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 simple harmonic motion of a spring can be used to give the required result. The formula for position can be substituted in the potential energy and by differentiating the position, we get the velocity, which we substitute in the kinetic energy formula.
Formula Used:
The displacement x of a particle performing the simple harmonic motion is given as:
$x = a \cos \omega t$
Time average of any quantity P is expressed as:
$< P >$ = $\dfrac{ \int_0^T Pdt }{\int_0^T dt}$

Complete answer:
For understanding this better, we consider an example of simple harmonic motion, we start with considering the case of horizontal oscillations of a block of mass m connected to a spring of force constant k. The displacement is given as:
$x = a \cos \omega t$
The velocity is obtained by differentiating this with respect to time, we get:
$v = -a \omega \sin \omega t$.
The formula for kinetic energy is written as:
$\dfrac{1}{2} m v^2 = \dfrac{1}{2} m (a \omega \sin \omega t )^2$.
The time average can be calculated as:
$K.E. = \dfrac{\dfrac{1}{2} m a^2 \omega^2 \int_0^T \sin ^2 \omega t dt}{\int_0^T dt} = \dfrac{1}{4} m a^2 \omega^2$.
The potential energy for the case of spring is written as:
$P.E. = \dfrac{1}{2} k x^2$
The time average potential energy is written as:
$P.E. = \dfrac{\dfrac{1}{2} k a^2 \int_0^T \cos^2 \omega t dt }{\int_0^T dt} = \dfrac{1}{4} m \omega^2 a^2$
Here, we have used the relation for frequency:
$\omega^2 = \dfrac{k}{m}$
Therefore, we are in position to conclude that the time averaged kinetic and potential energies are equal.

Note:
The value of the time average of square of sine or cosine is always 1/2. This integration is performed by first taking replacing the square of sines and cosines by using the cos2A formula. We know that the integral of cosine or sine is zero on one complete cycle. This is not the case for the squared terms. The squared sine/cosine terms always give an average of 1/2 over one complete cycle.