The maximum speed of a particle in S.H.M. is V. The average speed is

The correct option is (D).
Concept: For SHM, \(\mathrm{a}=-\left(\omega^2\right) \mathrm{x}\) is the necessary condition.
A general solution for SHM can be written as, \(\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi)\)
where, \(\mathrm{x}\) is the displacement from mean position, \(\mathrm{A}\) is the amplitude, \(\omega\) is the angular frequency and \(\phi\) is the phase angle. On taking the first derivative, velocity can be written as
\(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=(\mathrm{A} \omega) \cos (\omega \mathrm{t}+\phi)\)
So, the maximum speed is \((\mathrm{A} \omega)=\mathrm{V}\). For average speed, over one-time interval \(\mathrm{T}\), the body undergoing SHM moves by distance 4A. Thus, the average speed is given by
\(\langle|v|\rangle=\frac{4 \mathrm{~A}}{\mathrm{~T}}=\frac{4 \mathrm{~A}}{\left\{\frac{2 \pi}{\left(\frac{\mathrm{v}}{\mathrm{A}}\right)}\right\}}=\frac{2 \mathrm{~V}}{\pi}\)