A simple pendulum of length L is constructed | vedclass

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Question

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.When the pendulum swings from one side to the other,it hits the peg

Vedclass · Accepted Answer

$\pi \sqrt {\frac{L}{g}} \left( {1 + \frac{1}{{\sqrt 3 }}} \right)$

Vedclass · Answer

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.When the pendulum swings from one side to the other,it hits the peg.For the first half of the swing (from the extreme position to the vertical),the effective length is $L$. The time taken for this quarter oscillation is $t_1 = \frac{1}{4} T_1 = \frac{1}{4} (2\pi \sqrt{\frac{L}{g}}) = \frac{\pi}{2} \sqrt{\frac{L}{g}}$.After passing the vertical,the string hits the peg,and the effective length becomes $L' = L - \frac{2L}{3} = \frac{L}{3}$.The time taken for this quarter oscillation is $t_2 = \frac{1}{4} T_2 = \frac{1}{4} (2\pi \sqrt{\frac{L/3}{g}}) = \frac{\pi}{2} \sqrt{\frac{L}{3g}} = \frac{\pi}{2\sqrt{3}} \sqrt{\frac{L}{g}}$.The total time to return to the starting position is $T_{total} = 2(t_1 + t_2) = 2 \left( \frac{\pi}{2} \sqrt{\frac{L}{g}} + \frac{\pi}{2\sqrt{3}} \sqrt{\frac{L}{g}} \right) = \pi \sqrt{\frac{L}{g}} \left( 1 + \frac{1}{\sqrt{3}} \right)$.

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