 Solution

The wave equation on a string gives $c_j=\sqrt{T/{\rho }_j}$ and $\omega =c_j{k}_j$, henceThe tension is common across the join, and the frequency is unchanged by reflection and transmission at the stationary boundary.

Use the complex time convention $e^{i\omega t}$. The physical displacement is the real part ofThe first term is incident from the left, the second reflected to the left, and the right-side term transmitted to the right. Continuity of displacement gives $A+B=C$. At a massless join, continuity of the transverse force $T{y}_x$ gives $k_1(A-B)=k_{2}C$. Solving these two equations yieldsThese are displacement-amplitude coefficients for reflection at a density step on a string, not power fractions. Since the mean incident power is $T\omega k_1|A{|}^2/2$, the reflected and transmitted power fractions are $|r{|}^2$ and $(k_2/k_1)|t{|}^2$. They sum to one, as expected for the lossless join.