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