If the maximum velocity in S.H.M. is 'V' then the average velocity...

Maximum velocity in S.H.M. is given by $V = A\omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.

The displacement of the particle from one extreme position to the other extreme position is $2A$.

The time taken to travel from one extreme to the other is half of the time period, which is $t = \dfrac{T}{2} = \dfrac{\pi}{\omega}$.

Average velocity is the ratio of total displacement to the total time taken.

$v_{avg} = \dfrac{2A}{\dfrac{\pi}{\omega}} = \dfrac{2A\omega}{\pi}$

Substituting $V = A\omega$, we get:

$v_{avg} = \dfrac{2V}{\pi}$

Answer: $\dfrac{2V}{\pi}$