# Average Value of Potential Energy , Kinetic Energy of Harmonic Oscillator

The average value of P.E. for complete cycle is given by

$\displaystyle U_{avg}= \frac{1}{T}\int_{0}^{T}U dt$

$\displaystyle U_{avg}= \frac{1}{T}\int_{0}^{T}\frac{1}{2}m\omega^2 A^2 sin^2(\omega t +\phi) dt$

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

The average value of K.E. for complete cycle

$\displaystyle K_{avg}= \frac{1}{T}\int_{0}^{T}K dt$

$\displaystyle K_{avg}= \frac{1}{T}\int_{0}^{T}\frac{1}{2}m\omega^2 A^2 cos^2(\omega t +\phi)dt$

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

Thus average values of K.E. and P.E. of harmonic oscillator are equal and each equal to half of the total energy

Kaverage = Uaverage

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

Exercise : The amplitude of an SHM is doubled. Find the corresponding change in

(a) time period

(b) maximum velocity

(c) maximum acceleration

(d) total energy

Illustration : A particle executes SHM .

(a) What fraction of total energy is kinetic and what fraction is potential when displacement is one half of the amplitude;

(b) At what displacement the kinetic and potential energies are same?

Solution : (a)

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

For x = A/2

$\displaystyle K = \frac{1}{2}m\omega^2 (A^2 – x^2) = \frac{1}{2}m\omega^2 \frac{3}{4}A^2$

$\displaystyle \frac{K}{E} = \frac{3}{4}$

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

$\displaystyle \frac{U}{E} = \frac{1}{4}$

(b) K = U

$\displaystyle \frac{1}{2}m\omega^2 (A^2 – x^2) = \frac{1}{2}m\omega^2 x^2$

=> x = 0.707A