Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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?
PhysicsOscillationsNEETNEET 2008 (Screening)
Options:
  • A $\frac{T}{8}$
  • B $\frac{T}{6}$
  • C $\frac{T}{3}$
  • D $\frac{T}{12}$
Solution:
2252 Upvotes Verified Answer
The correct answer is: $\frac{T}{12}$
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.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.