= Solution
For a sinusoidal drive of variable positive angular frequency, the average absorbed <power> is proportional to
$$
\frac{\omega^2}{(\Omega^2-\omega^2)^2+4\gamma^2\omega^2}.
$$
It peaks at $\omega=\Omega$. The <absorbed-power bandwidth of a viscously damped oscillator> is the separation of its half-power frequencies: solving $|\Omega^2-\omega^2|=2\gamma\omega$ gives
$$
\omega_\pm=\sqrt{\Omega^2+\gamma^2}\pm\gamma,\qquad\boxed{\Delta\omega=2\gamma.}
$$
The usual <quality factor> is $\boxed{Q=\Omega/(2\gamma)=\Omega/\Delta\omega}$. Large $Q$ means a narrow response band relative to the resonant frequency. The post-drive motion has amplitude envelope $e^{-\gamma t}$, connecting a narrow <bandwidth> to a long <ringdown>. For weak damping, the cycle-averaged energy decays as $e^{-2\gamma t}$ and $Q$ is also $2\pi$ times stored energy divided by the energy lost per cycle. The exact instantaneous energy has an oscillatory factor, so this cycle interpretation uses the weak-damping approximation.
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