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

Options

A. `0.1 cos(4t + (pi)/(4))`
B. `0.1 sin(4t + (pi)/(4))`
C. `0.4 sin(t + (pi)/(4))`
D. `0.2 sin ((pi)/(2) + 2t)`

Answer

B

Explanation

From equilibrium
`(1)/(2)momega^(2)A^(2) = 8 xx 10^(-3) rArr (1)/(2) xx 0.1 xx omega^(2) xx (0.1)^(2) = 8 xx 10^(-3) rArr omega = 4`
So, equation of `SHM` is `x = A sin(omegat + phi) = 0.1 sin (4t + (pi)/(4))`.