If maximum speed and acceleration of a particle executing SHM be `10 cm //s` and `100 pi cm//s^(2)` respectively ,then find its time period.
Hint `:v_("max") =A omega = 10`
`a_(max) = A omega^(2) = 100pi`
Solve to get `omega`
`T = ( 2pi)/( omega)`

Explanation

`v_(max)= A omega = 10 ` ....(i)
`a_(max) = A omega^(2) = 100 pi` ....(ii)
Divid equation (ii) by equation (i) , `10pi `
Time period , `T= ( 2pi)/( omega) = ( 2pi)/( 10 pi) = 0.2 s`