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A membrane of mass per area $m$ and tension $T$ separating two identical acoustic half-spaces has evanescent modes withThe second term is the added mass of an evanescent fluid layer contributed by both sides. Hence the phase speed is below the vacuum membrane speed $\sqrt{T/m}$, and for $k\to 0$ it obeys $c\sim \sqrt{Tk/(2{\rho }_0)}\to 0$. Pressure and normal velocity are in quadrature, so the mean normal acoustic energy flux vanishes.

With no incoming wave, both fluids obey the outgoing acoustic square-root branch. Their surface pressure amplitudes areSubstitution in the sheet's force balance yields the dispersion relation for an acoustic membrane wave:The equivalent denominator $\Delta =\gamma (T{k}^2-m{\omega }^2)-2{\rho }_0{\omega }^2$ 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 formThe normal fields decay away from the sheet; continued by analytic continuation roots may describe radiating or leaky modes.

On $D=0$, $T{k}^2-m{\omega }^2=2{\rho }_0{\omega }^2/\gamma$, so the impedance from the preceding solution isThus 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.

Use the Fourier transform convention $\hat{p}(k,y)={\int }_{\mathbb{R}}{e}^{ikx}p(x,y)dx$. Let $\gamma (k{)}^2=k^2-k_0^2$, choosing $\mathrm{Re}\gamma >0$ on the real contour with $\mathrm{Im}\omega <0$. For time dependence $e^{i\omega t}$, this is the decaying continuation of the outgoing Sommerfeld radiation condition.

The transformed Helmholtz equation has upper and lower solutions $P(k)e^{-\gamma y}$ and $Q(k)e^{\gamma y}$. The two kinematic traces give $-\gamma P={\rho }_0{\omega }^2\hat{\eta }=\gamma Q$, henceThe elastic membrane equation gives $[\hat{p}]=(m{\omega }^2-T{k}^2)\hat{\eta }$. Therefore the acoustic wave on a tensioned massive membrane has dispersion relationEquivalently, $T{k}^2={\omega }^2(m+2{\rho }_0/\gamma )$. The fluid on both sides supplies a positive added mass of an evanescent fluid layer, which is a useful independent check on the sign. In particular, the incident pressure isThe symbol $m$ here denotes elastic membrane mass per area, rather than the fluctuating Mach number used in Question 1.

Takeand define the positive normal decay rateThe decaying solutions of the acoustic wave equation that satisfy the kinematic conditions areTheir pressures at the membrane areSubstitution into the dynamic boundary condition givesand hence the acoustic wave on a tensioned massive membrane dispersion relation

Writing the phase speed as $c=\omega /k$, the positive added-inertia term givesPhysically, each evanescent acoustic surface wave accelerates a layer of fluid on both sides of the membrane, so the inertia per unit area exceeds $m$. Moreover,and the added mass of an evanescent fluid layer $2{\rho }_0/q$ diverges as $k\to 0$. In that limit $c\ll c_0$, so $q\sim k$ and