The motion of the particle is described by
$\begin{align*}x(t) = A \cos \left(2 \pi f t + \delta \right)\end{align*}$
So, the particle's velocity is
$\begin{align*}x'(t) = -2 \pi f A \sin \left(2 \pi f t + \delta \right)\end{align*}$
When x(t) = A/2 , $\cos \left(2 \pi f t + \delta \right) = 1/2$ , so $\sin \left(2 \pi f t + \delta \right) = \sqrt{3}/2$ or $-\sqrt{3}/2$ . So, the particle's speed is
$\begin{align*}\left|x'(t) \right| = 2 \pi f A \cdot \frac{\sqrt{3}}{2} = \boxed{\sqrt{3} \pi f A}.\end{align*}$
Hence, answer (B) is correct.

Solution to 1986 Problem 77