A particle performing simple harmonic motion undergoes displacement of $\frac{A}{2}$ (where $A$ is the amplitude of simple harmonic motion) during first second. At $t=0$, the particle may be at the extreme position or mean position. The period of the simple harmonic motion can be

The correct options are
A $6\text{s}$
C $12\text{s}$
At $t=0,$ when particle is at extreme position, the situation is as shown in Fig. 1

From the figure, $\cos \theta =\frac{\frac{A}{2}}{A}=\frac{1}{2}$
$\theta =\frac{\pi }{3}\Rightarrow \frac{\pi }{3}=\omega t=\frac{2\pi }{T}\times 1\Rightarrow T=6\text{s}$

At $t=0,$ when particle is at mean position, the situation is as shown in Fig. 2

$\theta =\omega t$
From the figure, $\sin \theta =\frac{\frac{A}{2}}{A}$
$\theta =\frac{\pi }{6},$
$\Rightarrow \frac{\pi }{6}=\omega t=\frac{2\pi }{T}\times 1\Rightarrow T=12\text{s}$
If initially the particle is located somewhere else, then time period comes out to be differnet, A reverse question can also be formed on the same concept.

Why this question ? Simple harmonic motion can be visualized as the projection of uniform circular motion onto one axis. The phase angle $\omega t$ in SHM corresponds to the real angle $\omega t$ through which the ball has moved in circular motion