If the pendulum were not accelerating, the period would be
$\begin{align*}T = 2 \pi \sqrt{\frac{l}{g}}\end{align*}$
In the accelerating case, the effective gravitational acceleration is $g+a$ , so the period is
$\begin{align*}\boxed{T = 2 \pi \sqrt{\frac{l}{g+a}}}\end{align*}$
Therefore, answer (C) is correct.

Solution to 2008 Problem 59