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