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^(@)`.

Explanation

Given that amplitude `a = 0.1 m, m = 0.1 kg phi = 45^(@) = (pi//4)` rad , so the wquatude of SHM will be
`y = 0.1 sin [(omega t + (pi//4)]` (i)
Now as in SHM, `KE` is given by `K = (1)/(2) m omega^(2) (A^(2) - y^(2))`
which according to the given problem is `8 xx 10^(-3) = (1)/(2) xx 0.1 xx omega^(2) (0.1^(2) - 0^(2))`, i.e.
`omega = 4 ras//s` (ii)
Sumstituting the value of `omega` from Eq (ii) in Eq (i), we get `y = 0.1 sin [4 t + (pi//4)]`