2.9. Freely traveling cylindrical wave

Sound pressure

A solution to the cylindrical wave equation (2.23) is
p˜(w)=p˜+H0(2)(kw)+p˜H0(1)(kw),
image (2.125)
where p˜+ image is the amplitude of the sound pressure in the outgoing wave at unit distance from the axis of symmetry and p˜ image is the same for the reflected wave. H 0 (1) (x) and H 0 (2) (x) are Hankel functions defined by
H0(1)(x)=J0(x)+jY0(x),
image (2.126)
H0(2)(x)=J0(x)jY0(x),
image (2.127)
where J 0(x) and Y 0(x) are Bessel functions of the first and second kind respectively, as plotted in Fig. 2.10. The “2” in parentheses denotes an outgoing cylindrical wave and the “1” denotes an incoming one. In the far field
p˜(w)|w=2πkw(p˜+ej(kwπ/4)+p˜ej(kwπ/4)).
image (2.128)
image
Figure 2.10 Bessel functions of the first (black curve) and second (gray curve) kind.
We can see from Fig. 2.10 that cylindrical waves, which are essentially two-dimensional because of the lack of axial dependency, differ from plane ones in two respects: firstly the radial wavelength is longer nearer the axis of symmetry than in the far field; secondly they decay in amplitude as they spread out, adopting an inverse square-root law in the far field. The latter makes sense when we consider that the area of the wave front is proportional to the radial distance w. The radiated power is the intensity multiplied by the area, where the intensity is given by Eq. (1.12). The intensity, in turn, is proportional to the square of the pressure and therefore inversely proportional to the radial distance. Hence the power remains constant. The same kind of wave deformation can be seen if you drop a pebble in a pond. Note the singularity in the Y(x) function when x   =   0. If there are no reflecting surfaces in the medium, only the first term of Eq. (2.125) is needed, i.e.,
p˜(w)=p˜+H0(2)(kw).
image (2.129)

Particle velocity

With the aid of Eq. (2.4b), we solve for the particle velocity in the w direction:
u˜(w)=1jkρ0cwp˜(w)=jp˜+ρ0cH1(2)(kw).
image (2.130)
In the far field
u˜(w)=jp˜+ρ0c2πkwej(kw3π/4)=p˜+ρ0c2πkwej(kwπ/4).
image (2.131)

Specific acoustic impedance

The specific acoustic impedance is found from Eq. (2.129) divided by Eq. (2.130):
Zs=p˜(w)u˜(w)=jρ0cH0(2)(kw)H1(2)(kw)rayls.
image (2.132)
Plots of the magnitude and phase angle of the impedance as a function of kw are given in Fig. 2.11 and Fig.2.12 respectively.
For large values of kw, that is, for large distances or for high frequencies, Eq. (2.132) becomes, approximately,
Zsρ0crayls.
image (2.133)
image
Figure 2.11 Plot of the magnitude of the specific acoustic-impedance ratio |Z s |/(ρ 0 c) in a cylindrical freely traveling wave as a function of kw, where k is the wave number equal to ω/c or 2π/λ and w is the distance from the axis of symmetry. |Z s | is the magnitude of the ratio of pressure to particle velocity in a cylindrical free-traveling wave, and ρ 0 c is the characteristic impedance of air.
image
Figure 2.12 Plot of the phase angle, in degrees, of the specific acoustic-impedance ratio |Z s |/ρ 0 c in a cylindrical wave as a function of kw, where k is the wave number ω/c or 2π/λ and w is the distance from the axis of symmetry.
The impedance here is nearly purely resistive and approximately equal to the characteristic impedance for a plane freely traveling wave. In other words, the specific acoustic impedance for a large distance from a cylindrical source in free space is nearly equal to that in a tube in which no reflections occur from the end opposite the source.
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