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An acoustic wavelength measurement makes an invisible spatial pattern readable with a ruler. The authored guide Interesting and inexpensive experiments for high school physics by Joe Wolfe places a loudspeaker near a pipe's open end, drives it with an oscillator, and slides a microphone attached to a metre rule through the pipe. An oscilloscope or computer displays the microphone signal. Its oscillation amplitude rises and falls with position; adjacent minima locate pressure nodes of a standing wave a distance $d=\lambda /2$ apart. A cap gives a pressure antinode at the closed end; an open end has a pressure node just outside it. These are pressure patterns, distinct from displacement patterns. The guide advises wavelengths longer than the pipe diameter, sufficient pipe length to locate both a node and an antinode, and frequency spectrum analysis to distinguish the intended tone from other sounds.^[ref]

The inference follows from superposition of two oppositely directed acoustic plane waves. In an ideal uniform, lossless pipe, taking equal pressure amplitudes $P$ and a rigid cap at $x=0$ givesThe time-varying pressure perturbation $p^{\prime }$ has local oscillation amplitude $2|P\cos (kx)|$. Its zeros are separated by $\pi /|k|=\lambda /2$. Unequal incident and reflected amplitudes leave nonzero minima, but their spacing remains $\lambda /2$ in this ideal model. A microphone detects that pressure variation through a changing electrical signal, rather than watching a wave crest move. Record the tone's frequency too, then infer phase speed $v_p=2fd$.

For proposed controls, keep the drive frequency and amplitude fixed during a scan, verify the frequency on the recorded signal, and repeat the scan in both directions. Fit several minimum positions $x_j=a+jd$ using linear regression to reduce dependence on one pair of positions. Check the ruler scale and use one consistent reference point on the microphone. Compare with the source switched off to locate background measurement noise; use frequency spectrum analysis to check for additional tones, and reduce drive if the trace ceases to resemble a sinusoid. Record air temperature and avoid strong air motion. These are suggested controls, not additional procedures or observations attributed to the guide.

The pipe must support approximately plane waves over the measured region: transverse structure at shorter wavelengths, the microphone's finite size, and disturbance by the ruler can compromise the interpretation. Weak reflection gives shallow minima, while damping and other reflections can shift the extrema of the simple two-wave model. Measure interior spacings rather than assuming the first minimum is exactly at the pipe lip. A short scan may not include enough minima to estimate $d$ reliably, and drift during sequential scanning can imitate spatial variation.

For an explicitly hypothetical calculation, $f=1000\mathrm{Hz}$ and $d=0.1715\mathrm{m}$ give $\lambda =0.3430\mathrm{m}$, $K\simeq 18.3183\mathrm{rad}{\mathrm{m}}^{-1}$, and $v_p=343.0\mathrm{m}{\mathrm{s}}^{-1}$. No such measurements are claimed here. If independent small measurement errors in $f$ and fitted $d$ have standard deviations ${\sigma }_f$ and ${\sigma }_d$, first-order error propagation gives ${\sigma }_{v_p}/v_p\simeq \sqrt{({\sigma }_f/f{)}^2+({\sigma }_d/d{)}^2}$. A systematic scale error or changing air conditions requires separate assessment.

Work in the material's rest reference frame, with constant speed of sound $c>0$. For one-dimensional propagation, choose the positive $x$ direction along the travelling wave from its emission point toward the receiver. Define $u_s=d{x}_s/dt$ and $u_o=d{x}_o/dt$ as the signed source and receiver velocities along that same axis. Both use the same sign convention: positive $u_s$ moves the source toward the receiver ahead, whereas positive $u_o$ moves that receiver away from the incoming wave. Assume $|u_s|<c$ and $|u_o|<c$, constant velocities, and no source-receiver crossing during the interval considered. Then

To derive this, let $T_s=1/f_s$ be the emission wave period. During $T_s$, one wave crest travels $c{T}_s$ while the source advances $u_s{T}_s$. Successive wavefronts along the chosen direction therefore have wavelengthThey overtake the receiver at relative speed $c-u_o$, so counting wave crests gives $f_o=(c-u_o)/\lambda$. This separates the source's effect on wavelength from the receiver's effect on crossing rate. Even with a moving source, a stationary receiver obeys $f_o\lambda =c$ because the wave speed through the material remains $c$.

