The period of oscillation of a simple pendulum is given by T = 2 π L g , where L is the length of the pendulum and g is the acceleration due to gravity. The length is measured using a meter scale which has 500 divisions. If the measured value L is 50 cm , the accuracy in the determination of g is 1.1 % and the time taken for 100 oscillations is 100 seconds, what should be the possible error in measurement of the cloc

Step-by-step solution

Given, T = 2 π l g or T 2 = 4 π 2 l g ∴     2 ∆ T T = ∆ l l + ∆ g g ...(i) Now, l = 50   cm ,   ∆ l = 2   mm = 0.2   cm ∆ g g = 1.1 % = 1.1 100 Put these values in Equation. (i), then we get ∆ T T = 1 2 0.2 50 + 1.1 100 = 7.5 × 10 - 3   s or 7. 5   ms ∵ In 100 s, resolution of the clock is 7 .5   ms . ∴ In 60 s resolution of the clock is 7 .5 × 60 100 ≈ 5   ms