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An acoustically rigid boundary has zero perturbation normal velocity. For a reflected acoustic plane wave, pressure amplitudes add in phase and the reflection coefficient is $+1$. It is the infinite-surface acoustic impedance limit. A pressure-release boundary instead fixes the pressure perturbation to zero and reflects with coefficient $-1$.

For a progressive acoustic plane wave in a homogeneous fluid, the pressure-to-particle-speed ratio is $Z=\rho c$, where $\rho$ is mass density and $c$ is the speed of sound. Its units are $\mathrm{Pa}s/m$. This specific impedance differs from pressure divided by a volume flux, whose value also depends on the port area.

At angle $\theta$ from an interface normal, the normal particle-velocity amplitude is $U_n=U\cos \theta$. Therefore the normal acoustic impedance isFor two lossless fluids with propagating incident and transmitted waves, continuity of pressure and normal velocity givesand hence the pressure-amplitude reflection coefficientZero reflection requires equal normal acoustic impedances; equal bulk values $\rho c$ alone suffice only at normal incidence. Using particle displacement rather than velocity multiplies both normal impedances by the same frequency factor and does not replace $Z/\cos \theta$ by $Z\cos \theta$.^[ref]

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.

With pressure amplitude $P=ZV$ and velocity $V$ directed into a boundary, the mean absorbed acoustic energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. A passive boundary cannot supply net energy, so $\mathrm{Re}Z\ge 0$ at a real frequency. Its reflection from a propagating acoustic plane wave has magnitude at most one. A pole continued to a complex incidence angle or frequency must be interpreted separately from real-frequency passive scattering.

Take $0<{\theta }_0<\pi$, put $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0>0$, and suppress the common factor $e^{i\omega t}$. The upper-half-plane incident acoustic plane wave is $e^{iax+iby}$. A rigid zero-thickness screen imposes the same normal-derivative cancellation on both faces; the scattered velocity potential is consequently odd in $y$. Across the open part $x>0$ it is continuous, so its trace there is zero. These complementary half-line data are what make the Wiener-Hopf method applicable.

Use the Fourier transform $\hat{\varphi }(\kappa ,y)={\int }_{\mathbb{R}}\varphi (x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing Helmholtz equation solution is $\hat{\varphi }=F^{-}(\kappa )e^{-\gamma y}$ for $y>0$, where $\gamma =\sqrt{{\kappa }^2-k_0^2}$ has $\gamma =i\sqrt{k_0^2-{\kappa }^2}$ on the propagating interval and positive real part on evanescent components. The limiting absorption principle supplies the contour prescription. The transform $F^{-}$ of the trace supported on $x<0$ is analytic below the contour. Let $G^{+}$ be the analytic upper transform of the unknown derivative on $x>0$. The prescribed derivative on $x<0$ has transformSince the upper normal derivative is $-\gamma F^{-}$, the Wiener-Hopf equation isFor the Wiener-Hopf factorization, choose ${\gamma }_{+}=(\kappa -k_0{)}^{1/2}$ analytic above and ${\gamma }_{-}=(\kappa +k_0{)}^{1/2}$ analytic below. Their branch cuts go into the opposite half-planes, with ${\gamma }_{+}(-a)=i\sqrt{k_0+a}$. Dividing by ${\gamma }_{+}$ and performing pole subtraction in a Wiener-Hopf equation givesThe forcing pole is allocated to the lower analytic part; the second quotient has a removable pole and is upper analytic. Subtracting these parts leaves a common entire function. The finite-energy edge condition excludes an inverse-square-root singularity in the potential: its odd scattered trace is $O(\sqrt{|x|})$, so $F^{-}=O(|\kappa {|}^{-3/2})$. The normal derivative on the open side can be $O(x^{-1/2})$, giving $G^{+}=O(|\kappa {|}^{-1/2})$. Both sides of the entire remainder tend to zero. The Liouville theorem sets that remainder to zero, giving the explicit Wiener-Hopf solution of rigid half-plane diffraction:The scattered field on either side of the screen is thereforeIts forcing-pole residue at $\kappa =-a$ is $-i$. Deforming to the outgoing steepest descent contour crosses this pole precisely when $|\theta |>\pi -{\theta }_0$, where $x=r\cos \theta$, $y=r\sin \theta$ and $-\pi <\theta <\pi$. The residue contributes $\mathrm{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered geometrical optics field isAdding the incident acoustic plane wave gives incident plus unit specular reflection in the upper reflected sector, just the incident wave in the remaining illuminated sectors, and zero in the lower shadow sector. The geometrical shadow boundaries also follow directly by tracing a straight ray to $y=0$: the reflected ray hits the screen at $x+y\cot {\theta }_0<0$, and the transmitted incident ray crosses the open part at $x-y\cot {\theta }_0>0$.

