Question

Show that for a particle in linear $SHM$ the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer 1

The equation of displacement of a particle executing $SHM$ at an instant $t$ is given by

$x=A\sin \omega t$
Where
$A$=Amplitude of oscillation
$\omega$=Angular frequency
Now,

$\omega =\sqrt{\frac{k}{m}}\Rightarrow k=m{\omega }^2$
The velocity of the particle is,

$v=\frac{dx}{dt}=A\omega \cos \omega t$
Instantaneous kinetic energy of the particle is,
$K.E=\frac{1}{2}m{v}^2$
$=\frac{1}{2}m(A\omega \cos \omega t{)}^2$
$=\frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega t$
For time period $T$, the average kinetic energy over a single cycle is given by,

$E_k=\frac{1}{T}{\int }_0^T\frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega tdt$
$=\frac{1}{2T}m{A}^2{\omega }^2{\int }_0^T\frac{1+\cos 2\omega t}{2}dt$

$=\frac{1}{4T}m{A}^2{\omega }^2{[t+\frac{\sin 2\omega t}{2\omega }]}_0^T$

$=\frac{1}{4T}m{A}^2{\omega }^2[T-0+\frac{1}{2\omega }(\sin 2\omega T-\sin 0\left)\right]$

$=\frac{1}{4T}m{A}^2{\omega }^2\left[T\right]$ $[Using\omega =\frac{2\pi }{T}]$

$E_k=\frac{1}{4}m{A}^2{\omega }^2$ ...$\left(i\right)$


Potential energy of the particle is
$U=\frac{1}{2}k{x}^2=\frac{1}{2}m{A}^2{\omega }^2{\sin }^2\omega t$ $[\because k=m{\omega }^2\text{and}x=A\sin \omega t]$

and, average potential energy cycle is given as:

$E_p=\frac{1}{T}{\int }_0^T\frac{1}{2}m{\omega }^{2}A{\sin }^2\omega tdt$

$=\frac{1}{2T}m{A}^2{\omega }^2{\int }_0^T\left(\frac{1-\cos 2\omega t}{2}\right)dt$

$=\frac{1}{4T}m{A}^2{\omega }^2{[t-\frac{\sin 2\omega t}{2\omega }]}_0^T$

$=\frac{1}{4T}m{A}^2{\omega }^2\times [T-0+\frac{1}{2\omega }(\sin 2\omega T-\sin 0\left)\right]$

$\Rightarrow E_p=\frac{1}{4T}m{A}^2{\omega }^2\left[T\right]=\frac{1}{4}m{A}^2{\omega }^2$ ...$\left(ii\right)$

$E_k=\frac{1}{4}m{A}^2{\omega }^2$ ...$\left(i\right)$ $\Rightarrow E_p=\frac{1}{4}m{A}^2{\omega }^2$ ...$\left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, we can see that the average kinetic energy for a time period is equal to the average potential energy for the same time period.