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).
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For periodic motion of small amplitude A,the time period (T) of thes particle is proportional to.

Options

A. (a) Asqrtm/a`
B. (b) 1/Asqrtm/a`
C. (c ) Asqrta/m`
D. (d) `1/Asqrta/m`.

Answer

B

Explanation

(b) We can get the answer of this question with the help of dimensional analysis.
Given potential energy `=ax^(4)`
`a(Potential energy)/x^(4) =(ML^(2)T^(-2))L^(4) =[Ml^(-2)t^(-2)].
Now 1/Asqrtm/a=1/Lsq^(-2)T^(-2))=T`
Therefore potion (b) is correct.