Question

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

• A. T/3
• B. T/12
• C. T/8
• D. T/6

Answer 1

Answer: (B)

$x=a\sin (\omega t+\pi /6)$
$\frac{dx}{dt}a\omega \cos (\omega t+\pi /6)$
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 $=\frac{2\pi }{T}\cdot t+\pi /6$
$\frac{2\pi }{T}.t=\frac{2\pi }{6}-\frac{2\pi }{6}=+\frac{2\pi }{6}$
$\therefore t=+\frac{\pi }{6}\times \frac{T}{2\pi }=\left|+\frac{T}{12}\right|$