Two simple pendulums of length 5 m 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.
Text Solution
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Frequency of the pendulum $v_{1-5} = \frac{1}{2 \pi} \sqrt{\frac{g}{5}}$
$V_{1-20} = \frac{1}{2 \pi} \sqrt{\frac{g}{20}}$ $\therefore \frac{V_{1-5}}{V_{1-20}} = \sqrt{\frac{20}{5}} = 2$ $\Rightarrow V_{1-5} = 2 V_{1-20}$
As shorter length pendulum has frequency double the larger length pendulum.
Therefore shorter pendulum should complete 2 oscillations before they will be again in phase.
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