Question

Average Kinetic energy in one time period in an SHM is (Where $m,\omega$ and $y_0$ are mass of particle, angular velocity and maximum displacement respectively)

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

Answer 1

Answer: (D)

$\frac{m{\omega }^2{y}_0^2}{4}$

$K.E.=\frac{1}{2}m{\omega }^2{y}_o^2{\cos }^2\omega t$
$=\frac{{\int }_0^{T}K.E.dt}{{\int }_0^{T}dt}=\frac{{\int }_0^T\frac{1}{2}m{\omega }^2{y}_o^2{\cos }^2\omega tdt}{T}$
$K.E.=\frac{m{\omega }^2{y}_o^2}{2T}{\int }_0^T{\cos }^2\omega tdt=\frac{m{\omega }^2{y}_o^2}{4}$; $\because {\int }_0^T{\cos }^2\omega tdt=\frac{T}{2}$