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?

Question

Aicharya · Accepted Answer

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. /

A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=a \sin…

Solution