Simple Harmonic Motion — Easy JEE physics MCQ
A simple pendulum of length 1 m oscillates on Earth (g = 10 m/s²). Its time period is approximately:
- A. 1 s
- B. 2 s
- C. π s
- D. 2π s
Solution
Time period T = 2π√(L/g) = 2π√(1/10) = 2π/√10 ≈ 2 s.
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A simple pendulum of length 1 m oscillates on Earth (g = 10 m/s²). Its time period is approximately:
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