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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 $Z{\omega }^2|A{|}^2/2$.

An eigenfrequency is the natural frequency associated with a free normal mode. Always specify whether it means the angular frequency $\omega$ or the frequency $f=\omega /(2\pi )$ in hertz. For a uniform fixed-end wave equation on a string, ${\omega }_n=n\pi \sqrt{\tau /\mu }/L$ and $f_n=n\sqrt{\tau /\mu }/(2L)$; the amplitude does not change either within the linear model. In a generalized eigenvalue problem for small oscillations, the eigenvalue is ${\omega }^2$, rather than $\omega$ itself.

For a first-order system $\dot{z}=iAz$, a real eigenvalue $\lambda$ of $A$ instead gives a signed eigenfrequency in $e^{i\lambda t}$. A positive imaginary part of $\lambda$ gives exponential decay with this sign convention. Thus a complex damped eigenfrequency encodes both oscillation and decay; it is not just the number of cycles per second.

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 the ideal uniform wave equation on a string, the wave speed is $c=\sqrt{\tau /\mu }$, where $\tau$ is constant tension and $\mu$ is linear mass density. Assume small transverse slopes, negligible bending stiffness and no damping. Fix both ends at zero displacement, giving the homogeneous Dirichlet boundary conditions $y(0,t)=y(L,t)=0$. Our standing wave already satisfies the first. For a nonzero amplitude, the second requires $\sin (kL)=0$, henceThe value $n=0$ gives zero displacement in this sine family, not a vibrating normal mode. Each positive $n$ fits $n$ half-wavelengths into the string: it has $n$ antinodes, $n-1$ interior nodes of a standing wave, and two endpoint nodes of a standing wave. For $L=1\mathrm{m}$ and $c=20\mathrm{m}{\mathrm{s}}^{-1}$, $f_n=10n\mathrm{Hz}$; the third normal mode has $f_3=30\mathrm{Hz}$ and wavelength $2/3\mathrm{m}$.

The equally spaced eigenfrequencies follow from this string's uniform wave speed and fixed-end boundary conditions. They are not a universal spectrum for every vibrating object. Changing an endpoint constraint changes the allowed mode shapes; changing tension changes wave speed. At fixed $L$ and $\mu$, quadrupling $\tau$ doubles every $f_n$. The more general construction of arbitrary initial motion belongs to normal mode and wave equation on a string.

A slender flexible body modeled by a moving curve that transmits tension while its bending stiffness is neglected. For a uniform stretched string with constant tension $T$ and linear mass density $\mu$, small transverse oscillations obey $\mu y_{tt}=T{y}_{xx}$; see the wave equation on a string.

A node of an ideal standing wave is a position whose oscillation amplitude is zero: its displacement stays zero for all time. For a nontrivial normal mode $u(x,t)=\varphi (x)q(t)$, this means $\varphi (x)=0$. A momentary zero caused by $q(t)=0$ does not make every position a node. In more dimensions the zeros of a mode shape can form curves or surfaces.

For the $n$th normal mode of a uniform fixed-end wave equation on a string, the nodes of a standing wave areThere are $n+1$ nodes of a standing wave including the two endpoints, and $n-1$ interior ones. For $n=3$ and $L=1\mathrm{m}$, they lie at $0$, $1/3$, $2/3$, and $1\mathrm{m}$. Each remains stationary through every snapshot in the normal mode diagram; an antinode instead marks a maximum oscillation amplitude. ^[ref]

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

For one sine mode, $y_{xx}=-k_n^2{A}_n\sin (k_{n}x)\cos ({\omega }_{n}t)$ and $y_{tt}=-{\omega }_n^2{A}_n\sin (k_{n}x)\cos ({\omega }_{n}t)$. The wave equation on a string therefore holds whenThe superposition principle gives the same equation for the sum, assuming enough regularity to differentiate termwise, or interpreting finite-energy data in the usual modal sense. Fixed endpoints require $\sin (k_{n}L)=0$, so $k_n=n\pi /L$. Every cosine time factor has zero time derivative at $t=0$, giving the held-at-rest initial condition.

