Two simple pendulums of length 1 m and 16 m respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed n oscillations. The value of n is

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$\frac{2}{3}$

$\frac{4}{3}$

$\frac{1}{3}$

answer is D.

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Detailed Solution

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

at any time t, phase of pendulums are:

$\emptyset_{1} = ω_{1} t = \frac{2 π}{T_{1}} t, \emptyset_{2} = ω_{2} t = \frac{2 π}{T_{2}} t$

First pendulum is faster. Both will be in same phase again when faster pendulum has completed one oscillation more than slower pendulum

$\emptyset_{1} - \emptyset_{2} = 2 π$

$\Rightarrow \frac{2 π}{T_{1}} t - \frac{2 π}{4 T_{1}} t = 2 π \Rightarrow t = \frac{4 T_{1}}{3}$

No. of oscillations completed by shorter pendulum in time t:

$n = \frac{t}{T_{1}} = \frac{4}{3}$

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