For a stationary receiver and a source approaching at speed $v$, set $u_o=0$ and $u_s=v$: $f_o=f_{s}c/(c-v)>f_s$. For recession choose the positive axis along the departing wave toward the receiver again, so the source has $u_s=-v$: $f_o=f_{s}c/(c+v)<f_s$. For the deliberately chosen values $c=343\mathrm{m}{\mathrm{s}}^{-1}$, $v=20\mathrm{m}{\mathrm{s}}^{-1}$ and $f_s=500\mathrm{Hz}$, the two predictions areThe corresponding wavelengths are $0.646\mathrm{m}$ and $0.726\mathrm{m}$, compared with $c/f_s=0.686\mathrm{m}$ for a stationary source. These are calculations for chosen parameters, not reported measurements.

A receiver approaching a stationary source at $20\mathrm{m}{\mathrm{s}}^{-1}$ instead has $u_s=0$, $u_o=-20\mathrm{m}{\mathrm{s}}^{-1}$, giving $f_o=500(363/343)\mathrm{Hz}\simeq 529.15\mathrm{Hz}$. Moving away gives $f_o=500(323/343)\mathrm{Hz}\simeq 470.85\mathrm{Hz}$. The equal closing speeds do not make these equal to the moving-source values. If both source and receiver move together with $u_s=u_o$, however, the numerator and denominator cancel and $f_o=f_s$. For speeds small compared with $c$, the first-order shift is $(f_o-f_s)/f_s\simeq (u_s-u_o)/c$; the difference between moving-source and moving-receiver experiments appears beyond this first order.

The wave crest and wave trough are respectively the highest and lowest displacements in a wave.

The Doppler effect is a change between emitted and received frequency caused by motion of the source or receiver. An approaching siren can send more wave crests per second to a microphone than the source emits per second; a receding siren sends fewer. The source need not change its own oscillation. The Doppler shift is $\Delta f=f_o-f_s$, where $f_s>0$ is emitted frequency and $f_o>0$ is received frequency, both measured in hertz.

For acoustic waves, motion must be specified relative to the material carrying the wave. The speed of sound is measured in that material's rest reference frame. Moving the source changes the spacing of successive wavefronts; moving the receiver changes how quickly it crosses those wavefronts. These two mechanisms give different exact factors in the classical Doppler effect for sound, even for the same closing relative velocity.

This makes a microphone a motion probe: a stable tone supplies a repeating clock, and changes in arrival spacing reveal the source's motion toward or away from the receiver. An acoustic Doppler measurement can test the prediction directly. A passing source also illustrates why geometry matters: its speed can stay constant while the received frequency changes continuously, as in Doppler frequency along a trajectory.

The following sound model assumes linear acoustics in uniform stationary material with isotropy, constant speed of sound and negligible wave dispersion. Source and receiver speeds are below the speed of sound, and there is a single direct propagation path. Air motion requires transforming source and receiver velocities into the air's rest reference frame; a nonuniform flow requires a more general propagation model. At the sonic threshold the simple periodic-arrival picture fails; moving-point acoustic wavefronts and the Mach cone describe the relevant geometry, while finite shock waves require further physics. Light in vacuum instead requires the relativistic Doppler effect: the acoustic formula cannot be converted into its exact optical counterpart just by replacing the speed of sound with the speed of light.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

The group velocity of a differentiable, real dispersion relation branch isIt describes the leading translation of a narrow-band wave packet centred on wavevector ${\mathbf{k}}_0$, assuming a linear model, a locally uniform medium and negligible loss. Narrow-band means that the appreciably excited wavevectors occupy a small neighbourhood on which one smooth branch can be linearized. Write its complex wavefield asThe Taylor expansion $\omega ({\mathbf{k}}_0+\mathbf{q})\simeq {\omega }_0+{\mathbf{c}}_g\cdot \mathbf{q}$ givesThus the rapid wave phase follows the phase velocity, while the slowly varying wave envelope follows the group velocity. This Fourier analysis argument is an outline of the approximation, rather than a statement that every wave packet translates rigidly forever. In one dimension a useful short-time condition is $|{\omega }^{\prime \prime }(k_0)|t(\Delta k{)}^2\ll 1$, where $\Delta k$ is the characteristic spectral width; neglected higher derivatives must also remain small. Curvature of the dispersion relation eventually changes the wave envelope.^[ref]

