We use phasor analysis. The impedance of the resistor is $\mathbf{Z}_R= R$ . The impedance of the capacitor is $\mathbf{Z}_C = 1/(i \omega C)$ . The total impedance is
$\begin{align*}\mathbf{Z} = \mathbf{Z}_R + \mathbf{Z}_C = R + \frac{1}{i \omega C}\end{align*}$
The phasor input voltage is
$\begin{align*}\mathbf{V}_i = V_i\end{align*}$
Therefore, the steady state phasor current is
$\begin{align*}\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{V_i}{R + \frac{1}{i\omega C}}\end{align*}$
Therefore, the phasor output voltage is
$\begin{align*}\mathbf{V}_o = \mathbf{I} \mathbf{Z}_C = \frac{V_i}{R + \frac{1}{i\omega C}} \cdot \frac{1}{i \omega C} = \frac...$
The magnitude of the phasor output voltage is therefore,
$\begin{align*}V_o = \left|\mathbf{V}_o \right| = \frac{V_i}{\sqrt{1 + R^2 C^2 \omega^2}}\end{align*}$
From this we can see that answer (D) is correct.

Solution to 2008 Problem 69