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)

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