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Reflection at the fixed edges combines waves with opposite $k_y$ into a transverse standing wave while retaining their common longitudinal propagation factor. The membrane waveguide scalar transverse displacement satisfies the wave equation and vanishes at both edges:The $n=0$ sine gives no displacement. separation of variables givesFor a propagating branch with $k_x>0$,Their product is $v_m^2$. The phase velocity can exceed the material wave speed; energy transport follows the group velocity.

Here the cutoff sequence is $10n\mathrm{kHz}$. At $15\mathrm{kHz}$ only $n=1$ propagates, with $k_x=175.6{\mathrm{m}}^{-1}$ for travel toward positive $x$.

For a fixed transverse normal mode, define its cutoff angular frequencyThe longitudinal phase velocity and group velocity, for propagation in $+x$, areThus $v_p{v}_g=v^2$, $v_p>v$ and $v_g<v$. The large longitudinal phase velocity reflects the geometry of the transverse standing wave and does not imply faster transport of energy than the membrane wave speed.

The cutoff frequency of a membrane waveguide for mode $n$ isAt cutoff $k_x=0$ and $v_g=0$; the longitudinal phase velocity has no finite value. Below cutoff,and a bounded field excited at $x=0$ decays as $e^{-{\gamma }_{n}x}$. It is an evanescent wave, rather than a freely propagating mode carrying net time-averaged energy flux down an indefinitely long uniform strip. A finite section can still transmit through coupling of evanescent fields.

Twice differentiating the proposed normal mode givesIt therefore solves the two-dimensional wave equation precisely whenThe fixed-edge boundary conditions require $z(x,0,t)=z(x,b,t)=0$. The first is automatic, and the second requires $\sin (k_{y}b)=0$. The nonzero normal modes are thereforeThe apparent $n=0$ choice gives identically zero displacement for this sine solution. There are no imposed end conditions in $x$, so $k_x$ is continuous rather than quantized by a second length. For a propagating mode it is real; the plus sign in the stated $\cos (\omega t-k_{x}x)$ convention describes propagation in $+x$. Each positive integer $n$ labels a transverse standing wave across the strip.

A transmission line stores electric and magnetic energy along its length, so voltage and current depend on both position and time. A distributed circuit uses inductance $L^{\prime }$ and capacitance $C^{\prime }$ per length, together with resistance $R^{\prime }$ and leakage conductance $G^{\prime }$. Applying voltage and charge balance to a short segment gives the telegrapher's equations,For a lossless line, both variables obey wave equations with speed $v=1/\sqrt{L^{\prime }{C}^{\prime }}$ and characteristic impedance $Z_0=\sqrt{L^{\prime }/C^{\prime }}$. Forward and backward waves obey $I^{+}=V^{+}/Z_0$ and $I^{-}=-V^{-}/Z_0$. Matching a termination $Z_L$ gives the voltage reflection coefficientA matched load absorbs without reflection; an open circuit has ${\Gamma }_L=1$, and a short has ${\Gamma }_L=-1$. Reflections can produce standing waves. A coaxial line confines the fields between conductors and, for a uniform dielectric TEM wave, has $v=1/\sqrt{\mu ϵ}$. Finite loss attenuates waves; frequency-dependent losses or material properties can also distort pulses. A long line is a wave-carrying distributed system, so it cannot generally be replaced by one lumped capacitor.

irrotational flow has $\nabla \times \mathbf{v}=0$, so locally $\mathbf{v}=\nabla \Phi$; the simply connected fluid domain here permits a globally single-valued velocity potential. Incompressibility gives $\nabla \cdot \mathbf{v}=0$, hence ${\nabla }^2\Phi =0$. In a multiply connected domain, zero local vorticity alone would not rule out circulation and a multivalued potential.

