A particle executes S.H.M. between x = -A and x = +A . The time...

The equation of simple harmonic motion starting from the mean position is given by $x = A \sin(\omega t)$.

For the particle to travel from the mean position $x = 0$ to $x = \dfrac{A}{2}$, the time taken is $T_1$.

$\dfrac{A}{2} = A \sin(\omega T_1)$

$\sin(\omega T_1) = \dfrac{1}{2} \Rightarrow \omega T_1 = \dfrac{\pi}{6}$

$T_1 = \dfrac{\pi}{6\omega}$

For the particle to travel from the mean position to $x = A$, the total time taken is $T_1 + T_2$.

$A = A \sin(\omega (T_1 + T_2))$

$\sin(\omega (T_1 + T_2)) = 1 \Rightarrow \omega (T_1 + T_2) = \dfrac{\pi}{2}$

$T_1 + T_2 = \dfrac{\pi}{2\omega}$

Substituting the value of $T_1$:

$T_2 = \dfrac{\pi}{2\omega} - \dfrac{\pi}{6\omega} = \dfrac{3\pi - \pi}{6\omega} = \dfrac{2\pi}{6\omega} = \dfrac{\pi}{3\omega}$

Comparing $T_1$ and $T_2$, we get $T_2 = 2T_1$.

Therefore, $T_1 < T_2$.

Answer: $T_1 < T_2$