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For a right-moving sinusoidal string displacement $y=A\cos (kx-\omega t)$, the instantaneous energy flux of a transverse string wave is $P(x,t)=-T{y}_x{y}_t=T{A}^{2}k\omega {\sin }^2(kx-\omega t)$. Averaging over one period gives $⟨{\sin }^2⟩=1/2$. Since $c=\omega /k$ and $T=\rho c^2$,The displayed power is a cycle average, not the time-dependent instantaneous power. Here $\rho$ denotes linear mass density, rather than the volume density used in the surface-wave calculation.

At the same tension, $k\propto \sqrt{\rho }$, so $k_2/k_1=3$, $t=1/2$ and $r=-1/2$. The transmitted power fraction is not $t^2$, because the two strings have different wave impedances. Using the energy flux of a transverse string wave,The reflected fraction is $\mathcal{R}=|r{|}^2=1/4$, so $\mathcal{R}+\mathcal{T}=1$. The transmitted displacement amplitude is $2.5\mathrm{mm}$ and its wavelength is $0.5/3\mathrm{m}$. The negative $r$ indicates a reflected phase reversal; it does not mean negative reflected energy.

Work in the small-slope model of a transverse wave on a taut string. Let $T$ be the constant tension, $\rho$ the linear mass density, and take a right-moving sinusoidal displacement $y=A\cos (\kappa x-\omega t)$, with wave number $\kappa =\omega /v$. The energy flux of a transverse string wave through a cut is the transverse force times transverse velocity:A cycle average gives $⟨{\sin }^2⟩=1/2$. Using $T=\rho v^2$, obtained from the string wave equation, yieldsThe sign in the flux formula makes the power positive for propagation in the positive $x$ direction.

The energy per unit equilibrium length is the sum of the kinetic and elastic terms:For a segment of mass $M=\rho \ell$ containing many wavelengths, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq \frac{1}{2}M{\omega }^2{A}^2$. Convert the given quantities to SI units: $M=0.080\mathrm{kg}$, $A=1.6\times 10^{-4}\mathrm{m}$, and $\omega =2\pi (120){\mathrm{s}}^{-1}$. ThenThe many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of wavelengths.

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.

For a flexible uniform string with linear mass density $\rho$ and tension $T$, dimensional analysis gives $[T/\rho ]=L^2{T}^{-2}$ and hence $c\propto \sqrt{T/\rho }$. The small-slope force balance $\rho {\psi }_{tt}=T{\psi }_{xx}$ fixes the factor to one:Here $\rho$ has units kg/m, not the volume-density units used for a bulk fluid. The phase moves along the string at $c$, while a material point moves transversely with ${\psi }_t=A\omega \sin (kx-\omega t)$. Its maximum speed is $A\omega$, which is not the phase velocity. Small slope $Ak\ll 1$ makes $A\omega \ll c$ in this model.

The kinetic energy and extra elastic energy of a segment areThe second expression follows from extra length $\sqrt{1+{\psi }_x^2}dx-dx\simeq {\psi }_x^{2}dx/2$ against the pre-existing constant tension. Time averaging a harmonic wave gives $⟨{\mathcal{E}}_K⟩=\rho A^2{\omega }^2/4$ and $⟨{\mathcal{E}}_U⟩=T{A}^2{k}^2/4=\rho A^2{\omega }^2/4$. Therefore

At the driven endpoint the string pulls along its local tangent into $x>0$, with vertical component $T{\psi }_x$ for small slope. The point exerts the opposite force on the string, so the driven-string endpoint power isFor the outgoing wave $\psi =A\cos (kx-\omega t)$, at $x=0$ the derivatives are ${\psi }_x=Ak\sin \omega t$ and ${\psi }_t=-A\omega \sin \omega t$. Thus $P=T{A}^{2}k\omega {\sin }^2\omega t$ andThe energy flux of a transverse string wave follows by combining the energy density $\mathcal{E}=(\rho {\psi }_t^2+T{\psi }_x^2)/2$ and flux $P=-T{\psi }_x{\psi }_t$, which satisfy ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a pure right-going nondispersive wave, $P=c\mathcal{E}$ even instantaneously. energy is transported with the wave; a material string element merely oscillates about its unperturbed position.