A particle executes $S.H.M$ between $x =\, -A$ to $x =\, +A$ . The time taken for it in going from $0$ to $A/2$ is $T_1$ and from $A/2$ to $A$ is $T_2$. Then
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$\mathrm{As}, \boldsymbol{y}=A / 2 \therefore \frac{A}{2}=A \sin \omega T_{2} \therefore \sin \omega T_{1}=\sin \left(\frac{\pi}{6}\right)$
or $T_{1}=\frac{\pi}{6 \omega}$ Also $A=A \sin \omega\left(T_{1}+T_{2}\right)$
or $\sin \omega\left(T_{1}+T_{2}\right)=1=\sin \left(\frac{\pi}{2}\right)$ or $\omega\left(T_{1}+T_{2}\right)=\frac{\pi}{2}$
$T_{1}+T_{2}=\frac{\pi}{2 \omega}$ or $T_{2}=\frac{\pi}{2 \omega}-\frac{\pi}{6 \omega}=2 T_{1}$
$\therefore T_{1} < T_{2}$
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