Question

A particle is executing SHM according to the equation, $x=A\cos \omega t$. Average speed of the particle during the interval $0\le t\le \frac{\pi }{6\omega }$ is

• A. $\frac{\sqrt{3}A\omega }{2}$
• B. $\frac{\sqrt{3}A\omega }{4}$
• C. $\frac{3A\omega }{\pi }$
• D. $\frac{3A\omega }{\pi }(2-\sqrt{3})$

Answer 1

The correct option is D $\frac{3A\omega }{\pi }(2-\sqrt{3})$

Given:
$x=A\cos \omega t$
$\therefore v=\frac{dx}{dt}=-A\omega \sin \omega t$
We know that average of any function $y=f\left(x\right)$ is given by
$\overline{y}=\frac{\underset{x_1}{\overset{x_2}{\int }}f\left(x\right)dx}{\underset{x_1}{\overset{x_2}{\int }}dx}$
So, average velocity, $\overline{v}=\frac{\underset{0}{\overset{\frac{\pi }{6\omega }}{\int }}-A\omega \sin \omega tdt}{\underset{0}{\overset{\frac{\pi }{6\omega }}{\int }}dt}$
$\overline{v}=\frac{-A\omega {\left[\frac{-\cos \omega t}{\omega }\right]}_0^{\frac{\pi }{6\omega }}}{{\left[t\right]}_0^{\frac{\pi }{6\omega }}}$

$\overline{v}=\frac{3A\omega (\sqrt{3}-2)}{\pi }$
Now, average speed $=\left|\overline{v}\right|=\frac{3A\omega }{\pi }(2-\sqrt{3})$