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Two pendulum having lengths $1 m$ and $16 m$ are both provided small displacements in the same direction at the same instant. They will again be in phase at the mean position after the shorter pendulum completes

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$\frac{1}{4} t h$ oscillation

$4$ oscillation

$5$ oscillation

$16$ oscillation

answer is B.

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Detailed Solution

$T_{1} = 2 π \sqrt{\frac{1}{g}}$ and $T_{2} = 2 π \sqrt{\frac{16}{g}} = 4 T_{1}$

They will be again in phase, if shorter one complete one more oscillation, so

$(n + 1) \times T_{1} = n T_{2}$

or $(n + 1) \times T_{1} = n \times 4 T_{1}$

$∴ n = \frac{1}{3}$

and $n + 1 = \frac{4}{3}$

For whole number multiply $n$ and $(n + 1)$ by $3$ , so we get $1$ and $4$ .

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