Q. Maximum speed of a particle in SHM is v$_{max.}$ Then average speed of a particle in SHM is equal to
A) $\frac{v_{\max}}{2}$
B) $\frac{\pi v_{\max}}{2}$
C) $\frac{v_{\max}}{2\pi}$
D) $\frac{2v_{\max}}{\pi}$
Explanation
(d) Let $r$ be the amplitude of oscillation and $T$ be the time period in SHM. Then total distance travelled in time $T = 4r$
$\therefore$ Average velocity, $v_{av} = \frac{4t}{T} = \frac{4r}{2\pi/\omega}$
$= \frac{2r\omega}{\pi} = \frac{2v_{\max}}{\pi}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.