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Two simple pendulums of lengths 16I and I are in phase at the mean position at a certain time. If T is the time period of shorter pendulum, the minimum time after which they will be again in phase

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$\frac{3}{2} T$

$\frac{1}{3} T$

$\frac{2}{3} T$

$\frac{4}{3} T$

answer is D.

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Detailed Solution

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{l_{1}}{l_{2}}}$
since pendulam of greater the period makes one oscillation less than the other,
$\frac{T_{2} - T_{1}}{T_{2}} = \sqrt{\frac{l}{16 l}} = \frac{1}{4} \Rightarrow T_{2} = \frac{4}{3} T_{1}$

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