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(D) The equation for displacement in $SHM$ starting from the mean position is $y = A \sin(\omega t)$.Given $A = 4 \, cm$ and $T = 12 \, s$,the angular

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$1/2$

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(D) The equation for displacement in $SHM$ starting from the mean position is $y = A \sin(\omega t)$.Given $A = 4 \, cm$ and $T = 12 \, s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6} \, rad/s$.For the particle to reach $y = 2 \, cm$ from the mean position $(t_1)$:$2 = 4 \sin(\frac{\pi}{6} t_1) \implies \sin(\frac{\pi}{6} t_1) = \frac{1}{2} \implies \frac{\pi}{6} t_1 = \frac{\pi}{6} \implies t_1 = 1 \, s$.For the particle to reach the extreme position $(y = 4 \, cm)$ from the mean position $(t_2)$:$4 = 4 \sin(\frac{\pi}{6} t_2) \implies \sin(\frac{\pi}{6} t_2) = 1 \implies \frac{\pi}{6} t_2 = \frac{\pi}{2} \implies t_2 = 3 \, s$.The time taken to travel from $2 \, cm$ to the extreme position is $\Delta t = t_2 - t_1 = 3 - 1 = 2 \, s$.The required ratio is $\frac{t_1}{\Delta t} = \frac{1}{2}$.

$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

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