b
(b) Time taken by particle to move from $x=0$ (mean position) to $x = 4$ (extreme position) $ = \frac{T}{4} = \frac{{1.2}}{4} = 0.3\;s$
Let $t$ be the time taken by the particle to move from $x=0$ to $x=2 \,cm$
$y = a\sin \omega t$
$\Rightarrow 2 = 4\sin \frac{{2\pi }}{T}t$
$ \Rightarrow \frac{1}{2} = \sin \frac{{2\pi }}{{1.2}}t$
$ \Rightarrow \frac{\pi }{6} = \frac{{2\pi }}{{1.2}}t $
$\Rightarrow t = 0.1\;s$.
Hence time to move from $x = 2$ to $x = 4$ will be equal to $0.3 -0.1 = 0.2 s$
Hence total time to move from $x = 2$ to $x = 4$ and back again $ = 2 \times 0.2 = 0.4\sec $