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