= Solution
In the chosen <phasor> convention, output <phase> minus input <phase> is
$$
\boxed{\phi(\omega)=-\arctan\!\left[\frac{\omega RC-(\omega RC)^{-1}}3\right],\qquad \omega>0.}
$$
The real part of the denominator is positive, so this principal <arctangent> is continuous throughout the positive-<frequency> range. Below $\omega_0$, the output leads the input, approaching $+\pi/2$ at low <frequency>. Above $\omega_0$, it lags, approaching $-\pi/2$ at high <frequency>. At the maximum,
$$
\boxed{\phi(\omega_0)=0.}
$$
The input and output are in <phase> there, with the positive one-third <voltage> ratio found above.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-resit-rc-band-pass.png]
{title=Amplitude and phase response of the series-parallel RC band-pass filter}
{description=Original curves against $\omega RC$ use the exact unloaded transfer function. The positive output polarity gives a peak ratio of $1/3$ and zero relative <phase> at $\omega RC=1$.}
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