Momentum Conservation in Nonuniform Tubes
Newton's second law ``force equals mass times acceleration'' implies that the pressure gradient in a gas is proportional to the acceleration of a differential volume element in the gas. Let denote the area of the surface of constant phase at radial coordinate in the tube. Then the total force acting on the surface due to pressure is , as shown in Fig.C.45.
The net force to the right across the volume element
between and is then
where, when time and/or position arguments have been dropped, as in the last line above, they are all understood to be and , respectively. To apply Newton's second law equating net force to mass times acceleration, we need the mass of the volume element
where denotes air density.
or, dividing through by ,
In terms of the logarithmic derivative of , this can be written
Note that denotes small-signal acoustic pressure, while denotes the full gas density (not just an acoustic perturbation in the density). We may therefore treat as a constant.
In the case of cylindrical tubes, the logarithmic derivative of the area variation, ln, is zero, and Eq.(C.148) reduces to the usual momentum conservation equation encountered when deriving the wave equation for plane waves [318,349,47]. The present case reduces to the cylindrical case when
If we look at sinusoidal spatial waves, and , then and , and the condition for cylindrical-wave behavior becomes , i.e., the spatial frequency of the wall variation must be much less than that of the wave. Another way to say this is that the wall must be approximately flat across a wavelength. This is true for smooth horns/bores at sufficiently high wave frequencies.
Wave Impedance in a Cone
Conical Acoustic Tubes