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Eliminating the velocity and density perturbations from the linear homentropic acoustic equations gives

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:

Define $k_0=\omega /c_0$ and use the outgoing acoustic square-root branchFor positive real frequency reached from below, $\gamma =i\sqrt{k_0^2-k^2}$ on the propagating interval and is positive real for $|k|>k_0$. The outgoing field in the lower fluid has pressure amplitude $B{e}^{\gamma y}$, since $y<0$. If $\eta =\hat{\eta }{e}^{i\omega t-ikx}$, the shared normal velocity is $i\omega \hat{\eta }$. The linear homentropic acoustic equations therefore giveThis lower-fluid wave is outgoing; no additional incoming sound is included in defining the impedance seen by the upper fluid.

Put $K_s=T{k}^2-m{\omega }^2$. The sheet's force balance gives $K_s\hat{\eta }=-P+B$, hence $P=({\rho }_0{\omega }^2/\gamma -K_s)\hat{\eta }$. Its prescribed downward velocity amplitude is $V=-i\omega \hat{\eta }$. Thus the tensioned-sheet acoustic impedance isThe first term is the lower fluid's normal acoustic impedance, and the second is the sheet's inertial and elastic-sheet tension response. For a real propagating angle, $Z_f={\rho }_0{c}_0/\sin \theta$.

At fixed nonzero frequency and fixed wavenumber, $m\to \infty$ gives $|Z|\to \infty$, a zero-velocity, in-phase reflecting boundary. At fixed $k\ne 0$, $T\to \infty$ gives the same reflection limit, but there is an important exception: elastic-sheet tension does not resist the spatially uniform mode $k=0$. At normal incidence, the impedance remains ${\rho }_0{c}_0+im\omega$ however large the elastic-sheet tension is. With $m=0$ this mode is transparent. By contrast, arbitrarily large mass resists even a spatially uniform oscillation. These fixed-frequency limits exclude a simultaneously tuned structural resonance.

If $m=T=0$, $Z=Z_f$ and $R=0$. The identical fluids are effectively joined across a massless, untensioned interface: pressure and normal velocity continue without reflection. This is the matched case, rather than the pressure-release case.

Write the density and pressure as ${\rho }_0+{\rho }^{\prime }$ and $p_0+p^{\prime }$, and let the small fluid velocity be the gradient of an acoustic velocity potential, $\mathbf{u}=\nabla \varphi$. The linear homentropic acoustic equations arewhere the adiabatic sound speed of a perfect gas is $c_0^2=\gamma p_0/{\rho }_0$. Absorbing a function of time into $\varphi$, the momentum equation givesEliminating $p^{\prime }$ and ${\rho }^{\prime }$ gives the acoustic wave equation

Let the shell displacement be $a(t)-a_0=\mathrm{Re}(\eta e^{-i\omega t})$ and put $k=\omega /c_0$, $\theta =k{a}_0$. The spherically symmetric solution regular at the centre isThe kinematic boundary condition ${\varphi }_r(a_0,t)=\dot{a}(t)$ givesSince $p^{\prime }=-{\rho }_0{\varphi }_t$, the pressure amplitude on the shell is thereforeThe radial Newton's second law per unit area isso its complex amplitude satisfiesFor a nonzero mode, cancellation of $\eta$ and multiplication by $a_0^2/(m{c}_0^2)$ yield the frequency equation for spherically symmetric vibration of a gas-filled elastic shell:The spectrum is discrete and contains infinitely many positive solutions. Indeed, for sufficiently large $n$, $\theta \cot \theta -1$ runs from $+\infty$ to $-\infty$ on $(n\pi ,(n+1)\pi )$, forcing at least one root there. These high modes are predominantly standing acoustic modes of the gas, while the lowest mode is predominantly the shell's radial breathing mode; the coupling shifts both families.

The linear homentropic acoustic equations areDifferentiate the first equation in time, take the divergence of the second, and use the third. This eliminates ${\rho }^{\prime }$ and $\mathbf{u}$ and gives the acoustic pressure wave equation

Write the density and pressure as $\rho ={\rho }_0+{\rho }^{\prime }$ and $p=p_0+p^{\prime }$, and let $\mathbf{u}$ be the small fluid velocity. The linear homentropic acoustic equations arewhere the adiabatic sound speed isIf the flow is irrotational, write $\mathbf{u}=\nabla \varphi$. The momentum equation integrates spatially toa purely time-dependent integration term having been absorbed into $\varphi$. Substitution into continuity gives the wave equationso the wave speed is $c_0$.

To obtain energy conservation, dot the momentum equation with $\mathbf{u}$ and use the product rule:The continuity and pressure-density relations giveTherefore the acoustic energy conservation law iswith acoustic energy density and intensity

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.