1.

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 beA. `(2pibeta)/(alpha)`B. `(beta^(2))/(alpha^(2))`C. `(alpha)/(beta)`D. `(beta^(2))/(alpha)`

Answer» Correct Answer - 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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