Past exam of the physics 2 paper of the University of Oxford / 2021 / 10 / d / Solution

Put $x=\omega /{\omega }_0$. Maximizing the output wave amplitude is equivalent to minimizingIts derivative is $D^{\prime }(x)=2x[2x^2-2+Q^{-2}]$. An interior minimum at positive frequency therefore requires $Q>1/\sqrt{2}$, and occurs atIndeed $D^{\prime \prime }(x_{\mathrm{res}})=8x_{\mathrm{res}}^2>0$, and the peak value is $|H{|}_{\max }=Q/\sqrt{1-1/(4Q^2)}$. For $0<Q\le 1/\sqrt{2}$, $D$ increases for $x>0$ and the largest output is the DC value $|H(0)|=1$. At equality the putative peak has reached zero frequency; there is no distinct nonzero-frequency resonance. This threshold concerns capacitor voltage peaking: the series RLC circuit's electric current still has its maximum at ${\omega }_0$ for every positive $R$.