Question

Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer 1

Kinetic energy in SHM ,
KE = 1/2 mv²
= (1/T ).{\int(1/2mv²).dt}
For displacement x = Asinwt
V = dx/dt = wAcoswt
(KE)av = (1/T) {\int(mw²A²cos²(wt).dt}
= (1/4T)mw²A²{\int(1 + cos2wt).dt}
=( 1/4T) × mw²A²(T - 0)
= 1/4mw²A2

(PE)av = 1/2Kx²
= 1/2T{\int(mw²A²sin²wt).dt}
=(1/4T)mw²A²{\int(1 -cos2wt).dt}
=1/4T × mw²A² × ( T - 0)
= 1/4 mw²A²

(PE)av = (KE)av