A particle executes `SHM` in a straight line path .The amplitude of oscillation is `2cm`. When the displacement of the particle from the mean position is `1cm` the numerical value of acceleration is equal to the numerical value of velocity. Then find the frequency of `SHM`.

Options

A. `sqrt(3)//2pi`
B. `3//2 pi`
C. `3//sqrt(2)pi`
D. `sqrt(3)//pi`

Answer

A

Explanation

The speed of particle at a distance `x= 1` from meqan piosition is
`v = omega sqrt(A^(2) - x^(2)) = omega sqrt(2^(2) - 1^(2)) = sqrt(3s)`…(i)
The magnitude of acceleration at `x = 1` is
`a = omega^(2)x = omega^(2)`…(ii)
From equation (1) and (2)
`omega^(2) = sqrt(3)omega rArr omega = sqrt(3)`
`or f = (omega)/(2pi) = sqrt(3)/(2pi)`