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(D) For a particle executing simple harmonic motion starting from the mean position $(x=0)$,the displacement is given by $x(t) = A \sin(\omega t)$.For

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$T_1 = 2T_2$

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(D) For a particle executing simple harmonic motion starting from the mean position $(x=0)$,the displacement is given by $x(t) = A \sin(\omega t)$.For $x = 0$,$\omega t_0 = 0 \implies t_0 = 0$.For $x = \frac{A}{2}$,$\frac{A}{2} = A \sin(\omega t_1) \implies \sin(\omega t_1) = \frac{1}{2} \implies \omega t_1 = \frac{\pi}{6} \implies t_1 = \frac{\pi}{6\omega}$.For $x = \frac{A}{\sqrt{2}}$,$\frac{A}{\sqrt{2}} = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{1}{\sqrt{2}} \implies \omega t_2 = \frac{\pi}{4} \implies t_2 = \frac{\pi}{4\omega}$.The time taken to move from $x=0$ to $x=\frac{A}{2}$ is $T_1 = t_1 - t_0 = \frac{\pi}{6\omega}$.The time taken to move from $x=\frac{A}{2}$ to $x=\frac{A}{\sqrt{2}}$ is $T_2 = t_2 - t_1 = \frac{\pi}{4\omega} - \frac{\pi}{6\omega} = \frac{3\pi - 2\pi}{12\omega} = \frac{\pi}{12\omega}$.Comparing $T_1$ and $T_2$,we see that $T_1 = \frac{\pi}{6\omega} = \frac{2\pi}{12\omega} = 2 	imes \frac{\pi}{12\omega} = 2T_2$.Thus,$T_1 > T_2$,which implies $T_1$ is greater than $T_2$. Looking at the options,the relationship $T_1 > T_2$ is not explicitly listed as an option,but checking the provided options,$T_1 = 2T_2$ is the correct mathematical result.

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

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