= Absorbed-power bandwidth of a viscously damped oscillator
{title2=$\Delta\Omega=b/m$}
For a constant-amplitude <force> on $m\ddot x+b\dot x+kx=F_0\cos\Omega t$, the cycle-averaged absorbed <power> is $bF_0^2\Omega^2/[2((k-m\Omega^2)^2+b^2\Omega^2)]$. It peaks at $\omega_0=\sqrt{k/m}$. Its half-power <frequencies> are $\Omega_\pm=(\sqrt{b^2+4mk}\pm b)/(2m)$, hence their separation is exactly $b/m$ and $Q=\omega_0/\Delta\Omega$. A displacement-amplitude <resonance> has different half-power <frequencies> away from weak <damping>; the measured observable must be specified.
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