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InterviewSolution
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The time period of a simple pendulum is given by `T = 2 pi sqrt(L//g)`, where `L` is length and `g` acceleration due to gravity. Measured value of length is `10 cm` known to `1 mm` accuracy and time for 50 oscillations of the pendulum is 80 s using a wrist watch of 1 s resloution. What is the accuracy in the determination of `g`? |
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Answer» `T = 2 pi sqrt((L)/(g))` `g = (4 pi^(2) L)/(T^(2))` Here `T = (t)/(n), t:` total time, `n` : number of oscillaiton `Delta T = (Delta t)/(n)` `(Delta T)/(T) = (Delta t)/(t)` Maximum relative errors in `g` `(Delta g)/(g) = (Delta L)/(L) + 2 (Delta T)/(T)` Maximum percentage error is `g` `(Delta g)/(g) xx 100 = (Delta L)/(L) xx 100 + 2 (Delta T)/(T) xx 100` `= (0.1)/(10) xx 100 + 2 (1)/(80) xx 100` `= 1 + 2.3 = 3.5%` |
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