The acceleration of a simple harmonic oscillator is `1 m//s^(2)` when its displacement from mean position is 0.5 m. Then its frequency of oscillation is

Options

A. `sqrt(2)pi Hz`
B. `pi//sqrt(2)Hz`
C. `(1)/(sqrt(2)pi)Hz`
D. `(sqrt(2))/(pi)Hz`

Answer

C

Explanation

`a=omega^(2)x therefore omega^(2)=(a)/(x)=(1)/(0.5)`
`omega^(2)=2 therefore omega=sqrt(2)`
`2pin=sqrt(2)therefore n=(sqrt(2))/(2pi)=(1)/(sqrt(2))pi`