question_answer 4)

The potential energy function for a particle executing linear simple harmonic motion is given by  where k is the force constant of the oscillator. For  the graph of  versus x is shown in Fig. 6.47. Show that a particle of total energy 1 J moving under this potential must "turn back" when it reaches .    

Answer:

At any instant, the energy of the oscillator is partly kinetic and partly potential. Its total energy is   or            An oscillating particle turns back at the point where its instantaneous velocity is zero i.e., the particle will turn back at such a point x where .             But                or or