Question 7.1: A membrane of mass per unit area ms separates two gases but ...
A membrane of mass per unit area m_s separates two gases but it exerts no force of pressure on them (Figure 7.2). Write the boundary conditions on the membrane. Deduce the reflection coefficient and the transmission coefficient. Write down the expressions of the real waves of displacement in the gases and the expression of displacement of the membrane.
In this case, the displacement \xi(t) of the membrane is the same as that of the gases on both sides of it and the gases exert two forces of pressure on it. Thus, we have the boundary conditions
\underline{\xi } (t)= \underline{u} (t,0)+ \underline{u^\prime} (t,0)=\underline{u^{\prime \prime}} (t,0) \ \ \ \ \ \ \ m_s\underline{\ddot{\xi } } =\underline{p_a}(t,0)+\underline{p_a} ^\prime(t,0)+\underline{p_a} ^{\prime \prime}(t,0) .
Using expressions [7.12] and [7.13], these equations may be written as
\underline{u} (x,t)=\underline{u} _me^{i(ωt-kx)} \ \ \ , \ \ \ \underline{u^\prime}(x,t)=\underline{u^\prime}_me^{i(ω^\prime t+k^\prime x)} \ \ \ \, \ \ \ \ \underline{u^{\prime \prime}} (x,t)=\underline{u^{\prime \prime}} _me^{i(ω^{\prime \prime}t-k^{\prime \prime}x)}, [7.12]
\underline{p_a} (x,t)=iωZ_1\underline{u} _me^{i(ωt-kx)}, \ \ \ \ \underline{p_a} ^\prime(x,t)=-iωZ_1\underline{u^\prime} _me^{i(ω^\prime t +k^\prime x)} \\ \underline{p_a} ''(x,t)=iωZ_2\underline{u''} _me^{i(ω''t-k''x)}. [7.13]
\underline{\xi } (t)= (\underline{u}_m+\underline{u}^\prime _m )e^{iωt}=\underline{u} _m \ ^{\prime \prime} e^{iωt} \ \ \ , \ \ \ \ \ \ \ m_s\underline{\ddot{\xi } } iωZ_1(\underline{u}_m-\underline{u}_m^\prime )e^{iωt}-iωZ_2\underline{u} ^{\prime \prime}_me^{iwt}We deduce that
R_u\equiv \frac{\underline{u^\prime} _m}{\underline{u}_m }=\frac{Z_1-Z_2-iωm_s}{Z_1+Z_2+iωm_s} , \ \ ꭋ_u\equiv \frac{\underline{u^{\prime \prime}}_m }{\underline{u}_m } =\frac{2Z_1}{Z_1+Z_2+iωm_s} , \ \ \underline{\xi } (t)=\frac{2Z_1\underline{u}_m }{Z_1+Z_2+iωm_s} e^{iωt}Assuming that the incident wave is u=u_m \ e^{i(ωt-k_1x)} and taking the real parts, the reflected wave, the transmitted wave and the displacement \xi (t) may be written as
u^\prime=u^\prime_m \cos(ωt-k_1x-\phi ^–\phi ^+), \ u^{\prime \prime}= u^{\prime \prime}_m \cos(ωt-k_2x-\phi ^+) \ \text {and} \ \xi (t)=\xi _m \cos(ωt-\phi ^+).The amplitudes are u^\prime _m=u_m Z^-/Z^+ \ \text {and} \ u^{\prime \prime}_m=\xi _m=2u_mZ_1/Z^+ , where we have set Z^\pm =\sqrt{(Z_1\pm Z_2)^2+ω^2m_s^2} \ \text {and} \ \phi ^\pm = \text {Arctan}\left[ωm_s/(Z_1\pm Z_2)\right] (0\lt \phi ^\pm \lt \pi ).