Acoustic compliance, inertance and impedance

For this very special case, we define the characteristic acoustic impedance Z₀, where

Z₀  =  p/U  =  p/Au  =  z/A     so

Z₀  =  ρv/A

Of course, there usually are reflections from the other end of the pipe, whether open or closed. So there is a sum of waves travelling to the right and left and quite often these give strong resonances, which is after all how musical wind instruments work. For ducts however, the characteristic acoustic impedance Z₀, is the dimensional quantity that scales the acoustic impedance. To see what the impedance spectra of simple ducts and musical instruments look like, see What is acoustic impedance?

It is possible to have no reflections, however. If the pipe length L is very long, then it takes a long time for the reflection to return. So, for the time 2L/v, the duct is effectively infinite. Further, when the echos do eventualy return, they are much attenuated by losses at the wall. (I mention this because, in our research laboratory, we use acoustically infinite acoustical waveguides as calibrations for acoustic measurements. See this web site for the research lab, this page for measurement techniques or this scientific paper for some technical information, if you’re interested.)

Acoustic inertance

Z  =  jω.ρ.L/A ,    where j²  =  –1. 

Let's check the units: Z is pressure over flow, so (kg.m.s⁻²/m²)/(m³/s) = kg.m⁻⁴s⁻¹. On the right hand side, ωρL/A has units (s⁻¹).(kg.m⁻³).m/m⁻² = kg.m⁻⁴s⁻¹. The inertance is just L = ρL/A.

Here, the pressure is required because the mass of the air being accelerated has mass. That's why Z is proportional to ω: vibration at higher frequency requires greater acceleration and therefore greater acoustic pressure. For that reason, an impedance in which the pressure leads the flow by 90° is called an inertive impedance. You can remember the phase difference with this simplified argument: you need to apply the pressure for a while so that the acceleration produces a flow, so the pressure is ahead of the flow.

(You may wonder why the inertance has area in the denominator, when it would be in the numerator for inertia. I did, when I first encountered it. Yes, the mass has A in the numerator. But there are two factors giving A in the denominator: first, our numerator for impedance is pressure, which has A in the denominator, and our denominator is U = uA, so another A in the denominator.)

Acoustic Compliance

Now let's consider a small volume V, again with dimensions << λ. This time, the applied pressure will compress it, so let's imagine a cylinder of air in a duct with cross section A, that is closed at the right hand end. We apply an acoustic pressure p at the left hand end. This compresses it with an oscillating displacement, a the left end only, y  =  y_m cos ωt. At the instant of my sketch, p is positive and ΔV negative. The volume is changed over time by

ΔV  =  y_mcos ωt  =  (ΔV)_m cos ωt.

The flow U into the volume is – ∂V/∂t, so

U  =   – ∂/∂t (ΔV)_m cos ωt  =  ω (ΔV)_m sin ωt

How much pressure is required to compress a volume of air by ΔV/V?  The answer is complicated because, when you compress a gas, its temperature rises, particularly if you compress it so fast that there is insufficient time for heat to be conducted away. In the section on the

p  =  – γ P ΔV/V

Here, the fractional change in volume is (ΔV)_m cos ωt /V. Now the pressure at equilibrium is atmospheric pressure P_A. So the fractional change in pressure is

p_m/P_A  =  – γ (ΔV)_m cos ωt/V      so       p_m =  – γ P_A (ΔV)_m cos ωt/V

So, again, when defining the ratio of Δ p to U, we must include this phase difference. The

magnitude of the ratio is just

Δp/U  =  magnitude of (– γP_A (ΔV)_m cos ωt /V)/(ω(Δ V)_m sin ωt)  =  γP_A/ωV.

Z  =  γP_A/ωV  =  1/ωC      and has a negative phase of 90°

Z  =  – jγP_A /ωV  =  γP_A/jωV  =  1/jωC

The air in the confined volume acts like a spring. In analogy with mechanical compliance, a compact enclosed volume is called a compliance and an acoustical impedance in which the pressure lags the flow by 90° is called a compliant impedance. You can remember the phase difference with this simplified argument: you need the flow for a while before the pressure increases much, so the pressure must lag the flow.

You've possibly already thought that a compliance and an inertance could produce resonant oscillations, just like a capacitor and inductance in AC circuits. A simple example of such an acoustic resonance is a Helmholtz resonance.

The AC electricity analogies

One warning about this analogy: Kirchoff's law for electricity should be applied with caution: the fact that air is compressible and has non-negligible mass means that one almost always has acoustic compliance and inertance. So an acoustical duct is analogous to an electrical delay or a coaxial cable: both have inertive and compliant (inductive and capacitative) terms distributed continuously along their length.

