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For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.

Let $T$ be the common tension and ${\mu }_j$ the linear mass densities. The characteristic impedance of a stretched string is $Z_j=T/c_j=\sqrt{T{\mu }_j}$. Continuity of displacement and transverse force at a massless junction gives $1+r=t$ and $Z_1(1-r)=Z_{2}t$. ConsequentlyThe reflected power is $7.18\%$. The minus sign of $r$ represents a phase reversal, not negative power. Both incident and reflected waves travel in the first string, so their power ratio is simply the squared displacement-amplitude ratio. As a check, $\mathcal{T}=(Z_2/Z_1)|t{|}^2=0.92820$ and $\mathcal{R}+\mathcal{T}=1$.

The characteristic impedance of a stretched string is $Z=T/c=\rho c$, since its wave speed obeys $c^2=T/\rho$. Put the endpoint at $x=0$ and use displacement complex amplitudes $A$ incident and $B$ reflected:At the massless end, the string's transverse force $-T{\Psi }_x$ balances the damper's force $-\alpha {\Psi }_t$. Thus $T{\Psi }_x=-\alpha {\Psi }_t$, or $Z(A-B)=\alpha (A+B)$. This viscous termination of a stretched string hasHalf the incident mean power is reflected when $r=\pm 1/\sqrt{2}$, giving two positive solutions:They are approximately $0.1716\rho c$ and $5.828\rho c$, in $\mathrm{kg}{\mathrm{s}}^{-1}$. The smaller coefficient produces no reflected displacement-phase reversal; the larger produces a $\pi$ reversal. Both dissipate the remaining half of the incident power. Choosing the impedance-matched value $\alpha =\rho c$ would instead absorb all of it.

A characteristic wave impedance relates an effort-like quantity to its conjugate flow-like quantity in a single travelling wave: for a mechanical wave these are force and velocity, and their product is power. The sign depends on the propagation and force conventions. Use the transverse force transmitted towards increasing $x$, $F=-T{\Psi }_x$. A rightgoing travelling wave $\Psi =f(x-vt)$ has transverse velocity $u={\Psi }_t=-v{\Psi }_x$, soThis is the positive characteristic impedance of a stretched string. A leftgoing component has $F/u=-Z$ with the same force orientation. Its units differ from pressure-based acoustic impedance because it uses force, rather than force per area.

Assume a massless junction and common tension $T$. The characteristic impedance of a stretched string is $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ is its linear mass density. The displacement reflection coefficient is $r=(Z_1-Z_2)/(Z_1+Z_2)$, so the reflected fraction of power isSet $\mathcal{R}=1/2$. The positive solutions are $z=3\pm 2\sqrt{2}$, which are reciprocals. Thus the string density ratio for prescribed reflected power isThe impedance ratio is about $5.83$, rather than the required density ratio. Reversing which string is on the incident side changes the reflected displacement sign but not the reflected power fraction.

Use transverse displacement $y(x,t)$, small slopes and constant tension $T$. For a short string element of length $dx$, Newton's second law gives ${\rho }_{L}dx{y}_{tt}=T[y_x(x+dx)-y_x(x)]$. Hence the wave equation on a string iswhere ${\rho }_L$ is the linear mass density, in $\mathrm{kg}{\mathrm{m}}^{-1}$.

The mechanical impedance of a rightgoing travelling wave is the transverse driving force divided by the transverse velocity at the driven section. Choose the force exerted by the material to the left on the material to the right, $F_y=-T{y}_x$. For $y=f(x-c_{L}t)$, the transverse velocity is $v_y=y_t=-c_L{y}_x$. Thus the positive characteristic impedance of a stretched string isA leftgoing component has $F_y/v_y=-Z_L$ with this fixed orientation of force. Specifying the orientation avoids confusing that propagation sign with a negative material mechanical impedance.

A lossless intermediate section of positive real characteristic impedance $Z_M$ matches unequal real exterior characteristic impedances $Z_L,Z_R$ when $Z_M^2=Z_L{Z}_R$ and its propagation wave phase is an odd multiple of $\pi /2$. The length is an odd quarter of the internal wavelength. This principle also applies to the characteristic impedance of a stretched string.

For a uniform intermediate string of length $l$, put $\delta =k_{M}l$. Continuity of displacement and transverse force gives the displacement reflection coefficientAt real positive mechanical impedances, the denominator never vanishes. Its numerator vanishes only if $(Z_L-Z_R)\cos \delta =0$ and $(Z_M^2-Z_L{Z}_R)\sin \delta =0$. Thus unequal exterior strings require the quarter-wave condition. Equal exterior strings with a different middle string instead transmit perfectly at integer half-wavelengths; if all three mechanical impedances agree, every frequency is matched.

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.