Mechanical wave reflection at a boundary

[Toby_phys]

Homework Statement

Homework Equations

The right hand section (A) has an incident and reflected wave
$$y_1=Ae^{i(kx+\omega t)} +A'e^{i(-kx+\omega t)} $$

The middle section (B) has a transmission reflected wave

$$y_2=Be^{i(k_2x+\omega t)} +B'e^{i(-k_2x+\omega t)}$$

Section (C) just has the transmission wave:
$$y_3=Ce^{i(kx+\omega t)}$$

where $k=2\pi / \lambda$ and $\omega = 2\pi \upsilon$. The actual wave is the real part of the complex exponential.

We have the boundary conditions:

$$(1)y_3(0,t)=y_2(0,t) \text{ and } (2)\frac{\partial y_3(0,t)}{\partial x}=\frac{\partial y_2(0,t)}{\partial x}$$
and
$$(3) y_1(a,t)=y_2(a,t) \text{ and } (4) \frac{\partial y_1(a,t)}{\partial x}=\frac{\partial y_2(a,t)}{\partial x}$$

The Attempt at a Solution

by applying condition 1:
$$C=B+B'$$
condition 2:
$$Ck=(B-B')k_1 $$
condition 3:
$$Ae^{i(2\pi n k/k_1)}+A'e^{i(2\pi n k/k_1)}=B+B'$$
condition 4:
$$Ake^{i(2\pi n k/k_1)}-A'ke^{i(2\pi n k/k_1)}=k_1(B-B')$$We get $A'=0$ but that is the only progression I can make.

Please help, thank you x

 

[haruspex]

We get A'=0 but that is the only progression I can make.

Two of your equations have B+B' on the right hand side. Match those up.