= Solution
The original diagram's upper arm is a series <resistor> and <capacitor>; its lower arm is a parallel <resistor> and <capacitor>. The output is the upper-minus-lower <voltage> across that parallel arm. Assume an ideal input voltage source and an unloaded output. Their <electrical impedances> are
$$
Z_s=R+\frac1{i\omega C},\qquad Z_p=\frac{R}{1+i\omega RC}.
$$
For this <series-parallel RC band-pass filter>, the <voltage divider> gives, with $w=\omega RC$,
$$
\boxed{H(\omega)=\frac{\widetilde V_{\rm out}}{\widetilde V_{\rm in}}
=\frac{Z_p}{Z_s+Z_p}
=\frac1{3+i(w-w^{-1})}
=\frac{iw}{1-w^2+3iw}.}
$$
Taking the real part of $V_0H e^{i\omega t}$ yields the steady output
$$
\boxed{V_{\rm out}(t)=\frac{V_0}{\sqrt{9+(w-w^{-1})^2}}
\cos\!\left[\omega t-\arctan\!\left(\frac{w-w^{-1}}3\right)\right].}
$$
Initial charging contributes decaying transients if the source is switched on at a finite time; the displayed expression is the sinusoidal steady state. The output polarity matters: reversing the indicated output <voltage> would multiply $H$ by $-1$.
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