A particle executes simple harmonic oscillation with an amplitude `a`. The period of oscillations is `T`. The minimum time taken by the particle to travel half to the amplitude from the equliibrium position is

Options

A. `T/4`
B. `T/8`
C. `T/(12)`
D. `T/2`

Answer

C

Explanation

For SHM-
`y=Asinomegat`
`therefore (A)/(2)=Asin omegat`
`implies sin omegat=(1)/(2)`
`implies (2pi)/(T)t=(pi)/(6)implies t=(T)/(12)` sec.