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)

$x=a\text{sin}\left(\omega t+\frac{\pi }{6}\right)$
$\frac{dx}{dt}=a\omega \text{cos}\left(\omega t+\frac{\pi }{6}\right)$
Maximum velocity $=\text{a}\omega$
$\therefore \frac{a\omega }{2}=a\omega \text{cos}\left(\omega t+\frac{\pi }{6}\right)$
$\therefore \text{cos}\left(\omega t+\frac{\pi }{6}\right)=\frac{1}{2}$
$\omega t+\frac{\pi }{6}=60^o=\frac{2\pi }{6}\text{rad}$
$\omega t=\frac{2\pi }{6}-\frac{\pi }{6}=\frac{\pi }{6}$
$\therefore t=\frac{\pi }{6\omega }=\frac{\pi }{6}\times \frac{T}{2\pi }=\frac{T}{12}\left(\because \omega =\frac{2\pi }{T}\right)$