Q: A particle is executing SHM according to the equation x = Acos ωt . Average speed of the particle during the interval 0 ≤ t ≤ π/6ω

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

(B) $\frac{\sqrt{3}\omega A}{4}$

(C) $\frac{3 \omega A}{\pi}$

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

**Click to See Solution : **

$v_{avg} = \frac{\int_{0}^{t}dx}{t}$

$= \frac{x(t)-x(0)}{t} $

$ = \frac{A(cos\pi/6 – 1)}{\pi/6 \omega} $

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

since particle does not change it’s direction in the given interval , average speed

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