Question

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

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

Answer 1

Answer: (A)

${\pi }^{2}m{a}^2{v}^2$

The average kinetic energy during its motion from the position of equilibrium to the end will be given by:
$=\frac{{\int }_0^{T/4}\frac{1}{2}m{V}^2}{T/4}$
Substitute $V=a\omega cos\omega t$
$K.E_{avg}=\frac{{\int }_0^{T/4}\frac{1}{2}m(a\omega cos\omega t{)}^2}{T/4}$
After solving,
$K.E_{avg}=\frac{1}{4}m{a}^2{\omega }^2$
Now, substitute $\omega =2\pi v$
$K.E_{avg}=\frac{1}{4}m{a}^2(2\pi v{)}^2$
$K.E_{avg}={\pi }^{2}m{a}^2{v}^2$