The potential is zero at the origin, so the particle will have kinetic energy equal to $E$ there. So, its velocity will be $\sqrt{2 E/m}$ . If the particle is moving to the right, then its acceleration will be $-g$ . The time it will take for the particle to come to rest is $\sqrt{2 E/m}/g$ . Then the particle will start moving backward toward the origin. The time it will take to do that is also $\sqrt{2 E/m}/g$ . So, the total time the particle spends on the right side of the $y$ -axis is $4 \sqrt{2 E/m}/g$ .

Now, assume the particle is at the origin and is moving to the left with velocity $\sqrt{2 E/m}$ . If the potential were $0.5 k x^2$ for $-\infty < x < \infty$ (instead of only $-\infty < x$ ), would also be at the origin and moving to the left with speed $\sqrt{2 E/m}$ at some point in time during its period. The period for the $0.5 k x^2$ potential for $-\infty < x < \infty$ is

$\begin{align*}T = 2 \pi \frac{m}{k}\end{align*}$
By symmetry, the particle spends half of its time to the left of the y-axis and half of its time to the right of the y-axis. So, the amount of time it spends to the left of the axis is
$\begin{align*}\pi \frac{m}{k}\end{align*}$
We add the amount of time the particle spends to the left of the y-axis to the amount of time the particle spends to the right of the $y$ -axis to obtain the period:
$\begin{align*}T = \pi \frac{m}{k} + 4 \sqrt{2 E/m}/g\end{align*}$
Therefore, answer (D) is correct.

Solution to 1996 Problem 93