Question

A particle of mass $m$ executes simple harmonic motion with amplitude a and frequency $n$. The average kinetic energy during its motion from the position of equilibrium to the end is.

• A. $2\pi^2 ma^2 v^2$
• B. $\pi ^2 ma^2 v^2 $
• C. $\frac{1}{4} ma^2 v^2$
• D. $4\pi ^2 ma^2 v^2 $

Answer 1

Answer: (B)

The average kinetic energy during its motion from the position of equilibrium to the end will be given by:
$=\frac{\int\limits_{0}^{T / 4} \frac{1}{2} m V^{2}}{T / 4}$
Substitute $V = a\omega \cos \omega t$
$K . E _{\text {avg }}=\frac{\int\limits_{0}^{ T / 4} \frac{1}{2} m ( a \omega \cos \omega t )^{2}}{ T / 4}$
After solving,
$K \cdot E _{ avg }=\frac{1}{4} ma ^{2} \omega^{2}$
Now, substitute $\omega=2 \pi v$
$ K \cdot E _{ avg }=\frac{1}{4} ma ^{2}(2 \pi v )^{2} $
$ K \cdot E _{ avg }=\pi^{2} ma ^{2} v ^{2}$