# Normal modes of the conical duct

Solution of the wave equation for the narrow conical duct in low freqency approximation (i.e. 1D spherical wave) is well-known (only a spatial part):

$$p(r) = A\frac{\cos kr}{r} + B\frac{\sin kr}{r}$$

But I haven't been lucky so far in googling what are the normal modes of this duct for say a mixed boundary conditions:

$$p(r=L) = 0$$ $$\frac{\partial p(r=0)}{\partial r} = 0$$

This must be a solved problem. I even know how to get a reasonable approximation of eigenfrequencies using input impedance calculation, but I am interested in spatial pattern.

• Just plug your boundary conditions into your equation for $p(r)$ and solve for $A$ and $B$. There will only be non-zero solutions for certain values of $k$ - i.e. the eigenfrequencies. Then plot the function $p(r)$ for those values of $k$. It's not obvious (to me) which part of this is "hard". Aug 24, 2016 at 22:16
• @alephzero: always easy when you know it! ;-)
– Gert
Aug 24, 2016 at 22:36

I'm not sure where you get:

$$p(r) = A\frac{\cos kr}{r} + B\frac{\sin kr}{r}$$ ... from.

The wave equation in cylindrical coordinates is:

$$\frac{1}{c^2}u_{tt}=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta}+u_{zz}$$

With Ansatz:

$$u(r,\theta,z,t)=R(r)\Theta(\theta)Z(z)T(t)$$ Separation: $$\frac{1}{c^2}R\Theta ZT''=\Theta Z T R''+\frac{1}{r}\Theta ZTR'+\frac{1}{r^2}RZT\Theta''+R\Theta TZ''$$ $$\frac{1}{c^2}\frac{T''}{T}=\frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{\Theta''}{\Theta}+\frac{Z''}{Z}=-m^2$$ $$\frac{1}{c^2}\frac{T''}{T}=-m^2$$ $$\frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{\Theta''}{\Theta}+\frac{Z''}{Z}=-m^2$$ $$\frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{\Theta''}{\Theta}=-m^2-\frac{Z''}{Z}=-n^2$$ $$\frac{R''}{R}+\frac{1}{r}\frac{R'}{R}+\frac{1}{r^2}\frac{\Theta''}{\Theta}=-n^2$$ $$r^2\frac{R''}{R}+r\frac{R'}{R}+\frac{\Theta''}{\Theta}=-n^2r^2$$ $$r^2\frac{R''}{R}+r\frac{R'}{R}+n^2r^2=-\frac{\Theta''}{\Theta}=+k^2$$

Note that here the separation constant has to be positive, in order go give good solutions to: $$-\frac{\Theta''}{\Theta}=+k^2\implies \Theta''-k^2\Theta=0$$ Which has the solutions:

$$\Theta(\theta)=c_3\cos k\theta$$ Where the eigenvalues $k=1.2,3,...$.

So the first spatial ODE is: $$r^2R''+rR'+(n^2r^2+k^2)R=0$$ Which has solutions: $$R(r)=c_1J_k(nr)+c_2Y_k(nr)$$ Where $J_k$ is the Bessel Function and $Y_k$ is the Modified Bessel Function. The eigenvalues $n$ are the roots of: $$R(R_0)=c_1J_k(nR_0)+c_2Y_k(nR_0)=0$$

• "I'm not sure where you get ...from": In a narrow conical duct, you can make the approximation that the solution is a plane wave (i.e. a function of $r$ only) not a spherical wave. The OP's formula is the standard solution used in acoustics. (Or if you want to do it the hard way, approximate the Bessel functions by trig functions....) Aug 24, 2016 at 22:16
• @alephzero: thanks, will definitely look into that! :-)
– Gert
Aug 24, 2016 at 22:36

Conical waveguides are described by the wave equation in spherical coordinates, with the origin at the apex of the cone. If the field is radially symmetric, that is, if pressure $p$ depends only on radial distance from the apex and time, then $r_p$ is a sum of two functions of the characteristic variables $t-r/c$ and $t+r/c$.

This representation will identically satisfy the conditions at the side wall if it is rigid. Conditions at the far end, where $r$ is finite, can be satisfied with a combination of the functions. If your model is a complete cone, which means $r=0$ is in the domain, then a boundary condition is replaced by a finiteness condition on $p$. A truncated cone, like a megaphone, can have boundary conditions at both ends. All of these topics are covered in my recently published books: J.H. Ginsberg, "Acoustics-A Textbook for Engineers and Scientists, Volumes 1 and 2", Springer, 2017. See Chapters 6 and 9 for discussions of spherical waves and conical waveguides. Hope this helps.