Answer
$\dfrac{\pi}{2} \text{ seconds OR approximately }1.57\text{ seconds}
$
Work Step by Step
Substituting $
l=2
$ in the given equation, $
P=2\pi\sqrt{\dfrac{l}{32}}
,$ results to
\begin{array}{l}\require{cancel}
P=2\pi\sqrt{\dfrac{2}{32}}
\\
P=2\pi\sqrt{\dfrac{1}{16}}
\\
P=2\pi\sqrt{\left(\dfrac{1}{4}\right)^2}
\\
P=2\pi\left(\dfrac{1}{4}\right)
\\
P=\dfrac{\pi}{2}
\\\text{OR}\\
P\approx1.57
.\end{array}
Hence, the period, $P,$ is $
\dfrac{\pi}{2} \text{ seconds OR approximately }1.57\text{ seconds}
.$
