Q.

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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a

β2α

b

2πβα

c

β2α2

d

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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