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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.

An antinode is a position of maximum oscillation amplitude in a standing wave. For $u_n=A\sin (n\pi x/L)\cos ({\omega }_{n}t+\phi )$, the antinodes occur atwhere the amplitude is $|A|$. They sit halfway between consecutive nodes of a standing wave. For the first normal mode, the sole antinode is $x=L/2$. It reaches zero displacement at a quarter oscillation period while moving fastest, which illustrates why an antinode is defined by oscillation amplitude rather than instantaneous height.

For a rectangular membrane mode with peak antinode displacement A, the integrals of each squared sine factor are half the corresponding side length. Integrating local kinetic energy density $\rho ({\partial }_t\psi {)}^2/2$ gives $K=\rho ab{\omega }^2{A}^2{\sin }^2(\omega t+\varphi )/8$. The total energy is $\rho ab{\omega }^2{A}^2/8$, and potential energy has the complementary cosine-squared factor. Its time-averaged kinetic energy is $\rho ab{\omega }^2{A}^2/16$. At equal peak displacement, frequency and density, such mean energies are proportional to area.

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 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 standing wave has a fixed spatial amplitude profile and oscillates in time, with stationary nodes of a standing wave. It arises from equal-amplitude waves of the same frequency travelling in opposite directions. For a uniform string with linear mass density $\mu$, the wave speed is $c=\sqrt{T/\mu }$ and a separated mode has $Y^{\prime \prime }+k^{2}Y=0$.

The fixed boundary requires $Y(0)=0$, while the free boundary requires $T{Y}^{\prime }(L)=0$. Thus $Y=A\sin kx$ and $\cos kL=0$, givingThe fixed end is a node of a standing wave and the free end an antinode. Arbitrary small motions are superpositions of these normal modes; the tension alone determines their frequencies only when $\mu$ is also specified.

For two equal-amplitude waves with the same frequency and wavenumber, choose their phases so thatThe spatial amplitude and temporal oscillation separate. Positions with $\cos kx=0$ are fixed nodes of a standing wave, while positions with $|\cos kx|=1$ are antinodes. Different initial phases merely shift this pattern in space and time.

A travelling wave has a phase profile that moves at its phase velocity; a standing wave has stationary nodes and each point oscillates with its fixed local amplitude. The two counterpropagating waves carry equal opposite time-averaged energy flux, so their superposition has zero net time-averaged energy transport. Instantaneous energy can still move within the standing pattern.

Free ends give $Y^{\prime }(0)=Y^{\prime }(L)=0$, so the oscillatory normal modes areThere is also the zero mode $k_0=0$, with uniform displacement $y_0(x,t)=a+bt$: a free string can translate without a restoring force. It is not a positive-frequency oscillation. The first three oscillatory shapes are $\cos (\pi x/L)$, $\cos (2\pi x/L)$ and $\cos (3\pi x/L)$, with an antinode at each end. The figure includes the uniform mode too, so that the zero-frequency limit is explicit.

A free end has zero transverse force, hence $T{Y}^{\prime }(L)=0$, rather than zero displacement. With $Y(0)=0$, write $Y=A\sin kx$. The remaining condition is $\cos kL=0$, givingThe first three shapes have one, three and five quarter-waves. Each has a node of a standing wave at the fixed end and an antinode at the free end; the mode frequencies are in the ratio $1:3:5$.

Substitution into the wave equation gives ${\partial }_x^2\psi +{\partial }_y^2\psi =-(k_x^2+k_y^2)\psi$ and ${\partial }_t^2\psi =-{\omega }^2\psi$. Therefore ${\omega }^2=c^2(k_x^2+k_y^2)$. The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges requireNeither integer may vanish, because that would make the entire displacement zero.

The element's mass is $\rho dxdy$, and its speed is $-\omega A\sin (k_{x}x)\sin (k_{y}y)\sin (\omega t)$. Squaring and multiplying by half its mass gives the requested energy density. The two sine-squared spatial integrals are $a/2$ and $b/2$, soThere is no dissipation. The total mechanical energy equals the maximum kinetic energy, givingEquivalently, integrate the small-slope tension energy $\rho c^2|\nabla \psi {|}^2/2$ and use the dispersion relation.

For $b=4a/5$, the two lowest modes are $(p,q)=(1,1)$ and $(2,1)$:The fundamental has one antinode at $(a/2,b/2)$ and no interior nodal line. The next mode has a node along $x=a/2$ and opposite-sign antinodes at $(a/4,b/2)$ and $(3a/4,b/2)$. Every boundary is a node.

The square fundamental has $f=c\sqrt{2}/(2\beta a)$. Equating it to $f_{11}$ givesAt the same peak displacement amplitude and frequency, time-averaged kinetic energy is proportional to membrane area. ThereforeThe reciprocal ratio is $41/40$ if the rectangle is placed first.

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.

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.

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.

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.

A stationary mode shape does not imply absence of energy. Write $A=2a$ and $y=A\sin (kx)\cos (\omega t)$. For the ideal wave equation on a string, the kinetic energy and elastic potential energy per unit reference length add toHere $P$ is signed power crossing a position on the string, positive to the right; $e$ is an energy density per length, not per volume. Differentiation and $\mu y_{tt}=\tau y_{xx}$ give the local conservation of energy identity ${\partial }_{t}e+{\partial }_{x}P=0$. Substitution into $P$ givesIt can be nonzero away from a node of a standing wave or an antinode, but its time average over an oscillation period is zero. Energy flows locally back and forth even though this pure standing wave transports no net energy across a section over a complete cycle. At a node of a standing wave, $y_t=0$, so kinetic energy density is zero, but $y_x$ need not vanish: elastic potential energy density can be positive there.

For a fixed-end normal mode, integration over $0\le x\le L$ givesThe last equality uses $\tau k^2=\mu {\omega }^2$. At a straight-string snapshot, $U=0$ while $K=E$; at an extreme snapshot, $K=0$ while $U=E$. This is normal-mode energy exchange. A driven real string with damping requires continuing input power to replace lost energy, so a visually stationary pattern does not establish zero average supplied power.