Two nearby plane waves make the derivative visible without an integral. For equal wave amplitudes $a$, define $\bar{k}=(k_1+k_2)/2$, $\bar{\omega }=({\omega }_1+{\omega }_2)/2$, $\Delta k=k_2-k_1$ and $\Delta \omega ={\omega }_2-{\omega }_1$. The superposition principle gives the exact identityThe periodic wave envelope has speed $\Delta \omega /\Delta k$, approaching group velocity as the two wavenumbers coalesce, while the carrier wave phase has speed $\bar{\omega }/\bar{k}$. Two tones alone form an infinite beat train, not a localized wave packet. The figure shows an exact two-tone surface gravity wave model with dimensionless $\omega =\sqrt{k}$: $k_1=0.9$, $k_2=1.1$. Its carrier phase velocity is approximately $0.99875$, its finite-difference envelope velocity approximately $0.50063$, and ${\omega }^{\prime }(1)=0.5$ is the limiting group velocity. The marked carrier phase line need not coincide exactly with a maximum of the full modulated wave.

The deep-water surface gravity wave therefore has $c_g=c_p/2$, whereas the deep-water capillary wave limit has $c_g=3c_p/2$: crests can drift either forward or backward relative to a wave envelope. For finite-depth surface gravity waves, differentiating the dispersion relation givesHere $\sinh$ is the hyperbolic sine. The shallow-water limit has $c_g\simeq c_p\simeq \sqrt{gh}$; the deep-water limit has $c_g\simeq c_p/2$.^[ref]

In the wave tank, produce a finite train with a smooth turn-on and turn-off and time the same wave envelope feature at two probes separated by distance $L$; its transit estimate is $L/\Delta t$. Independently follow wave crests through that train. Techet's 2003 lab asks students to observe a deep-water train's foremost crest disappearing as the group moves forward, and to compare deep- and shallow-water cases. The interpretation is that a wave crest overtakes the slower wave envelope, dwindling at its front while other crests become visible at its rear. This is a qualitative expected observation in the teaching protocol, not a published set of measured curves.^[ref]

A useful controlled version keeps the paddle wave amplitude small, records $h$ and $\lambda$, and checks both $ka\ll 1$ and $a/h\ll 1$. Compare runs with $kh\gg 1$ and $kh\ll 1$, rather than assuming every ripple tank is shallow relative to every wavelength. Increase train duration at fixed central frequency to narrow its spectrum, and check that the envelope transit estimate stabilizes. Calibrate both probes, sample fast enough to resolve the carrier, and compare records before reflected waves return; an absorbing beach reduces this contamination. A current changes laboratory velocities, large wave amplitudes introduce nonlinear effects, and very short wavelengths require surface tension. Strong damping or noticeable reshaping makes the chosen threshold or peak an unreliable marker of simple group velocity. These are proposed controls and model limits, not claims that the cited lab measured every correction.

Three velocities should consequently be kept separate. The group velocity predicts narrow-band wave envelope motion. An energy-transport velocity is energy flux divided by energy density, using the appropriate time average; it agrees with group velocity in standard lossless linear wave models when all stored energy is included.^[ref] A signal-front velocity concerns the earliest response to a new disturbance and requires a causal model over its full spectrum. A strictly band-limited wave packet has no sharply bounded onset, and a mathematical two-tone envelope cannot test front arrival. In strongly absorbing or broadband propagation, a single group velocity need not describe the observed peak, energy transport, or first detectable response. A complex $\omega$ or $k$ also prevents interpreting its derivative as an ordinary real transport velocity without further analysis.

In particular, an unusually large or negative group velocity from pulse reshaping does not establish superluminal information transmission. For the causal dielectric model analysed by Fitzpatrick, the field ahead of the front vanishes for $t<x/c$, even when its phase velocity exceeds the vacuum speed of light $c$. That result concerns the electromagnetic model's causality; the ideal water-wave diagrams above contain no relativistic signal-front claim.^[ref]

Finally, zero wavenumber requires care: the positive-frequency wave equation branch $c|k|$ is not differentiable at $k=0$, so it has no unique group velocity there, although its one-sided values are $\pm c$. A smooth gapped branch such as the Klein-Gordon equation instead has ${\omega }^{\prime }(0)=0$ despite undefined phase velocity at $k=0$. The same signed branch must be used in both derivatives and ratios; a negative group velocity means an envelope moving in the negative coordinate direction, not a negative elapsed time. The practical value of group velocity is precisely this translation between a local slope on a dispersion diagram and a measurable packet transit, with its assumptions made explicit.

