A body of mass 0.1 kg is executing a simple harmonic motion of amplitude 0.2 m. When the body passes through the mean position, its kinetic energy is `8xx10^(-3)` J. The initial phase of oscillation is 60. What is the equation of motion of the body ?

Options

A. `y=0.2 "sin" ((1)/(4)+(pi)/(3))`
B. `y=0.2 "sin" (4t+(pi)/(3))`
C. `y=0.2 "sin" ((1)/(2)+(pi)/(3))`
D. `y=0.2 "sin" (4t-(pi)/(3))`

Answer

B

Explanation

For the body executing a S.H.M.
a =0.2 m, m =0.1 kg and initial phase `alpha =60^(@)=(pi)/(3)`
`therefore` When the body passes through the mean position, its K.E. is maximum and Maximum K.E.`= (1)/(2) m omega^(2)a^(2)`
`therefore 8xx10^(-3)=(1)/(2)xx0.1xx omega^(2)xx(10^(-1))^(2)`
`16xx10^(-3)=omega^(2)xx10^(-3)`
`therefore omega^(2)=16 or omega =4 rad//s`
`therefore` The equation of S.H.M. is
`y=a "sin" (omega t+ alpha)=0.2 "sin" [4 t+(pi)/(3)]`