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

Answer

D

Explanation

Kinetic energy at mean position `= (1)/(2) omega^(2)a^(2)= 8 xx 10^(-3)`
or `omega=((2 xx 8 xx 10^(-3))/(ma^(2)))=[(2 xx 8 xx 10^(-3))/(0.1 xx (0.1)^(2))]^(1//2) = 4`
Equation of SHM, `y = a sin (omega t + theta) = 0.1 sin (4 t + (pi)/(4))`