Take $k>0$ so the disturbance decays at depth. Differentiating the given candidate potential yieldsThe $x$ and $z$ second derivatives of $\Phi$ cancel in the Laplacian. At $z=0$, $v_z=h_0\omega \cos (\omega t)\sin (kx)={\partial }_{t}Z$, so the linearized kinematic free-surface condition holds. This uses $k{h}_0\ll 1$; advection of the displaced surface is higher order.

Since ${\cos }^2(kx)+{\sin }^2(kx)=1$, the local squared speed is $h_0^2{\omega }^2{\cos }^2(\omega t)e^{2kz}$. The depth integral is thereforeIt is independent of horizontal position; no further spatial average is needed for this kinetic energy. Replacing the upper boundary $Z$ by zero changes it only beyond quadratic order in amplitude.

Relative to the flat surface, the gravitational energy at each horizontal position changes by $\rho g{\int }_0^{Z}zdz=\rho g{Z}^2/2$. Averaging over one wavelength givesThe energy of a deep-water standing gravity wave, averaged over wavelength, is $T+⟨U{⟩}_x$ and must be constant for a free inviscid standing wave, which requires equal coefficients of the complementary squared sine and cosine:As an independent check, the linearized dynamic boundary condition ${\partial }_t\Phi +gZ=0$ at $z=0$ gives exactly $g-{\omega }^2/k=0$. Infinite depth, negligible surface tension and no dissipative forcing are part of this gravity-wave model.

A transmission line must be treated as a distributed circuit when its propagation time is significant on the signal timescale. Let $R^{\prime },L^{\prime },G^{\prime },C^{\prime }$ denote resistance, inductance, leakage conductance and capacitance per unit length. Applying Kirchhoff's voltage law and Kirchhoff's current law to a short length $dx$ gives the telegrapher's equationsFor a lossless line, $R^{\prime }=G^{\prime }=0$. Differentiating again gives wave equations withThe travelling solutions can be written $V=V_{+}(t-x/v)+V_{-}(t+x/v)$ and $I=[V_{+}(t-x/v)-V_{-}(t+x/v)]/Z_0$. The minus sign for the backward wave is essential to power transport.

At a load $Z_L$, continuity of voltage and current gives the voltage reflection coefficientA matched termination $Z_L=Z_0$ absorbs the incident signal without reflection. An open circuit has ${\Gamma }_L=+1$ and a short circuit has ${\Gamma }_L=-1$. Superposed incident and reflected waves produce standing waves. A sinusoidal lossless line of length $\ell$ transforms the termination intofor the $e^{i\omega t}$ convention. At a quarter wavelength, $Z_{\mathrm{in}}=Z_0^2/Z_L$, allowing impedance transformation. A coaxial line supports a transverse electromagnetic mode whose ideal $L^{\prime },C^{\prime }$ are calculated in A4. Its energy travels in the fields between conductors; the characteristic impedance is not the wire's DC resistance. Losses introduce attenuation and make the characteristic impedance and propagation constant complex.

A cubic reflecting cavity supports standing waves with positive integer mode indices and wavevector components $k_i=\pi n_i/L$. A cell in mode space has volume $(\pi /L{)}^3$. In the continuum approximation, a spherical shell of radius $k$ and thickness $dk$ has one-eighth of its volume in the positive octant. There are two independent transverse photon polarizations for a generic wavevector. HenceThe photon energy is $ϵ=\hslash \omega =\hslash ck$, so $dk=dϵ/(\hslash c)$. ThereforeThis is the leading bulk density of states, valid when the cavity is large compared with the relevant wavelengths and many modes lie in the counting interval. A finite cavity has a discrete mode spectrum, with boundary corrections beyond this approximation. Modes with vanishing indices require separate boundary counting but do not change the leading $L^3$ term. Here $\hslash$ is the reduced Planck constant, as printed in the original PDF; replacing it by $h$ would introduce an erroneous factor of $(2\pi {)}^3$.

A travelling wave, for example $A\cos (kx-\omega t)$, has moving constant-phase surfaces. A standing wave, for example $2A\sin (kx)\cos (\omega t)$, has stationary nodes and can be formed from two equal counterpropagating waves.

