A particle of mass `10g` executes a linear `SHM` of amplitude `5 cm` with a pariod of `2s`. Find the `PE` and `KE,(1)/(6) s` after crossing the mean position.

Explanation

Given `m = 10g = 10^(-2) kg, T = 2s`,
`omega = (2pi)/(T) = (2pi)/(2) = pi rad//s`
`A = 5cm = 5xx 10^(-2)m, KE = (1)/(2) mA^(2) omega^(2) cos^(2) omegat`
At `t = (1)/(6)s, KE = (1)/(2) xx 10^(-2) xx (5 xx 10^(-2))^(2) (pi^(2)) cos^(2). (pi)/(6)`
`= (25 xx 10^(-6))/(2) xx pi^(2) xx ((sqrt(3))/(2))^(2) = 9.25 xx 10^(-5)J`
`PE = (1)/(2) mA^(2) omega^(2) sin^(2) omegat`
`= (1)/(2) xx 10^(-2) xx (5xx10^(-2))^(2) pi^(2) sin^(2).(pi)/(6)`
`= (25 xx 10^(-6))/(2) xx pi^(2) xx ((1)/(2))^(2) = 3.085 xx 10^(-5)J`