A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be

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β2α

2πβα

β2α2

αβ

answer is B.

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Detailed Solution

If A and ω be amplitude and angular frequency of vibration, thenα=ω2A . . . . (i) and β=ωA . . . . .(ii)Diving eqn. (i) by eqn. (ii), we getαβ=ω2AωA=ω∴ Time period of vibration is T=2πω=2π(α/β)=2πβα

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