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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.

Take $\omega >0$, with time convention $e^{-i\omega t}$, and use the Fourier transform pair $\hat{h}(k)=\int h(x)e^{-ikx}dx$, $h(x)=(2\pi {)}^{-1}\int \hat{h}(k)e^{ikx}dk$. For an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$ and $p=-{\rho }_0{\varphi }_t$. Its harmonic amplitude solves the Helmholtz equation. The outgoing Fourier component has the formThe outgoing acoustic square-root branch is fixed by decay for evanescent components and upward radiation for propagating components:A precise causal prescription is to replace $\omega$ by $\omega +i\sigma$, $\sigma >0$, select $\mathrm{Re}\gamma >0$ for real $k$, invert on the real $k$ axis, and take $\sigma \downarrow 0$. It corresponds to switching on the forcing from the remote past. The limiting inversion contour $C$ passes below the positive outgoing surface-wave pole and above the negative one. The branch points are approached with the same causal prescription; analytic continuation during contour deformation must remain on this sheet. A principal real square root with positive imaginary part inside the acoustic interval would produce incoming sound and the wrong solution.

The elastic membrane's normal velocity equals the fluid velocity at the boundary, so $-\gamma A=-i\omega \hat{\eta }$. ConsequentlyThe elastic membrane balance therefore givesWriting $D=(T{k}^2-m{\omega }^2)\gamma -{\rho }_0{\omega }^2$, the one-sided fluid-loaded membrane radiation formulas areThe sign of the fluid term describes positive added mass of an evanescent fluid layer, not negative inertia.

For $0<\theta <\pi$, the acoustic saddle point is $k_s=k_0\cos \theta$, with ${\gamma }_s=-i{k}_0\sin \theta$. Applying the stated method of steepest descent result givesSince $p=i{\rho }_0\omega \varphi$ in amplitude notation, the angular acoustic directivity of the pressure, apart from angle-independent factors, isThe intensity directivity is proportional to ${\mathcal{D}}^2$. This cylindrical radiating field has amplitude of order $r^{-1/2}$. The formula is for fixed interior angles; grazing-angle limits require a separate uniform approximation.

For real $k>k_0$, $\gamma$ is positive, $D\to -{\rho }_0{\omega }^2$ as $k\downarrow k_0$, and $D\to +\infty$ as $k\to \infty$. A root exists. A root also requires $T{k}^2-m{\omega }^2>0$. In that range,so the root is unique. Evenness supplies the second root, $-k_{\ast }$. There are no real roots inside $|k|<k_0$: the fluid-loading term is real and nonzero while the other term is purely imaginary. Thus there are exactly two real physical-sheet roots, $\pm k_{\ast }$, for nonzero forcing frequency, positive elastic-sheet tension and positive fluid mass density. The zero-frequency static limit is not a two-distinct-root statement.

These roots are counterpropagating evanescent acoustic surface waves coupled to the membrane:Their normal profiles decay as $e^{-{\gamma }_{\ast }y}$, and their phase speed is below both $c_0$ and the vacuum membrane speed. They matter for the elastic membrane motion and for observations near its surface, but are not automatically part of the leading far-field radiation. A residue is picked up only when the causal contour deformation crosses its pole; at fixed $y/r>0$ its magnitude is exponentially small, $e^{-{\gamma }_{\ast }r\sin \theta }$.

For example, in the right half-plane put $k=k_0\cos \zeta$, $\gamma =-i{k}_0\sin \zeta$ and $k_{\ast }=k_0\cosh a$. The pole is at $\zeta =ia$ and the saddle at $\zeta =\theta$. The steepest-descent path has $\mathrm{Re}\cos (\zeta -\theta )=1$. At the pole height its left branch has real coordinate $\theta -\arccos (\mathrm{sech}a)$; comparing it with the original contour just to the right of the pole shows that its residue is crossed forThe analogous sector lies near $\theta =\pi$ for the negative pole. This is why the existence of real dispersion zeros alone does not make their residues relevant in every observation direction.

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