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A particle performs simple harmonic motion with period of 3 second. The time taken by it to cover a distance equal to half the amplitude from mean position is $\left[\sin 30^{\circ}=0.5\right]$
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The correct answer is:
$\frac{1}{4} \mathrm{~S}$
(D)
T $=3 \sec$
$y=A \sin \omega t$
$\frac{A}{2}=A \sin \frac{2 \pi}{T} t$
$\sin \frac{\pi}{6}=\sin \frac{2 \pi}{3} t$
$\frac{\pi}{6}=\frac{2 \pi}{3} t \quad \therefore t=\frac{\pi}{6} \times \frac{3}{2 \pi}=\frac{1}{4} s$
T $=3 \sec$
$y=A \sin \omega t$
$\frac{A}{2}=A \sin \frac{2 \pi}{T} t$
$\sin \frac{\pi}{6}=\sin \frac{2 \pi}{3} t$
$\frac{\pi}{6}=\frac{2 \pi}{3} t \quad \therefore t=\frac{\pi}{6} \times \frac{3}{2 \pi}=\frac{1}{4} s$
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