The Gaussian envelope in a Klein-Gordon medium supplies a quantitative next step: the first derivative translates the wave envelope, the second derivative spreads it, and higher derivatives limit that approximation. At finite spectral width, a peak's measured speed need not equal the group velocity at the central wavenumber. Neither a peak nor a fixed fraction of its height generally identifies front velocity.

A useful stipulated dispersive medium obeys the Klein-Gordon equation $u_{tt}-c^2{u}_{xx}+{\omega }_0^{2}u=0$, with $c>0$ and ${\omega }_0>0$. Its positive-angular frequency dispersion relation isHere ${\omega }_0$ is the gap at zero wavenumber, not the central angular frequency ${\omega }_c=\omega (k_0)$ of a wave packet; the parameter in the preceding formula is $A={\omega }_0/c$. For $k>0$, the phase velocity satisfies $v_p>c$, the group velocity satisfies $0<v_g<c$, and $v_p{v}_g=c^2$. In this stipulated medium $c$ is the limiting propagation speed of the model; it need not mean the vacuum speed of light. The same square-root dispersion relation appears in waveguide phase and group velocities, with a waveguide cutoff frequency playing the role of ${\omega }_0$.^[ref]

For the original dimensionless choice $c={\omega }_0=k_0=1$, a carrier wave has phase velocity $v_p=\sqrt{2}$, while a narrow wave packet has leading group velocity $v_g=1/\sqrt{2}$. After time $t=8$, the carrier's zero-wave phase reference has moved to $x=8\sqrt{2}\simeq 11.31$, whereas the leading wave envelope centre is at $x=8/\sqrt{2}\simeq 5.66$. A carrier wave crest therefore moves through the wave envelope; it does not carry an identifiable piece of the envelope with it.

The origin-to-curve secant on the dispersion diagram gives phase velocity, while its local tangent gives group velocity. As $k\to 0^{+}$, $v_p\to \infty$ and $v_g\to 0$; at $k=0$ the phase velocity itself is undefined. As $k\to \infty$, both approach $c$, from opposite sides. On the same positive-angular frequency branch with $k<0$, both signed velocities are negative and their product remains $c^2$. These conventions distinguish direction from phase speed. Reversing both $k$ and $\omega$ represents the same real plane wave; changing just one sign changes its direction.

For an original finite-bandwidth check, take equal-amplitude plane waves at $k_1=0.8$ and $k_2=1.2$, still with $c={\omega }_0=1$. The exact two-tone wave envelope speed iswhile its carrier wave has phase velocity $(\sqrt{1+0.8^2}+\sqrt{1+1.2^2})/2\simeq 1.421337$. Their product is exactly $1$: rationalizing the difference gives $v_{\mathrm{beat}}=2/({\omega }_1+{\omega }_2)$. Nevertheless $v_{\mathrm{beat}}\ne {\omega }^{\prime }(1)$; the derivative describes the limit of vanishing separation, not an arbitrary pair of tones. The two-tone wave envelope repeats indefinitely, so it has no isolated first arrival and cannot measure front velocity.

For the local phase $S$, define $k_i={\partial }_{i}S$ and $\omega =-{\partial }_{t}S$. The dispersion relation is the Hamilton-Jacobi equation $S_t+\Omega (\nabla S,\mathbf{x},t)=0$. Differentiate it in space and follow a ray with group velocity ${\dot{x}}_i={\Omega }_{k_i}$; symmetry of the phase Hessian cancels the terms involving ${\partial }_i{k}_j$. Thus the ray-tracing equations areHere $d/dt={\partial }_t+\dot{\mathbf{x}}\cdot \nabla$ means differentiation along the moving ray; subscripts on $\Omega$ mean partial derivatives holding its other independent arguments fixed.