For the half-line finite well with $E>V_0$, define $k=\sqrt{2mE}/\hslash$ and $q=\sqrt{2m(E-V_0)}/\hslash$. The hard wall requires $\psi =0$ for $x<0$ and $\psi (0)=0$, so the general allowed forms areAt the finite step both $\psi$ and ${\psi }^{\prime }$ are continuous:There is no requirement to match the derivative to zero across the infinite wall.

For a bound state $0<E<V_0$, put $\kappa =\sqrt{2m(V_0-E)}/\hslash >0$. Normalizability excludes the growing exponential, leaving $A\sin (kx)$ inside and $D{e}^{-\kappa (x-a)}$ outside. Matching their logarithmic derivatives givesThe exterior expression is equivalent to $C{e}^{-\kappa x}$ by rescaling its constant. Negative cotangent first occurs for $\pi /2<ka<\pi$; all higher allowed intervals also have $ka>\pi /2$. Thus ${\lambda }_1=2\pi /k<4a$. Since $k<\sqrt{2m{V}_0}/\hslash$, a normalizable bound state requiresThis is also sufficient: on the lowest negative-cotangent branch, the two matching sides cross whenever $a\sqrt{2m{V}_0}/\hslash >\pi /2$. The printed non-strict inequality includes the threshold limit. At equality $E=V_0$ and the exterior is not square-integrable, so it is not itself a bound state.

For the helium mass and energy scale given, the limiting width isActual binding requires a width just larger than this threshold in the stated ideal potential.

The ground-state energies are $1.50\mathrm{eV}$ and $0.376\mathrm{eV}$, respectively. For a particle in a box of width $L$, the wave function must vanish at both infinite walls. The lowest nonzero standing wave therefore has wavelength $2L$, giving the kinetic energy eigenvalueUsing the electron mass and converting joules to electronvolts givesDoubling the confinement length divides this kinetic energy by four. A zero potential energy inside the box does not imply zero total energy: a constant, zero-momentum wave function cannot satisfy both wall conditions unless it vanishes everywhere. This is a concrete manifestation of quantum confinement and the Heisenberg uncertainty principle.

Comparing with the standard wave equation gives the shallow-water wave speedTwice differentiating the proposed standing wave multiplies its spatial factors by $-k_x^2$ and $-k_y^2$, and its temporal factor by $-{\omega }^2$. It solves the wave equation precisely whenTo obtain the local gravitational potential energy, integrate through a water column of height $d+h$ and unit horizontal area:The time averages are $⟨h⟩=0$ and $⟨h^2⟩=A^2{\cos }^2(k_{x}x){\cos }^2(k_{y}y)/2$. ThusThe constant is retained because the specified zero of gravitational potential is the tank bottom. This is local time-averaged potential energy, not the sum of potential energy and kinetic energy.

The no-normal-flow wall conditions are Neumann boundary conditions. Differentiating the cosine factors at the opposite walls givesNegative integers give the same cosine modes. The formal $m=n=0$ case has $\omega =0$: it is a uniform change of the mean level, not a finite-frequency oscillation. In a closed tank with fixed water volume even that mean-level change is excluded from the oscillatory disturbance.

Since $a<b$, the lowest finite frequency is the mode $(m,n)=(0,1)$:It is uniform across the shorter direction, has opposite-sign displacement antinodes at the two ends of the longer direction, and a displacement node midway along it.

For the open-ended Bay of Fundy model, approximate the ocean mouth as a displacement node and the closed head as an antinode. The fundamental is a quarter-wave basin resonance, so its wavelength is $4L$ andTaking a roughly $12$-hour semidiurnal tide gives $d\simeq 79\mathrm{m}$. Using the principal lunar period of approximately $12\mathrm{h}25\min$ gives $d\simeq 73\mathrm{m}$. The idealized estimate is therefore about $7\text{–}8\times 10^1\mathrm{m}$. The lunar period is half of a roughly $24$-hour $50$-minute lunar day.^[ref]

This estimate treats the mouth as an ideal open boundary and replaces the real basin by a uniform depth. The supplied large tidal height motivates resonance; it is not enough by itself to determine damping or an amplification factor in this model.

