 Solution

For a sinusoidal drive of variable positive angular frequency, the average absorbed power is proportional toIt peaks at $\omega =\Omega$. The absorbed-power bandwidth of a viscously damped oscillator is the separation of its half-power frequencies: solving $|{\Omega }^2-{\omega }^2|=2\gamma \omega$ givesThe 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.