 Solution

Take incidence from A into B and first specify 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 givesSolving the two linear equations gives the acoustic reflection and transmission at an interface coefficientsFor 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 $Z_B/Z_A=\sqrt{{\rho }_B/{\rho }_A}=\sqrt{\delta }$, givingThe word amplitude is otherwise ambiguous. For fluid velocity, or displacement at the same frequency, the corresponding answer isThese 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 $\mathcal{T}=4\sqrt{\delta }/(1+\sqrt{\delta }{)}^2\le 1$, 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.