Question

A point performs simple harmonic oscillation of period T and the equation of motion is given by

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. $\frac{T}{3}$
• B. $\frac{T}{12}$
• C. $\frac{T}{8}$
• D. $\frac{T}{6}$

Answer 1

The correct option is B

$\frac{T}{12}$

Step 1: Given the equation of motion:

$X=a\sin \left(\omega t+\frac{\mathrm{\pi }}{6}\right)$

Where $X$=displacement, $a$=amplitude, $\omega$=angular velocity, $t$=time,

Step 2: Differentiate with respect to time to get velocity:

$V=\frac{dX}{dt}=a\omega \cos \left(\omega t+\frac{\mathrm{\pi }}{6}\right)$…….$\left(i\right)$

Also, it is given that $V=\frac{a\omega }{2}$…………..$\left(ii\right)$

Hence equating $\left(i\right)$ and, we get

$\frac{a\omega }{2}=a\omega \cos \left(\omega t+\frac{\mathrm{\pi }}{6}\right)$

or $\frac{1}{2}=\cos \left(\omega t+\frac{\mathrm{\pi }}{6}\right)$

$$\begin{gathered} {\cos }^{-1}\left(\frac{1}{2}\right)=\omega t+\frac{\mathrm{\pi }}{6} \\ \frac{\mathrm{\pi }}{3}=\omega t+\frac{\mathrm{\pi }}{6} \\ \omega t=\frac{\mathrm{\pi }}{6} \end{gathered}$$

$$\begin{gathered} t=\frac{\mathrm{\pi }}{6\omega },Now\omega =\frac{2\mathrm{\pi }}{T} \\ Hencet=\frac{\mathrm{\pi T}}{6x2\mathrm{\pi }} \\ Givest=\frac{T}{12} \end{gathered}$$

At $t=\frac{T}{12}$, the velocity of the point will be equal to half of its maximum velocity.

Hence Option B is correct.