Let the motion be $x=A\sin (\omega t)$.

• The total period is $T=2\pi /\omega$.
• To reach $x=A/2$: $A/2=A\sin (\omega t)⟹\omega t=\pi /6$.
• To reach $x=-A/2$: $\omega t=-\pi /6$.
• Time spent in this range within one cycle: $\Delta t=2\times \frac{\pi /6}{\omega }=\frac{\pi /3}{\omega }$ (one crossing) $\times 2$ (one full period involves two crossings) = $\frac{2\pi }{3\omega }$.
• Fraction $=\frac{2\pi /3\omega }{2\pi /\omega }=\frac{1}{3}$.
• Correction: The question asks for the region between $-A/2$ and $A/2$. In one full period, the particle crosses this region twice. Total time in region is $1/6+1/6=1/3$ of the cycle. Wait, the option says $1/3$ is correct. Option '2/3' was previously listed - let's re-verify. Actually, the phase angle for $-A/2$ to $A/2$ is $60^{\circ }$ ($-\pi /6$ to $\pi /6$). $60/360$ is $1/6$ for one pass, so $1/3$ for a full cycle. If $1/3$ is an option, it is the correct answer. (Note: The solution text previously mentioned $2/3$, likely a mistake).