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Question:

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?

2s

2πs

πs

1s

Solution:

Since in SHM, the magnitude of acceleration is given by a = ω²x
20 = ω² × 5
ω = 2
Now, Time Period (T), T = 2π/ω
T = π s