= Solution
Use complex <phasors> with time dependence $e^{i\omega t}$ and assume an unloaded output. The <capacitor> has <electrical impedance> $Z_C=1/(i\omega C)$. The <voltage divider> formed with the <resistor> gives the <transfer function>
$$
\boxed{H(\omega)=\frac{\widetilde V_o}{\widetilde V_i}=\frac{Z_C}{R+Z_C}=\frac1{1+i\omega RC}.}
$$
For input $V_i(t)=V_{i0}\cos\omega t$, the output <amplitude> and <phase shift> are
$$
|\widetilde V_o|=\frac{V_{i0}}{\sqrt{1+(\omega RC)^2}},\qquad
\phi_o-\phi_i=-\tan^{-1}(\omega RC),
$$
so $V_o(t)=|\widetilde V_o|\cos(\omega t-\tan^{-1}(\omega RC))$.
Here $\tau=RC=16.0\,\mu\mathrm s$, and the <cutoff frequency> is $f_c=1/(2\pi RC)=9.947\,\mathrm{kHz}$. This is an <RC low-pass filter>: at small <frequency> the <capacitor> has large <electrical impedance> and the output nearly equals the input; at large <frequency> it increasingly shunts the signal to ground. The <amplitude> then falls as $1/f$, and the <phase shift> approaches $-90^\circ$.
|| <Frequency> $f$ | Output <amplitude> divided by $V_{i0}$ | Output <phase shift> \\
| $100\,\mathrm{Hz}$ | $0.99995$ | $-0.576^\circ$ \\
| $10^4\,\mathrm{Hz}$ | $0.70523$ | $-45.152^\circ$ \\
| $5\times10^4\,\mathrm{Hz}$ | $0.19512$ | $-78.748^\circ$
No numerical input <amplitude> is supplied, so the output voltages are specified as fractions of $V_{i0}$. An application is suppression of high-<frequency> <electrical noise> on a slowly varying signal. A finite load changes the <transfer function> and must be included if its <electrical impedance> is comparable with the circuit's output <electrical impedance>.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019-low-pass-filter-responses.png]
{title=Passive and active low-pass amplitude and phase responses}
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