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 maximum velocity
B. velocity will be one fourth its maximum energy
C. velocity will be equal to its maximum velocity
D. none of the above

Answer

A

Explanation

`V=V_(0) cos omegat=V_(0) cos.(2pit)/(T)`
at `t=T/6`
`V=V_(0) cos.(2pi)/T .T/6=(V_(0))/2 .... (1)`
`a=a_(0) sin omegat=a_(0) sin ((2pit)/T)rArr t=T/6`
`a=a_(0) sin ((2pi)/(T)T//6) =(sqrt(3) a_(0))/2=0.86 a_(0) ...(3)`