NEET 2017 · PhysicsPrevious Year Question
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity equals the magnitude of its acceleration. Its time period (in seconds) is
Answer: (C) (4π)/√5. Correct Answer (C): (4π)/√5 Step-by-step solution: 1.
- A.5/(2π)
- B.(5√2)/π
- C.(4π)/√5✓
- D.(2π)/√3
Correct Answer
(C) (4π)/√5
Solution & Explanation
Correct Answer (C): (4π)/√5 Step-by-step solution: 1. In SHM, |v| = ω√(A² − x²) and |a| = ω²x. 2. Set |v| = |a|: ω√(A² − x²) = ω²x → √(A² − x²) = ωx. 3. ω = √(A² − x²)/x. With A = 3 cm, x = 2 cm: ω = √(9 − 4)/2 = √5/2 rad/s. 4. T = 2π/ω = 2π·(2/√5) = 4π/√5 s.
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