= Solution
Use the <Fourier transform> convention $\widehat p(k,y)=\int_{\mathbb R}e^{ikx}p(x,y)dx$. Let $\gamma(k)^2=k^2-k_0^2$, choosing $\operatorname{Re}\gamma>0$ on the real contour with $\operatorname{Im}\omega<0$. 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 $-\gamma P=\rho_0\omega^2\widehat\eta=\gamma Q$, hence
$$
Q=-P,\qquad P=-\frac{\rho_0\omega^2}{\gamma}\widehat\eta,\qquad[\widehat p]=-\frac{2\rho_0\omega^2}{\gamma}\widehat\eta.
$$
The <elastic membrane> equation gives $[\widehat p]=(m\omega^2-Tk^2)\widehat\eta$. Therefore the <acoustic wave on a tensioned massive membrane> has <dispersion relation>
$$
\boxed{D(\omega,k)=m\omega^2-Tk^2+\frac{2\rho_0\omega^2}{\gamma(k)}=0.}
$$
Equivalently, $Tk^2=\omega^2(m+2\rho_0/\gamma)$. 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> is
$$
p_I(x,y)=-\operatorname{sgn}(y)\frac{\rho_0\omega^2}{\gamma_I}e^{-ik_Ix-\gamma_I|y|},\qquad\gamma_I=\gamma(k_I).
$$
The symbol $m$ here denotes <elastic membrane> mass per area, rather than the fluctuating <Mach number> used in Question 1.
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