Answer
$3.24$ feet
Work Step by Step
Given \begin{equation}
T(L)=2 \pi \sqrt{\frac{L}{32}}
\end{equation} We are given that $T = 2$ seconds and want to find the length of the clock's pendulum. Set $T(L) = 2$ into the formula and solve for $L$. \begin{equation}
\begin{aligned}
T(L) & =2 \pi \sqrt{\frac{L}{32}} \\
2 & =2 \pi \sqrt{\frac{L}{32}} \\
\frac{2}{2 \pi} & =\frac{2 \pi \sqrt{\frac{L}{32}}}{2 \pi} \\
\frac{1}{ \pi} & =\sqrt{\frac{L}{32}} \\
\left(\frac{1}{\pi}\right)^2 & =\left(\sqrt{\frac{L}{32}}\right)^2 \\
\frac{1}{ \pi^2} & =\frac{L}{32} \\
32\left(\frac{1}{ \pi^2}\right) & =32\left(\frac{L}{32}\right) \\
\frac{32}{\pi^2} & =L \\L& \approx 3.24.
\end{aligned}
\end{equation} The length of the pendulum with period of $1$ second is about $3.24$ feet.
