A particle executes simple harmonic motion with an amplitude A. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude form the equilibrium position is………..

Options

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

Answer

D

Explanation

Displacement at mean position
`y= a sin ((2pi)/(T))t`
For `y= (a)/(2), (a)/(2)= a sin ((2pi)/(T))t`
`therefore (1)/(2)= sin (pi)/(6) = sin [((2pi)/(T))t]`
`therefore t((2pi)/(T))= (pi)/(6)`
`therefore t= (T)/(12)` is minimum.