Two simple pendulums of length 5m and 20 m respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed.... oscillations

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Detailed Solution

If t is the time taken by pendulums to come in same phase again first time after $t = 0$ and

$N_{S}$ = Number of oscillations made by shorter length pendulum with time period $T_{S}$

$N_{L} =$ Number of oscillations made by longer length pendulum with time period $T_{L}$

Then $t = N_{S} T_{S} = N_{L} T_{L}$

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