Radiation impedance

Now we can consider this piston as approximating the flow at an open pipe. So, for a pipe ending in a large flange, there is an inertance due to the volume of air, with length δ ~ 0.85a, that moves with the wave at the end of the pipe. For a simple, narrow, open pipe, δ is 0.6 a. These are called end effects. For frequencies low enough tht δ << λ, these pipes behave approximately like ideally open pipes, but the radiation inertance at the open end gives them an extra length δ.

Reflection

So, what happens to a wave when it gets to the end of a pipe? Suppose that our pipe is very long, but ends at x = 0, as sketched below. The pipe (x < 0) has a characteristic impedance Z₀, and the radiation impedance at the end is Z_rad. So the impedance at x = 0 depends on which way one 'looks'. 'Looking out' (at x = 0, but looking at x > 0), the impedance at the end of the pipe is Z_rad. 'Looking in' (at x = 0, but looking at x < 0), the impedance at the end of the pipe is Z₀. (As we've seen above, Z₀ is real: the impedance of a very long pipe is resistive, while Z_rad is largely intertive: its impedance is nearly completely imaginary.)

Let's suppose that we send a sine wave along the pipe towards the right. Let the pressure due to this wave be p = p_>sin (kx − ωt) , where the subscript in the amplitude is to remind us that the wave is travelling to the right (See

p_>/U_> = Z₀.

Outside the pipe (at x = 0 but looking out, which we'll write as x = 0₊), there is sound radiation away from the pipe. Let's write that the pressure* looking into the radiation field (at x = 0₊) is p(0₊) = p_radsin (− ωt).g(x) , and the flow due out of the pipe due to this wave is U(0₊) = U_radsin (− ωt).h(x) .   As we also wrote above

p_rad/U_rad = Z_rad ,

remembering that Z_rad is almost entirely imaginary, so that p_rad is almost 90° ahead of U_rad in phase—so at least one of them has to be a complex number.

* We can't write a simple one dimensional wave equation here, because the radiation diverges. The geometry very near the pipe end is actually somewhat complicated, but it's not important to this argument. Away from the end, i.e. for kr >> 1, where r is the distance for the end of the pipe, the amplitudes of both p_rad and U_rad go like 1/r and they are in phase: the local geometry approaches that of a plane wave.

Continuity of pressure. Now exactly at x = 0, at the end of the pipe, the pressure just inside the pipe must equal that just outside. (If we didn't satisfy continuity of pressure, we'd have an infinite pressure gradient and so infinite force.) We also have

Continuity of flow: the flow coming out of the pipe (at x = 0₋) must equal the flow radiating outwards (at x = 0₊). (Continuity of pressure is required because we are not creating or destroying air.) Of course, we can't satisfy those conditions with these two waves: the magnitude of Z₀ is much greater than that of Z_rad so, if we set the flows equal, the pressures can't be equal.

What happens is no surprise: there is a reflection at the open end: let's write the reflected wave, which travels to the left, as p = p_<sin (kx + ωt) and the flow due to this wave is U = U_<sin (kx + ωt). So, at x = 0, we have to satisfy the continuity of pressure and flow conditions:

p_> + p_< = p_rad     and    U_> − U_< = U_rad .

(Why the minus sign? Because U_> and U_rad are flowing out of the pipe while U_< is flowing in.) We can use the equations above for Z₀ and Z_rad to rewrite the equation above right as     p_>/Z₀ − p_</Z₀ = p_rad/Z_rad. This and the equation above left are two equations that can be solved to give the reflection coefficent at the end:

Reflection coefficient  =  p_</p_>  =  (Z_rad − Z₀)/(Z_rad + Z₀)  =  − (1 − Z_rad/Z₀)/(1 + Z_rad/Z₀)

where we remember that the reflection coefficient is in general a complex number (Z_rad is almost completely imaginary). Now, except at very high frequencies, |Z_rad| << Z₀, so the reflection coefficient is almost −1: we have reflection with a change in phase of π in pressure, but no change in phase in the flow. This gives us a pressure node (two waves out of phase add approximately to zero) and flow antinode at an open end (two waves in phase add to give a maximum).

So, we have a reflection at the open end of a pipe: very useful the standing waves in musical instruments, the voice and elsewhere. Examples, explanations and animations are given in Open and Closed Pipes and Pipes and Harmonics.

In the sketch, a wavetrain p_> in red is shown reflecting at the end of a pipe: p_< is in blue and their sum, the standing wave, in black. The radiated wave is purple.

Reflection at a closed end is somewhat easier. If the pipe is ideally closed, then the conservation of flow must mean that U_> and U_< add to zero, which means that they are out of phase by π. Their pressures are in phase, so the closed end of a pipe has an antinode in pressure and a node in flow.

More information

• This is a background page to the multimedia chapter Quantifying Sound.
• The wave equation for sound sets up the physical background.
• Acoustic impedance, intensity and power introduces specific acoustic impedance.
• What is acoustic impedance? gives examples of simple pipes and also musical instruments.
• Further explanations and animations in Open and Closed Pipes
• and in Pipes and Harmonics.