The acceleration of a particle performing SHM is`12 cm //s^(2)` at a distance of 3 cm from the mean position . Calculate its time - period .
Hint `: a = omega^(2) x`
`a = 12 cm//s^(2) , x=3cm`
Find `omega`
`T= (pi)/(omega)`

Explanation

If in SHMs , at any instant the displacement of a particle from the mean position is x, then at this instant the acceleration is given by ltbr.`a= omega^(2) x`
`a= 12 cm//s^(2)`
`x= 3 cm`
` :. omega^(2) = (a)/( y) = (12)/ (3) = 4`
`omega = 2 rad s^(-1)`
Time period `T = ( 2pi)/(omega) = ( 2xx 3.14 )/(2) `
`= 3.14 s`