If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is

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KollegeApply · Accepted Answer

1% Let l be the length if the pendulum and T be the period. \text { Also, let ∆ l be the error in the length and ∆ T be the error in the period } . \text { We have } ∆ l/l $\times$ 100 = 2 \Rightarrow dl/l $\times$ 100 = 2 \text { Now,} T = 2$\pi$$\sqrt$(l/g) Taking log on both sides, we get \log T = \log 2$\pi$ + $\frac{1}{2}$\log l - $\frac{1}{2}$\log g \text { Differentiating both sides w . r . t . x, we get } 1/TdT/dl = $\frac{1}{2}$l \Rightarrow dT/dl = T/2l \Rightarrow dl/l $\times$ 100 = 2dT/T $\times$ 100 \Rightarrow dT/T $\times$ 100 = $\frac{2}{2}$ \Rightarrow ∆ T/T $\times$ 100 = 1 \text { Hence, there is an error of 1 % in calculating the period of the pendulum } .

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