Explanation

(a) In order to find the time taken by the particle from -12.5 cm to +12.5 cm on either side of mean position, we will find the time taken by particle to go from $x = - 12.5\ cm\ to\ x = 0\ and\ to\ go\ from\ x = 0\ to\ x = + 12.5\ cm.$
Let the equation of motion be $x = A\sin{\omega t}$
First, the particle moves from
$x = - 12.5\ cm\ to\ x = 0$
$\therefore\ \ 12.5 = 25\sin{\omega t}$ $\because A = 25\ cm$
$\Rightarrow \ \ \ \ \ \frac{1}{2} = \sin{\omega t}$
$\Rightarrow \ \ \ \ \ \omega t = \frac{\pi}{6}$
$\therefore\ \ \ \ \ \ \ t = \frac{\pi}{6\omega}$
Similarly to go from $x = 0\ to\ x = 12.5\ cm$
$\omega t = \frac{\pi}{6}\ \Rightarrow t = \frac{\pi}{6\omega}$
$\therefore$ Total time taken from $x = - 12.5\ cm\ to\ x = 12.5\ cm$
$t^{'} = \frac{\pi}{6\omega} + \frac{\pi}{6\omega} = \frac{\pi}{3\omega}$
$= \ \frac{\pi}{3(\frac{2\pi}{T})} = \frac{T}{6} = \frac{3}{6} = 0.5$ s