A point particle if mass `0.1 kg` is executing SHM of amplitude `0.1 m`. When the particle passes through the mean position, its kinetic energy is `8 xx 10^(-3)J`. Write down the equation of motion of this particle when the initial phase of oscillation is `45^(@)`.

Answer

A, D

Explanation

`1/2 mv_(max)^(2) = 8 xx 10^(-3)`
`rArr 1/8 momega^(2)a^(2) = 8 xx 10^(-3)`
`rArr omega = 4rad//s`
Therefore equation of SHM
`x = 0.1 sin (4t + (pi)/(4))`