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(A) The equation for displacement starting from the mean position is $x(t) = A \sin(\omega t)$.Given $T = 6 \pi 	ext{ s}$,the angular frequency is $\

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$2$

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(A) The equation for displacement starting from the mean position is $x(t) = A \sin(\omega t)$.Given $T = 6 \pi 	ext{ s}$,the angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{6 \pi} = \frac{1}{3} 	ext{ rad/s}$.At $x = A$,the particle is at the extreme position. The time taken to reach the mean position $(x=0)$ from $x=A$ is $T/4 = (6 \pi)/4 = 1.5 \pi 	ext{ s}$.However,the question asks for the time to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$.Using the phasor diagram,the position $x = A \sin(	heta)$ corresponds to the projection on the vertical axis.At $x = A$,the phase angle $	heta_1 = \frac{\pi}{2}$.At $x = \frac{\sqrt{3}}{2} A$,$\sin(	heta_2) = \frac{\sqrt{3}}{2}$,so $	heta_2 = \frac{\pi}{3}$.The change in phase is $\Delta 	heta = 	heta_1 - 	heta_2 = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.Since $\Delta 	heta = \omega \Delta t$,we have $\frac{\pi}{6} = \frac{1}{3} \Delta t$.Therefore,$\Delta t = \frac{3 \pi}{6} = \frac{\pi}{2} 	ext{ s}$.Comparing this with $\frac{\pi}{x} 	ext{ s}$,we get $x = 2$.

$A$ simple harmonic oscillator has an amplitude $A$ and time period $6 \pi \text{ s}$. Assuming the oscillation starts from its mean position,the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x} \text{ s}$,where $x=$ . . . . . . .

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