Question
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
A.
$$\frac{T}{4}$$
B.
$$\frac{T}{8}$$
C.
$$\frac{T}{12}$$
D.
$$\frac{T}{2}$$
Answer :
$$\frac{T}{12}$$
Solution :
Let displacement equation of particle executing $$SHM$$ is
$$x = a\sin \omega t$$
As particle travels half of the amplitude from the equilibrium position, so
$$x = \frac{a}{2}$$
$$\eqalign{
& {\text{Therefore,}}\,\,\frac{a}{2} = a\sin \omega t \cr
& {\text{or}}\,\,\sin \omega t = \frac{1}{2} = \sin \frac{\pi }{6} \cr
& {\text{or}}\,\,\omega t = \frac{\pi }{6}\,\,{\text{or}}\,\,t = \frac{\pi }{{6\omega }} \cr
& {\text{or}}\,\,t = \frac{\pi }{{6\left( {\frac{{2\pi }}{T}} \right)}}\,\,\left( {{\text{as,}}\,\,\omega = \frac{{2\pi }}{T}} \right) \cr
& {\text{or}}\,\,t = \frac{T}{{12}} \cr} $$
Hence, the particle travels half of the amplitude from the equilibrium in $$\frac{T}{{12}}s.$$