= Solution
Use an <RC low-pass filter>: put the <resistor> in series with the input and the <capacitor> between the output node and the return conductor. Measure the output <voltage> across the <capacitor>, with negligible output loading. The <voltage divider> then has <transfer function>
$$
H(\omega)=\frac{1/(i\omega C)}{R+1/(i\omega C)}=\frac1{1+i\omega RC},\qquad |H|=\frac1{\sqrt{1+(\omega RC)^2}}.
$$
At the high <frequency>, the attenuation requirement is
$$
2\pi f_hRC>\sqrt{500^2-1},\qquad C>\frac{\sqrt{500^2-1}}{2\pi(20000)(330)}\simeq12.1\ \mu\mathrm F.
$$
At the low <frequency>, approximately unchanged <amplitude> means $2\pi f_lRC\ll1$, so
$$
\boxed{12.1\ \mu\mathrm F<C\ll482\ \mu\mathrm F.}
$$
The upper relation is an asymptotic design condition, because “approximately” specifies no numerical tolerance. For example, retaining at least $99\%$ of the low-<frequency> <amplitude> would sharpen the upper bound to approximately $68.7\ \mu\mathrm F$.
\b[A practical illustrative choice is $C=22\ \mu\mathrm F$.] Its <voltage gain> is approximately $0.99896$ at $1\ \mathrm{Hz}$ and $0.001096$ at $20\ \mathrm{kHz}$: the latter <amplitude> is reduced by a factor of about $912$. Its <cutoff frequency> is $1/(2\pi RC)\simeq21.9\ \mathrm{Hz}$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-low-pass.png]
{title=An original RC low-pass circuit and its amplitude response for a 22 microfarad capacitor}
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