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(B) The time taken by the particle to move from the mean position $(x=0)$ to the extreme position $(x=4 \, cm)$ is $T/4 = 1.2/4 = 0.3 \, s$.Let $t_1$

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$0.4$

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(B) The time taken by the particle to move from the mean position $(x=0)$ to the extreme position $(x=4 \, cm)$ is $T/4 = 1.2/4 = 0.3 \, s$.Let $t_1$ be the time taken to move from $x=0$ to $x=2 \, cm$. The equation of motion is $x = A \sin(\omega t)$.Substituting the values: $2 = 4 \sin(\frac{2\pi}{T} t_1) \Rightarrow 1/2 = \sin(\frac{2\pi}{1.2} t_1)$.This gives $\frac{\pi}{6} = \frac{2\pi}{1.2} t_1$,which simplifies to $t_1 = 0.1 \, s$.The time taken to move from $x=2 \, cm$ to $x=4 \, cm$ is $t_2 = (T/4) - t_1 = 0.3 - 0.1 = 0.2 \, s$.The total time to move from $x=2 \, cm$ to $x=4 \, cm$ and back again is $2 	imes t_2 = 2 	imes 0.2 = 0.4 \, s$.

$A$ particle is performing simple harmonic motion along the $x$-axis with an amplitude of $4 \, cm$ and a time period of $1.2 \, s$. The minimum time taken by the particle to move from $x = 2 \, cm$ to $x = +4 \, cm$ and back again is given by .... $s$.

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