(n)}
=\frac{n\pi\bar c_sc_s}{h\sqrt{c_s^2-\bar c_s^2}}}.
$$
The $n=1$ dispersion curve begins on the line $\omega=c_sk$ at $(k_c^{(1)},\omega_c^{(1)})$, bends below it, and approaches the layer shear-wave line $\omega=\bar c_sk$ from above as $k\to\infty$.
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