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Questions  

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/s2 at a distance of 5 m from the mean position. The time period of oscillation is

a
l s
b
2 π s
c
2 s
d
π s

detailed solution

Correct option is D

In case of SHM a→ = -ω2y→Given a = 20 m/s2 at y = 5 mHence |a→| = ω2|y→|20 = ω2×5or ω2 = 4  ⇒ ω = 2 rad/secBut ω = 2πT2= 2πT    or T = π sec

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