Take the threshold from the above-step side used in the preceding parts, $E\to V_0^{+}$. Then $k_2\to 0$ while $k_1\to \sqrt{2m{V}_0}/\hslash$ remains finite. The reflected probability amplitude approaches $B=A$, and the transmitted amplitude approaches $C=2A$. At each fixed position, the potential-step threshold wavefunction is thereforeIts probability density is

The left-hand standing wave has zero-density nodes at $k_{1}x=-\pi /2,-3\pi /2,\dots$ and maxima of height $4|A{|}^2$. Its density period is $\pi /k_1$, half the incident de Broglie wavelength. At the origin the density is a maximum, continuous with the constant value on the right; its first derivative is also zero on either side there.

The nonzero right-hand density does not imply transmitted flux: $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m\to 0$, so $R\to 1$ and $T\to 0$. The exact threshold solution with no incoming component from the right is the limiting constant solution of ${\psi }^{\prime \prime }=0$ in the raised region. These stationary scattering states are not normalized over the whole real line; finite-interval density ratios are nevertheless well defined.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, as a basis near their Bragg degeneracy. The potential has Fourier coefficients $V_{\pm \mathbf{G}}$; its zero Fourier coefficient vanishes. All other plane waves are off resonance and are neglected in the leading nearly-free electron model. With ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor the real cosine potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.

At the Brillouin-zone boundary, $\mathbf{k}\cdot \mathbf{G}=G^2/2$, the free energies are equal. The two standing waves have different probability densities relative to the lattice potential and are split byRepeating this avoided crossing at the reciprocal lattice boundaries joins smooth branches into energy bands. In one dimension a nonzero Fourier coefficient opens a gap at the corresponding crossing. In higher dimensions a direct gap at one wavevector need not give a common forbidden energy interval at every wavevector: different bands may still overlap in energy. The present two-dimensional example remains metallic even though the crossings are locally gapped.

For the rectangular lattice, choose real-space primitive vectors $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe occupied disk has precisely the area of the first Brillouin zone, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two electrons per cell do not imply a filled first band in the absence of a separating band gap.

Fold the protruding caps by $\pm {\mathbf{G}}_x$. The original circle and the translated circle intersect at $k_x=K$, $k_y=\pm \sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an avoided crossing. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

The largest pocket in this weak-potential construction is the lens-shaped intersection of two disks enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta =\arccos (K/k_F)$, two sectors minus their two triangles give the lens bound on a reconstructed Fermi pocket:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. An orbit on this closed contour can return to itself under the Lorentz force and occupy quantized Landau levels. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.

For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ this reproduces the lens bound on a reconstructed Fermi pocket; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\mathrm{eV}$. Numerical integration gives $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis estimate neglects remote plane-wave admixtures and other Fourier coefficients. It also holds the chemical potential fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.

For real $x=ka$, taking the modulus squared of the amplitude gives the half-line delta-well amplitude enhancementThe supplied resonance poles have real parts $x_n\simeq n\pi (1+\lambda /2+{\lambda }^2/4)$ and negative imaginary parts $-{\gamma }_n\simeq -(\lambda n\pi /2{)}^2$ for fixed low $n$ at $\lambda \ll 1$.^[ref] Linearizing the denominator at a simple pole givesThus the sketch has narrow, tall peaks with a locally Lorentzian line shape, just above positive integer multiples of $\pi$. Their half-widths are ${\gamma }_n$, and the leading peak heights are $4/({\lambda }^2{n}^2{\pi }^2)$. In this fixed-low-mode approximation, increasing $n$ broadens and lowers the peaks. Exactly at $x=n\pi \ne 0$, the value is one. At zero it has the finite limit ${\lambda }^2/(2-\lambda {)}^2$ for $\lambda \ne 2$.

