Two simple pendulums whose lengths are 1,m an | vedclass

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.Let the length of the longer pendulum be $\ell_L = 121\,cm = 1.2

Vedclass · Accepted Answer

$10$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.Let the length of the longer pendulum be $\ell_L = 121\,cm = 1.21\,m$ and the shorter pendulum be $\ell_S = 1\,m$.Let $N$ be the number of oscillations of the longer pendulum and $(N+1)$ be the number of oscillations of the shorter pendulum when they are in phase again.Since the total time taken is the same,we have $N \cdot T_L = (N+1) \cdot T_S$.Substituting the formula for $T$,we get $N \cdot 2\pi \sqrt{\frac{\ell_L}{g}} = (N+1) \cdot 2\pi \sqrt{\frac{\ell_S}{g}}$.$N \sqrt{\ell_L} = (N+1) \sqrt{\ell_S}$.Substituting the values: $N \sqrt{1.21} = (N+1) \sqrt{1}$.$1.1N = N + 1$.$0.1N = 1$.$N = 10$.Thus,the longer pendulum completes $10$ oscillations to be in phase with the shorter one.

Two simple pendulums whose lengths are $1\,m$ and $121\,cm$ are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum will the two be in phase again?

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