Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 37B / Solution

Linearized momentum balance is ${\rho }_0{\partial }_t\tilde{u}=-\nabla \tilde{p}$. For a progressive acoustic plane wave $\tilde{p}=P(\hat{k}\cdot x-c_{0}t)$, integration givesThe 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 $⟨I⟩$. For complex harmonic pressure amplitude $P$, $⟨I⟩=|P{|}^2\hat{k}/(2{\rho }_0{c}_0)$.

Use phase convention $e^{i(k_{y}y-\omega t)}$, with $\omega =k{c}_0$, $k_y=k\sin \theta$, $k_x=k\cos \theta >0$. Let incident, reflected and transmitted pressure amplitudes be $1,R,T_p$. The two normal velocities at the membrane are $(1-R)\cos \theta /({\rho }_0{c}_0)$ and $T_p\cos \theta /({\rho }_0{c}_0)$, and both equal $-i\omega X_0$. Hence $T_p=1-R$ and $X_0=i{T}_p\cos \theta /({\rho }_0{c}_0\omega )$. Only normal velocity is matched for an inviscid fluid; tangential velocities need not agree.

Put $D_m=k^2(m{c}_0^2-T{\sin }^2\theta )$. The dynamic condition is $-D_m{X}_0+T_p-(1+R)=0$, or $-D_m{X}_0-2R=0$. Since $D_m\cos \theta /({\rho }_0{c}_0\omega )=1/\alpha$, elimination givesComplex 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 $\alpha \to 0$, giving almost total reflection.

For real $\alpha$, $|R{|}^2+|T_p{|}^2=1$. Thus the incoming net normal energy flux is $\cos \theta (1-|R{|}^2)/(2{\rho }_0{c}_0)$, equal to the transmitted flux $\cos \theta |T_p{|}^2/(2{\rho }_0{c}_0)$. Time-averaged energy flux is conserved: the lossless membrane stores and returns energy but has no mean dissipation.