We can approximate the acoustical behaviour of the mouthpiece of a typical brass instrument by the simplified Helmholtz resonator model sketched in Fig. 1. The cup has volume $V$, and the neck and backbore are idealised as a cylinder of length $L$ and cross-sectional area $S$. The player injects a volume flow rate $qe^{i \omega t}$ via their lips. The corresponding flow rate through the exit tube is $q_1e^{i \omega t}$. From now on, for clarity I will omit the factors $e^{i \omega t}$ on everything. The pressure inside the cup is $p_1$, the pressure just outside the neck, at the entry to the instrument tube, is $p_2$, and the instrument tube presents an input impedance

$$Z(\omega)=p_2/q_1 \tag{1}$$

at that point. Our aim is to determine the modified input impedance

$$Z_{in}(\omega)=p_1/q \tag{2}$$

at the player’s lips, on the dotted line in the figure.

The net mass flow into the volume $V$ is $\rho_0 (q-q_1)$, where $\rho_0$ is the mean density of air. This is balanced by a density perturbation $\rho’$ inside the cup. For our harmonic signal, this gives

$$\rho_0(q-q_1)=i \omega V \rho’ = i \omega V \dfrac{p_1}{c^2} \tag{3}$$

using the relation between density and pressure via the speed of sound $c$. Now we apply Newton’s law to the mass of air in the exit tube, treating it as if it were rigid just as we did when we looked at the Helmholtz resonator in section 4.2.1:

$$(p_1-p_2) S = L S \rho_0 \dfrac{\partial u}{\partial t} \tag{4}$$

where $u$ is the flow speed. But $q_1 = Su$, so for harmonic response

$$L \rho_0 i \omega \dfrac{q_1}{S}=p_1-p_2 = p_1 -Zq_1 \tag{5}$$

using equation (1).

Eliminating $q_1$ using equation (3) we obtain

$$p_1=\left[ Z+\dfrac{i \omega L \rho_0}{S} \right] \left[ q-\dfrac{i \omega V p_1}{\rho_0 c^2} \right] \tag{6}$$

and so

$$Z_{in} =\dfrac{p_1}{q}=\dfrac{Z+i \omega L \rho_0/S}{1+ \dfrac{i \omega V Z}{\rho_0 c^2} – \omega^2 \dfrac{LV}{Sc^2}} . \tag{7}$$

Now we can recall from section 4.2.1 that the resonance frequency $\omega_m$ of an isolated Helmholtz resonator is given by

$$\omega_m^2 =\dfrac{c^2S}{VL} . \tag{8}$$

If we introduce non-dimensionalised versions of the impedances defined by

$$X=\dfrac{i \omega V}{\rho_0 c^2}Z \mathrm{~~~and~~~} X_{in}=\dfrac{i \omega V}{\rho_0 c^2}Z_{in} \tag{9}$$

we can simplify equation (7) into the form

$$X_{in}=\dfrac{X-\omega^2/\omega_m^2}{X+1-\omega^2/\omega_m^2}. \tag{10}$$

Finally, recognising the term $1-\omega^2/\omega_m^2$ in the denominator as a form of the usual resonant denominator for the response of a single degree of freedom oscillator, we can insert a factor to take account of the damping of the Helmholtz resonator. If the Q-factor of the mouthpiece alone is $Q_m$, we can write

$$X_{in} \approx \dfrac{X-\omega^2/\omega_m^2}{X+1+i \omega/\omega_m Q_m-\omega^2/\omega_m^2}. \tag{11}$$

We can get an idea of what equation (10) or (11) predicts about resonance frequencies if we think about the undamped version, for which $X(\omega)$ is a purely real function. The original resonances of the tube are given by the poles of $X$, while the new ones are given by the zeros of the denominator of equation (10). In other words, they occur at frequencies satisfying

$$X(\omega)=\dfrac{\omega^2}{\omega_m^2}-1. \tag{12}$$

We can represent this equation graphically, as indicated schematically in Fig. 2.

The graph of $X$, shown in blue, looks a bit like an upside-down version of the tangent function. The original resonance frequencies are marked by the vertical dotted lines, where $X$ goes off to infinity. Now we superimpose a graph of the function $\omega^2/\omega_m^2 – 1$, in red. This curve crosses the blue curve at the points circled in green. It is immediately obvious that there must be one of these green circles in every gap between an adjacent pair of tube resonances.

For the purposes of the simulation results shown in section 11.5, parameter values for a particular trombone mouthpiece were used. According to Campbell et al. [1], the Denis Wick 7AL mouthpiece has a cup volume $V=12 \times 10^3\mathrm{~mm^3}$, and a direct measurement of the mouthpiece in isolation gives a Helmholtz resonance frequency 535 Hz and a Q-factor of 23. However, when the mouthpiece is plugged into the trombone tube the end correction will be bigger than in free space, and a better fit to the impedance was given by assuming that the relevant Helmholtz resonator frequency was reduced to 460 Hz.

[1] Murray Campbell, Joël Gilbert and Arnold Myers, “The science of brass instruments”, ASA Press/Springer (2021)