Wave equation on a string ($y_{tt}=(T/\mu )y_{xx},c=\sqrt{T/\mu }$)

An ideal uniform, flexible string under constant tension $T>0$ has transverse displacement $y(x,t)$ satisfyingwhere $\mu >0$ is linear mass density, meaning mass per unit length of the straight reference string. The approximation assumes small slope $|y_x|\ll 1$, transverse motion, negligible bending stiffness, negligible damping, and no distributed external force. Constant tension and uniform linear mass density make $c$ constant. Some texts use $\rho$ for this same linear mass density, giving $c=\sqrt{T/\rho }$; here $\rho$ must have units $\mathrm{kg}{\mathrm{m}}^{-1}$, rather than units of volume mass density. ^[ref]

The units provide a useful check: $T/\mu$ has units $(\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2})/(\mathrm{kg}{\mathrm{m}}^{-1})={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For $T=100\mathrm{N}$ and $\mu =0.010\mathrm{kg}{\mathrm{m}}^{-1}$, the wave speed is $100\mathrm{m}{\mathrm{s}}^{-1}$, so a pulse crosses a one-metre string in $0.010\mathrm{s}$. Quadrupling tension doubles wave speed; quadrupling linear mass density halves it. These square-root laws explain how material and loading set the time scale of the vibration.

For a length-$L$ string with fixed endpoints, the homogeneous Dirichlet boundary conditions are $y(0,t)=y(L,t)=0$. Its nonzero normal modes haveThus $cn\pi /L$ is an angular frequency, in radians per second, whereas $cn/(2L)$ is the frequency in hertz. The wavelength is $2L/n$, and each node of a standing wave lies at $x=jL/n$ for some $j=0,\dots ,n$. For the numerical string above with $L=1\mathrm{m}$, the first three frequencies are $50$, $100$ and $150\mathrm{Hz}$, with angular frequencies $100\pi$, $200\pi$ and $300\pi {\mathrm{s}}^{-1}$. ^[ref]

Each normal mode is a standing wave, but it also has a travelling wave description:At a fixed endpoint an incident displacement pulse reflects with reversed sign, allowing incident and reflected displacements to cancel at the endpoint. Periodic reflection for a fixed-end string implements both endpoints through odd, $2L$-periodic extensions of the initial data. This is the bridge between moving pulses and the stationary Fourier sine basis. ^[ref]

A short derivation comes from Newton's second law. The transverse force on a small segment is approximately $T[y_x(x+\delta x,t)-y_x(x,t)]$, and its mass is $\mu \delta x$. Dividing the force balance by $\delta x$ and taking the limit gives $\mu y_{tt}=T{y}_{xx}$. Separation of variables then gives ${\varphi }^{\prime \prime }+k^2\varphi =0$; the fixed-end boundary conditions select $k=n\pi /L$, while the temporal equation is a harmonic oscillator equation with $\omega =ck$.

The second-order central difference approximation to the fixed-end wave equation on $n$ interior points has normal modesHere $h=L/(n+1)$ and the endpoint displacements are zero. The eigenvectors form an orthonormal basis in the ordinary Euclidean norm, without an extra grid-spacing weight. They sample the continuous standing wave shapes exactly, while their angular frequencies differ from $k\pi c/L$ because of spatial discretization. For fixed $k$, the difference tends to zero as $h\to 0$.

For a uniform ideal fixed-end string, a real normal mode of displacement amplitude $A_n$ has total mechanical energy $T{\pi }^2{n}^2{A}_n^2/(4L)$. This follows by integrating the squared sine velocity and cosine slope over the length; each squared trigonometric function integrates to $L/2$. Thus finitely many excited modes with $|A_n|$ proportional to $1/n$ have equal total modal energies. Their instantaneous kinetic energies need not agree, because their time factors oscillate at different frequencies.

At a lossless string junction with displacement reflection coefficient $r$ from the incident side, reverse incidence has coefficient $-r$ and the product of transmission coefficients is $1-r^2$. Reflection from a fixed far end contributes $-1$. Summing all round trips with phase $q=e^{2ikL}$ gives $R=r-(1-r^2)q/(1-rq)=(r-q)/(1-rq)$. For real $r$ with $|r|<1$ and $|q|=1$, numerator and denominator have equal absolute value, so $|R|=1$: the finite segment returns all incoming energy while adding a frequency-dependent phase. Short, nonoverlapping pulses can instead be counted as a geometric series of returning pulse energies; overlapping echoes must be added as amplitudes before their energy is evaluated.

For zero initial velocity and initial displacement $h(x)$ on $x\ge 0$ with $h(0)=0$, let $H$ be its odd extension to the real line. The displayed D'Alembert's formula solves the wave equation with $y(0,t)=0$, since $H(-ct)+H(ct)=0$. A pulse travelling toward the endpoint reflects with reversed displacement. For piecewise linear data the equation holds classically between moving corners and in the sense of weak derivatives globally.

