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)

Let the displacement of the particle executing $SHM$ at any instant of time $t$ from its equilibrium position is given by
$x=Acos\left(\omega t+\varphi \right)$
Velocity, $v=\frac{dx}{dt}=-\omega Asin\left(\omega t+\varphi \right)$
Kinetic energy of the particle is
$K=\frac{1}{2}m{v}^2$
$=\frac{1}{2}m{\omega }^2{A}^{2}si{n}^2\left(\omega t+\varphi \right)$
Potential energy of the particle is
$U=\frac{1}{2}m{\omega }^2{x}^2$
$=\frac{1}{2}m{\omega }^2{A}^{2}co{s}^2\left(\omega t+\varphi \right)$
Average value of kinetic energy over a period is
$\left(K\right)=\frac{1}{T}\underset{0}{\overset{T}{\int }}Kdt$
$=\frac{1}{T}\underset{0}{\overset{T}{\int }}\frac{1}{2}m{\omega }^2{A}^{2}si{n}^2\left(\omega t+\varphi \right)dt$
$=\frac{1}{2T}m{\omega }^2{A}^2\underset{0}{\overset{T}{\int }}\left[\frac{1-cos2\left(\omega t+\varphi \right)}{2}\right]dt$
Since the average value of both a sine and a cosine function for a complete cycle or over a time period $T$ is $0$.
$\therefore \left(K\right)=\frac{1}{4T}m{\omega }^2{A}^2{\left[t\right]}_0^T$
$=\frac{1}{4T}m{\omega }^2{A}^{2}T$
$=\frac{1}{4}m{\omega }^2{A}^2$
Average value of potential energy over a period is
$\left(U\right)=\frac{1}{T}\underset{0}{\overset{T}{\int }}\frac{1}{2}m{\omega }^2{A}^{2}co{s}^2\left(\omega t+\varphi \right)dt$
$=\frac{1}{2T}m{\omega }^2{A}^2\underset{0}{\overset{T}{\int }}\left[\frac{1+cos2\left(\omega t+\varphi \right)}{2}\right]dt$
Since the average value of both a sine and a cosine function for a complete cycle or over a time period $T$ is zero.
$\therefore \left(U\right)=\frac{1}{4T}m{\omega }^2{A}^2{\left[t\right]}_0^T$
$=\frac{1}{4T}m{\omega }^2{A}^{2}T$
$=\frac{1}{4}m{\omega }^2{A}^2$