= Solution
Write the SH displacement as $u=U(z)e^{i(kx-\omega t)}$ with $c=\omega/k$. A trapped mode requires
$$
\bar c_s<c<c_s.
$$
In the substrate $U=Ae^{qz}$, where
$$
q=k\sqrt{1-c^2/c_s^2},
$$
and the stress-free condition at $z=h$ makes the layer solution $U=B\cos(m(z-h))$, where
$$
m=k\sqrt{c^2/\bar c_s^2-1}.
$$
Continuity of displacement and shear stress at $z=0$ eliminates $A,B$ and gives
$$
\boxed{\tan(mh)=\frac{\mu}{\bar\mu}\frac qm
=\frac{\mu}{\bar\mu}
\frac{\sqrt{1-c^2/c_s^2}}{\sqrt{c^2/\bar c_s^2-1}}}.
$$
The bars lost from the converted statement are required dimensionally and by the stated ordering of wave speeds.
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