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

Options

A. `y=0.1sin(+-4t+(pi)/(4))`
B. `y=0.2sin(+-4t+(pi)/(4))`
C. `y=0.1sin(+-2t+(pi)/(4))`
D. `y=0.2sin(+-2t+(pi)/(4))`

Answer

A

Explanation

The displacement of a particle in S.H.M. is given by
`y=a sin(omega t+phi)` velocity `=(dy)/(dt)=omega a cos(omega t+phi)`
The velocity is maximum when the particle passes through the mean position i.e.,
`((dy)/(dt))_("max")=omega a`
The kinetic energy at this instant is given by
`(1)/(2)m((dy)/(dt))^(2)_("max")=(1)/(2)m omega^(2)a^(2)=8xx10^(-3)` joule
or `(1)/(2)xx(0.1)omega^(2)xx(0.1)^(2)=8xx10^(-3)`
Solving we get `omega=+-4`
Substituting the values of a, `omega` and `phi` in the equation of S.H.M., we get
`y=0.1sin(+-4t+pi//4)` metre.