The coefficients must represent the actual held profile $f(x)$ through its Fourier sine series, $A_n=(2/L){\int }_0^{L}f(x)\sin (n\pi x/L)dx$ for $n\ge 1$. Thus the series accommodates the displacement initial condition once these coefficients are chosen; it does not select an unspecified initial shape by itself. The printed $n=0$ term is identically zero because $\sin 0=0$, so the nontrivial fixed-end modes start at $n=1$.

For an element $dx$, the vertical components of the approximately constant tension are $T{y}_x$ at its two ends. Their difference is $T{y}_{xx}dx$ to first order in the small slope. Its inertial response is $\rho dx{y}_{tt}$. Thus Newton's second law givesHere $\rho$ is mass per unit length. Small slopes, negligible gravity, uniform tension and negligible bending stiffness or dissipation are the assumptions in this wave equation on a string.

Interpret “held” and then “released” as zero initial velocity. In the small-slope wave equation on a string model, the elastic energy is $U=(T/2){\int }_0^L{y}_x^{2}dx$ and the initial kinetic energy is zero. The given profile hasThe two cosine functions are orthogonal on $[0,L]$ and each has squared integral $L/2$. ThusThe implicit leading displacement amplitude is one in the source's chosen length units. If that amplitude is written explicitly as $A$, all energies here and below acquire a factor $A^2$.

Substituting $y(x,t)=f(x)g(t)$ into the wave equation on a string gives $f^{\prime \prime }g=f{g}^{\prime \prime }/c^2$. Wherever both factors are nonzero,The left side depends only on $x$ and the right side only on $t$, so both equal a separation constant $\alpha$. ThereforeThe resulting differential equations extend through zeros of $f$ or $g$ by continuity; division at a node is unnecessary.

The wave equation on a string gives $c_j=\sqrt{T/{\rho }_j}$ and $\omega =c_j{k}_j$, henceThe tension is common across the join, and the frequency is unchanged by reflection and transmission at the stationary boundary.

Use the complex time convention $e^{i\omega t}$. The physical displacement is the real part ofThe first term is incident from the left, the second reflected to the left, and the right-side term transmitted to the right. Continuity of displacement gives $A+B=C$. At a massless join, continuity of the transverse force $T{y}_x$ gives $k_1(A-B)=k_{2}C$. Solving these two equations yieldsThese are displacement-amplitude coefficients for reflection at a density step on a string, not power fractions. Since the mean incident power is $T\omega k_1|A{|}^2/2$, the reflected and transmitted power fractions are $|r{|}^2$ and $(k_2/k_1)|t{|}^2$. They sum to one, as expected for the lossless join.

Put $\xi =x+vt$ and $F(\xi )=A{e}^{-{\xi }^2/x_0^2}$, with $x_0\ne 0$. The chain rule givesFor a nonzero pulse the wave equation on a string therefore requiresIt is not a solution for arbitrary $v$; the zero pulse $A=0$ is the trivial exception. Here $\mu$ is linear mass density, and $T>0$ is tension. With $v>0$, the maximum lies at $x=-vt$, so the Gaussian function translates towards decreasing $x$ by $v(t_2-t_1)$ between the two times. Its amplitude and width stay constant.

Assume $0<a<b<l$ for the interior pulse; endpoint variants use the same coefficient formula. Orthogonality of the Fourier sine basis givesThe Fourier sine series equals $f$ at its interior continuity points and takes value $1/2$ at the jumps $a,b$; the sine series is zero at both endpoints. Thus it does not retain an arbitrarily assigned value at a jump.

For the string, write $y(x,t)={\sum }_{n\ge 1}{q}_n(t)\sin (n\pi x/l)$. The wave equation on a string givesThe last equality is the Fourier sine series of the Dirac delta at $l/4$. Solving the modal ordinary differential equations givesThis impulsively struck fixed-end string is understood as a distributional weak solution; the initial velocity is a distribution, not a classical continuous function.

