Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 37B / Solution

The linear acoustic momentum and continuity equations are $\rho u_t=-p_x$ and $p_t=-\rho c^2{u}_x$. For a right-moving plane wave depending on $x-ct$, integration gives $\boxed{p=\rho cu}$, with zero perturbation constants. For a left-moving wave the sign is reversed. The factor $Z=\rho c$ is the acoustic impedance. Proportionality applies to a traveling wave; a superposition of opposite directions generally has no single such constant.

Use time dependence $e^{-i\omega t}$ and set $Z_j={\rho }_j{c}_j$. In the outer regions write pressures $I{e}^{i{k}_{0}x}+R{e}^{-i{k}_{0}x}$ and $TI{e}^{i{k}_{0}x}$, taking $I=1$ for convenience. The corresponding velocities are $(e^{i{k}_{0}x}-R{e}^{-i{k}_{0}x})/Z_0$ and $T{e}^{i{k}_{0}x}/Z_0$. Pressure and normal velocity are continuous at both layer boundaries. Within the cooled layer a decomposition into right and left waves gives, with $\delta =k_{1}L$,This matrix follows by writing $p=A{e}^{i{k}_{1}x}+B{e}^{-i{k}_{1}x}$ and $Z_{1}u=A{e}^{i{k}_{1}x}-B{e}^{-i{k}_{1}x}$. Applying its inverse and using $p(0)+Z_{0}u(0)=2$ eliminates $R$, givingConsequentlyFor $\lambda \gg 1$, transmission is weak except near $k_{1}L=n\pi$, where $|T|=1$. The cooled layer acts as a resonant filter: an integer number of half-wavelengths fits within it at each resonance. The half-power width in $\delta$ is asymptotically $2/\lambda$ on either side. Since the two outer impedances agree, the transmitted power fraction is $|T{|}^2$.