= Solution
Take incidence from A into B and first specify \b[<pressure> amplitude]. For a plane <acoustic wave> traveling normally to the interface, the <acoustic impedance> is $Z=\rho c_s$. A right-moving wave has <pressure> amplitude $P=ZU$, where $U$ is its fluid-velocity amplitude; a left-moving reflected wave has $P=-ZU$.
Let $P_i,P_r,P_t$ denote incident, reflected, and transmitted <pressure> amplitudes. Continuity of <pressure> and normal <velocity> at the interface gives
$$
P_i+P_r=P_t,\qquad \frac{P_i-P_r}{Z_A}=\frac{P_t}{Z_B}.
$$
Solving the two linear equations gives the <acoustic reflection and transmission at an interface> coefficients
$$
\frac{P_t}{P_i}=\frac{2Z_B}{Z_A+Z_B},\qquad
\frac{P_r}{P_i}=\frac{Z_B-Z_A}{Z_A+Z_B}.
$$
For the same perfect gas, take a common <adiabatic index> $\gamma$. The <adiabatic sound speed> satisfies $c_s^2=\gamma p/\rho$. The equal equilibrium <pressure> therefore implies $Z_B/Z_A=\sqrt{\rho_B/\rho_A}=\sqrt\delta$, giving
$$
\boxed{\frac{P_t}{P_i}=\frac{2\sqrt\delta}{1+\sqrt\delta}.}
$$
The word amplitude is otherwise ambiguous. For fluid <velocity>, or <displacement> at the same frequency, the corresponding answer is
$$
\boxed{\frac{U_t}{U_i}=\frac{\xi_t}{\xi_i}=\frac{2}{1+\sqrt\delta}.}
$$
These are consistent because <pressure> and <velocity> amplitudes differ by the impedance. As an energy check, the time-averaged <acoustic intensity> is $|P|^2/(2Z)$, so its transmission coefficient is $\mathcal T=4\sqrt\delta/(1+\sqrt\delta)^2\leq1$, even when the transmitted <pressure> amplitude exceeds the incident one.
Equal <mean molecular weight> alone does not logically guarantee equal $\gamma$. If the gases have different values of the <adiabatic index>, replace $\sqrt\delta$ in these formulas by $\sqrt{\gamma_B\delta/\gamma_A}$; a density-only answer requires the common-$\gamma$ assumption.
Back to article page
