A particle of mass $m$ is moving in a potential field $U(x)=a{x}^2-b{x}^4$, where $a$ and $b$ are positive constants. For what range of energy $E$ can the particle undergo bounded motion?

Solution

1. The potential $U(x)=a{x}^2-b{x}^4$ has local maxima where $$\begin{array}{rl}dU/dx& =2ax-4b{x}^3\\ & =0\end{array}$$.
2. The extrema are at $x=0$ and $x^2=a/(2b)$.
3. The value at the local maxima is $$\begin{array}{rl}{U}_{max}& =a(a/2b)-b(a/2b{)}^2\\ & =a^2/(2b)-a^2/(4b)\\ & =a^2/(4b)\end{array}$$.
4. For bounded motion, the energy must be less than the potential barrier height $U_{max}$ and greater than the minimum $U(0)=0$.
5. Hence the answer is (A).