To see how to express the input impedance of a tube in terms of modal parameters, we start with the “Webster horn equation” from section 4.2.3. We need to augment the version of the equation derived there, by including the term describing forcing input in the form of a volume flow source $q$ per unit length. The result, given in section 7.2.1 of Chaigne and Kergomard [1], is
$$\dfrac{S}{c^2} \dfrac{\partial^2 p}{\partial t^2} = \dfrac{\partial}{\partial x}\left(S\dfrac{\partial p}{\partial x} \right) + \rho_0 \dfrac{\partial q}{\partial t} \tag{1}$$
where $p$ is the pressure, $S(x)$ is the cross-sectional area of the tube, $c$ is the speed of sound and $\rho_0$ is the density of air.
We are interested in the pressure modes $p_n(x) e^{i \omega_n t}$ when the tube has a closed end at $x=0$. Substituting in equation (1) without the source term, these modes must satisfy
$$-\frac{S \omega_n^2}{c^2}p_n = \dfrac{\partial}{\partial x}\left(S\dfrac{\partial p_n}{\partial x} \right) . \tag{2}$$
As usual, the mode shapes are orthogonal with respect to an integral deduced from the potential energy. From section 4.1.3, the potential energy can be expressed in the form
$$V=\dfrac{1}{2} \dfrac{1}{\rho_0 c^2} \int_0^L{S p^2 dx} \tag{3}$$
so we can deduce that
$$\dfrac{1}{2} \dfrac{1}{\rho_0 c^2} \int_0^L{S p_n(x) p_m(x) dx} = 0 \mathrm{~~~for~~~} n \ne m . \tag{4}$$
We can choose to normalise the modes using the same expression, so that
$$\dfrac{1}{\rho_0 c^2} \int_0^L{S p_n^2(x) dx} = 1. \tag{5}$$
Now go back to equation (1), and include a source term $q e^{i \omega t} \delta(x)$ describing a concentrated sinusoidal source at $x=0$ (using the Dirac delta function — see section 2.2.8). The pressure $p(x) e^{i \omega t}$ must satisfy
$$-\dfrac{S}{c^2} \omega^2 p = \dfrac{\partial}{\partial x}\left(S\dfrac{\partial p}{\partial x} \right) + \rho_0 i \omega q \delta(x). \tag{6}$$
Now we seek a solution of this equation in the form of a modal expansion
$$p(x) = \sum_n{a_n p_n(x)} \tag{7}$$
with complex coefficients $a_n$. If we multiply equation (6) by $p_m(x)$ and integrate over the range $0 \rightarrow L$ we find
$$-\dfrac{\omega^2}{c^2} \sum_n{a_n \int_0^L{S p_n p_m dx}}=$$
$$\sum_n{a_n \int_0^L{p_m \dfrac{\partial}{\partial x}\left(S\dfrac{\partial p_n}{\partial x} \right) dx}} + i \omega \rho_0 q p_m(0) \tag{8}$$
where the properties of the delta function have been used to simplify the final term. Now we can use equations (4), (5) and (6) to deduce
$$-\dfrac{\omega^2}{c_2}a_m \rho_0 c^2 = -\dfrac{\omega_m^2}{c_2}a_m \rho_0 c^2 + i \omega \rho_0 q p_m(0) \tag{9}$$
so that
$$a_m (\omega_m^2 – \omega^2) = i \omega q p_m(0). \tag{10}$$
Substituting in equation (7) gives
$$p(x,\omega)= i \omega q \sum_m{\dfrac{p_m(0) p_m(x)}{\omega_m^2 – \omega^2}} \tag{11}$$
so that the input impedance we want is
$$Z(\omega) = \dfrac{p(0)}{q} = i \omega \sum_m{\dfrac{p_m^2(0)}{\omega_m^2 – \omega^2}} . \tag{12}$$
In the usual way (see section 2.2.7) we can tweak this undamped result to allow for small modal damping with Q-factor $Q_m$ for mode $m$, to give
$$Z(\omega) \approx i \omega \sum_m{\dfrac{p_m^2(0)}{\omega_m^2 + i \omega \omega_m/Q_m – \omega^2}} . \tag{13}$$
This has exactly the same form as the modal expression for the input admittance of a mechanical system: see for example equation (12) of section 2.2.7 (which would need to be multiplied by a factor $i \omega$ to convert it to admittance).
[1] Antoine Chaigne and Jean Kergomard; “Acoustics of musical instruments”, Springer/ASA press (2013)