The depth is stationary and independent of $y$, so $\omega ={\omega }_{\infty }$ and $l$ are constant. Choose the incident direction with positive $l$; the other sign reflects the picture. In the far deep water,Since $\Omega$ depends isotropically on the horizontal wavevector, the group-velocity ratio is $l/k$. Therefore the ray equation isNear shore, $\kappa h\to 0$, so the shallow-water dispersion relation gives $\kappa \sim {\omega }_{\infty }/(\sqrt{g\alpha }{x}^{p/2})$ and $k\sim -\kappa$. Integrating the ray slope yieldsFor $p<2$, the near-shore phase is $S\sim ly-{\omega }_{\infty }t-{\omega }_{\infty }{x}^{1-p/2}/[\sqrt{g\alpha }(1-p/2)]$. A constant-phase wave crest thus hasIts tangent becomes parallel to the shoreline, while rays approach perpendicular to it. The exponent is between zero and one, giving the curved crests sketched below for $p=1$.

These are the formal geometrical optics asymptotics. For fixed $\alpha$ and $p<2$, $\kappa x\to 0$ arbitrarily close to the shore, so the slowly varying wave approximation eventually fails there; the sketch describes the ray-theory regime before that final inner region.

A wave crest moves at the phase velocity, while the envelope of a narrow wave packet moves at the group velocity. Since $c_g>c_p$ for every nonzero wavenumber, the crests move more slowly than the packet.

The phase velocity is the motion of a constant-wave phase surface. For a one-dimensional plane wave $u=a\cos (kx-\omega t+{\varphi }_0)$ with real $k\ne 0$ and real $\omega$, fixing $kx-\omega t+{\varphi }_0$ givesEach wave crest of this monochromatic travelling wave advances at that signed phase velocity. Its magnitude is the phase speed $|c_p|=\lambda /T$, where $\lambda =2\pi /|k|$ is the wavelength and $T=2\pi /|\omega |$ is the temporal period when $\omega \ne 0$. Choosing $\omega >0$ makes positive $k$ mean rightward phase velocity and negative $k$ mean leftward phase velocity. Reversing both signs leaves the real plane wave and its phase velocity unchanged. At $k=0$ there are no spatial crests to track, so $\omega /k$ is undefined; a finite limit as $k\to 0$ is a limiting phase speed, not the speed of a spatially uniform oscillation.

In several dimensions, a plane wave has wave phase $\mathbf{k}\cdot \mathbf{x}-\omega t+{\varphi }_0$. Constant-wave phase planes obey $\mathbf{k}\cdot \mathbf{v}=\omega$. Their normal phase velocity isTangential motion within a plane does not move that wavefront, which is why the normal direction fixes the vector convention. In an anisotropic medium the group velocity need not point along this wavevector. Likewise the horizontal phase velocity of a tilted wavefront is its crossing speed along a horizontal line, rather than a component of the normal phase velocity.

For the nondispersive wave branch $\omega =ck$, a wave crest and a wave envelope both move at $c$. For deep-water surface gravity waves, $\omega =\sqrt{gk}$ gives $c_p=\sqrt{g/k}$ and group velocity $c_g=c_p/2$. With the illustrative choices $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\lambda =1\mathrm{m}$, the model predicts $c_p\simeq 1.25\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.625\mathrm{m}{\mathrm{s}}^{-1}$. These are calculated values for negligible surface tension and sufficiently large $kh$, not measurements. A wave crest moves forward through the wave packet, so timing a crest does not time the packet's journey.

There is a direct laboratory meaning to this distinction. MIT's 2005 surface-wave lab uses a long water tank, a controllable hydraulic paddle, and an absorbing beach. Resistance probes convert submerged length into voltage; calibration converts their readouts into free surface elevation versus time. Two probes a known distance apart, backed by a ruler and stopwatch, let students time a wave crest and separately time a feature of the wave packet. The lab records depth and varies frequency and wave amplitude.^[ref]

For a clean measurement of phase velocity, fit the periodic portion of both records or follow one identifiable wave crest. A phase difference is only known modulo $2\pi$: probe spacing below half the shortest relevant wavelength, or continuity while sweeping frequency, avoids assigning the wrong whole-cycle delay. Then compare the timed phase velocity with $\lambda /T$. Tracking pepper grains instead measures fluid-particle motion, which has a different meaning from either phase velocity or group velocity.

The Klein-Gordon packet velocities example makes the distinction especially sharp: its phase velocity exceeds $c$, yet its narrow wave envelope moves below $c$. A carrier wave labels successive oscillations; its speed alone does not establish information velocity. The front velocity depends on the full causal model, including angular frequencies far outside the carrier's narrow range.

