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Acoustic Energy Density

The two forms of energy in a wave are kinetic and potential. Denoting them at a particular time $ t$ and position $ \underline{x}$ by $ w_v(t,\underline{x})$ and $ w_p(t,\underline{x})$, respectively, we can write them in terms of velocity $ v$ and wave impedance $ R=\rho c$ as follows:

\begin{eqnarray*}
w_v &=& \frac{1}{2} \rho v^2 \eqsp \frac{1}{2c} R v^2 \quad\le...
...ad\left(\frac{\mbox{\small Energy}}{\mbox{\small Volume}}\right)
\end{eqnarray*}

More specifically, $ w_v$ and $ w_p$ may be called the acoustic kinetic energy density and the acoustic potential energy density, respectively.

At each point in a plane wave, we have $ p(t,\underline{x})=R\,v(t,\underline{x})$ (pressure equals wave-impedance times velocity), and so

\begin{eqnarray*}
w_v &=& \frac{1}{2c} R v^2 = \frac{1}{2}\cdot \frac{I}{c}\\
w_p &=& \frac{1}{2c} \frac{p^2}{R} = \frac{1}{2} \cdot \frac{I}{c},
\end{eqnarray*}

where $ I(t,\underline{x})\isdef p(t,\underline{x})\,v(t,\underline{x})$ denotes the acoustic intensity (pressure times velocity) at time $ t$ and position $ \underline{x}$. Thus, half of the acoustic intensity $ I$ in a plane wave is kinetic, and the other half is potential:B.30

$\displaystyle \frac{I}{c} = w = w_v+w_p = 2w_v = 2w_p
$

Note that acoustic intensity $ I$ has units of energy per unit area per unit time while the acoustic energy density $ w=I/c$ has units of energy per unit volume.


Previous: Acoustic Intensity
Next: Energy Decay through Lossy Boundaries

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About the Author: Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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