Question

A particle executing SHM with time period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation is then :

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

Answer 1

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

Let the displacement of the particle executing SHM at any instant $t$ is

$x=Asin\omega t$

Velocity, $v=\frac{dx}{dt}=\frac{d}{dt}Asin\omega t=A\omega cos\omega t$

The mean velocity of the particle averaged over quarter oscillation is

$v_{0\Rightarrow \frac{T}{4}}=\frac{{\int }_0^{\frac{T}{4}}vdt}{{\int }_0^{\frac{T}{4}}dt}=\frac{{\int }_0^{\frac{T}{4}}A\omega cos\omega dt}{\frac{t}{4}}$

$\frac{\frac{A\omega }{\omega }[sin\omega t{]}_0^{\frac{T}{4}}}{\frac{T}{4}}=\frac{4A}{T}$

Thus the mean velocity of the particle averaged over quarter oscillation is $\frac{4A}{T}$