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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 is equal to that of its acceleration. Then its time period in seconds is

1. √5 / 2π
2. 4π / √5
3. 2π / √3
4. √5 / π

Correct answer: √5 / 2π

Solution

In SHM, velocity and acceleration are given by v = ω√(A² - x²) and a = ω²x, respectively. Equating their magnitudes, ω√(A² - x²) = ω²x. Simplifying, we get A² = 2x². Substituting A = 3 cm and x = 2 cm, we find ω = √5. The time period T = 2π/ω = √5 / 2π seconds.

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