The peaks represent long-lived standing waves between the wall and the attractive delta potential. Despite the source's transmission-coefficient terminology, $|A{|}^2$ is an internal amplitude factor and can exceed one. The interior probability current is zero, and the reflected probability current equals the incident magnitude; the hard wall passes no transmitted flux.

Each supplied eigenvector gives relative displacements at $x=1,2,3$ in a standing wave. Insert zero at $x=0,4$ to describe the whole sampled string. A real normal mode is $u_j(t)=C{v}_j\cos (\omega t+\varphi )$: its shape stays fixed while its amplitude oscillates.

These semidiscrete fixed-string normal modes are samples of $\sin (k\pi x/4)$, with $k=3,2,1$ in the order listed. More precisely, their interior vectors are $v_j^{(k)}=2^{-1/2}\sin (k\pi j/4)$, which have unit Euclidean norm and form an orthonormal basis of ${\mathbb{R}}^3$. Their eigenvalues and angular frequencies obeyThus larger absolute value of a negative eigenvalue means faster oscillation, rather than faster decay. The numerical angular frequencies approximate the continuous fixed-string values $k\pi c/L$; the finite difference spatial approximation changes them. Only three independent interior normal modes exist on this grid, and any interior displacement or initial velocity vector can be expanded in this orthonormal basis.

Consider a uniform tube $0\le x\le L$ with rigid ends, constant undisturbed mass density ${\rho }_0$ and speed of sound $c$. Neglect losses and transverse structure. A free normal mode hasand a consistent choice of sound pressure, acoustic particle velocity and acoustic displacement isHere $P>0$ is a peak sound pressure. Direct differentiation verifies ${\rho }_0{\partial }_{t}u=-{\partial }_x{p}^{\prime }$ and ${\partial }_t{p}^{\prime }=-{\rho }_0{c}^2{\partial }_{x}u$, the one-dimensional linear homentropic acoustic equations. The rigid ends enforce $u=0$ there. A driven experiment only approximates this free-mode geometry; the formulas do not predict a finite resonant amplitude for an undamped driven tube.

The maxima of sound pressure amplitude occur at $kx=j\pi$, where acoustic particle velocity vanishes. The maxima of acoustic particle velocity amplitude instead occur at $kx=(j+1/2)\pi$, where sound pressure vanishes. The two families are displaced by $\lambda /4$, but each family repeats after $\lambda /2$. At a position where both are nonzero, the time dependence is also a quarter cycle apart. Using consecutive maxima of sound pressure amplitude therefore givesThe relation also applies to consecutive minima of sound pressure amplitude, provided the same family is used throughout. Mixing maxima and minima would halve the spacing and give an erroneous speed of sound.

The local acoustic energy density is $p^{\prime 2}/(2{\rho }_0{c}^2)+{\rho }_0{u}^2/2$. Averaging over one wave period gives $P^2/(4{\rho }_0{c}^2)$, independent of $x$. Likewise $⟨p^{\prime }u⟩=0$: this ideal standing wave has no mean longitudinal acoustic energy flux, although its pressure and fluid velocity oscillate. Energy transport into a real lossy tube is a different question from the free-mode pattern.

An ideal uniform shallow basin, closed at one end and held at zero displacement at its open end, has its lowest standing wave with a displacement node at the opening and an antinode at the closed end. Its wavelength is four times its length. Real mouths need not satisfy this perfectly reflecting idealization.

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 $h\to 0$.

A slotted line samples an electromagnetic field along a guide with a movable probe introduced through a longitudinal slot. Positions of standing wave minima determine the guided wavelength; a detector response need not be a calibrated field amplitude.

