The maximum speed of a particle in S.H.M. is V. The average speed is
The maximum speed of a particle in S.H.M. is V. The average speed is
$\frac{\mathrm{V}}{\pi}$
$\frac{3 V}{\pi}$
$\frac{4 \mathrm{~V}}{\pi}$
$\frac{2 \mathrm{~V}}{\pi}$
Solution
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
$\langle|v|\rangle=\frac{4 \mathrm{~A}}{\mathrm{~T}}=\frac{4 \mathrm{~A}}{\left\{\frac{2 \pi}{\left(\frac{\mathrm{v}}{\mathrm{A}}\right)}\right\}}=\frac{2 \mathrm{~V}}{\pi}$
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