A point particle of mass `0.1kg` is executing SHM of amplitude `0.1m`. When the particle passes through the mean position, its kinetic energy is `8xx10^-3J`. Obtain the equation of motion of this particle if the initial phase of oscillation is `45^@`.

Explanation

Equation of SHM
`y = a sin (omega t + phi)`
`a = 0.1 m " "phi = pi//4m = 0.1 kg`
K.E. at mean position
`8 xx 10^(3) J = (1)/(2) m omega^(2) a^(2) rArr omega^(2) = (16 xx 10^(-3))/(ma^(2))`
`omega^(2) = (16 xx 10^(-3))/(0.1 xx (0.1)^(2))`
`omega^(2) = 16," "omega = 4 rad//s`
Hence equation of SHM
y = 0.1 (m) sin `(4t + pi//4)`