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(D) The equation for displacement in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$,where $\omega = \frac{2

Vedclass · Accepted Answer

$\frac{1}{3} \,s$

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(D) The equation for displacement in simple harmonic motion starting from the mean position is given by $y = A \sin(\omega t)$,where $\omega = \frac{2\pi}{T}$.Given $T = 4 \,s$,we have $\omega = \frac{2\pi}{4} = \frac{\pi}{2} \,rad/s$.We need to find the time $t$ when the displacement $y = \frac{A}{2}$.Substituting the values into the equation:$\frac{A}{2} = A \sin\left(\frac{\pi}{2} t\right)$$\frac{1}{2} = \sin\left(\frac{\pi}{2} t\right)$Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we equate the angles:$\frac{\pi}{2} t = \frac{\pi}{6}$$t = \frac{2}{6} = \frac{1}{3} \,s$.Therefore,the time taken is $\frac{1}{3} \,s$.

The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

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