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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

`(alpha)/(beta)`

B

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

C

`(2pibeta)/(alpha)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
`omega^(2)A=alpha`
`omegaA=beta`
`implies omega=(alpha)/(beta)implies T=(2pi)/(omega)=(2pibeta)/(alpha)`.
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