A point particle of mass `0.1kg` is executing SHM of amplitude `0.1m`. When the particle passes through the mean position, its kinetic energy is `8xx10^-3J`. Obtain the equation of motion of this particle if the initial phase of oscillation is `45^@`.

Options

A. `y=0.4 "sin" (6t+ pi//4)`
B. `y=0.1 "sin" (4 t + pi//4)`
C. `y=0.1 "sin" (2t + pi//4)`
D. `y=0.4 "sin" (3t+ pi//4)`

Answer

B

Explanation

`K.E.=(1)/(2)mv^(2) =(1)/(2)m omega^(2)A^(2)`
`therefore omega^(2)=(2K)/(mA^(2)) therefore omega =sqrt((2K)/(mA^(2))`
`therefore omega= sqrt((2xx8xx10^(-3))/(0.1xx(0.1)^(2)))=4`
`therefore` The equation of motion of the particle is
`y=A "sin" (omega t+ alpha) " where " alpha=45^(@)=pi//4`
`therefore y=0.1 "sin" (4t + pi//4)`