= Solution
Let $\mu=\pi/H$. The end-wall forcing has transverse factor $\cos(\mu y)$, which already satisfies both fixed-wall <Neumann boundary conditions>. By <linearity> and <separation of variables>, seek its periodic response as
$$
\phi(x,y,t)=\operatorname{Re}\left\{\Phi(x)\cos(\mu y)e^{-i\omega t}\right\}.
$$
Substitution into the <wave equation> and the moving-wall condition gives
$$
\Phi''+\left(\frac{\omega^2}{c^2}-\mu^2\right)\Phi=0,\qquad
\Phi'(0)=\epsilon\omega.
$$
The <acoustic waveguide cutoff> is therefore
$$
\boxed{\omega_c=c\mu=\frac{\pi c}H.}
$$
It is the lowest nonzero transverse-mode frequency. The uniform transverse mode has no cutoff, but is not excited here because $\int_0^H\cos(\pi y/H)\,dy=0$.
For $0<\omega<\omega_c$, put $\alpha=\sqrt{\mu^2-\omega^2/c^2}>0$. The axial equation has solutions $e^{\pm\alpha x}$. Boundedness removes the growing exponential, and $\Phi'(0)=\epsilon\omega$ fixes the coefficient. Thus
$$
\boxed{\phi=-\frac{\epsilon\omega}{\alpha}e^{-\alpha x}\cos\frac{\pi y}H\cos(\omega t)
\qquad(0<\omega<\omega_c).}
$$
This <evanescent wave> oscillates in time but decays with penetration length $1/\alpha$. Its axial <velocity> is proportional to $\cos(\omega t)$, while its excess <pressure> $p'=-\rho_0\phi_t$ is proportional to $\sin(\omega t)$. They are in <phase quadrature>, so their mean product, the axial <acoustic energy flux>, is zero.
For $\omega>\omega_c$, put $q=\sqrt{\omega^2/c^2-\mu^2}>0$. The axial solutions are $e^{\pm iqx}$. The outgoing condition selects $e^{iqx}$ for the chosen time convention, and
$$
iq\Phi(0)=\epsilon\omega,\qquad
\Phi(x)=\frac{\epsilon\omega}{iq}e^{iqx}.
$$
Taking the real part gives
$$
\boxed{\phi=\frac{\epsilon\omega}{q}\cos\frac{\pi y}H\sin(qx-\omega t)
\qquad(\omega>\omega_c).}
$$
Its axial <velocity> is $\epsilon\omega\cos(\pi y/H)\cos(qx-\omega t)$, so the moving-wall condition is satisfied with the correct sign. This is a traveling <acoustic waveguide mode> carrying <energy> toward positive $x$. Its <dispersion relation> is
$$
\omega^2=c^2(q^2+\mu^2),\qquad
c_{\rm phase}=\frac\omega q>c,\qquad
c_{\rm group}=\frac{d\omega}{dq}=\frac{c^2q}\omega<c.
$$
The larger axial <phase velocity> does not imply faster energy transport: the <group velocity> controls that transport.
At $\omega=\omega_c$, the axial equation becomes $\Phi''=0$. The imposed nonzero derivative forces $\Phi=\epsilon\omega_c x+$ a constant, which is unbounded. Thus \b[there is no bounded periodic response at exact cutoff] for this infinite, undamped, velocity-driven model. The divergent coefficients as $q\to0$ or $\alpha\to0$ also show why the assumed small perturbations fail sufficiently close to cutoff at fixed forcing. Initial conditions can add free transients; the displayed formulas give the selected periodic forced response.
\Image[/past-exam-of-the-fluids-and-waves-course-of-the-university-of-oxford/2015-acoustic-channel-regimes.png]
{title=An oscillating end wall excites either an evanescent disturbance or an outgoing acoustic wave}
{description=The transverse shape is $\cos(\pi y/H)$. Below cutoff, the normalized axial potential decays exponentially; above cutoff it oscillates downstream with an outgoing phase. The right panel shows the common cutoff and the propagating dispersion branch.}
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