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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).
.
The acceleration of this partile for `|x|gtX_(0)` is
(a) proprtional to `V_(0)`
(b) proportional to.

A

proportional to `V_0`

B

proportional to `(V_0)/(m X_0)`

C

proportional to `sqrt((V_0)/(m X_0))`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
D
As potential energy is constant for `|x| gt X_(0)`, the force on the particle is zero hence acceleration is zero.
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