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………

Options

A. `(2pi beta)/(alpha)`
B. `(beta^2)/(alpha^2)`
C. `(alpha)/(beta)`
D. `(beta^2)/(alpha)`

Answer

A

Explanation

Maximum acceleration in SHM
`a_("max")= A omega^(2)`
`therefore alpha = A omega^(2)` and
Maximum velocity `v_("max")= A omega`
`therefore beta = A omega`
`therefore (a_("max"))/(v_("max"))= (A omega^(2))/(A omega)`
`(alpha)/(beta)= omega`
`therefore (alpha)/(beta)= (2pi)/(T)`
`therefore T= 2pi (beta)/(alpha)`.