For the diffracted contribution, the outgoing phase $-i\kappa r\cos \theta -\gamma r|\sin \theta |$ has saddle ${\kappa }_s=k_0\cos \theta$, value $-i{k}_{0}r$, and second derivative $ir/(k_0{\sin }^2\theta )$. Stationary phase therefore supplies $\sqrt{2\pi k_0/r}|\sin \theta |e^{-i{k}_{0}r+i\pi /4}$. Substituting $F^{-}({\kappa }_s)$, and using $\mathrm{sgn}(y)|\sin \theta |/\sqrt{1+\cos \theta }=\sqrt{2}\sin (\theta /2)$, givesThe full asymptotic scattered field is ${\varphi }_{\mathrm{GO}}+{\varphi }_d$. The angular edge-wave expression is not uniform on the geometrical boundaries. There is also a sign inconsistency in the supplied saddle formula: with its printed inverse exponential $e^{+i\kappa x}$, the outgoing saddle is $-k_0\cos \theta$, not $+k_0\cos \theta$. The consistent inverse convention used above gives the requested outgoing field without that ambiguity.

Take a cut-on mode with $k>0$, put $\lambda =n\pi /h$ and $\alpha ={\tan }^{-1}(\lambda /k)$, so $k=k_0\cos \alpha$, $\lambda =k_0\sin \alpha$, and initially take $n>0$. The acoustic waveguide mode splits into two acoustic plane waves:At the upper edge $y=h$, the upward-going constituent is incident from below. Use the reflected local coordinate $Y_{+}=h-y$; its incident potential becomes $(-1{)}^n{e}^{-ikx+i\lambda Y_{+}}/2$, with local incidence angle ${\theta }_0=\pi -\alpha$. At the lower edge use $Y_{-}=y+h$ and the downward-going constituent; it has the same local incident amplitude and incidence angle. At each wall the other constituent is the geometrical optics reflection of this incident half-amplitude, so it must not be counted again as an independent incident wave at the same edge.

Let $D(r)=\sqrt{2/(\pi k_{0}r)}{e}^{-i{k}_{0}r-i\pi /4}$. For a far observer, the upper and lower local observation angles are $-\theta$ and $\theta$, and their distances are $r_{+}=r-h\sin \theta +O(h^2/r)$ and $r_{-}=r+h\sin \theta +O(h^2/r)$. Applying the Wiener-Hopf solution of rigid half-plane diffraction separately givesThe sign difference is the reversal of the upper-edge transverse coordinate. Their sum is the leading independent-edge radiation from an open planar duct:Restore $e^{i\omega t}$ if a time-dependent potential is desired. This is the leading large-$k_{0}h$ approximation, neglecting further diffraction between edges, and assumes a far-field distance $r\gg k_0{h}^2$ as well as $r\gg h$. It is not a uniform approximation for a mode approaching cutoff. The zero mode follows as the grazing-incidence limit $\alpha =0$; its beam-direction limit is treated separately below.

