Q. Average value of KE and PE over entire time period is
A) $0,\frac{1}{2}m\ \omega^{2}A^{2}$
B) $\frac{1}{2}m\ \omega^{2}A^{2},0$
C) $\frac{1}{2}m\ \omega^{2}A^{2},\ \frac{1}{2}m\ \omega^{2}A^{2}$
D) $\frac{1}{4}m\ \omega^{2}A^{2},\frac{1}{4}m\ \omega^{2}A^{2}$
Explanation
(d) Maximum KE$= \frac{1}{2}m\omega^{2}A^{2}$; minimum KE=0
Average KE$= \frac{0 + \frac{1}{2}m\omega^{2}\ A^{2}}{2} = \frac{1}{4}m\omega^{2}A^{2}$
Similarly average PE$= \left( \frac{0 + \frac{1}{2}m\omega^{2}\ A^{2}}{2} \right)/2$
$= \frac{1}{4}m\omega^{2}A^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.