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Question

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

A
2πβα
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B
β2α2
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C
αβ
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D
β2α
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Solution

The correct option is A 2πβα
For SHM, maximum acceleration, amax=ω2A=α, where A is the amplitude of SHM.
Maximum velocity, vmax=ωA=β
So, αβ=ω2AωA=ω
or αβ=2πT
or T=2πβα

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