The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of `(T)/(6)` it passes its mean position ,then at t=0 its,

Options

A. velocity will be one half its mean position, its
B. displacement will be one half its amplitude
C. acceleration will be nearly `85%` of its maximum acceleration
D. `KE=PE`

Answer

A, C

Explanation

`V=V_0cosomegat=V_0cos((2pit)/(T))`
`V=V_0cos((2pi)/(T))(T)/(6)=(V_0)/(2)` (A)
`a=a_0sinomegat=a_0sin((2pit)/(T))`
`=a_0sin((2pi)/(T)(T)/(6))=(sqrt3a_0)/(2)=86a_0`. (C )