A potential is given by V ( x ) = k ( x + a ) 2 / 2 for x < 0 and V ( x ) = k ( x - a ) 2 / 2 for x > 0 . The schematic variation of oscillation period T for a particle performing periodic motion in this potential as a function of its energy E is

Step-by-step solution

Given, potential function for the oscillating particle is V x = k ( x + a ) 2 2 , x < 0 k ( x - a ) 2 2 , x > 0 So, potential energy of the particle (mass m ) is U x = k m ( x + a ) 2 2 , x < 0 k m ( x - a ) 2 2 , x < 0 d U d x = k m ( x + a ) , x < 0 k m ( x - a ) , x > 0 If d U d x = 0 , when x = ± a Now, d 2 U d x 2 = k m > 0 So, particle is in unstable equilibrium at x = ± a Hence, particle is unbounded for - a > x and x > a In region, - a ≤ x ≤ a , time period of particle reduces from a maximum. So, correct gr