Question

Average KE of a particle of mass $m$ performing SHM of amplitude $x_0$ and angular frequency $\omega$ in one complete oscillation with respect to position, is:

• A. $\frac{m{\omega }^2{x}_0^2}{2}$
• B. $\frac{m{\omega }^2{x}_0^2}{3}$
• C. $\frac{m{\omega }^2{x}_0^2}{4}$
• D. $\frac{m{\omega }^2{x}_0^2}{6}$

Answer 1

The correct option is B $\frac{m{\omega }^2{x}_0^2}{3}$

We know that kinetic energy of a particle during SHM is given by
$\text{KE}=\frac{1}{2}m{\omega }^2(x_0^2-x^2)$

Then, average $KE$ for one complete oscillation
${\text{KE}}_{av}=\frac{\underset{-x_0}{\overset{x_0}{\int }}KE\left(x\right)dx}{\underset{-x_0}{\overset{x_0}{\int }}dx}$
$\Rightarrow {\text{KE}}_{av}=\frac{1}{2}m{\omega }^2\frac{2\underset{0}{\overset{x_0}{\int }}(x_0^2-x^2)dx}{2\underset{0}{\overset{x_0}{\int }}dx}$
$\Rightarrow {\text{KE}}_{av}=\frac{1}{2}m{\omega }^2\frac{[x_0^3-\frac{x_0^3}{3}]}{x_0}=\frac{m{\omega }^2{x}_0^2}{3}$