Fix time dependence $e^{-i\omega t}$ and use pressure amplitude for $\psi$. Write the incident acoustic plane wave as ${\psi }_i=A{e}^{i\alpha x-i\beta z}$, where $\alpha =k\cos \theta$, $\beta =k\sin \theta >0$. “Perfectly reflecting” specifies zero absorbed energy but does not uniquely select a boundary condition: a rigid wall has a Neumann boundary condition, whereas a pressure-release surface has a Dirichlet boundary condition. We first give the rigid-wall result, and then the alternative pressure-release result.

For the rigid surface the normal-velocity condition, without an immaterial normalizing factor, isExpand $\psi ={\psi }^{[0]}+{\psi }^{[1]}+{\psi }^{[2]}+\cdots$ in the height amplitude, scaling a fixed differentiable profile. The flat total field is ${\psi }^{[0]}=A{e}^{i\alpha x}(e^{-i\beta z}+e^{i\beta z})$. Taylor expansion givesLet $\hat{h}(q)=\int h(x)e^{-iqx}dx$ and $\beta (q)=\sqrt{k^2-q^2}$ on the outgoing sheet: it is positive real for propagating modes and positive imaginary for evanescent modes. Fourier transforming this boundary derivative gives $2A(k^2-\alpha q)\hat{h}(q-\alpha )$. The outgoing angular spectrum therefore yields the small-height Neumann scattering fieldFor an infinite stationary surface the Fourier quantities may be interpreted as random distributions or through a finite illumination window and its limiting mean. The correction is linear in $h$ and $h^{\prime }$, so zero mean height and stationarity giveOnly the rough correction has zero mean at first order; the flat reflected wave remains.

The first nontrivial coherent correction depends on correlation, not just the r.m.s. height. Define $C_h(s)=⟨h(x+s)h(x)⟩$, its power spectrum of surface height $S_h(q)=\int C_h(s)e^{-iqs}ds$, and ${\sigma }^2=(2\pi {)}^{-1}\int S_h(q)dq$. At second order the flat-field third normal derivative and mixed derivative vanish at zero, leavingFor each first-order outgoing Fourier component, its derivative data are $g_1(q)=2A(k^2-\alpha q)\hat{h}(q-\alpha )$, its second normal derivative is $i\beta (q)g_1(q)$, and its tangential derivative is $q{g}_1(q)/\beta (q)$. Averaging their products with $h,h^{\prime }$ gives the mean derivativeDividing by the specular outgoing derivative $i\beta$ therefore gives oblique coherent reflection from a stationary rigid rough surface:This second-order expression requires the relevant spectral moments to exist and the boundary expansion to remain controlled. Small $|kh|$ alone does not control arbitrarily large slopes or high-frequency roughness. The given stationarity and $\sigma$ determine the first-order mean above, but do not determine this spectrum-weighted correction numerically.

For a pressure-release surface, instead impose $\psi (x,h)=0$. The flat reflected coefficient is $-1$, and Taylor expansion gives ${\psi }^{[1]}(x,0)=2i\beta Ah{e}^{i\alpha x}$. Thus small-height Dirichlet scattering givesThe zero-mean first correction again vanishes on averaging. At second order ${\psi }^{[2]}(x,0)=-h{\partial }_z{\psi }^{[1]}(x,0)$, givingIf the roughness spectrum is concentrated near $q=\alpha$ after the incident tangential shift, the additional long-correlation approximation yields $r_N\simeq 1-2{\beta }^2{\sigma }^2$ and $r_D\simeq -1+2{\beta }^2{\sigma }^2$. A local Gaussian random-height phase model would then give $r_N=e^{-2{\beta }^2{\sigma }^2}$ and $r_D=-e^{-2{\beta }^2{\sigma }^2}$, but Gaussian statistics and that local approximation were not supplied. They must not be inferred from stationarity and r.m.s. height alone.

