= Solution
For a harmonic field, the time-averaged energy densities at $r=a$ are
$$
\langle K\rangle
=\frac{\rho_0}{4}|\widehat u_r|^2
=\frac{\rho_0\omega^2\epsilon^2}{4},
$$
$$
\langle W\rangle
=\frac{|\widehat p|^2}{4\rho_0c_0^2}
=\frac{\rho_0\omega^2\epsilon^2}{4}
\frac{x^2}{1+x^2}.
$$
From part (ii),
$$
|\langle I_r\rangle|
=\frac{\rho_0\omega^2\epsilon^2c_0}{2}
\frac{x^2}{1+x^2}.
$$
Consequently
$$
\boxed{
\frac{c_0\langle K+W\rangle}{|\langle I_r\rangle|}
=\frac{1+2x^2}{2x^2}
=1+\frac{c_0^2}{2\omega^2a^2}
}.
$$
This is the <surface acoustic energy-to-flux ratio of a pulsating sphere>.
Back to article page
