To understand how the behaviour of a soundpost is affected by the various contact stiffnesses at its ends, it is useful to start with the simplest problem. Instead of looking at the interaction of the post with flexible top and back plates, we will see what happens when the post is held between two flat, parallel, rigid surfaces. For that problem, the axial motion of the post is decoupled from the bending motion, and we can consider them separately. The two cases are illustrated in Fig. 1. The linear springs in the left-hand sketch have stiffness $k_p$, while the rotational springs in the right-hand sketch have stiffness $K_p$.
We first consider axial motion, at frequency $\omega$. The displacement of a point at vertical position $z$ is $w_p(z) e^{i \omega t}$, and it satisfies the differential equation
$$E_p \dfrac{d^2 w_p}{d z^2}= -\rho_p \omega^2 w_p \tag{1}$$
where $E_p$ is the Young’s modulus and $\rho_p$ is the density. The post has length $L_p$ and radius $a_p$ so that the cross-section area is $A_p=\pi a_p^2$. The general solution to this equation takes the form
$$w_p = P \cos \alpha z + Q \sin \alpha z \tag{2}$$
where
$$\alpha = \omega \sqrt{\dfrac{\rho_p}{E_p}} \tag{3}$$
is the wavenumber.
The boundary condition at $z=0$ is
$$k_p w_p= E_p A_p \dfrac{d w_p}{d z} \tag{4}$$
so that
$$k_p P = Q E_p A_p \alpha . \tag{5}$$
The corresponding boundary condition at $z=L_p$ is
$$k_p w_p= – E_p A_p \dfrac{d w_p}{d z} \tag{6}$$
so that
$$P \cos \alpha L_p + Q \sin \alpha L_p $$
$$=-\dfrac{E_p A_p \alpha}{k_p}\left[ -P \sin \alpha L_p + Q \cos \alpha L_p \right] . \tag{7}$$
Equating the values of $P/Q$ given by equations (5) and (7) and simplifying leads to the condition
$$\tan \alpha L_p = \dfrac{2 \lambda \alpha}{\lambda^2 \alpha^2 – 1} \tag{8}$$
where
$$\lambda = \dfrac{E_p A_p}{k_p} . \tag{9}$$
The solutions for $\alpha$ of this equation give the natural frequencies of axial motion of the post, via equation (3).
Solving numerically for a wide range of values of $k_p$, using the parameter values given in section 7.6.1, gives the results shown in Fig. 2. Notice that this plot covers a wider frequency range than usual, in order to see the behaviour clearly. When the contact stiffness is very low, the lowest resonance involves the post bouncing as a rigid body between the two springs. But as the stiffness increases, the frequency of this resonance rises across the entire audible frequency range, until with very high stiffness the post is constrained between fixed ends. The mode shape then involves a half-wavelength of deformation between nodal points at the ends, with a frequency near 27 kHz.
This is exactly the same resonance frequency as that of the second mode when the stiffness is very low. The reason is that as the contact stiffness tends to zero, the post has free ends. The next mode then consists of a half-wavelength of deformation, but in the cosine phase rather than the sine phase of the first mode with high stiffness.
We can now look at bending motion of the soundpost, related to the right-hand sketch in Fig. 1. For bending in the $x-z$ plane, the horizontal displacement of the post is $u_p(z)e^{i \omega t}$, and it obeys the differential equation
$$E_p I_p \dfrac{\partial^4 u_p}{\partial z^4}= \rho_p A_p \omega^2 u_p \tag{10}$$
where $I_p=\pi a_p^4/4$ is the second moment of area. The general solution can be written
$$u_p=P \cos \alpha z + Q \sin \alpha z + R \cosh \alpha z + S \sinh \alpha z \tag{11}$$
where now the wavenumber $\alpha$ is given by
$$\alpha^4 = \omega^2 \dfrac{\rho_p A_p}{E_p I_p} . \tag{12}$$
We have two boundary conditions at each end of the post. At $z=0$ we have
$$u_p=0 \mathrm{~~and~~} E_p I_p \dfrac{d^2 u_p}{dz} = K_p \dfrac{d u_p}{dz} \tag{13}$$
which lead to
$$P+R=0 \tag{14}$$
and
$$\alpha^2 E_p I_p (-P+R)=K_p \alpha (Q+S) \tag{15}$$
so that
$$R=-P \mathrm{~~and~~} S=-2 \mu P – Q \tag{16}$$
where
$$\mu = \dfrac{E_p I_p \alpha}{K_p} . \tag{17}$$
At $z=L_p$ we have
$$u_p=0 \mathrm{~~and~~} E_p I_p \dfrac{d^2 u_p}{dz} = -K_p \dfrac{d u_p}{dz} . \tag{18}$$
Substituting and working through some rather tedious algebra, these four boundary conditions lead to the equation
$$\cos \alpha L_p \cosh \alpha L_p – \mu^2 \sin \alpha L_p \sinh \alpha L_p$$
$$+ 2 \mu \cos \alpha L_p \sinh \alpha L_p – 2 \mu \sin \alpha L_p \cosh \alpha L_p = 1 \tag{19}$$
which governs the natural frequencies.
Solving numerically for the roots $\alpha$ and then deducing frequencies from equation (12) leads to the results plotted in Fig. 3. When the torsional spring stiffness is very low the post behaves like a pinned-pinned beam, but when the stiffness becomes very large it behaves like a clamped-clamped beam. These two limits determine the pattern: each frequency rises from the pinned-pinned value to the clamped-clamped value as the stiffness increases.
There is one more useful calculation to do in connection with the torsional spring stiffness. It is very hard to have an intuitive feel for a torsional spring stiffness, and it is useful to link it to the axial stiffness. If we imagine a perfectly-fitted post, then the contact stiffness should be uniformly distributed over the cross-section of the post (on a length scale bigger than the individual asperity contacts). Suppose this stiffness is $s$ per unit area. The axial stiffness is then simply the aggregate over the section:
$$k_p=A_p s . \tag{20}$$
We can now calculate the torsional stiffness, by a suitable integration. Figure 4 shows a pair of strips of width $dx$, symmetrically distributed at a distance $x$ from the centre of the post. The half-length of each strip is $\sqrt{a_p^2-x^2}$. In response to a rotation by $\theta$ the compression at position $x$ is $x \theta$, so the combined moment from the two strips is
$$2x \times x \theta \times 2s\sqrt{a_p^2-x^2} dx \tag{21}$$
and the total torsional stiffness follows by integration:
$$K_p \theta = \theta \int_0^{a_p}{4sx^2 \sqrt{a_p^2-x^2} dx} . \tag{22}$$
Substituting $x=a_p \sin \phi$,
$$K_p=4sa_p^4 \int_0^{\pi/2}{ \sin^2 \phi \cos^2 \phi~d\phi} . \tag{23}$$
This integral is included in standard tables of integrals (e.g. [1]), so
$$K_p = 4s a_p^4 \left[-\dfrac{1}{8}\left(\dfrac{1}{4} \sin 4 \phi ~-~ \phi\right) \right]_0^{\pi/2}$$
$$=\dfrac{\pi s a_p^4}{4} = \dfrac{a_p^2 k_p}{4} . \tag{24}$$
Replotting the results from Figs. 2 and 3, but with $K_p$ expressed in terms of $k_p$ using this formula, gives Fig. 4. The transition for bending modes occurs at somewhat higher stiffness than the transition for axial modes, but they lie in a similar range.
[1] I. S. Gradshteyn and I. M. Ryzhik, “Table of integrals, series and products”, Academic Press (1994)