3.1 The impedance problem

Sound travels well in air because the air molecules can move easily. Sound travels well in water for the same reason at the molecular level, but for a different reason at the macroscopic level: water is dense and incompressible, so any compression propagates very rapidly. The relevant property is acoustic impedance, defined as the ratio of pressure to particle velocity in a propagating wave: $Z = p/v$ .

For air at room temperature, $Z_\text{air} \approx 415\,\text{Pa}\cdot\text{s/m}$ . For water — and for the perilymph of the cochlea, which is essentially salt water — $Z_\text{water} \approx 1.5 \times 10^6\,\text{Pa}\cdot\text{s/m}$ . The ratio is about 3500. This is enormous.

Consider what happens when a sound wave in air meets a water surface. The pressure has to be continuous across the interface, and so does the particle velocity. But $Z_\text{water}/Z_\text{air} = 3500$ means that for a given pressure, the water moves about 3500 times less than the air on the other side. The acoustic wave cannot push the water hard enough to keep itself going. Most of its energy reflects back.

You can compute exactly how much. The reflection coefficient at a boundary between two media is

R = \frac{Z_2 - Z_1}{Z_2 + Z_1}.

▶ Derivation: reflection at a boundary between two media Derivation

Consider an acoustic plane wave $p_i = p_0\, e^{i(\omega t - k_1 x)}$ traveling rightward in medium 1 (impedance $Z_1$ ) toward a boundary at $x = 0$ . At the boundary, some of the wave reflects ( $p_r$ ) and some transmits ( $p_t$ ) into medium 2 (impedance $Z_2$ ):

p_r = R\, p_0\, e^{i(\omega t + k_1 x)}, \quad p_t = T\, p_0\, e^{i(\omega t - k_2 x)}.

Two boundary conditions hold at $x = 0$ :

Pressure is continuous: $p_i + p_r = p_t$ at $x = 0$ , giving $1 + R = T$ .
Particle velocity is continuous: $v_i + v_r = v_t$ . Using $v = p/Z$ for a forward wave and $v = -p/Z$ for a backward wave: $(p_i - p_r)/Z_1 = p_t/Z_2$ , so $(1 - R)/Z_1 = T/Z_2$ .

From (1), $T = 1 + R$ . Substituting into (2):

\frac{1 - R}{Z_1} = \frac{1 + R}{Z_2}.

Cross-multiplying: $Z_2 (1 - R) = Z_1 (1 + R)$ , so $Z_2 - Z_1 = R(Z_1 + Z_2)$ , giving

R = \frac{Z_2 - Z_1}{Z_1 + Z_2}.

Plug in $Z_1 = Z_\text{air}$ and $Z_2 = Z_\text{water}$ : $R \approx 0.999$ . The intensity reflection coefficient is $|R|^2 \approx 0.9989$ , which means the transmitted power fraction is

T = 1 - |R|^2 \approx 0.0011,

or in decibels, about $-29.6$ dB. More than 99.9% of incoming sound power would reflect off the cochlear fluid if it were simply exposed to air. A sound at 60 dB SPL in air would arrive at the cochlea at the equivalent of 30 dB SPL — most of speech would fall below the threshold of audibility.

This is what the middle ear has to fix.

⏳ The history — Fletcher, Munson, and the equal-loudness contours

In 1933 Harvey Fletcher and Wilden Munson at Bell Telephone Laboratories published the first systematic measurement of equal-loudness contours — curves in the frequency-intensity plane along which tones of different frequencies sound equally loud. The work required thousands of loudness-matching judgments from trained listeners and produced the now-familiar family of curves showing that human hearing is most sensitive near 3—4 kHz (the ear-canal resonance) and falls off steeply at low and very high frequencies.

Fletcher and Munson’s contours were adopted as an international standard (ISO 226) and are the basis of the A-weighting filter used in environmental noise measurement. The ear-canal and middle-ear transfer functions — the physics developed in this chapter — directly explain the shape of the contours: the 3 kHz sensitivity peak is the quarter-wave resonance of the ear canal, and the low-frequency rolloff reflects the stiffness-dominated impedance of the middle ear.