Answer
$0.81$ feet
Work Step by Step
Given \begin{equation}
T(L)=2 \pi \sqrt{\frac{L}{32}}.
\end{equation} We are given that $T = 1$ seconds and want to find the length of the clock's pendulum. Set $T(L) = 1$ into the formula and solve for $L$. \begin{equation}
\begin{aligned}
T(L) & =2 \pi \sqrt{\frac{L}{32}} \\
1 & =2 \pi \sqrt{\frac{L}{32}} \\
\frac{1}{2 \pi} & =\sqrt{\frac{L}{32}} \\
\left(\frac{1}{2 \pi}\right)^2 & =\left(\sqrt{\frac{L}{32}}\right)^2 \\
\frac{1}{4 \pi^2} & =\frac{L}{32} \\
32\left(\frac{1}{4 \pi^2}\right) & =32\left(\frac{L}{32}\right) \\
\frac{8}{\pi^2} & =L \\
L & \approx 0.8106.
\end{aligned}
\end{equation} The length of the pendulum with period of $1$ second is about $0.81$ feet.
