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A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m s⁻² at a distance of 5 m from the mean position. The time period of oscillation is

Asked in NEET 2018 · v-x and a-x relations

Answer: (2) π s

Step-by-step solution

Given: |a|=20 m s⁻² at x=5 m from the mean position.

Idea: in SHM the acceleration and the displacement are tied by |a|=ω²x, whatever the system.

ω²=(|a|)/x=(20)/5=4 s⁻²

ω=2 rad s⁻¹

T=(2π)/ω=(2π)/2=π s

Why the other options are wrong

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