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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.

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    $\begingroup$ 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". $\endgroup$
    – alephzero
    Aug 24, 2016 at 22:16
  • $\begingroup$ @alephzero: always easy when you know it! ;-) $\endgroup$
    – Gert
    Aug 24, 2016 at 22:36

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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: $$R(R_0)=c_1J_k(nR_0)+c_2Y_k(nR_0)=0$$

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  • $\begingroup$ "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....) $\endgroup$
    – alephzero
    Aug 24, 2016 at 22:16
  • $\begingroup$ @alephzero: thanks, will definitely look into that! :-) $\endgroup$
    – Gert
    Aug 24, 2016 at 22:36
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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.

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