= Solution
For a vector $a$ described in axes rotating at constant <angular velocity> $\Omega$, the <rotating-axes theorem> is $(da/dt)_{\rm inertial}=(da/dt)_{\rm rotating}+\Omega\times a$. Inertial particle <velocity> is $u+\Omega\times x$. Applying the theorem once more gives inertial acceleration $Du/Dt+2\Omega\times u+\Omega\times(\Omega\times x)$. Thus the rotating momentum equation is
$$
\boxed{\frac{Du}{Dt}+2\Omega\times u+\Omega\times(\Omega\times x)=-\frac1\rho\nabla p+g.}
$$
If $g=-\nabla\Phi$ and <density> is constant, the centrifugal term can be absorbed into the <reduced pressure in a rotating fluid>:
$$
\boxed{P=p+\rho\Phi-\frac\rho2|\Omega\times x|^2,\qquad
u_t+(u\cdot\nabla)u+2\Omega\times u=-\nabla P/\rho.}
$$
For small disturbances with $\Omega=\Omega e_z$, linearize to $u_t-2\Omega v=-P_x/\rho$, $v_t+2\Omega u=-P_y/\rho$, $w_t=-P_z/\rho$, together with <incompressibility>. Put $\zeta=v_x-u_y$. Taking the vertical <curl> gives $\zeta_t=2\Omega w_z$. Taking divergence gives $\nabla^2P=2\rho\Omega\zeta$. Applying the Laplacian to the vertical momentum equation therefore gives $\partial_t\nabla^2w=-2\Omega\zeta_z$. Another time <derivative> yields
$$
\boxed{\partial_t^2\nabla^2w+4\Omega^2w_{zz}=0.}
$$
All three <velocity> components are retained, even for fields independent of $y$; rotation produces the transverse component.
Set $m=\pi/h$ and seek $w=ae^{-i\omega t}W(x)\sin mz$. The rigid-wall conditions are then automatic and the scalar equation gives
$$
W''+m^2\left(\frac{4\Omega^2}{\omega^2}-1\right)W=0,\qquad W(0)=1.
$$
For $0<|\omega|<2|\Omega|$, define $q=m\sqrt{4\Omega^2/\omega^2-1}$. The two possible oscillatory factors are distinguished by energy propagation. With horizontal <wavenumber> $k$, the <dispersion relation> is $\omega^2=4\Omega^2m^2/(k^2+m^2)$ and $c_{g,x}=-\omega k/(k^2+m^2)$. An outgoing wave in $x>0$ must have $\omega k<0$, giving
$$
\boxed{w=ae^{-i\omega t-i\operatorname{sgn}(\omega)qx}\sin(\pi z/h).}
$$
For $|\omega|>2|\Omega|$, define $\kappa=m\sqrt{1-4\Omega^2/\omega^2}$. Reject the exponentially growing branch, leaving the evanescent response
$$
\boxed{w=ae^{-i\omega t-\kappa x}\sin(\pi z/h).}
$$
The real part is the physical <velocity>. The radiation or bounded-decay condition is needed in addition to the wavemaker boundary value to pick a unique half-channel response. Steady forcing ($\omega=0$) is a singular limit: with nonzero rotation a steady <geostrophic balance> and impermeable top/bottom require $w=0$, incompatible with a nonzero imposed <sine> profile. At the cutoff equality, the bounded outgoing polarization also degenerates. These exceptional cases cannot be obtained by blindly substituting into the two strict-inequality formulas. The propagating sign is the <radiation sign for an inertial wave>.
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