To evaluate the requested time, take the odd $2l$-periodic extension $v$ of $\delta (x-l/4)$ and use the D'Alembert formula $y=(2c{)}^{-1}{\int }_{x-ct}^{x+ct}v(s)ds$. At $ct=l/2$, the interval contains the positive impulse at $l/4$. For $x<l/4$, it also contains the negative reflected impulse at $-l/4$, so the contributions cancel. For $l/4<x<3l/4$ only the positive impulse contributes. For $x>3l/4$ neither contributes. Thus the midpoint-valued Fourier series givesThe discontinuity values are a convention for the Fourier representative and do not affect the distribution. The sketch shows the reflected negative front cancelling the left portion of the initially generated plateau.

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely

Let the uniform string have constant tension $\tau >0$ and linear mass density $\mu >0$. Assume small transverse motion with $|u_x|\ll 1$, negligible longitudinal motion, and no external transverse force. If the string makes angle $\beta$ with the horizontal, then $\tan \beta =u_x$, and the small-angle approximation gives $\sin \beta \simeq u_x$ and $ds\simeq dx$ to first order.

The net upward force on the segment from $x$ to $x+\Delta x$ is thereforeBy Newton's second law, its mass $\mu \Delta x$ times its transverse acceleration equals this force, to the order retained in the small displacement model. Divide by $\Delta x$ and let $\Delta x\to 0$ to obtainThus the wave equation on a string iswhere $c$ is the wave speed. The horizontal force balance makes the horizontal component of tension constant; to first order it equals $\tau$.

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.

For a short string element of length $\Delta x$, the small-slope approximation makes the upward components of tension at its two ends $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$. Newton's second law givesDivide by $\Delta x$ and take the limit to obtain the wave equation on a string,Here $\rho$ is linear mass density, and the derivation assumes small transverse slopes, uniform tension, and a flexible string.

At the right endpoint the string pulls the ring toward the interior. Its vertical force is therefore $-T{y}_x(L,t)$ to first order; the frictionless wire supplies no vertical force. The ring obeysThis is the dynamic endpoint law for a string with an endpoint mass.

For a separated oscillatory solution, write $y=X(x)q(t)$ with $q^{\prime \prime }=-{\omega }^{2}q$ and $X^{\prime \prime }+(\omega /c{)}^{2}X=0$. The fixed left end removes the spatial cosine, so the nonzero normal modes haveThe endpoint law gives $M{\omega }_n^2\sin ({\omega }_{n}L/c)=T({\omega }_n/c)\cos ({\omega }_{n}L/c)$. Since $\sin ({\omega }_{n}L/c)=0$ would also force the cosine to vanish, it can be divided out for a nonzero mode. Using $T=\rho c^2$ givesFor physical $M\ge 0$, these give all positive modal frequencies. There is no nonzero static linear mode: the left value and right slope conditions would force it to vanish. Positivity also follows by integration by parts from

For $M=0$, the right endpoint is free, with the Neumann boundary condition $y_x(L,t)=0$. At $L=\pi$, the allowed wave numbers are $k_n=n+1/2$, $n\ge 0$, and ${\omega }_n=c{k}_n$. Zero initial displacement removes every cosine-in-time coefficient. The initial velocity contains only the modes $k=1/2$ and $k=5/2$, each with coefficient one. Dividing those velocity coefficients by their angular frequencies givesAt $x=0$ each term vanishes, at $x=\pi$ each spatial derivative vanishes, and at $t=0$ the displacement and its time derivative reproduce the two initial conditions.

The symbol $\rho$ here is linear mass density, with units $\mathrm{kg}{\mathrm{m}}^{-1}$. A short element $[x,x+\delta x]$ has transverse forcebecause small slopes give $\sin \theta \simeq \tan \theta ={\Psi }_x$. Its mass is $\rho \delta x$. Applying Newton's second law, dividing by $\delta x$, and taking the limit gives the wave equation on a string,The derivation assumes constant positive tension, small transverse displacements and slopes, a flexible string with negligible bending stiffness, and negligible damping. Variations of tension due to the motion enter beyond the linear approximation.