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PhysicsSHMMiscellaneousEasy2 minPYQ_2020
PhysicsEasysingle choice

A simple pendulum of length1 mis oscillating with an angular frequency10 rad s-1The support of the pendulum starts oscillating up and down with a small angular frequency of1 rad s-1and an amplitude of10-2 m.The relative change in the angular frequency of the pendulum is best given by:

Options:

Answer:
D
Solution:

Angular frequency of simple pendulum,

ω=gl

As the support oscillates up and down, effectivegchanges

∆g=2.ω02A=2×12×10-2 m s-2

=0.02 m s-2

Change in angular frequencydωof the pendulum is given by

dωω=-12⋅dgg

∆ωω=12∆gg

=12×0.0210

=0.001 rad s-1

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous
⏱ 2mℹ️ Source: PYQ_2020

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