Question

A boy performs an experiment in which he uses a simple pendulum to find the value of 'g' using the formula $g=\frac{4{\pi }^{2}l}{T^2}$ = Where l = 1 m. Error in measurements of length is $\Delta l=2cm$, human error in time measurement is 0.15sec and least count of stop watch $\Delta T=0.01s$, What will be the percentage error in determining value of $g$?

Answer 1

value of $g$ is dependent on time period of pendulum $\left(T\right)$ & as follows:

$g=\frac{4{\pi }^{2}L}{T^2}$......(1)

differentiating both sides give error relation:

$\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}$....(2)

given that, $\Delta L=0.02m$ and combined $\Delta T=0.15+0.01=0.165$ [Human error and least count error taken together]

When $l=1m$ taking $g=9.8m/s^2$ the correct time period will be:

$T=2\pi \sqrt{\frac{l}{g}}$ [From reassuring (1)]

$=2\pi \sqrt{\frac{1}{9.8}}\simeq 2s$

Thus, from (2) we have:

$\frac{\Delta g}{g}=\frac{0.01}{L}+2\times \frac{0.16}{2}=0.17$, or $\frac{\Delta g}{g}=17\%.$