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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 $ A(x)$ denote the area of the surface of constant phase at radial coordinate $ x$ in the tube. Then the total force acting on the surface due to pressure is $ f(t,x)=A(x)p(t,x)$ , as shown in Fig.C.45.

Figure C.45: Differential volume element for the conical acoustic tube.
\includegraphics[width=3in]{eps/fconic}

The net force $ df(t,x)$ to the right across the volume element between $ x$ and $ x+dx$ is then

$\displaystyle df(t,x) = f(t,x)-f(t,x+dx)$ $\displaystyle =$ $\displaystyle f(t,x)-[f(t,x) + dx \cdot f'(x)]$  
  $\displaystyle =$ $\displaystyle - dx \cdot f'(x)$  
  $\displaystyle =$ $\displaystyle - dx \cdot[A(x)p(t,x)]'$  
  $\displaystyle =$ $\displaystyle -dx \cdot[A' p + A'p],$  

where, when time and/or position arguments have been dropped, as in the last line above, they are all understood to be $ t$ and $ x$ , respectively. To apply Newton's second law equating net force to mass times acceleration, we need the mass of the volume element

\begin{eqnarray*}
dM(t,x) &=& \int_x^{x+dx} \rho(t,\xi) A(\xi)\,d\xi \\ [5pt]
&\approx& \int_x^{x+dx} \left[\rho(t,x)+\rho'(t,x)(\xi-x)][A(x)+A'(x)(\xi-x)\right]d\xi\\ [5pt]
&\approx& \rho(t,x) A(x)\,dx + \int_x^{x+dx} \left[\rho'(t,x)A(x)+\rho(t,x)A'(x)\right](\xi-x)\,d\xi\\ [5pt]
&=& \rho\,A\,dx + \left(\rho' A+\rho A' \right)\frac{(dx)^2}{2}\\ [5pt]
&\approx& \rho\, A\,dx,
\end{eqnarray*}

where $ \rho(t,x)$ denotes air density.

The center-of-mass acceleration of the volume element can be written as $ {\dot u}(t,x)$ where $ u(t,x)$ is particle velocity.C.16 Applying Newton's second law $ df = dM\cdot {\dot u}$ , we obtain

$\displaystyle -dx \cdot (A'p + Ap') \eqsp \rho\, A\,dx\, {\dot u} \protect$ (C.143)

or, dividing through by $ -A\,dx$ ,

$\displaystyle p' + p \, \frac{A'}{A} \eqsp - \rho\,{\dot u}. \protect$ (C.144)

In terms of the logarithmic derivative of $ A$ , this can be written

$\displaystyle p' + p \,$   ln$\displaystyle 'A \eqsp - \rho\,{\dot u}. \protect$ (C.145)

Note that $ p$ denotes small-signal acoustic pressure, while $ \rho$ denotes the full gas density (not just an acoustic perturbation in the density). We may therefore treat $ \rho$ as a constant.



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``Physical Audio Signal Processing'', by Julius O. Smith III, W3K Publishing, 2010, ISBN 978-0-9745607-2-4.
Copyright © 2014-03-23 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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