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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