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)

T/12

Answer 2

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