Question

The period of oscillation of a simple pendulum is $T =2 \pi \sqrt{\frac{ L }{ g }} .$ Measured value of $L$ is $20.0\, cm$ known to $1\, mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $80\, s$ using a wrist watch of $1\, s$ accuracy. What will be the accuracy (in $\%$) in the determination of $g$?

Answer 1

Time period, $T =2 \pi \sqrt{\frac{ L }{ g }}$
$\therefore g =4 \pi^{2} \frac{ L }{ T ^{2}}$
$\therefore \frac{\Delta g }{ g } \times 100 =\frac{\Delta L }{ L } \times 100+2 \frac{\Delta T }{ T } \times 100$
$=\frac{0.1}{20.0} \times 100+2 \times \frac{1}{80} \times 100$
$=\frac{10}{20}+\frac{200}{80}=\frac{1}{2}+\frac{20}{8}$
$=3 \%$