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 1

Answer: (C)

$\frac{ T }{12}$

Answer 2

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 $y=\frac{a}{2}$ Therefore, $\frac{a}{2}=a \sin \omega t$ or $\sin \omega t=\frac{1}{2}=\sin \frac{\pi}{6}$ or $\omega t=\frac{\pi}{6}$ or $t=\frac{\pi}{6 \omega}$ or $t= \frac{\pi}{6(\frac{2\pi}{T})}\left(\right.$ as $\left.\omega=\frac{2 \pi}{T}\right)$ or $t=\frac{T}{12}$ Hence, the particle travels half of the amplitude from the equilibrium in $\frac{T}{12} s$.