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 same period.

Explanation

The equation of displacement of a particle executing SHM at an instant t is given as:
`x=Asinomegat`
Where, A= Amplitude of oscillation
=`sqrt((K)/(M))`
omega = Angular frequency
The velocity of the particle is:
`v=(dx)/(dt)=Aomegacosomegat`
The kinetic energy of the particle is: `E_(K)=(1)/(2)Mv^(2)=(1)/(2)Momega^(2)A^(2)cos^(2)omegat`
The potential energy of the particle is:
`E_(P)=(1)/(2)Kx^(2)=(1)/(2)Momega^(2)A^(2)sin^(2)omegat`
For time period T, the average kinetic energy over a single cycle is given as:
`(E_(K))_(Avg)=(1)/(T)overset(T)underset(0)intE_(K)dt=(1)/(T)overset(T)underset(0)int(1)/(2)MA^(2)omega^(2)cos^(2)omegatdt=(1)/(2T)MA^(2)omega^(2)overset(T)underset(0)int((1+cos2omegat))/(2)dt=(1)/(4T)MA^(2)omega^(2)[t+(sin2omegat)/(2omega)]_(0)^(T)`
=`(1)/(4T)MA^(2)omega^(2)(T)=(1)/(4)MA^(2)omega^(2)`.....(i)
And,average potential energy over one cycle is given as:
`(E_(P))_(Avg)=(1)/(T)overset(T)underset(0)intE_(P)dt=(1)/(T)overset(T)underset(0)int(1)/(2)Momega^(2)A^(2)sin^(2)omegatdt=(1)/(2t)Momega^(2)A^(2)overset(T)underset(0)int((1-cos2omegat))/(2)dt`
`(1)/(4T)Momega^(2)A^(2)[t-(sin2omegat)/(2omega)]_(0)^(T)=(1)/(4T)Momega^(2)A^(2)(T)=(Momega^(2)A^(2))/(4)`.......(ii)
It can we inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.