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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
$\dfrac{T}{3}$
B
$\dfrac{T}{12}$
C
$\dfrac{T}{8}$
D
$\dfrac{T}{6}$
Detailed Solution
$x = a\sin\left(\omega t + \dfrac{\pi}{6}\right)$
Velocity: $v = \dfrac{dx}{dt} = a\omega\cos\left(\omega t + \dfrac{\pi}{6}\right)$
Maximum velocity: $v_{max} = a\omega$
Given $v = \dfrac{v_{max}}{2}$: $\dfrac{a\omega}{2} = a\omega\cos\left(\omega t + \dfrac{\pi}{6}\right)$
$\cos\left(\omega t + \dfrac{\pi}{6}\right) = \dfrac{1}{2}$
$\omega t + \dfrac{\pi}{6} = \dfrac{\pi}{3} \Rightarrow \omega t = \dfrac{\pi}{6}$
$t = \dfrac{\pi}{6\omega} = \dfrac{\pi}{6} \times \dfrac{T}{2\pi}$
$t = \dfrac{T}{12}$
Velocity: $v = \dfrac{dx}{dt} = a\omega\cos\left(\omega t + \dfrac{\pi}{6}\right)$
Maximum velocity: $v_{max} = a\omega$
Given $v = \dfrac{v_{max}}{2}$: $\dfrac{a\omega}{2} = a\omega\cos\left(\omega t + \dfrac{\pi}{6}\right)$
$\cos\left(\omega t + \dfrac{\pi}{6}\right) = \dfrac{1}{2}$
$\omega t + \dfrac{\pi}{6} = \dfrac{\pi}{3} \Rightarrow \omega t = \dfrac{\pi}{6}$
$t = \dfrac{\pi}{6\omega} = \dfrac{\pi}{6} \times \dfrac{T}{2\pi}$
$t = \dfrac{T}{12}$