Question

A particle performs SHM with period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation, starting from right extreme position is

• A. $0$
• B. $\frac{2A}{T}$
• C. $\frac{4A}{T}$
• D. $\frac{3A}{T}$

Answer 1

The correct option is C $\frac{4A}{T}$

Let the displacement of the particle executing SHM at any instant is $x=Asinwt$

Velocity $v=\frac{dx}{dt}=\frac{d}{dt}\left(Asinwt\right)=Awcoswt$

The mean velocity of the particle average over quarter oscillation is

${⟨v⟩}_{0\to T/4}=\frac{{\int }_0^{T/4}vdt}{{\int }_0^{T/4}dt}=\frac{{\int }_0^{T/4}Aw\cos dt}{T/4}$

$=\frac{\frac{Aw}{w}{\left[sinwt\right]}_0^{T/4}}{T/4}$

$=\frac{A{\left[sinwt\right]}_0^{T/4}}{T/4}$

$=\frac{4A}{T}$