Question 355: Oscillations

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 corresponding time period is proportional to $\sqrt {{m \over k}} $, as can be seen easily using dimensional analysis. 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 to infinity. Consider a particle of mass m moving on the x-axis. Its potential energy is V(x) = $\alpha$x
4
($\alpha$ > 0) for | x | near the origin and becomes a constant equal to V
0
for (see figure).
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l31km1oy/9062ceb7-1d06-4ac8-a560-9d5ec8300523/a86fc630-d127-11ec-b5bf-9d106fd05fc4/file-1l31km1oz.png?format=png)

Accepted Answer

The kinetic energy of the particle cannot be negative. The total energy E is the sum of the kinetic energy K(x) and potential energy V(x) i.e.,
E = K(x) + V(x). ....... (1)
From the given figure, V(x) $\ge$ 0 for all x. If E $\le$ 0 for some x then kinetic energy K(x) = E $-$ V(x) $\le$ 0 for those x and hence the motion of the particle is not allowed for those x. On the other hand, if E $\ge$ V
0
then K(x) = E $-$ V(x) $\ge$ 0 for all x and particle is allowed to move for all x, including infinity (in this case the particle will escape to infinity). Thus, for the particle to have a periodic motion, V
0
> E > 0. In this case, particle is allowed at points where
K(x) = E $-$ V(x) = E $-$ $\alpha$x
4
$\ge$ 0 i.e.,
| x | $\le$ (E/$\alpha$)
1/4
.
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l31krb4q/31a8bfd6-9679-4f0e-9b5b-7af2521de1ad/3ac5eeb0-d128-11ec-b5bf-9d106fd05fc4/file-1l31krb4r.png?format=png)

Answer

E < 0

Answer

E > 0

Answer

V
0
> E > 0

Answer

E > V
0

Oscillations

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