Q107 / 200 · NEET 2008 (Screening) · Physics

A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=a \sin (\omega t+ \frac{\pi}{6})$. 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
    $\frac{T}{8}$
  • B
    $\frac{T}{6}$
  • C
    $\frac{T}{3}$
  • Correct answer
    D
    $\frac{T}{12}$

Solution

Key Idea : Velocity is the time derivative of displacement. Writing the given equation of a point performing SHM $x=a \sin \left(\omega t+\frac{\pi}{6}\right)$ Differentiating Eq. (i), w.r.t. time, we obtain $y=\frac{d x}{d t}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)$ It is given that $y=\frac{a \omega}{2}$, so that or $\frac{a \omega}{2}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)$ $\frac{1}{2}=\cos \left(\omega t+\frac{\pi}{6}\right)$ or $\cos \frac{\pi}{3}=\cos \left(\omega t+\frac{\pi}{6}\right)$ $\begin{array}{ll} \text {or } & \omega t+\frac{\pi}{6}=\frac{\pi}{3} \\ \Rightarrow & \omega t=\frac{\pi}{6} \\ \text {or } & t=\frac{\pi}{6 \omega}=\frac{\pi \times T}{6 \times 2 \pi}=\frac{T}{12} \end{array}$ Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its maximum velocity.

Hence, the correct answer is option: 'D'.

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