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(B) The displacement equation for a particle executing simple harmonic motion starting from the mean position is given by $y = a \sin(\omega t)$, wher

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$1/2 \,s$

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(B) The displacement equation for a particle executing simple harmonic motion starting from the mean position is given by $y = a \sin(\omega t)$, where $\omega = \frac{2\pi}{T}$.Given that the particle moves from the mean position $(y = 0)$ to half the amplitude $(y = a/2)$, we substitute these values into the equation:$\frac{a}{2} = a \sin\left(\frac{2\pi}{T} t\right)$$\frac{1}{2} = \sin\left(\frac{2\pi}{T} t\right)$Since $\sin(\pi/6) = 1/2$, we have:$\frac{2\pi}{T} t = \frac{\pi}{6}$Solving for $t$:$t = \frac{T}{12}$Given the period $T = 6 \,s$:$t = \frac{6}{12} = 0.5 \,s = \frac{1}{2} \,s$.

$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

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