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^(@)`.

Options

A. `y = 0.1sin(pm 4t + (pi)/(4))`
B. `y = 0.2sin(pm 4t + (pi)/(4))`
C. `y = 0.1sin(pm2t + (pi)/(4))`
D. `y = 0.2sin(pm2t + (pi)/(4))`

Answer

A

Explanation

The displacement of a particle in S.H.M. is given by
`y a = sin (omega t + phi)`
velcoity `= (dt)/(dy) = omega a cos (omegat + phi)`
The velocityis 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))_(max)^(2)= (1)/(2) m omega^(2) a^(2) = 8 xx 10^(-3)` joule .
or `(1)/(2) xx (0.1) omega^(2) (0.1)^(2) = 8 xx 10^(-3)`