Boundary conditions for 2 ropes fixed to a massless ring with a damper

[Redwaves]

Homework Statement

The boundary conditions for 2 ropes fixed to a massless ring with a massless damper
A wave comes from the left (rope 1) to the right (rope 2).

Relevant Equations

$\psi_1 = Ae^{i(\omega t -k_1x)} + Re^{i(\omega t + k_1x)}$
$\psi_2 = Te^{i(\omega t -k_2x)}$

Hi,
I'm not quite sure if I'm correct. I need to find the boundary conditions for 2 ropes $T_1 \mu_1, T_2 \mu_2$ fixed at $x=0$ to a massless ring with a massless damper of force $F_d - -bv_y$

Here what I think, since the ring and the damper is massless $\sum F_y = 0$. Thus, $-T_1 \frac{\partial \psi_1}{\partial x} + T_2 \frac{\partial \psi_2}{\partial x} -b \frac{dy}{dt} = 0$

Then, the height must be the same for the first rope, the ring and the second rope at $x=0$, thus $y_d = \psi_1(0,t) = \psi_2(0,t)$
Which mean, A + R = T, since $Ae^{i(\omega t -k_1x)} + Re^{i(\omega t + k_1x)} = Te^{i(\omega t -k_2x)}$ at x=0

 

[berkeman]

Is there a diagram that goes with this problem? I'm having trouble getting all of the terms correct...

 

[Redwaves]

I did a diagram. Is it clear enough?

 

[berkeman]

I did a diagram. Is it clear enough?

Bwaaa. So the two ropes are co-linear, and there is a ring in the middle connecting them with some sort of damper? Is there really no diagram with the problem that you've been given?

 

[Redwaves]

It's exactly like that. I just drew it myself, but it's basically the same. Correct me if I'm wrong, but to find the boundary conditions, I don't think I need this diagram. Since, there are only 3 forces in the y direction and I know that the damper and the ring is massless.

The goal is the find that $b = Z_1 - Z_2$ to have $R = 0$. However, I found $R = -Z_1 + Z_2$. I'm pretty sure my boundary conditions is not correct. $Z_i$ are the impedance.

 

[Redwaves]

$-T_1 \frac{\partial \psi_1}{\partial x} + T_2 \frac{\partial \psi_2}{\partial x} -b \frac{dy}{dt} = 0$
$T = A + R$

With those conditions I get $R = \frac{Z_1A - Z_2B - bA}{Z_1 + Z_2 +b}$
if $b = Z_1 - Z_2$, $R = 0$
This is the correct answer. But are my boundary conditions correct?

 

[TSny]

$-T_1 \frac{\partial \psi_1}{\partial x} + T_2 \frac{\partial \psi_2}{\partial x} -b \frac{dy}{dt} = 0$
$T = A + R$

With those conditions I get $R = \frac{Z_1A - Z_2B - bA}{Z_1 + Z_2 +b}$
if $b = Z_1 - Z_2$, $R = 0$
This is the correct answer. But are my boundary conditions correct?

This looks good to me, except I think the B in the numerator of R should be A.

 

[Redwaves]

If B in the numerator of R is A, I didn't get 0 if $b = Z_1 - Z_2$

 

[TSny]

If B in the numerator of R is A, I didn't get 0 if $b = Z_1 - Z_2$

Can you show the individual steps of how you get $R = \frac{Z_1A - Z_2B - bA}{Z_1 + Z_2 +b}$ to reduce to $R = 0$ when $b = Z_1 - Z_2$?