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

The voltage divider relation for the capacitor output gives the transfer functionConsequently the relative wave amplitude isFor input $V_0\cos \omega t$, the output is $V_0|H|\cos (\omega t+\varphi )$, where the continuous phase shift isFor positive frequency and $R>0$, this phase shift goes from $0$ towards $-\pi$, passing through $-\pi /2$ at ${\omega }_0$. A one-argument arctangent without a quadrant correction would wrongly predict a positive phase shift above ${\omega }_0$. The capacitor follows a slowly varying input voltage, while its output falls as $1/({\omega }^{2}LC)$ at high frequency.