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A point performs simple harmonic oscillation of period $\mathrm{T}$ and the equation of motion is given by $x=a \sin (w t+\pi / 6)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
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Verified Answer
The correct answer is:
$\frac{\mathrm{T}}{12}$
$$
\begin{aligned}
& v=\omega \cos \left(\omega t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\omega a}{2}=\omega \cos \left(\frac{2 \pi}{T} t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\pi}{3}=\frac{2 \pi}{T} t+\frac{\pi}{6} \\
& \Rightarrow t=\frac{T}{12}
\end{aligned}
$$
\begin{aligned}
& v=\omega \cos \left(\omega t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\omega a}{2}=\omega \cos \left(\frac{2 \pi}{T} t+\frac{\pi}{6}\right) \\
& \Rightarrow \quad \frac{\pi}{3}=\frac{2 \pi}{T} t+\frac{\pi}{6} \\
& \Rightarrow t=\frac{T}{12}
\end{aligned}
$$
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