= Solution
Linearized momentum balance is $\rho_0\partial_t\tilde u=-\nabla\tilde p$. For a progressive <acoustic plane wave> $\tilde p=P(\hat k\cdot x-c_0t)$, integration gives
$$
\boxed{\tilde u=\frac{\hat k}{\rho_0c_0}\tilde p}.
$$
The integration constant is zero for the pure wave with no added uniform flow. The instantaneous <acoustic intensity> is $I=\tilde p\tilde u$, the pressure-work energy flux; time-averaged intensity is $\langle I\rangle$. For complex harmonic pressure amplitude $P$, $\langle I\rangle=|P|^2\hat k/(2\rho_0c_0)$.
Use phase convention $e^{i(k_y y-\omega t)}$, with $\omega=kc_0$, $k_y=k\sin\theta$, $k_x=k\cos\theta>0$. Let incident, reflected and transmitted pressure amplitudes be $1,R,T_p$. The two normal velocities at the membrane are $(1-R)\cos\theta/(\rho_0c_0)$ and $T_p\cos\theta/(\rho_0c_0)$, and both equal $-i\omega X_0$. Hence $T_p=1-R$ and $X_0=iT_p\cos\theta/(\rho_0c_0\omega)$. Only normal velocity is matched for an inviscid fluid; tangential velocities need not agree.
Put $D_m=k^2(mc_0^2-T\sin^2\theta)$. The dynamic condition is $-D_mX_0+T_p-(1+R)=0$, or $-D_mX_0-2R=0$. Since $D_m\cos\theta/(\rho_0c_0\omega)=1/\alpha$, elimination gives
$$
\boxed{R=\frac1{1+2i\alpha},\qquad T_p=\frac{2i\alpha}{1+2i\alpha}}.
$$
Complex conjugation gives the amplitudes for the opposite time-phase convention. At zero $D_m$, the formulas have the limiting values $R=0,T_p=1$ even though $\alpha$ itself is then infinite. A strongly inertial membrane has $\alpha\to0$, giving almost total reflection.
For real $\alpha$, $|R|^2+|T_p|^2=1$. Thus the incoming net normal energy flux is $\cos\theta(1-|R|^2)/(2\rho_0c_0)$, equal to the transmitted flux $\cos\theta|T_p|^2/(2\rho_0c_0)$. \b[Time-averaged energy flux is conserved]: the lossless membrane stores and returns energy but has no mean dissipation.
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