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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:

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)

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

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},

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.

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Energy Decay through Lossy Boundaries
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Acoustic Intensity