Question

A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \text{sin} \left(\omega t + \frac{\pi }{6}\right)$ . After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

• A. $\frac{T}{3}$
• B. $\frac{T}{12}$
• C. $\frac{T}{8}$
• D. $\frac{T}{6}$

Answer 1

Answer: (B)

$x = a \text{sin} \left(\omega t + \frac{\pi }{6}\right)$
$\frac{ d x}{ d ⁡ t} = a \omega \text{cos} \left(\omega t + \frac{\pi }{6}\right)$
Maximum velocity $= \textit{a} \omega $
$\therefore \frac{a \omega }{2}=a\omega \text{cos}\left(\omega t + \frac{\pi }{6}\right)$
$\therefore \text{cos}\left(\omega t + \frac{\pi }{6}\right)=\frac{1}{2}$
$\omega t+\frac{\pi }{6}=60^{ο}=\frac{2 \pi }{6}\text{rad}$
$\omega t = \frac{2 \pi }{6} - \frac{\pi }{6} = \frac{\pi }{6}$
$\therefore t=\frac{\pi }{6 \omega }=\frac{\pi }{6}\times \frac{T }{2 \pi }=\frac{T ⁡}{1 2}\left(\because \omega = \frac{2 \pi }{T ⁡}\right)$