A particle moves in a potential $V(x)=k{x}^4$. For small oscillations about the equilibrium position, what is the relationship between the period of oscillation $T$ and the amplitude $A$?

Solution

1. For a potential of the form $V(x)=c{x}^n$, the period of oscillation $T$ scales with the amplitude $A$ as $T\propto A^{1-n/2}$.
2. Given the potential $V(x)=k{x}^4$, we identify the exponent as $n=4$.
3. Substituting $n=4$ into the scaling relation yields $T\propto A^{1-4/2}=A^{-1}$.
4. Hence the answer is (C).