Question

A particle executes SHM in a straight line. The maximum speed of the particle during its motion is $v_0$. Its average speed in one complete oscillation is

• A. $\frac{\sqrt{2}{V}_0}{\pi }$
• B. $\frac{V_0}{2\pi }$
• C. $\frac{2V_0}{\pi }$
• D. $\frac{V_0}{\sqrt{2\pi }}$

Answer 1

The correct option is C $\frac{2V_0}{\pi }$

Average speed is $V_{av}=\frac{Totaldistance}{Totaltime}=\frac{4A}{T}\cdots \left(i\right)$
Maximum speed is $V_0$.
$\therefore V_0=A\omega =\frac{A.2\pi }{T}$
So, $\frac{A}{T}=\frac{V_0}{2\pi }$
Using this in Eq. (i), we get
$V_{av}=\frac{2v_0}{\pi }$