A particle executing SHM with time period T and amplitude A.... | PHYSICS - VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION | SolveeIt

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

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

$\frac{\text{2A}}{T}$

$\frac{\text{3A}}{T}$

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

Answer

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

Explanation

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

$x = A\sin\omega t$

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

The mean velocity of the particle averaged over quarter oscillation is

$< v >_{0 \rightarrow T/4} = \frac{\int_{0}^{T/4}{vdt}}{\int_{0}^{T/4}{dt}} = \frac{\int_{0}^{T/4}{A\omega\cos\omega tdt}}{T/4}$

$\frac{\frac{A\omega}{\omega}\lbrack\sin\omega t\rbrack_{0}^{T/4}}{T/4} = \frac{A\lbrack\sin\omega t\rbrack_{0}^{T/0}}{T/4} = \frac{4A}{T}$

Question: A particle executing SHM with time period T and amplitude A. The mean velocity of the particle avera...