To model the effect of inserting the soundpost into a violin body, we can make use of a very general theorem relating to the vibration of coupled systems. We already met the simplest example in section 5.3. If two subsystems are joined together at a single point, the admittance $Y_{coup}$ at the coupling point is related in a very simple way to the admittances $Y_1$ and $Y_2$ of the separate subsystems at that point:
$$\dfrac{1}{Y_{coup}}=\dfrac{1}{Y_1} + \dfrac{1}{Y_2} . \tag{1}$$
This formula expresses the fact that, after coupling, the motion of the two subsystems at the coupling point must be equal, while the force that must be applied there to cause the vibration is the sum of the forces needed to excite the separate subsystems.
If the two subsystems are coupled at several points, at $N$ points say, we need to consider $N \times N$ matrices of admittance. But the physical principle remains the same: the motion must match at each point, and the excitation force at each point must be the sum of the forces to drive the separate subsystems. The result is a straightforward generalisation of equation (1) involving the inverses of the matrices:
$$Y_{coup}^{-1} = Y_1^{-1} + Y_2^{-1} . \tag{2}$$
For the case of a soundpost in a violin body, if we only consider the axial motion of the post then we have $N=2$, and we need to consider $2 \times 2$ matrices of admittance between the positions of the two ends of the post. For the violin, we can measure this matrix directly.
For the soundpost, we need to calculate the corresponding matrix. The system is sketched in the upper diagram of Fig. 1. The post, of length $L_p$ and radius $a_p$, has contact springs of stiffness $k_p$ at both ends. The post can undergo axial vibration at frequency $\omega$, with displacement $w(x,t) = u(x)e^{i \omega t}$ where $x$ measures distance along the post and $t$ is time as usual.
The system is symmetrical, so we only need to consider the case in which a force is applied at one end. The lower diagram of Fig. 1 shows the system, with the spring separated from the end of the post. A force $F_1$ is applied at the left-hand end. The spring is considered to be massless, so the same force acts in an equal and opposite pair between the spring and the post. The displacements at various key points are denoted $u_1$, $u_2$ and $u_3$ as shown in the diagram. The spring at the right-hand end of the post plays no role in the calculation: there is no force at that end, so the massless spring is simply carried along by the vibrating post.
The governing equation for axial vibration is
$$E_p \dfrac{\partial^2 w}{\partial x^2} = \rho_p \dfrac{\partial^2 w}{\partial t^2} \tag{3}$$
where $E_p$ is the Young’s modulus of the post and $\rho_p$ is its density, so that
$$\dfrac{d^2 u}{dx^2}=-\dfrac{\rho_p \omega^2}{E_p} u . \tag{4}$$
The general solution of this equation is
$$u=P \cos \omega x/c_p + Q \sin \omega x/c_p \tag{5}$$
where $P$ and $Q$ are constants, and $c_p = \sqrt{E_p/\rho_p}$ is the speed of axial waves. Now we need to satisfy the relevant boundary conditions. At $x=0$ we have
$$F_1=-E_p A_p \dfrac{du}{dx} = – E_p A_p Q \omega/c_p \tag{6}$$
where $A_p = \pi a_p^2$ is the cross-sectional area of the post. At $x=L_p$ we have $du/dx = 0$, so that
$$-P \sin \omega L_p/c_p + Q \cos \omega L_p/c_p = 0 \tag{7}$$
and thus
$$P = Q \cot \omega L_p/c_p . \tag{8}$$
Combining with equation (6) we find
$$ Q = -\dfrac{F_1 c_p}{E_p A_p} \mathrm{~~~ and~~~} P = -\dfrac{F_1 c_p \cot \omega L_p/c_p}{E_p A_p} . \tag{9}$$
It follows that
$$u_2 = -\dfrac{F_1 \cot \omega L_p/c_p}{E_p A_p \omega /c_p} \tag{10}$$
and
$$u_3 = -\dfrac{F_1 c_p}{E_p A_p \omega} \left[ \cot \dfrac{\omega L_p} {c_p} \cos \dfrac{\omega L_p} {c_p} + \sin \dfrac{\omega L_p} {c_p} \right] $$
$$= -\dfrac{F_1 c_p}{E_p A_p \omega \sin \omega L_p/c_p} . \tag{11}$$
Finally, we can relate $u_1$ to $u_2$ via the stiffness of the spring:
$$F_1=k_p (u_1 – u_2) \tag{12}$$
so that
$$u_1 = F_1 \left[ \dfrac{1}{k_p} – \dfrac{F_1 c_p \cot \omega L_p/c_p}{E_p A_p} \right] . \tag{13}$$
The two admittances we require are thus
$$Y_{11} = \dfrac{i \omega u_1}{F_1} = \dfrac{i \omega}{k_p} -\dfrac{i c_p}{E_p A_p} \cot \dfrac{\omega L_p} {c_p} \tag{14}$$
and $$Y_{12} = -\dfrac{i \omega u_3}{F_1} = \dfrac{i c_p}{E_p A_p \sin \omega L_p/c_p} , \tag{15}$$
and, taking advantage of the symmetry, the $2 \times 2$ admittance matrix for the soundpost model is
$$Y_p = \begin{bmatrix} Y_{11} & Y_{12} \\Y_{12} & Y_{11}\end{bmatrix} . \tag{16}$$
To avoid any problems with infinities, it is useful to incorporate some damping into the model. A simple way to do this is to replace $\omega L_p/c_p$ in the trigonometric expressions in equations (14) and (15) by the complex form
$$\dfrac{\omega L_p}{c_p} \left[ 1 – \dfrac{i}{2Q_p} \right] \tag{17}$$
where $Q_p$ is the Q-factor of axial resonances of the post. For the purposes of computation, the value $Q_p = 300$ was used.