A simple pendulum of length L is constructed | vedclass

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Question

(D) The pendulum swings with two different effective lengths during its motion.When the pendulum is on the left side,the effective length is $L_1 = L$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The pendulum swings with two different effective lengths during its motion.When the pendulum is on the left side,the effective length is $L_1 = L$. The time taken for a quarter oscillation is $t_1 = \frac{1}{4} T_1 = \frac{1}{4} (2\pi \sqrt{\frac{L}{g}}) = \frac{\pi}{2} \sqrt{\frac{L}{g}}$.When the pendulum is on the right side,it hits the peg at a distance $2L/3$ from the pivot. The effective length becomes $L_2 = L - 2L/3 = L/3$. The time taken for a quarter oscillation is $t_2 = \frac{1}{4} T_2 = \frac{1}{4} (2\pi \sqrt{\frac{L/3}{g}}) = \frac{\pi}{2} \sqrt{\frac{L}{3g}}$.The total time to return to the starting position is $T' = t_1 + t_2 + t_2 + t_1 = 2(t_1 + t_2)$.$T' = 2 \left( \frac{\pi}{2} \sqrt{\frac{L}{g}} + \frac{\pi}{2} \sqrt{\frac{L}{3g}} \right) = \pi \sqrt{\frac{L}{g}} \left( 1 + \frac{1}{\sqrt{3}} \right)$.

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