For \(x = A/2,\;\;\sin \omega {T_1} = 1/2 \Rightarrow {T_1} = \frac{\pi }{{6\omega }}\)

For \(x = A,\;\sin \omega ({T_1} + {T_2}) = 1 \Rightarrow {T_1} + {T_2} = \frac{\pi }{{2\omega }}\)

\( \Rightarrow {T_2} = \frac{\pi }{{2\omega }} - {T_1} = \frac{\pi }{{2\omega }} - \frac{\pi }{{6\omega }} = \frac{\pi }{{3\omega }}i.e.\;{T_1} < {T_2}\)

Alternate method : In \(S.H.M\)., velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from \(0\) to \(\frac{A}{2}\) will be less than the time taken to go from \(\frac{A}{2}\) to \(A\).

Hence \({T_1} < {T_2}.\)