Characteristic impedance of a stretched string ($Z=\sqrt{T\mu }$)

For the lossless small-slope wave equation on a string, tension $T$ and linear mass density $\mu$ give wave speed $c=\sqrt{T/\mu }$. With transverse force $F_y=-T{y}_x$ exerted across a cut toward increasing $x$, a rightgoing displacement $y=f(x-ct)$ has velocity $v_y=-c{y}_x$. Its positive mechanical impedance is therefore $Z=F_y/v_y=T/c=\sqrt{T\mu }$. A leftgoing component has $F_y/v_y=-Z$ under the same fixed orientation. A rightgoing time-harmonic wave with peak displacement wave amplitude $|A|$ carries mean power $Z{\omega }^2|A{|}^2/2$.

For a semi-infinite ideal string on $x\le 0$ with tension $T$ and linear mass density $\mu$, let a free transverse endpoint mass be $m$. Use a time-harmonic wave $y=\mathrm{Re}\{A(e^{ikx}+R{e}^{-ikx})e^{-i\omega t}\}$, where $k=\omega \sqrt{\mu /T}>0$. The endpoint boundary condition from Newton's second law is $m{y}_{tt}=-T{y}_x$. It gives the title displacement reflection coefficient, with characteristic impedance of a stretched string $Z=\sqrt{T\mu }$. Since numerator and denominator have equal moduli, all mean incident power is reflected. The endpoint kinetic energy varies within a cycle, but its mean rate of change is zero. A free mass changes the reflected phase, and does not by itself provide dissipation. The limits $m\to 0$ and $m\to \infty$ give $R=+1$ and $R=-1$ respectively at fixed positive angular frequency.

A static lossless taper preserves frequency. If characteristic impedance varies slowly on the local wavelength scale, reflection is small and mean power $Z{\omega }^2{A}^2/2$ is conserved. At fixed tension this gives the displayed amplitude law, while wave speed and wavelength scale as ${\rho }^{-1/2}$. A long section containing an abrupt step is not automatically adiabatic.

A semi-infinite wave equation on a string with linear mass density $\rho$, wave speed $c$, and tension $T=\rho c^2$ has positive characteristic mechanical impedance $Z=\rho c$. At an end attached to a massless viscous damper of coefficient $\alpha >0$, displacement and transverse-force balance give the displacement reflection coefficientHere $\mathcal{A}$ is the fraction of mean incident power dissipated. A prescribed $0<\mathcal{R}<1$ gives two positive damper coefficients, $\alpha /Z=(1\pm \sqrt{\mathcal{R}})/(1\mp \sqrt{\mathcal{R}})$, with opposite reflected phases. Matching $\alpha =Z$ removes reflection.

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.

For two ideal strings under a common tension, the positive mechanical impedances satisfy $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ are their linear mass densities. The reflected fraction of incident power is $\mathcal{R}=[(Z_1-Z_2)/(Z_1+Z_2){]}^2$. Solving for the positive ratio $z=Z_2/Z_1$ gives $z=(1+\sqrt{\mathcal{R}})/(1-\sqrt{\mathcal{R}})$ or its reciprocal. Squaring yields the displayed linear mass density ratios. For $\mathcal{R}=1/2$, the heavier string has $17+12\sqrt{2}\simeq 33.97$ times the lighter string's linear mass density; the impedance ratio is only $3+2\sqrt{2}$.