MIT's documented slotted-line waveguide measurement shorts one end of a rectangular electromagnetic waveguide and excites it through a probe coupled by capacitance. A second such probe moves through a longitudinal slot, aligned with the wall electric current to reduce disturbance. Its output locates axial standing wave nulls. The book's standing-wave setup chooses a frequency above the first waveguide cutoff frequency and below higher-mode cutoffs. Lowering it below the first cutoff instead reveals spatial decay away from the source. The text supplies apparatus and an observable comparison, rather than a table of measured null positions.^[ref]

The inference can be checked without assuming that detector output is proportional to electric field amplitude. At fixed measured frequency, record several minima at coordinates $z_0,\dots ,z_N$; then estimate ${\lambda }_g=2(z_N-z_0)/N$ when the minima are consecutive. This uses positions alone. Compare the result with the guided wavelength prediction using independently measured $a$ and $f$. The $40\mathrm{mm}$ spacing above is a synthetic example, not a reading from the MIT apparatus.

Additional proposed controls are to keep probe insertion fixed, repeat the scan in both directions, recheck source frequency, and compare shallow probe settings for shifts in the minima. These test probe loading, mechanical backlash and drift. A deliberately reflecting termination supplies the standing wave; varying its position should translate the pattern without changing adjacent-null spacing in a uniform single-mode region. Source-adjacent evanescent waves, unintended reflections, additional propagating normal modes, detector noise and finite positional resolution can spoil that inference. A disappearing remote signal alone does not distinguish below-cutoff decay from loss or poor excitation, and a scan at one frequency does not measure a pulse's group velocity.

A standing wave results from the superposition principle applied to equal-frequency, equal-amplitude waves travelling in opposite directions. Its spatial nodes of a standing wave remain fixed while its displacement oscillates in time.

For a pure standing wave, the two travelling waves also have equal amplitudes: their interference leaves a fixed mode shape, rather than a profile moving along the string. Choose the position origin and time origin so that their transverse displacements areHere $a>0$ is each constituent's amplitude, $k=2\pi /\lambda$ is the wavenumber, $\lambda$ is the wavelength, and $\omega =2\pi f$ is the angular frequency corresponding to frequency $f$. Their wave speed is $c=\omega /k$. The superposition principle and the sine addition formula giveThe spatial factor stays put; the temporal factor changes its size and sign. Other constant wave phases shift the position or time origin. Unequal constituent amplitudes generally leave moving displacement at the would-be nodes of a standing wave, rather than this exact factorization.^[ref]

For an original ideal-string example, set $L=1\mathrm{m}$, $c=20\mathrm{m}{\mathrm{s}}^{-1}$, and choose the third normal mode. Then $k=3\pi {\mathrm{m}}^{-1}$, $\omega =60\pi {\mathrm{s}}^{-1}$, and $f=30\mathrm{Hz}$ by fixed-end standing wave frequencies. With each constituent amplitude $a=5\mathrm{mm}$,Its nodes of a standing wave are at $x=0,1/3,2/3,1\mathrm{m}$ and its antinodes at $x=1/6,1/2,5/6\mathrm{m}$. These are constructed model parameters, not experimental readings.

At $t=1/120\mathrm{s}$, every point has zero displacement, but most have nonzero transverse velocity: the string is crossing its straight configuration. The black dots alone mark permanent nodes of a standing wave. At an antinode the greatest transverse speed is $2a\omega =0.6\pi \mathrm{m}{\mathrm{s}}^{-1}$, much smaller than the chosen $20\mathrm{m}{\mathrm{s}}^{-1}$ wave speed. One is motion of string material; the other is propagation of the constituent profiles.

The beauty of a standing wave is that motion reveals geometry: fixed nodes of a standing wave turn a moving disturbance into a spatial ruler. Their separation measures wavelength, and boundary conditions turn the continuous choice of wavenumber into a discrete set of normal modes. This is useful both for interpreting string vibrations and for understanding how a bounded wave equation selects its eigenfrequencies.