= Solution
The <voltage divider> relation for the <capacitor> output gives the <transfer function>
$$
H(\omega)=\frac{V_{\rm out}}{V_{\rm in}}=\frac{1/(i\omega C)}{R+i\omega L+1/(i\omega C)}=\boxed{\frac1{1-\omega^2LC+i\omega RC}.}
$$
Consequently the relative <wave amplitude> is
$$
\boxed{|H|=\frac1{\sqrt{(1-\omega^2LC)^2+(\omega RC)^2}}.}
$$
For input $V_0\cos\omega t$, the output is $V_0|H|\cos(\omega t+\phi)$, where the continuous <phase shift> is
$$
\boxed{\phi=-\operatorname{atan2}(\omega RC,\,1-\omega^2LC).}
$$
For 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 $1/(\omega^2LC)$ at high <frequency>.
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