A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=Asin$$\left( {\omega t + \frac{\pi }{6}} \right)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
AIPMT 2008, Medium
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$\quad x=a \sin (\omega t+\pi / 6)$
$\frac{d x}{d t}=a \omega \cos (\omega t+\pi / 6)$
Max. velocity $=a \omega$
$\therefore \quad \frac{a \omega}{2}=a \omega \cos (\omega t+\pi / 6)$
$\therefore \quad \cos (\omega t+\pi / 6)=\frac{1}{2}$
$\Rightarrow 60^{\circ}$ or $\frac{2 \pi}{6}$ radian $=\frac{2 \pi}{T} \cdot t+\pi / 6$
$\Rightarrow \frac{2 \pi}{T} \cdot t=\frac{2 \pi}{6}-\frac{\pi}{6}=+\frac{\pi}{6}$
$\therefore \quad t=+\frac{\pi}{6} \times \frac{T}{2 \pi}=\left|+\frac{T}{12}\right|$
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