 Quality factor from ringdown

For an isolated weakly damped resonator mode, $Q=\omega E/P_{\mathrm{loss}}$. If $E(t)=E(0)e^{-t/{\tau }_E}$, differentiating gives $P_{\mathrm{loss}}=E/{\tau }_E$, so $Q=\omega {\tau }_E$. Since energy is quadratic in wave amplitude, the wave amplitude exponential relaxation time is ${\tau }_A=2{\tau }_E$. A decay by an energy factor $D$ in a measured interval $t_d$ gives $Q=\omega t_d/\ln D$. For a conventional reverberation time, $Q=2\pi f{T}_{60}/[6\ln (10)]$ under this exponential single-rate approximation.