A particle of mass 0.2 kg is excuting SHM of amplitude 0.2 m. When it passes through the mean position its kinetic energy is `64 xx 10^(-3) J` . Obtain the equation of motion of this particle if the initial phase of oscillation is `pi//4`.

Explanation

`A= 0.2 m, phi = (pi)/(4)`
`KE= (1)/(2) momega^(2) A^(2)`
`64 xx 10^(-3) = (1)/(2) xx 0.2 xx omega^(2) xx 0.2 ^(2)`
`omega^(2) = ( 128 xx 10^(-3))/( 0.2 xx 0.2 xx 0.2)`
or `omega = 4s^(-1)`
` x= A sin ( omega t + phi) = 0.2 sin ( 4 t+ (pi)/( 4))`