3.2 Acoustic impedance from first principles

Consider a plane wave $p(x, t) = p_0\, \exp[i(\omega t - k x)]$ propagating rightward through a medium of density $\rho$. The particle velocity $v$ is the time-derivative of the displacement of a small element of medium.

Newton’s second law, applied to a slab of the medium of thickness $dx$, gives

$$\rho \frac{\partial v}{\partial t} = -\frac{\partial p}{\partial x}.$$

For our travelling wave, $\partial p / \partial x = -i k p$ and $\partial v / \partial t = i\omega v$, so

$$\rho \cdot i\omega \cdot v = i k p,$$

giving $v = (k/\rho\omega)\, p$. The speed of sound is $c = \omega/k$, so

$$\frac{p}{v} = \frac{\rho\omega}{k} = \rho c \equiv Z.$$

Density times wave speed. ∎

For air at sea level: $\rho_\text{air} \approx 1.2\,\text{kg/m}^3$, $c_\text{air} \approx 343\,\text{m/s}$, so $Z_\text{air} \approx 412\,\text{Pa}\cdot\text{s/m}$.

For water: $\rho \approx 1000\,\text{kg/m}^3$, $c \approx 1480\,\text{m/s}$, so $Z \approx 1.48 \times 10^6$. The ratio is about $\rho_w c_w / (\rho_a c_a) \approx 3600$.