For a small-slope travelling wave $y=A\cos (kx-\omega t+\varphi )$ on a uniform string with linear mass density $\mu$ and tension $T$, the wave equation on a string requires ${\omega }^2=T{k}^2/\mu$. Its elastic energy of a small-slope string and kinetic energy density are equal, giving total energy density $u=\mu A^2{\omega }^2{\sin }^2(kx-\omega t+\varphi )$. Its spatial average is $\mu A^2{\omega }^2/2$, and integrating over a full wavelength $\lambda =2\pi /|k|$ gives the title formula. Omitting $\lambda$ produces an energy density, not an energy; units and a direct one-period integral detect the mistake independently.

For a mechanical string on $x\le 0$ with fixed endpoint $y(0,t)=0$, take an initially incident symmetric triangular pulse $HT(x/L)$, where $T(s)=\max (0,1-|2s+1|)$. The superposition principle gives the title formula. Both terms satisfy the wave equation away from the corners and satisfy it weakly everywhere; their values cancel at the endpoint. At $ct=L/2$ their displacements cancel on the whole half-line, but the velocity is nonzero on $-L/2<x<0$. Hence a flat instantaneous shape need not mean zero wave energy.

For a uniform fixed-end mechanical string of length $L$, linear mass density $\mu$ and wave speed $c$, let its $n$th normal mode be $y_n=[A_n\cos {\omega }_{n}t+B_n\sin {\omega }_{n}t]\sin (n\pi x/L)$, where ${\omega }_n=cn\pi /L$. The conserved energy is the sum of kinetic energy and elastic potential energy,The sine and cosine squared integrals are $L/2$, so substitution proves the title formula. Orthogonality makes energies of distinct normal modes add. For zero initial displacement and initial modal velocity $V_n$, $A_n=0$ and $B_n=V_n/{\omega }_n$, giving $E_n=\mu L{V}_n^2/4$.

For a uniform fixed-end mechanical string of length $l$ and tension $T$, a single sine normal mode of amplitude $A_n$ has conserved energy $E_n=n^2{\pi }^{2}T{A}_n^2/(4l)$. Hence amplitudes proportional to $1/n$ give equal energies, even though their displacement magnitudes differ. In particular $A_2=A_1/2$ makes the first two modal energies equal. Orthogonality removes cross terms in the integrated kinetic energy and potential energy, so the modal energies add.

A uniform string with fixed ends $x=\pm \ell$, tension $T$, linear mass density $\rho$, and attached central point mass $m>0$ obeys the wave equation away from zero, continuous displacement at zero, and $m{y}_{tt}(0,t)=T[y_x(0^{+},t)-y_x(0^{-},t)]$. The normal modes split into antisymmetric modes that leave the mass at rest and symmetric modes that move it.

To prove the spectral equations, use $X=A\sin (k(x+\ell ))$ on the left and $X=B\sin (k(\ell -x))$ on the right, with ${\omega }^2=(T/\rho )k^2$. Continuity gives $(A-B)\sin (k\ell )=0$ and the force law gives $Tk(A+B)\cos (k\ell )=m{\omega }^{2}A\sin (k\ell )$. If $\sin (k\ell )=0$, then $k\ell =n\pi$, $n\ge 1$, and $B=-A$. Otherwise $A=B$ and cancellation gives $\cot (k\ell )=mk/(2\rho )$. In dimensionless variables $z=k\ell$, $\alpha =m/(2\rho \ell )$, strict monotonicity gives one symmetric root in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$. The apparent zero root of the sine equation has only the zero profile and is excluded.

For $m/(\rho \ell )\gg 1$, the lowest root tends to zero. The Taylor series $\cot z=1/z-z/3+O(z^3)$ gives $z\sim {\alpha }^{-1/2}$ and hence $\omega \sim \sqrt{2T/(m\ell )}$. The two nearly straight half-strings supply a restoring force $-2TY/\ell$, explaining this harmonic oscillator limit.

For the wave equation on a string, prescribing $y(0,t)=f(t)$ and $y_x(0,t)=0$ selects equal incoming and outgoing profiles. Write D'Alembert's formula as $P(t-x/c)+Q(t+x/c)$. The two traces imply $P+Q=f$ and $P^{\prime }=Q^{\prime }$, proving the displayed expression. These data do not generally describe a disturbance launched from an everywhere resting string. For nonsmooth $f$, derivatives and boundary traces need their piecewise classical or weak derivative interpretation; an endpoint cusp may prevent a classical derivative at an isolated time.

For two-sided string data with zero slope, take $f(t)=\sin (\pi t/a)$ on $[-a,a]$ and zero outside. At $x=ca/2$, the displacement vanishes throughout $|t|<a/2$, whereas the spatial derivative is $y_x=-\pi \sin (\pi t/a)/(ca)$. Cancellation along a line in spacetime need not cancel the derivative transverse to that line. At $x=ca,t=0$, however, oddness gives $y(x,0)=0$ for every $x$, so the spatial derivative is zero even though its one-sided temporal limits there differ.

