11.1 The long-wavelength limit: from field to circuit

A sound field is described by pressure and velocity at every point. But when the object under study is much smaller than a wavelength, that spatial detail is wasted: the pressure barely varies from one end of the object to the other, and the field is captured by a few bulk numbers. This lesson makes “much smaller than a wavelength” precise, chooses the two variables that describe a small acoustic element, and derives the three elements every acoustic circuit is built from.

The lumped criterion

Sound of angular frequency $\omega$ has wavelength $\lambda = 2\pi c/\omega$ and wavenumber $k = \omega/c = 2\pi/\lambda$ . Across an object of size $L$ , the phase of the wave changes by $kL = 2\pi L/\lambda$ . If that phase change is small,

\boxed{\;kL \;=\; \frac{2\pi L}{\lambda} \;\ll\; 1,\;}

then the pressure is very nearly the same everywhere on the object at any instant. There is no room for a standing wave to form; the object cannot tell where the wave is in its cycle, only what the pressure currently is. This is the lumped-element regime (also called the acoustically compact regime). When it fails — when $kL$ approaches and exceeds 1 — spatial structure reappears and we are back to the full wave equation of Chapter 7.

where

$k = \omega/c$: wavenumber 1/m
$\lambda = 2\pi/k$: wavelength m
$L$: characteristic size of the object m
$kL$: phase change across the object — the dimensionless smallness parameter —

regime: borderline

f = 1000 Hz λ = 34.3 cm kL = 0.46

tube length L = 2.5 cm frequency f = 1000 Hz

The lumped-element picture is the low-frequency limit kL ≪ 1: when the object is far smaller than a wavelength, the pressure is essentially uniform across it and the field collapses to a handful of circuit numbers. Raise f (or L) until standing waves appear and the approximation breaks.

The interactive drives a short closed tube and plots the pressure pattern along it. At $kL \ll 1$ the pattern is flat — one pressure describes the whole tube, and it acts as a single lumped element. Push the frequency (or the length) up until $kL \sim 1$ and the curve bends into a standing wave; the lumped description has broken. For a 25 mm ear canal, $kL = 1$ falls near 2 kHz, which is why the canal is “lumped” through the low frequencies and “distributed” (a quarter-wave resonator) by the upper mid-range — a transition we return to in 11.4.

The two variables: pressure and volume velocity

To describe a lumped element we need one variable for the “effort” driving flow and one for the “flow” itself — the acoustic analogues of voltage and current. They are:

Pressure $p$ — the acoustic pressure (uniform across the element, by the lumped assumption). This is the across variable, the effort.
Volume velocity $U$ — the volume of air per second crossing a surface, $U = S\,v$ for a duct of area $S$ carrying particle velocity $v$ . This is the through variable, the flow.

where

$p$: acoustic pressure across the element Pa
$U = S v$: volume velocity — volumetric flow rate through the element m^3/s
$S$: cross-sectional area of the opening m^2
$v$: acoustic particle velocity in the opening m/s

Volume velocity is the right flow variable because it is what is conserved where ducts of different area meet: mass conservation for an incompressible plug of air means $U$ is continuous across a junction even when $v$ jumps. The ratio of pressure to volume velocity defines the acoustic impedance $Z_a = p/U$ (units $\text{Pa·s/m}^3$ ), the quantity the rest of the chapter computes for each element and combination.

Element 1 — the compliance of a cavity

Take a rigid-walled cavity of volume $V$ , small compared with a wavelength, with a single opening. Push a volume velocity $U$ in through the opening; the trapped air compresses, and its pressure rises. Because the cavity is compact, that pressure is uniform, and the air acts as a spring: pressure proportional to the volume pushed in.

▶ Acoustic compliance C_a = V/(ρ₀c²) Derivation

Pushing volume $\delta \mathcal V = \int U\,dt$ into a rigid cavity reduces the air’s volume available… more precisely, it adds mass and compresses the gas. For a compact cavity the density rises uniformly by $\rho'$ , and conservation of mass gives the added volume as $\delta\mathcal V = (\rho'/\rho_0)\,V$ (a fractional density rise $\rho'/\rho_0$ over the whole volume $V$ ).

The gas responds adiabatically (4.4): $p = c^2\rho'$ , so $\rho' = p/c^2$ and

\delta\mathcal V \;=\; \frac{V}{\rho_0 c^2}\; \rho_0 \cdot \frac{p}{\rho_0 c^2}\ \ \Longrightarrow\ \ \delta\mathcal V \;=\; \frac{V}{\rho_0 c^2}\,p.

