Answer to Question #102196 in Physics for Megha

Question

Show that the average kinetic energy associated with SHM is ka^2/4.

Answer 1

The virial theorem states

$$2<T>=n<U>$$

where $<T>$ is a mean kinetic energy, $<U>$ is a mean potential energy.

In the case of SHM $n=2,$ so

$$<T>=<U>$$

The total mean energy of SHM

$$<E>=<T>+<U>=2<T>=\frac{kA^2}{2}.$$

Hence

$$<T>=\frac{kA^2}{4}.$$