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

Let displacement equation of particle executing SHM is `y = a sin omegat`
As particle traveles half of the amplitude from the
Therefore (a)/(2) = a sin omega t`
or sin omega t = (1)/(2) = sin (pi)/(6)` or `omega t = (pi)/(6)`
or `t = (pi)/(6 omega)` or `t = (pi)/(6((2pi)/(T))) (as `omega = (2pi)/(T)`)
or `t = (T)/(12)` Hence the particle travel half of the amplitude from thhe equilibrium in `(T)/(12) sec`