A node of a standing wave must have $y(x,t)=0$ for every time, not merely at one snapshot. For $y=2a\sin (kx)\cos (\omega t)$, with $a,k,\omega >0$, set $t=0$ to see that this requires $\sin (kx)=0$. ThusAn antinode instead maximizes the oscillation amplitude $2a|\sin (kx)|$ and lies halfway between successive nodes of a standing wave, at $x=(2j+1)\lambda /4$. Neighbouring antinodes are also separated by $\lambda /2$; the distance from a node of a standing wave to its nearest antinode is $\lambda /4$. These distances refer to the string's longitudinal coordinate, not its curved arc length.

Within one interval between nodes of a standing wave, all moving points have the same oscillation phase. Adjacent intervals have opposite signs of $\sin (kx)$, so their phase difference is $\pi$: one loop rises while the next falls. The oscillation phase at an exact node of a standing wave is undefined because its amplitude is zero. A ruler measures the spacing of the fixed pattern; it does not track a moving wave crest.

For a propagating branch of an ideal vacuum-filled rectangular electromagnetic waveguide, the phase velocity tracks constant wave phase along the axis, whereas the group velocity describes a sufficiently narrow, weakly distorted wave packet. Using ${\omega }^2=c^2{\beta }^2+{\omega }_c^2$ givesThus $v_{\mathrm{ph}}>c$, $0<v_g<c$, and $v_{\mathrm{ph}}{v}_g=c^2$. Differentiating the dispersion relation proves the second formula; dividing it by ${\beta }^2$ proves the first. This is the same mathematical structure as Klein-Gordon packet velocities. In the stipulated $f=1.25f_c$ example,Over $0.30\mathrm{m}$, the corresponding narrow-envelope delay is $(0.50\mathrm{m})/c$, approximately $1.668\mathrm{ns}$. Timing a carrier wave crest would give $0.6004\mathrm{ns}$; it does not time the arrival of a newly introduced message.

There is an independent energy check. For the forward ${\mathrm{TE}}_{10}$ fields, the average axial Poynting vector is $⟨S_z⟩=\beta |E_y{|}^2/(2{\mu }_0\omega )$. Integrating across the guide gives average power $P$ and average electromagnetic energy per unit length $U^{\prime }$:The transverse sine and cosine squared each integrate to $a/2$; ${\beta }^2+(\pi /a{)}^2={\omega }^2/c^2$ then makes the integrated electric and magnetic contributions to $U^{\prime }$ equal. The axial energy transport therefore agrees with the group velocity in this ideal model.

As $\omega \downarrow {\omega }_c$, $v_g\to 0$ and $v_{\mathrm{ph}}\to \infty$; exactly at cutoff $\omega /\beta$ is undefined. Strong wave dispersion close to cutoff also limits the narrow-packet approximation. Neither a large phase velocity nor a pulse peak establishes superluminal information velocity; front velocity asks a separate causal question. In a uniform filling with constant positive permittivity $\epsilon$ and magnetic permeability $\mu$, replace $c$ by $u=1/\sqrt{\epsilon \mu }$, giving $v_{\mathrm{ph}}{v}_g=u^2$. Frequency-dependent material response or significant loss requires differentiating the actual dispersion relation, so this product is not universal.

The wavelength $\lambda$ is the spatial period of a monochromatic wave, measured at a fixed time along the wavevector. For a plane wave with a sinusoid as its profile, it is the distance between consecutive wave crests, or between any consecutive points with the same wave phase modulo $2\pi$:Here $\mathbf{k}$ uses angular wavenumber, $f$ is ordinary frequency, and $v_p$ is phase speed. A distance between arbitrary zero crossings need not be a wavelength: consecutive zeros of a sinusoid are only $\lambda /2$ apart, with opposite slopes. In a standing wave, consecutive nodes of a standing wave are also $\lambda /2$ apart; $\lambda$ refers to the constituent travelling waves, even though the nodes of a standing wave do not travel.

