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

`KE` at mean position
`K = (1)/(2)m v_(max)^(2) = (1)/(2)m omega^(2) A^(2)`
`:. omega = (sqrt((2K)/(m)))1/A`
`= (sqrt((2xx10^(-3)xx8)/(0.1))) (1)/(0.1) = 4 rad//s`
Now, `y = A sin (omega t + pi//4)` is the required equation.