Question

When a particle of mass m moves on the x-axis in a potential of the form `V(x) =kx^(2)` it performs simple harmonic motion. The correspondubing time period is proprtional to `sqrtm/h`, as can be seen easily using dimensional analusis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of `x=0` in a way different from `kx^(2)` and its total energy is such that the particle does not escape toin finity. Consider a particle of mass m moving on the x-axis. Its potential energy is `V(x)=ax^(4)(agt0)` for |x| neat the origin and becomes a constant equal to `V_(0)` for |x|impliesX_(0)` (see figure).

If total energy of the particle is E, it will perform perildic motion only if.
A. (a) `Elt0`
B. (b)`Egt0`
C. (c )`V_(0)gtEgt0`
D. (d)`EgtV_(0)`

Answer 1

Correct Answer - C
If the energy is zero, the particle will not perform oscillations. Therefore € should be greater than zero. Further if (c )`E=V_(0)`, the potential energy will become constant as depicted in the potential energy will become constant as depicted in the graph given. In this acse also the particle will not oscillate.
Therefore (E) should be than zero but less than `V_(0)`.