At a free termination of a taut string, zero transverse force imposes a Neumann boundary condition. Incident and reflected displacement amplitudes add with equal sign so their slopes cancel. The endpoint is a displacement antinode, all incident power is reflected, and the flux $-T{\psi }_x{\psi }_t$ through the boundary vanishes. This is a transverse-wave boundary condition, distinct from the quantum open-string usage of similar notation.

The small-slope energy density is $\mathcal{E}=(\rho y_t^2+T{y}_x^2)/2$ and power flux is $P=-T{y}_x{y}_t$. The wave equation gives ${\partial }_t\mathcal{E}+{\partial }_{x}P=0$. For a right-going wave, $P=c\mathcal{E}$; a harmonic wave has mean flux $\rho c{A}^2{\omega }^2/2$.

For a string occupying $x>0$, its small-slope force on the endpoint is $T{y}_x$. The driver force on the string has the opposite sign, so injected power is $-T{y}_x{y}_t$. A right-going traveling wave has nonnegative outgoing instantaneous power in this ideal model.

For a uniform string with tension $T>0$, linear mass density $\rho >0$, left end fixed, and a freely transverse-moving point mass $M\ge 0$ at $x=L$, the wave equation and endpoint law areThe last equation is Newton's second law: the string pulls the mass back toward the interior, giving the minus sign. With wave speed $c=\sqrt{T/\rho }$, a nonzero normal mode has spatial part $X(x)=\sin (\omega x/c)$ and satisfiesFor $M>0$, its dimensionless roots $q=\omega L/c$ occur once in each $(n\pi ,n\pi +\pi /2)$, $n\ge 0$, because $\cot q$ decreases from infinity to zero while $Mq/(\rho L)$ increases. For $M=0$, $q=(n+1/2)\pi$: the right end has a homogeneous Neumann boundary condition. There is no nonzero zero-frequency mode, since a linear spatial profile vanishing at the left end and with zero right slope must vanish everywhere.

Multiplying the spatial equation by $X$ and using integration by parts givesFor two different modes, the same calculation gives the weighted inner product orthogonality $\rho {\int }_0^L{X}_m{X}_{n}dx+M{X}_m(L)X_n(L)=0$. The endpoint mass therefore contributes to both inertia and mode orthogonality.

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 $3+2\sqrt{2}$.

For $0<\xi <l$, the wave equation $y_{tt}=c^2{y}_{xx}$ with zero endpoint displacement, zero initial displacement and initial velocity $\delta (x-\xi )$ has the distributional weak solutionExpand the Dirac delta in the Fourier sine basis and solve each modal ordinary differential equation. Equivalently take the odd, $2l$-periodic extension $v$ of the initial velocity and use $y=(2c{)}^{-1}{\int }_{x-ct}^{x+ct}v(s)ds$. The image impulses enforce the fixed endpoints. At jump fronts the Fourier series selects the midpoint of the two one-sided values.

A fixed-end string of length $L$ initially straight receives velocity $v$ on an interval of width $a$ centred at $l$, and zero elsewhere. Its sine-mode velocity coefficient is $g_n=4v\sin (n\pi l/L)\sin (n\pi a/(2L))/(n\pi )$. For linear density $\rho$, orthogonality gives mode energy $E_n=\rho L{g}_n^2/4$ and total energy $E=\rho a{v}^2/2$. Their ratio is the displayed formula. Zeros in either sine factor suppress a normal mode completely.

Zero values at both endpoints of a length-$L$ string allow the D'Alembert formula to be written with opposite travelling waves and a $2L$-periodic profile. This represents reflection with sign reversal at each fixed endpoint. For positive-time boundary conditions, periodicity concerns the arguments visited by the string solution; unused profile values can be defined by periodic extension.

With fixed endpoints, $y_{tt}-y_{xx}+y_t=0$ separates into sine modes whose amplitudes satisfy

For $y_{tt}+2k{y}_t=c^2{y}_{xx}$ with fixed endpoints, initial displacement zero and initial velocity a Dirac delta at $a\in (0,l)$, the Fourier sine series coefficient of that velocity is $2\sin (n\pi a/l)/l$. Each coefficient evolves by $T_n^{\prime \prime }+2k{T}_n^{\prime }+(n\pi c/l{)}^2{T}_n=0$. Here ${\omega }_n^2=(n\pi c/l{)}^2-k^2$. At critical damping, replace $\sin ({\omega }_{n}t)/{\omega }_n$ by $t$; for an overdamped mode replace it by $\sinh (\sqrt{k^2-(n\pi c/l{)}^2}t)/\sqrt{k^2-(n\pi c/l{)}^2}$. The initial conditions hold in the sense of distributions.

For initial sine coefficient $a_n$ and zero initial velocity, put ${\omega }_n^2=n^2{\pi }^2-1/4$. The modal amplitude is

For fixed endpoints andintegration by parts and the damped wave equation give