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

Here, A=0.2 m `phi =(pi)/(4)`
As we know , KE`=(1)/(2)momega^(2)A^(2)`
where , m=mass of particle =0.2 kg
`64 xx10^(-3) =(1)/(2)xx0.2xx omega^(2)xx0.2^(2)`
`omega^(2)=(128 xx10^(-3))/(0.2xx0.2xx0.2) "or" omega=4s^(-1)`
`therefore` Equation of motion can be written as
`x=A"sin"(omegat+phi)`
`=0.2 "sin"(4t+(pi)/(4))`