A particle executing SHM of amplitude 4 , cm | vedclass

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(B) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$. Assuming the particle starts from the mean position,$x = A \sin(\om

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$1/3 \, s$

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(B) The displacement equation for $SHM$ is given by $x = A \sin(\omega t + \phi)$. Assuming the particle starts from the mean position,$x = A \sin(\omega t)$.Given amplitude $A = 4 \, cm$ and time period $T = 4 \, s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \, rad/s$.For position $x_1 = 2 \, cm$:$2 = 4 \sin(\omega t_1) \Rightarrow \sin(\omega t_1) = 1/2 \Rightarrow \omega t_1 = \pi/6$.For position $x_2 = 2\sqrt{3} \, cm$:$2\sqrt{3} = 4 \sin(\omega t_2) \Rightarrow \sin(\omega t_2) = \sqrt{3}/2 \Rightarrow \omega t_2 = \pi/3$.The time taken $\Delta t = t_2 - t_1$ is given by:$\omega(t_2 - t_1) = \pi/3 - \pi/6 = \pi/6$.Substituting $\omega = \pi/2$:$(\pi/2) \Delta t = \pi/6 \Rightarrow \Delta t = \frac{\pi}{6} 	imes \frac{2}{\pi} = 1/3 \, s$.

$A$ particle executing $SHM$ of amplitude $4 \, cm$ and $T = 4 \, s$. The time taken by it to move from $+2 \, cm$ to $+2\sqrt{3} \, cm$ is

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