Potential energy of a particle executing SHM is U at an instant. Magnitude of  acceleration is a at the same instant. If x is the magnitude of displacement of the  particle

detailed solution

Correct option is D

For simple harmonic motion, the potential energy can be written as  U=12Kx2  ⇒U  α  x2---(1) . Differentiating with respect to x we get  dUdx=−Kx.  here K =force constant ; x = displacement; m=mass of the oscillatorHere dUdx  is the force due to simple harmonic motion which can be equated to  ma.    force F= ma=−Kx.   acceleration =a=−Kxm                                    ⇒a  α  x---(2)equations (1) and (2) are the reasons for the options chosen