Across a stationary interface between stationary linear media, the incident, reflected and transmitted monochromatic waves have the same frequency. The boundary conditions must hold at every time: terms oscillating with different angular frequencies cannot cancel throughout the entire time record. Each medium's dispersion relation then determines its own angular wavenumber and wavelength, soFor electromagnetic radiation in a transparent medium with refractive index $n$, $v_p=c_0/n$ and $\lambda ={\lambda }_0/n$, where $c_0$ is the vacuum speed of light. A synthetic example with ${\lambda }_0=600\mathrm{nm}$ and $n=1.5$ gives $\lambda =400\mathrm{nm}$ at the same frequency. This conservation argument assumes time-independent linear material response and a stationary interface in the chosen frame. A moving interface can produce a Doppler effect; time-dependent or nonlinear response can change or generate frequencies.^[ref]

A broad frequency spectrum generally supplies many component wavelengths, rather than one wavelength for an entire pulse. Even two tones have both their individual wavelengths and a different overall repetition distance: the $2$- and $8$-cycle-per-metre sum in spatial sampling of a wave repeats after $0.50\mathrm{m}$, but contains the finer $0.125\mathrm{m}$ structure. A narrow wave packet can have an approximate carrier wavelength; the wave envelope propagates approximately at group velocity, determined by the slope of the dispersion relation. Substituting that group velocity into $\lambda =v_p/f$ generally gives the wrong carrier wavelength. Where the medium varies slowly, local wavelength gives a more useful description.

A wavenumber measures how rapidly a wave phase changes with distance. Two conventions coexist: angular wavenumber counts radians per unit length, while ordinary spatial frequency counts cycles per unit length. For a monochromatic wave with wavelength $\lambda >0$,Here $K\ge 0$ is the magnitude of the angular wavenumber. The symbol $k$ can instead denote a signed one-dimensional angular wavenumber: a factor $e^{ikx}$ has wavelength $2\pi /|k|$ for real $k\ne 0$. Both conventions have dimensions of inverse length; writing $\mathrm{rad}{\mathrm{m}}^{-1}$ or “cycles per metre” exposes the convention. A quoted value in ${\mathrm{m}}^{-1}$ alone does not resolve the factor of $2\pi$. Spectroscopic wavenumber usually uses the cycles-per-length convention.^[ref]

For a real monochromatic plane wave, fix the conventionThe wavevector $\mathbf{k}$ supplies both the spatial wave phase slope and the normal to the wavefronts; its dot product with position makes the wave phase constant on parallel planes. The angular frequency is $\omega =2\pi f$, with $f>0$ the ordinary frequency. Moving a distance $\lambda$ along $\mathbf{k}$ changes the wave phase by $2\pi$. Holding a wave crest's wave phase fixed gives its phase velocity $\omega \mathbf{k}/K^2$ and phase speedIn one dimension the signed phase velocity is $\omega /k$. This identifies a moving pattern; it does not identify the motion of the medium's particles or a wave packet's group velocity.

For the synthetic plane wave below, choose $\lambda =0.40\mathrm{m}$ and $f=5\mathrm{Hz}$. Then $K=5\pi \mathrm{rad}{\mathrm{m}}^{-1}$, ${\nu }_x=2.5$ cycles per metre, $\omega =10\pi \mathrm{rad}{\mathrm{s}}^{-1}$, and $v_p=2\mathrm{m}{\mathrm{s}}^{-1}$. The wave crests are $0.40\mathrm{m}$ apart at one instant; after a quarter oscillation period, each has advanced $0.10\mathrm{m}$. The lower panel retains whole turns of wave phase, making its slope visible instead of reducing it modulo $2\pi$.

For the homogeneous one-dimensional wave equation $u_{tt}=c^2{u}_{xx}$ with constant $c>0$, wave speed means the speed at which a travelling wave profile translates. In $F(x-ct)$, a feature initially at $x=s$ follows $x=s+ct$, giving signed velocity $+c$; in $G(x+ct)$ it follows $x=s-ct$, giving signed velocity $-c$. The characteristic lines for the one-dimensional wave equation therefore have slopes $dx/dt=\pm c$. The speed of either profile is $c$, even though the displacement at a fixed point can change sign or pass through zero.^[ref]

The same $c$ bounds the causal cone for the constant-speed wave equation. For a wave equation on a string, $c=\sqrt{T/\mu }$ depends on tension and linear mass density; for small acoustic waves in stationary uniform air it is the speed of sound. This interpretation assumes the stated ideal wave equation. In media with wave dispersion, phase velocity, group velocity and the earliest signal arrival have different meanings: a measured wave crest speed need not be the speed of a wave envelope, and neither alone establishes a causal cone for the constant-speed wave equation.