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

Options

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

Answer

A

Explanation

let displacement equation of particle executing SHM is `y = a sin omega t`
As particle travels half of the amplitude from the equilibrium position, so `y = (a)/(2)`
Therefore, `(a)/(2) = a sin omega t`
or `sin omega t = (1)/(2) = "sin" (pi)/(2)`
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 travels half of the amplitude from the equlilibrium in `(T)/(12)` sec.