Physics - SHM Question with Solution | TestHub

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$=\left(\frac{0+\frac{1}{2}m{\omega }^2 A^2}{2}\right)$

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

(average of${\text{sin}}^2\theta$and${\text{cos}}^2\theta$in a time period is$\frac{1}{2}$)