Question

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

Answer 1

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, $L=100\, cm , T=2 \,s$
$\Delta T=\frac{1}{500}=0.002\, s$
$\Delta L=2 \,mm =0.2 \,cm$
$\Rightarrow\left|\frac{\Delta g}{g}\right|_{\max }=\frac{\Delta L}{L}+\frac{2 \Delta T}{T}$
$=\frac{0.2}{100}+2 \times \frac{0.002}{2}$
Thus, maximum percentage error
$\left|\frac{\Delta g}{g}\right|_{\max } \times 100=\left(\frac{0.2}{100} \times 100\right)+\left(\frac{2 \times 0.002}{2} \times 100\right)=0.4$