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Q. Average value of kinetic energy and potential energy over entire time period in a SHM is

AIIMSAIIMS 2013

Solution:

Maximum $KE =\frac{1}{2} m \omega^{2} A^{2}$
Minimum $K E=0$
Average $KE =\left(\frac{0+\frac{1}{2} m \omega^{2} A^{2}}{2}\right)$
$=\frac{1}{4} m \omega^{2} A ^{2} $
Similarly average $ PE =\left(\frac{0+\frac{1}{2} m \omega^{2} A^{2}}{2}\right)$
$=\frac{1}{4} m \omega^{2} A^{2}$