11.1.4 Bore perturbation and Rayleigh’s principle

We will use Rayleigh’s principle to find an expression for the effect on natural frequencies of a small perturbation to the bore profile of a cylindrical tube. This is a problem that Rayleigh himself addressed [1], and I will follow his treatment closely. As in previous sections, we consider a tube of length $L$ which is open at both ends. We now allow the cross-sectional area to be

$$S(x) = S_0 +S'(x) \tag{1}$$

with $S_0$ a constant, and $|S’| \ll S_0$.

In order to apply Rayleigh’s principle, we need suitable expressions for the potential and kinetic energies. It is simplest to formulate these in terms of what Rayleigh described as “the total transfer of fluid at time $t$ across the section at $x$, reckoned from the equilibrium condition”, and following his lead we denote this by $X(x,t)$. In terms of the particle displacement $\xi(x,t)$ which we used earlier,

$$X=S \xi . \tag{2}$$

The kinetic energy is straightforward: it is given by

$$T=\dfrac{1}{2} \rho_0 \int{S \dot{\xi}^2 dx} = \dfrac{1}{2} \rho_0 \int{\dfrac{1}{S} \dot{X}^2 dx}. \tag{3}$$

For the potential energy, we start from an expression derived in section 4.1.3:

$$P=\dfrac{1}{2} \dfrac{c^2}{\rho_0} \int{\rho’^2 dV}=\dfrac{1}{2} \dfrac{c^2}{\rho_0} \int{S\rho’^2 dx} \tag{4}$$

in terms of the acoustic density fluctuation $\rho’$, where the first integral is taken over the volume inside the tube, and the second integral uses the fact that $dV=S dx$.

Figure 1. A small element of the acoustic wavefield in a tube with varying cross-section, showing particle displacements.

Now we need to express $\rho’$ in terms of $X$. Figure 1 shows a small section of the tube between positions $x$ and $x+\delta x$. The particle displacements are $\xi$ and $\xi + \delta \xi$, and the cross-sectional areas are $S$ and $S+\delta S$. The initial volume of this fluid element is $V_0 \approx S \delta x$, and after taking the particle displacements into account it increases by an amount

$$\delta V \approx (S+\delta S)(\xi + \delta \xi) -S \xi \approx S \delta \xi + \xi \delta S =\delta(S \xi) . \tag{5}$$

Now the mass of the fluid element remains constant, so the perturbed density $\rho’$, which we are trying to find, satisfies

$$\dfrac{\rho’}{\rho_0}=-\dfrac{\delta V}{V_0} \tag{6}$$

so that

$$\dfrac{S \delta x}{\rho_0}\rho’ \approx -\delta(S \xi) . \tag{7}$$

Dividing by $\delta x$ and taking the usual calculus limit gives

$$\rho’=-\dfrac{\rho_0}{S}\dfrac{\partial}{\partial x}(S \xi) \tag{8}$$

so that our potential energy is

$$P=\dfrac{1}{2} \rho_0 c^2 \int{\dfrac{1}{S}\left[\dfrac{\partial}{\partial x}(S \xi)\right]^2 dx}=\dfrac{1}{2} \rho_0 c^2 \int{\dfrac{1}{S}\left[\dfrac{\partial X}{\partial x}\right]^2 dx} . \tag{9}$$

Now we can apply Rayleigh’s principle to our original problem of making a small perturbation to the bore profile. We solved a similar problem back in section 3.3.1, relating to the tuning of a marimba bar. The argument is the same this time: we evaluate the Rayleigh quotient using the corrected energy expressions (including the perturbed bore), but we use a mode shape from the original, unperturbed problem. Errors arising from this incorrect mode shape will only be of second order, so that the predicted natural frequency will be correct to leading order in the small perturbation.

We are dealing with a cylindrical tube, open at both ends. The pressure mode shapes are thus of the form $\sin n \pi x/L$ (as in Fig. 11 of section 4.2). But for this formulation, we need the mode shapes expressed in terms of particle displacements and the variable $X$. By virtue of equation (7) from section 4.1.1, we know that the answer for the $n\mathrm{th}$ mode is

$$X_n=\cos \dfrac{n \pi x}{L} . \tag{10}$$

Noting that

$$\dfrac{1}{S}=\dfrac{1}{S_0+S’} \approx \dfrac{1}{S_0}\left[1-\dfrac{S’}{S_0}\right] \tag{11}$$

by the binomial theorem, our Rayleigh quotient reads

$$\omega_n^2 \approx \dfrac{c^2 \int{(1-S’/S_0)(n \pi/L)^2 \sin^2(n \pi x/L) dx}}{\int{(1-S’/S_0) \cos^2(n \pi x/L) dx}} \tag{12}$$

using equations (3) and (9). This takes the form

$$\omega_n^2 \approx \dfrac{P_0 + \delta P}{T_0 + \delta T} \tag{13}$$


$$P_0=c^2 \left(\dfrac{n \pi}{L}\right)^2 \int{ \sin^2 \dfrac{n \pi x}{L} dx}=c^2 \left(\dfrac{n \pi}{L}\right)^2 \dfrac{L}{2} \tag{14}$$

$$\delta P=-\dfrac{c^2}{S_0} \left(\dfrac{n \pi}{L}\right)^2 \int{S'(x) \sin^2 \dfrac{n \pi x}{L} dx} \tag{15}$$

$$T_0=\int{ \cos^2 \dfrac{n \pi x}{L} dx} =\dfrac{L}{2} \tag{16}$$


$$\delta T=-\dfrac{1}{S_0} \int{S'(x) \cos^2 \dfrac{n \pi x}{L} dx} . \tag{17}$$

It follows that

$$\omega_n^2 \approx \dfrac{P_0}{T_0}\left[1 + \dfrac{\delta P}{P_0}\right] \left[1 – \dfrac{\delta T}{T_0}\right]\approx \dfrac{P_0}{T_0}\left[1 + \dfrac{\delta P}{P_0} – \dfrac{\delta T}{T_0}\right]$$

$$=\left(\dfrac{c n \pi}{L}\right)^2 \left[1+\dfrac{2}{S_0L} \int{S'(x) \left(\cos^2\dfrac{n \pi x}{L}-\sin^2 \dfrac{n \pi x}{L} \right) dx}\right]$$

$$=\left(\dfrac{c n \pi}{L}\right)^2 \left[1+\dfrac{2}{S_0L} \int{S'(x) \cos\dfrac{2n \pi x}{L} dx}\right] . \tag{18}$$

Finally, Rayleigh expressed the result as an equivalent length correction to the cylindrical tube to achieve the perturbed resonance frequency: the extra length required is

$$\Delta L \approx -\dfrac{1}{S_0L} \int{S'(x) \cos\dfrac{2n \pi x}{L} dx} . \tag{19}$$

Apart from a multiplicative factor, this rather simple form is a Fourier series coefficient of the bore perturbation $S'(x)$.

The frequency is raised by an enlargement of the bore near the ends, or near any other nodal point of pressure (which is an antinodal point of particle displacement or $X$). Conversely frequency is reduced by a bore enlargement near an antinode of pressure (or a node of particle displacement). As Rayleigh points out, the frequencies are rather insensitive to a slight conicity of the tube, because that would give a linear form for $S'(x)$, and such a linear form is orthogonal to the cosine function in the integral.

[1] J. W. S. Rayleigh:The Theory of Sound (1877, reprinted by Dover, New York 1945). This calculation appears in Volume 2, section 265.