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PhysicsSHMMiscellaneousEasy2 minPYQ_2022
PhysicsEasysingle choice

Two pendulums of length121 cmand100 cmstart vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position isNEET 2022 (Phase 1)

Options:

Answer:
D
Solution:

Time period of a pendulum,

T∝l

⇒T1T2=l1l2

If n is the number of vibrations after which the pendulums are again in phase, the number of vibration of the longer pendulum will be n-1. Therefore,

T1×n-1T2×n=1 ⇒121n-1100n=1

⇒11n-11=10n ⇒n=11

Stream:NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous
⏱ 2mℹ️ Source: PYQ_2022

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