Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation equals the average potential energy over the period.

Show that for a particle in linear S.H.M., the average kinetic energy

1.	Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation equals the average potential energy over the period.
Answer» Solution :Consider a particle of mass m EXECUTING S.H.M. with period T . The displacement of the particle at an instant t, when time period is noted from the MEAN position is given by y`=A sin omega t` `therefore ` VELOCITY , `V= (dy)/(dt) = A omega cos omega t`, KINETIC energy `E_K = 1/2 mV^2 = 1/2 m A^2 omega^2 cos^2 omega t` POTENTIAL energy `E_P =1/2ky^2 = 1/2 m omega^2 A^2 sin^2 omega t` `(because K = m omega^2) therefore ` Average K.E. over one cycle. `E_(k_(av)) = 1/T int_0^T E_k dt = 1/T int_0^2 1/2 mA^2 omega^2 cos^2 omega t dt` `=1/(2T) m omega^2 A^2 int_0^T ((1+ cos 2 omegat))/2 dt = 1/(4T) m A^2 omega^2 [k+(sin 2omega t)/(2omega)]_0^T = 1/(4T) mA^2 omega^2 (T) = 1/4 mA^2 omega^2` Average P.E. over one cycle is `E_(k_(av)) = 1/T int_0^T E_k dt = 1/T * 1/T int_0^T 1/2 mA^2 omega^2 sin^2 omega t dt` `=1/(2T) m omega^2 A int_0^T ((1-cos 2omegat))/2 = 1/(4T) m omega^2 A^2 [1-(sin2 omega t)/(2omega)]_0^T` `=1/(4T) m omega^2 A^2 [T] =1/4 A^2 omega^2` From (i) and (ii) , `E_(K_(aÅ))= E_(P_(aÅ))`