The potential energy of a particle Ux executing SHM is given by
Ux=K2a−x2
Ux=K1x+K2x2+K3x3
Ux=Ae−bx
Ux=a constant
PE of body in SHM at an instant PE=U=12mω2y2=12Ky2
m is mass of oscillator, ω is angular velocity, y is displacement of oscillator, k is force constantIf the displacement , y=a−x then U=12Ka−x2=12Kx−a2