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 \( = 2 \times 0.2 = 0.4\sec \)