Wave behavior across two semi-infinite membranes with a special boundary

[Miles123K]

Homework Statement

This is a problem from the Physics of Waves by Howard Georgi. Consider a boundary between two semi-infinite membranes stretched in the x-y plane connected at $x=0$. The membrane in the area $x<0$ has surface tension $T_s$ and surface density $\rho_s$ and the membrane in $x>0$ has surface tension $T_s$ and surface density $\rho_s '$. There is some device that cause friction on the boundary $x=0$ such that the force on a small chunk of the boundary stretching from the point $(0,y)$ to $(0,y + dy)$ is $dF = - dy \gamma \frac{\partial \psi}{\partial t} (0,y,t)$. For $x<0$, there is a wave of the form:
$\psi(x,y,t) = A e^{i(k \cos(\theta)x + k \sin(\theta)y - \omega t)}$
For $x>0$, there is a wave of the form:
$\psi(x,y,t) = A e^{i(k' \cos(\theta')x + k' \sin(\theta ') y- \omega t)}$
The membrane have dispersion relations $\omega^2 =\frac{T_s}{\rho_s} k^2$ and $\psi$ denotes the displacement in the z-direction. Find $k', \theta', \gamma$. The hint says $\gamma$ should approach zero as $\rho'$ approach $\rho$, and I'm also asked to explain why.

Relevant Equations

$\frac{\partial^2 \psi}{\partial t^2} = v^2 \frac{\partial^2 \psi}{\partial x^2}$

Since the membrane doesn't break, the wave is continuous at $x=0$ such that
$\psi_{-}(0,y,t) = \psi_{+}(0,y,t)$
$A e^{i(k \cos(\theta)x + k \sin(\theta)y - \omega t)} = A e^{i(k' \sin(\theta ') y- \omega t)}$
Which is only true when $k' \sin(\theta ') = k \sin(\theta)$.
From the dispersion relations and the provided $\omega$, we may find $k'$ as
$k' = \omega \sqrt{\frac{\rho_s '}{T_s}}$
Thus, we may also find $\theta '$ as:
$\sin(\theta ' ) = \frac{k}{k'} \sin(\theta) = \sqrt{\frac{\rho_s}{\rho_s '}} \sin(\theta)$
I think up until now things should be correct(?), but here comes problem number 1.
On that small chunk of boundary from $(0,y)$ to $(0,y + dy)$, I wrote the force as the following. Is this correct?
$dF = - dy \gamma \frac{\partial \psi}{\partial t} = T_s( \frac{\partial \psi_{+}}{\partial x} - \frac{\partial \psi_{-}}{\partial x} ) dy$
In the previous text, it is mentioned that if the boundary is a massive string with linear density $\rho_L$ and $T_L$, the force will take the form
$dF = \rho_L dy \frac{\partial^2 \psi}{\partial t^2} = T_s( \frac{\partial \psi_{+}}{\partial x} - \frac{\partial \psi_{-}}{\partial x} ) dy + T_L dy \frac{\partial^2 \psi}{\partial y^2}$
While we treat the boundary as being connected by something massless, implying $\rho_L = 0$, do we also treat it as something without linear tension? Does the surface tension of the membrane not contribute to the force of this small chunk because the x-width of the considered "mass" is zero?
Now moving on with the assumption that things haven't gone wrong up until now (I suspect that the above equation above is correct, or else I don't see how I could achieve the results of the hint). I just solved the above equations to get:
$\gamma \omega = T_s ( k' \cos(\theta ') - k \cos(\theta) )$
Now here is question number 2. Is this correct? It seems to fit the conditions in the hint, and the units work out, but I want to be sure. Anyway, now assuming everything is done correctly, I still have the question of why $\gamma$ would vanish as $\rho_s '$ approach $\rho$. From what I understood, $\gamma$ is a given associated with that friction machine, but somehow it became a variable here. What happened? (There's actually a note on the textbook from the author himself saying he doesn't know what's going on with the "device," are there just no explanations at all?)

 

[member 553083]

Thank you in advance!A:The idea is that if the surface tension goes to zero, then the wave can't be "reflected" off of the membrane. The energy of the wave just passes through it, and this causes the frequency of the wave to change.The force equation you wrote is correct, since we are considering a massless string. The surface tension does not contribute to the force because the width of the "mass" is zero.The equation you wrote for $\gamma\omega$ is correct, and the reason why $\gamma$ vanishes as $\rho'_s \rightarrow \rho_s$ is because as the surface tension goes to zero, the wave is not reflected off of the membrane. This means the wave will not experience any additional friction, so $\gamma = 0$.