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

$\text{ Average velocity }\begin{array}{c}\overline{v}=\frac{{\int }_0^t \frac{dx}{dt}\cdot dt}{t}=\frac{{\int }_0^t dx}{t}=\frac{x\left(t\right)-x\left(0\right)}{t}\\ =\frac{A(\cos \pi /6-1)}{\pi /6\omega }=\frac{3A\omega }{\pi }(\sqrt{3}-2)\end{array}$

Since particle does not change its direction in the given interval, average speed becomes 

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