Define the acoustic compliance as the constant of proportionality between the volume pushed in and the pressure it produces, $\delta\mathcal V = C_a\,p$ :

\boxed{\;C_a \;=\; \frac{V}{\rho_0 c^2}.\;}

Differentiating $\delta\mathcal V = C_a p$ in time gives the element’s law in flow form, $U = C_a\,\dfrac{dp}{dt}$ — the acoustic capacitor. A bigger cavity is a softer spring (larger $C_a$ ): easier to push volume in for a given pressure.

where

$C_a = V/\rho_0 c^2$: acoustic compliance — the 'springiness' of trapped air m^3/Pa
$V$: cavity volume m^3

This is the same trapped-air response an audiologist measures directly: the equivalent-ear-canal volume of a tympanogram is a reading of $C_a$ for the air in front of the probe (Tools of Audiology 4.1).

Element 2 — the inertance of a short duct

Now the opposite element: a short open duct (a neck, a port) of area $S$ and length $\ell$ , small compared with a wavelength. The air inside it is too short to compress appreciably, but it has mass, and to drive volume velocity through it you must accelerate that mass. The duct is an inertia.

▶ Acoustic inertance M_a = ρ₀ℓ/S Derivation

The plug of air in the duct has mass $m = \rho_0\,\ell\,S$ . A pressure difference $p$ across the duct exerts a net force $F = p\,S$ on the plug (pressure times area). Newton’s second law for the plug moving at particle velocity $v$ :

\rho_0\,\ell\,S\;\frac{dv}{dt} \;=\; p\,S.

Write it in terms of volume velocity $U = Sv$ , so $v = U/S$ and $dv/dt = (1/S)\,dU/dt$ :

p \;=\; \frac{\rho_0\,\ell}{S}\;\frac{dU}{dt} \;\equiv\; M_a\,\frac{dU}{dt}, \qquad \boxed{\;M_a \;=\; \frac{\rho_0\,\ell}{S}.\;}

The acoustic inertance (or acoustic mass) $M_a$ is the constant relating pressure to the rate of change of volume velocity — the acoustic inductor. A longer, narrower neck has more inertance: more air to accelerate, forced through a smaller aperture.

where

$M_a = \rho_0\ell/S$: acoustic inertance (acoustic mass) — the inertia of a plug of air kg/m^4
$\ell$: length of the duct (with end correction — see 11.3) m

A subtlety we take up in 11.3: the air just outside each end of the duct also gets dragged along, so the effective length is a little longer than the geometric $\ell$ . This end correction matters for any real port.

Element 3 — resistance

The third element dissipates rather than stores. Two mechanisms give an acoustic resistance $R_a$ , the element for which pressure is in phase with flow, $p = R_a\,U$ :

Viscous resistance — air dragging against the walls of a narrow duct loses energy to viscosity (Chapter 10.1). It dominates in thin slits and meshes, and it is how the damping screens in a hearing aid or the acoustic mesh over a microphone port are engineered.
Radiation resistance — an opening that faces free space launches sound away, and that radiated power looks, from the circuit’s side, like a resistance draining energy. It is the useful “loss” of any resonator that is meant to be heard.

Both give $Z_a = R_a$ , a real, frequency-flat (to first approximation) impedance. Unlike the compliance and inertance, a resistance stores no energy and introduces no $90^\circ$ phase shift — it is where acoustic energy leaves the circuit, either as heat or as radiated sound.

The three elements together

That is the entire toolkit — three elements, each a pressure–flow law and its single-frequency impedance:

Compliance $C_a = V/\rho_0 c^2$ — a compressed cavity, the spring. Law $U = C_a\,\dot p$ ; impedance $Z_a = 1/(i\omega C_a)$ .
Inertance $M_a = \rho_0\ell/S$ — an accelerated plug, the mass. Law $p = M_a\,\dot U$ ; impedance $Z_a = i\omega M_a$ .
Resistance $R_a$ — viscous or radiation loss. Law $p = R_a\,U$ ; impedance $Z_a = R_a$ .

The impedances are read off by substituting a single-frequency drive $e^{i\omega t}$ , exactly as for the RLC circuit of Foundations 3.3: $\dot p \to i\omega p$ turns the compliance law into $U = i\omega C_a p$ , i.e. $Z_a = p/U = 1/(i\omega C_a)$ , and likewise $M_a\dot U \to i\omega M_a U$ gives $Z_a = i\omega M_a$ . The next lesson makes the correspondence with electrical circuits explicit and shows how to combine elements; 11.3 puts an inertance and a compliance in series to build the Helmholtz resonator.