Question

The length of a simple pendulum is about $100cm$ known to an accuracy of $2mm$. Its period of oscillation is $2s$ determined by measuring the time for $500$ oscillations using a clock of $1s$ resolution. What is the accuracy (in percentage) in the determined value of $g$ ?

Answer 1

Answer: 0.40

Time period of a simple pendulum,
$T=2\pi \sqrt{\frac{L}{g}}$
or $g=\frac{4{\pi }^{2}L}{T^2}$ ...(i)
Differentiating (i), we have
$\frac{\Delta g}{g}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}$ ...(ii)
Given, $L=100cm,T=2s$
$\Delta T=\frac{1}{500}=0.002s$
$\Delta L=2mm=0.2cm$
$\Rightarrow {\left|\frac{\Delta g}{g}\right|}_{\max }=\frac{\Delta L}{L}+\frac{2\Delta T}{T}$
$=\frac{0.2}{100}+2\times \frac{0.002}{2}$
Thus, maximum percentage error
${\left|\frac{\Delta g}{g}\right|}_{\max }\times 100=\left(\frac{0.2}{100}\times 100\right)+\left(\frac{2\times 0.002}{2}\times 100\right)=0.4$