A particle executes simple harmonic motion (amplitude = A) between $\mathbf{x} = -\mathbf{A}$ and $\mathbf{x} = +\mathbf{A}$ . The time taken for it to go from 0 to A/2 is $T_1$ and to go from A/2 to A is $T_2$ . Then
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Using $x = A \sin \omega t$
$For x = A/2, \sin \omega T_1 = 1/2 \Rightarrow T_1 = \frac{\pi}{6 \omega} For x = A, \sin \omega (T_1 + T_2) = 1 \Rightarrow T_1 + T_2 = \frac{\pi}{2 \omega} \Rightarrow T_2 = \frac{\pi}{2 \omega} - T_1 = \frac{\pi}{2 \omega} - \frac{\pi}{6 \omega} = \frac{\pi}{3 \omega} \text{ i.e. } T_1 < T_2$
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