Two pendulums having lengths 1.0 m and 1.21 | vedclass

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.For the first pendulum $(L_1 = 1.0 \ m)$: $T_1 = 2\pi \sqrt{\frac{1

Vedclass · Accepted Answer

$11$

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.For the first pendulum $(L_1 = 1.0 \ m)$: $T_1 = 2\pi \sqrt{\frac{1.0}{g}} = 2\pi \sqrt{\frac{1}{g}}$.For the second pendulum $(L_2 = 1.21 \ m)$: $T_2 = 2\pi \sqrt{\frac{1.21}{g}} = 1.1 	imes 2\pi \sqrt{\frac{1}{g}} = 1.1 T_1$.Let $n_1$ be the number of oscillations of the smaller pendulum and $n_2$ be the number of oscillations of the larger pendulum when they are in the same phase again.At this time,$n_1 T_1 = n_2 T_2$.Substituting $T_2 = 1.1 T_1$,we get $n_1 T_1 = n_2 (1.1 T_1)$,which simplifies to $n_1 = 1.1 n_2$ or $n_1 = \frac{11}{10} n_2$.For $n_1$ to be an integer,the smallest value for $n_2$ is $10$,which gives $n_1 = 11$.Therefore,the smaller pendulum completes $11$ oscillations.

Two pendulums having lengths $1.0 \ m$ and $1.21 \ m$ start oscillating simultaneously in the same phase. They will again come in the same phase after how many oscillations of the smaller pendulum?

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