Reflection at a density step on a string

For a massless join between two ideal strings with a common tension $T$, require continuous transverse displacement and continuous transverse force $T{\partial }_x\Psi$. With $Z_j=\sqrt{T{\rho }_j}$ for linear mass densities ${\rho }_j$, the displacement reflection coefficient and transmission amplitude areThe reflected and transmitted mean-power fractions are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(Z_2/Z_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. Reflection from the heavier side reverses phase; reflection from the lighter side does not. Total reflection occurs in the limits $Z_2/Z_1\to \infty$ or zero, while a finite positive impedance mismatch transmits some energy. Equal mechanical impedances remove reflection.

Two semi-infinite strings with common tension $T$ and a point mass $M$ at their join have continuous transverse displacement, $A+B=C$. With time factor $e^{i\omega t}$ and a right-moving incident factor $e^{-i{k}_{1}x}$, Newton's second law at the join gives $k_{1}T(A-B)=(k_{2}T+iM{\omega }^2)C$. Solving gives the displayed reflection coefficient and $t=C/A=2k_{1}T/[T(k_1+k_2)+iM{\omega }^2]$. Their power fractions obey $|r{|}^2+(k_2/k_1)|t{|}^2=1$. Reversing the time convention complex-conjugates the coefficients. The limits $M=0$ and $M\to \infty$ recover a massless join and a fixed boundary, respectively.