InterviewSolution
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InterviewSolution
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The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` and is known to `1mm` accuracy . The period of oscillation is about `0.5 s`. The time of 100 oscillation is measured with a wrist watch of `1 s` resolution . What is the accuracy in the determination of `g` ? |
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Answer» Correct Answer - `5%` `T = 2 pi sqrt((L)/(g))` or `T^(2) = 4 pi^(2) (L)/( g)` or `g = ( 4 pi^(2) L)/( T^(2))` Now `(Delta g)/( g) = (Delta L) /( L) + 2 xx(Delta T)/(T)` IN terms of percentage , `(Delta g)/(g) xx 100 = (Delta L)/(L) xx 100 + 2 xx (Delta T)/(T) xx 100` Percentage error in `L = 100 xx ( Delta L)/(L) = 100 xx (0.1)/(10) = 1%` Percentage error in `T = 100 xx ( Delta T%)/(T) = 100 xx (1)/(50 ) = 2%` Percentage error in `g = 100 xx ( Delta g)/(g) = 1% + 2 xx 2% = 5%` |
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