Answer
$0.3 \%$
Work Step by Step
The period $\mathrm{T}$ of a simple pendulum with oscillations is given as:
\[
(2 \pi)(\sqrt{\frac{L}{g}})=T
\]
Thus, the total differential $d T$ of $\mathrm{T}$ is given by
\[
\begin{aligned}
&\frac{\pi}{\sqrt{L g}} d L-\frac{\pi \sqrt{L}}{g \sqrt{g}} d g =d T \\
&-\frac{2 \pi \sqrt{L}}{2 g \sqrt{g}} d g+\frac{2 \pi L}{2 L \sqrt{L g}} d L =d T \\
d T &=\frac{T}{2 L} d L-\frac{T}{2 g} d g \\
\frac{d T}{T} &=\frac{d L}{2 L}-\frac{d g}{2 g}
\end{aligned}
\]
,
\[
\left|\frac{\Delta L}{L}\right| \leq 0.005
\]
\[
\left|\frac{\Delta g}{g}\right| \leq 0.001
\]
\[
\begin{aligned}
\frac{\Delta T}{T_{0}} \approx & \frac{d T}{T}=\left|\frac{1}{2} \frac{\Delta L}{\Delta L}-\frac{1}{2} \frac{\Delta g}{g}\right| \\
& \frac{0.001}{2}+\frac{0.005}{2}=\frac{\Delta T}{T_{0}} \\
&0.003= \frac{\Delta T}{T_{0}}
\end{aligned}
\]
