Lumped Acoustic Elements

The fundamental wavelength of sound produced by a musical instrument is generally much larger than the dimensions of certain of its component parts, such as toneholes and mouthpieces. The behavior of large wavelength sound waves within small structures is generally well approximated by assuming uniform pressure throughout the volume of interest (Fletcher and Rossing, 1991). In this way, we can analyze small acoustic components in terms of ideal lumped elements.

An Acoustic Mass (Inertance)

Figure 4: A short acoustic tube.

• In the low-frequency limit, the air within a short, open tube will be displaced by equal amounts at both its ends when subjected to an external pressure at one end only. If the length and cross section of the tube are given by $L$ and $A$, respectively, then the enclosed air has a mass of $\rho L A$, where $\rho$ is the mass density of air.

• The acoustic response of this system is analyzed by assuming an applied sinusoidal pressure of the form $P(\omega)$. Using Newton's second law (force = mass $\times$ acceleration),
    $$P(\omega) A = \rho L A \, \frac{dV(\omega)}{dt} = \rho L \, \frac{dU(\omega)}{dt},$$ (7)
  where $U(\omega) = A V(\omega)$ is the acoustic volume velocity of the air mass.

• The volume velocity response to the applied pressure will also vary sinusoidally with frequency $\omega,$ so that Eq. (7) reduces to
    $$P(\omega) = \frac{j \omega \rho L}{A} U(\omega).$$ (8)

• The acoustic impedance of the tube is then given by
    $$Z(\omega) = \frac{P(\omega)}{U(\omega)} = \left(\frac{\rho L}{A}\right) j \omega$$ (9)
  (or $Z(s) = (\rho L / A) s$ when expressed in terms of the Laplace transform), where initial conditions are assumed equal to zero.

• In the low-frequency limit, the open tube is called an acoustic inductance or an inertance and it has a direct analogy to the inductance in electrical circuit analysis or the mass in mechanical system analysis.

• The impedance of a mechanical mass is equal to $m s,$ and thus the open tube has an equivalent acoustic “mass” equal to $\rho L / A.$ An inertance is also sometimes referred to as a constriction (Morse, 1981, p. 234).

An Acoustic Spring (Cavity)

• The acoustic analog of the electrical capacitor or the mechanical spring is a cavity, or a tank (Morse, 1981, p. 234).

• In the low-frequency limit, an applied external pressure will compress the enclosed air, which then acts like a spring because of its elasticity. Assuming a cavity volume $Q,$ an increase in applied pressure $P$ will decrease this volume by an amount $-dQ$.

• The ratio of change in volume to original volume is called volume strain or dilation and is given by $\epsilon = dQ/Q.$ Using Hooke's law, we find
    $$P = -K \, \frac{dQ}{Q} = - K \epsilon,$$ (10)
  where $K$ is the bulk modulus as discussed earlier.

• Using Eq. (10) and writing the change in cavity volume, $-dQ$, in terms of the sinusoidal volume velocity $U$ as
    $$-dQ = \int U dt = \frac{U}{j \omega} \:,$$ (11)
  the acoustic impedance of the cavity is given in terms of the Laplace transform by
    $$Z(s) = \left(\frac{\rho c^{2}}{Q}\right) \frac{1}{s} \:.$$ (12)

• The impedance of a mechanical spring is equal to $k/s,$ so by analogy the acoustic cavity has an equivalent “spring constant” equal to $\rho c^{2}/Q.$

The Helmholtz Resonator

Figure 5: The Helmholtz resonator and its mechanical correlate.

• In the “low-frequency limit”, an open tube is a direct acoustic correlate to the mechanical mass.

• In the “low-frequency limit”, a cavity is a direct acoustic correlate to the mechanical spring.

• Using Newton's Second Law to model the air mass in the tube and Hooke's Law for fluids to model the compressibility of the air cavity, a sinusoidal solution can be found with natural frequency $\omega_{0} = c \sqrt{A/(L Q)}$, where $c$ is the speed of sound in air, $A$ is the cross-sectional area of the tube, $L$ is the length of the tube, and $Q$ is the volume of the cavity.