Solution to 2008 Problem 59

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.

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