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

Answer: (D)

$\frac{3A\omega }{\pi }(2-\sqrt{3})$

$t=\frac{\pi }{6\omega }$
$\therefore$ $\omega t=\frac{\pi }{6}$
According to the equation,
$X=A\cos \omega t$
$X=A$ at $t=0$ and $X=\frac{\sqrt{3}A}{2}$ at $\omega t=\frac{\pi }{6}$
$\therefore$ Distance traveled
$=A-\frac{\sqrt{3}}{2}A$
$=(2-\sqrt{3})\frac{A}{2}$
Average speed$=\frac{distance}{time}$
$=\frac{(2-\sqrt{3})A/2}{\left(\pi /6\omega \right)}$
$=\frac{3\omega A}{\pi }(2-\sqrt{3})$