Question

A particle of mass m executing SHM with amplitude A and angular frequency $\omega$. The average value of the kinetic energy and potential energy over a period is

• A. $0,\frac{1}{2}m{\omega }^2{A}^2$
• B. $\frac{1}{2}m{\omega }^2{A}^2,0$
• C. $\frac{1}{2}m{\omega }^2{A}^2,\frac{1}{2}m{\omega }^2{A}^2$
• D. $\frac{1}{4}m{\omega }^2{A}^2,\frac{1}{4}m{\omega }^2{A}^2$

Answer 1

Answer: (D)

$\frac{1}{4}m{\omega }^2{A}^2,\frac{1}{4}m{\omega }^2{A}^2$

$x=A\sin (\omega t+\varphi )$

$v=\frac{dx}{dt}=A\omega \cos (\omega t+\varphi )$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )$

$<KE>=\frac{1}{2}m{A}^2{\omega }^2<{\cos }^2(\omega t+\varphi )>$

$=\frac{1}{4}m{A}^2{\omega }^2$ [Avg. value of ${\sin }^2\theta$ and ${\cos }^2\theta$ in one time-period$=\frac{1}{2}]$

$PE=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2(\omega t+\varphi )$

$=\frac{1}{4}m{\omega }^2{A}^2$