Two pendulums of length 1.21 ,m and 1.0 ,m st | vedclass

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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.For the first pendulum of length $L_1 = 1.21 \,m$,the time period

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$10$

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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.For the first pendulum of length $L_1 = 1.21 \,m$,the time period is $T_1 = 2 \pi \sqrt{\frac{1.21}{g}} = 1.1 	imes 2 \pi \sqrt{\frac{1}{g}}$.For the second pendulum of length $L_2 = 1.0 \,m$,the time period is $T_2 = 2 \pi \sqrt{\frac{1.0}{g}}$.Thus,$T_1 = 1.1 T_2 = \frac{11}{10} T_2$,which implies $10 T_1 = 11 T_2$.Let $n_1$ be the number of vibrations of the longer pendulum and $n_2$ be the number of vibrations of the shorter pendulum when they are in phase again.The condition for being in phase is $n_1 T_1 = n_2 T_2$.Substituting $T_1 = \frac{11}{10} T_2$,we get $n_1 (\frac{11}{10} T_2) = n_2 T_2$,which simplifies to $11 n_1 = 10 n_2$.For the smallest integer values,$n_1 = 10$ and $n_2 = 11$.Therefore,the two pendulums will be in phase again after $10$ vibrations of the longer pendulum.

Two pendulums of length $1.21 \,m$ and $1.0 \,m$ start vibrating. At some instant,the two are in the mean position in the same phase. After how many vibrations of the longer pendulum will the two be in phase again?

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