Let a particle of mass #m # is executing SHM with amplitude #a# and its displacement x at time t is given by the following equation
#x= asinomegat......(1)#,
where #omega# is the angular velocity of the reference point in the reference circle associated with the SHM.
Differentiating (1) w.r to t we get the velocity v of the particle at t th sec
#v=(dx)/(dt)=aomegacosomegat......(2)#
Again differentiating (2) w.r to t we get the aceeleration f of the particle at t th sec
#f=(d^2x)/(dt^2)=-aomega^2sinomegat=-omega^2x.....(3)#
So KE of the particle at displacement x
#E_k=1/2mv^2=1/2ma^2omega^2cos^2omegat=1/2momega^2(a^2-a^2sin^2omegat)#
#E_k=1/2momega^2(a^2-x^2)...........(4)#
Th PE of the particle at displacement x
#E_p=-int_0^xmfdx#
#=>E_p=-m int_0^x(-omega^2x)dx=1/2momega^2x^2#
So total energy at any position x
#E=E_k+E_p=1/2momega^2(a^2-x^2)+1/2momega^2x^2#
#E=1/2m omega^2a^2color(red)(...................[5])#
when #x=a/2 # then KE by equation (4)
#(E_k)_(x=a/2)=1/2momega^2(a^2-(a/2)^2)#
#(E_k)_(x=a/2)=3/4xx1/2momega^2a^2=3/4E#
