Two pendulums of length 121 cm and 100 cm s | vedclass

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(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.Let $L_1 = 121 \ cm = 1.21 \ m$ and $L_2 = 100 \ cm = 1.0 \ m$.Let

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$11$

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(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.Let $L_1 = 121 \ cm = 1.21 \ m$ and $L_2 = 100 \ cm = 1.0 \ m$.Let $n_1$ be the number of oscillations of the longer pendulum and $n_2$ be the number of oscillations of the shorter pendulum.For them to be in phase at the mean position again,the total time must be equal: $n_1 T_1 = n_2 T_2$.$n_1 (2 \pi \sqrt{\frac{1.21}{g}}) = n_2 (2 \pi \sqrt{\frac{1.0}{g}})$.$n_1 (1.1) = n_2 (1.0)$.$1.1 n_1 = n_2$,which implies $\frac{n_2}{n_1} = \frac{1.1}{1} = \frac{11}{10}$.Since $n_2$ and $n_1$ must be integers,the minimum number of vibrations for the shorter pendulum $(n_2)$ is $11$ and for the longer pendulum $(n_1)$ is $10$.

Two pendulums of length $121 \ cm$ and $100 \ cm$ start vibrating in phase. At some instant,the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

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