The instantaneous K.E of a mass attached to the end of an elastic spring is:

Explanation

In summary, the correct option representing the instantaneous kinetic energy of a mass attached to the end of an elastic spring is a) 1/2 k (xo2 - x2). This option represents the instantaneous kinetic energy of the mass-spring system. Here, 'k' is the spring constant, 'xo' is the equilibrium position (the position where the spring is neither stretched nor compressed), and 'x' is the displacement of the mass from its equilibrium position at any given time. The formula 1/2 k (xo2 - x2) represents the difference between the potential energy (P.E) at equilibrium position (1/2 k xo2) and the potential energy at the displaced position (1/2 k x2) converted into kinetic energy.