= Solution
With no incoming wave, both fluids obey the <outgoing acoustic square-root branch>. Their surface <pressure> amplitudes are
$$
\widehat p_+=-\frac{\rho_0\omega^2}{\gamma}\widehat\eta,\qquad
\widehat p_-=+\frac{\rho_0\omega^2}{\gamma}\widehat\eta.
$$
Substitution in the sheet's <force balance> yields the <dispersion relation> for an <acoustic membrane wave>:
$$
\boxed{D(k,\omega)=Tk^2-m\omega^2-\frac{2\rho_0\omega^2}{\gamma}=0.}
$$
The equivalent denominator $\Delta=\gamma(Tk^2-m\omega^2)-2\rho_0\omega^2$ is useful away from the branch points $\gamma=0$. For real guided modes with $|k|>k_0$, this is the familiar <added mass of an evanescent fluid layer> form
$$
\left(m+\frac{2\rho_0}{\gamma}\right)\omega^2=Tk^2.
$$
The normal fields decay away from the sheet; continued by <analytic continuation> roots may describe radiating or leaky modes.
On $D=0$, $Tk^2-m\omega^2=2\rho_0\omega^2/\gamma$, so the impedance from the preceding solution is
$$
\boxed{Z=-\frac{i\rho_0\omega}{\gamma}=-Z_f.}
$$
Thus the divergence of the <reflection coefficient> corresponds to a <reflection pole of a fluid-loaded membrane>. The prescribed incoming amplitude is zero, but a homogeneous fluid-sheet mode can have nonzero amplitude. This is a resonance or guided-mode pole of the analytic scattering problem, not arbitrarily large passive reflection at a real propagating angle. In particular, an undamped guided mode has an evanescent normal field and therefore a complex incidence angle in the plane-wave continuation. For real propagating incidence, $Z_f>0$ and the finite-mass, finite-tension sheet has a purely imaginary structural impedance, so its passive reflection remains bounded.
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