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