A particle executes simple harmonic oscillation with an amplitude 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:

Let displacement equation of particle executing SHM is $\mathrm{y} = \mathrm{a} \sin \mathrm{\omega t}$

As particle travels half of the amplitude from the equilibrium position, so y = $\frac{\mathrm{a}}{2}$

Therefore, $\frac{\mathrm{a}}{2} = \mathrm{a} \sin \mathrm{\omega t}$

or $\sin \mathrm{\omega t} = \frac{1}{2}= \sin \frac{\mathrm{\pi }}{6} \mathrm{or} \mathrm{\omega t} = \frac{\mathrm{\pi }}{6}$

or $\mathrm{t} = \frac{\mathrm{\pi }}{6\mathrm{\omega }} \mathrm{or} \mathrm{t} = \frac{\mathrm{\pi }}{6\left({\displaystyle \frac{2\mathrm{\pi }}{\mathrm{T}}}\right)}(\mathrm{as} \mathrm{\omega } = \frac{2\mathrm{\pi }}{\mathrm{T}})$

or t = $\frac{\mathrm{T}}{12}$

Hence, the particle travels half of the amplitude from the equilibrium in $\frac{\mathrm{T}}{12}\sec$