Use the prescribed harmonic convention $e^{i\omega t}$. For a propagating acoustic plane wave, let $k=(\omega /c_0)\cos \theta$ and $\ell =(\omega /c_0)\sin \theta >0$. The incident and reflected pressure amplitudes in the upper half-space have vertical factors $e^{i\ell y}$ and $e^{-i\ell y}$ respectively:The linear homentropic acoustic equations imply $i\omega {\rho }_0{v}_y=-{\partial }_y{p}^{\prime }$. At the surface, the normal velocity is therefore $\ell (R-A)/({\rho }_0\omega )=-V$, while the pressure amplitude is $P=A+R$. The surface acoustic impedance condition $P=ZV$ givesHere $Z_f$ is the normal acoustic impedance; the angle in this question is measured from the horizontal, not the normal. For a passive acoustic impedance, the mean power absorbed per unit area is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2\ge 0$. The four limiting cases have distinct meanings:

For the incident acoustic plane wave, set $q_0=k\sin \theta$ and ${\beta }_0=k\cos \theta >0$. The flat reflected field and total field areTherefore ${\partial }_z{\psi }^{[0]}(x,0)=-2i{\beta }_0{e}^{i{q}_{0}x}$ and the first-order rough-surface scattered field isIt is linear in the height. Since $⟨h(x)⟩=0$, its mean vanishes, with the expectation interpreted through finite windows or stationary spectral distributions when needed. HenceThe coherent first-order reflection is the flat-surface reflection. If the field symbol is instead used only for the rough correction, its first-order mean is zero. Stationarity ensures the coherent reflection retains the incident horizontal wavenumber, but zero mean height already explains the vanishing linear correction. The nonzero root mean square height does not enter this mean at first order; it does enter the fluctuating reflected field and its intensity.

Linearized momentum balance is ${\rho }_0{\partial }_t\tilde{u}=-\nabla \tilde{p}$. For a progressive acoustic plane wave $\tilde{p}=P(\hat{k}\cdot x-c_{0}t)$, integration givesThe integration constant is zero for the pure wave with no added uniform flow. The instantaneous acoustic intensity is $I=\tilde{p}\tilde{u}$, the pressure-work energy flux; time-averaged intensity is $⟨I⟩$. For complex harmonic pressure amplitude $P$, $⟨I⟩=|P{|}^2\hat{k}/(2{\rho }_0{c}_0)$.

Use phase convention $e^{i(k_{y}y-\omega t)}$, with $\omega =k{c}_0$, $k_y=k\sin \theta$, $k_x=k\cos \theta >0$. Let incident, reflected and transmitted pressure amplitudes be $1,R,T_p$. The two normal velocities at the membrane are $(1-R)\cos \theta /({\rho }_0{c}_0)$ and $T_p\cos \theta /({\rho }_0{c}_0)$, and both equal $-i\omega X_0$. Hence $T_p=1-R$ and $X_0=i{T}_p\cos \theta /({\rho }_0{c}_0\omega )$. Only normal velocity is matched for an inviscid fluid; tangential velocities need not agree.

Put $D_m=k^2(m{c}_0^2-T{\sin }^2\theta )$. The dynamic condition is $-D_m{X}_0+T_p-(1+R)=0$, or $-D_m{X}_0-2R=0$. Since $D_m\cos \theta /({\rho }_0{c}_0\omega )=1/\alpha$, elimination givesComplex conjugation gives the amplitudes for the opposite time-phase convention. At zero $D_m$, the formulas have the limiting values $R=0,T_p=1$ even though $\alpha$ itself is then infinite. A strongly inertial membrane has $\alpha \to 0$, giving almost total reflection.

For real $\alpha$, $|R{|}^2+|T_p{|}^2=1$. Thus the incoming net normal energy flux is $\cos \theta (1-|R{|}^2)/(2{\rho }_0{c}_0)$, equal to the transmitted flux $\cos \theta |T_p{|}^2/(2{\rho }_0{c}_0)$. Time-averaged energy flux is conserved: the lossless membrane stores and returns energy but has no mean dissipation.