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A particle is executing a simple harmoni...

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

A

`(2pibeta)/(alpha)`

B

`(beta^(2))/(alpha^(2))`

C

`(alpha)/(beta)`

D

`(beta^(2))/(alpha)`

Text Solution

Verified by Experts

The correct Answer is:
A
Maximum acceleration
`alpha=Aomega^(2)` . . (1)
Maximum velocity
`beta=A omega`
`(Eq^(n)(1))/(Eq^(n)(2))implies(alpha)/(beta)implies(alpha)/(beta)=(2pi)/(T)`
`T=2pi(beta)/(alpha)`
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