A particle performing `SHM` with amplitude `10cm`. At What distance from mean position the kinetic energy of the particle is thrice of its potential energy ?

Answer

`+- 5 cm`

Explanation

According to question
`(1)/(2)mv^(2) = 3 xx(1)/(2)momega^(2)x^(2)`
`rArr (1)/(2)momega^(2)(A^(2) - x^(2)) = (3)/(2)momega^(2)x^(2) rArr A^(2) = 4x^(2) rArr x = +- (A)/(2) +- (10cm)/(2) = +- 5cm`.