Past exam of the mathematical methods I course of the University of Oxford / 2016 resit / 11 / ii / Solution

The weak-damping mechanical energy is proportional to the squared oscillation amplitude, so it decays as $e^{-\gamma t}$. Over one period $P=2\pi /{\omega }_d$, the fractional loss is $1-e^{-\gamma P}\simeq \gamma P$. HenceFor the release-from-rest solution, comparison at successive corresponding turning points gives the more precise ratio $E/\Delta E=(1-e^{-\gamma P}{)}^{-1}$. The quality factor is $Q\simeq 2\pi E/\Delta E={\omega }_0/\gamma$; the requested energy-to-loss ratio itself has the extra factor $1/(2\pi )$.