Physics

# 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 $\Delta l = 2\space cm$, human error in time measurement is 0.15sec and least count of stop watch $\Delta T = 0.01 s$, What will be the percentage error in determining value of $g$?

##### SOLUTION
value of $g$ is dependent on time period of pendulum $(T)$ & as follows:
$g=\dfrac{4\pi^2L}{T^2}$......(1)
differentiating both sides give error relation:
$\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}$....(2)
given that, $\Delta L=0.02\ m$ and combined $\Delta T=0.15+0.01=0.165$ [Human error and least count error taken together]
When $l=1\ m$ taking $g=9.8\ m/s^2$ the correct time period will be:
$T=2\pi \sqrt{\dfrac{l}{g}}$        [From reassuring (1)]
$=2\pi \sqrt{\dfrac{1}{9.8}}\simeq 2s$
Thus, from (2) we have:
$\dfrac{\Delta g}{g}=\dfrac{0.01}{L}+2\times \dfrac{0.16}{2}=0.17$, or $\dfrac{\Delta g}{g}=17\%.$

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Subjective Medium Published on 18th 08, 2020
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