= Solution
The linear acoustic momentum and <continuity> equations are $\rho u_t=-p_x$ and $p_t=-\rho c^2u_x$. For a right-moving <plane wave> depending on $x-ct$, integration gives $\boxed{p=\rho c\,u}$, 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 $Z_j=\rho_jc_j$. In the outer regions write pressures $I e^{ik_0x}+R e^{-ik_0x}$ and $T I e^{ik_0x}$, taking $I=1$ for convenience. The corresponding velocities are $(e^{ik_0x}-R e^{-ik_0x})/Z_0$ and $Te^{ik_0x}/Z_0$. Pressure and normal velocity are continuous at both layer boundaries. Within the cooled layer a decomposition into right and left waves gives, with $\delta=k_1L$,
$$
\begin{pmatrix}p(L)\\u(L)\end{pmatrix}=
\begin{pmatrix}\cos\delta&iZ_1\sin\delta\\ i\sin\delta/Z_1&\cos\delta\end{pmatrix}
\begin{pmatrix}p(0)\\u(0)\end{pmatrix}.
$$
This matrix follows by writing $p=Ae^{ik_1x}+Be^{-ik_1x}$ and $Z_1u=Ae^{ik_1x}-Be^{-ik_1x}$. Applying its inverse and using $p(0)+Z_0u(0)=2$ eliminates $R$, giving
$$
T=\frac{e^{-ik_0L}}{\cos\delta-\frac i2(\lambda+\lambda^{-1})\sin\delta},\qquad \lambda=Z_1/Z_0.
$$
Consequently
$$
\boxed{|T|=\left[\cos^2(k_1L)+\frac14(\lambda+\lambda^{-1})^2\sin^2(k_1L)\right]^{-1/2}.}
$$
For $\lambda\gg1$, transmission is weak except near $k_1L=n\pi$, 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 $|T|^2$.
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