The potential energy function of a particle of mass \(m\) is \(V=-\frac{1}{2} c\left(x^{2}-\right.\) \(\left.a^{2}\right)^{2}\), where \(c\) and \(a\) are positive constants. Sketch this function, and describe the possible types of motion in the three cases (a) \(E>0\), (b) \(E<-\frac{1}{2} c a^{4}\), and \((\mathrm{c})-\frac{1}{2} c a^{4}

To sketch the function \(V=-\frac{1}{2}c(x^2 - a^2)^2\), first note that it is even, meaning \(V(x) = V(-x)\). The function is also a quartic polynomial, where the leading term is negative. Hence, the graph has two minima at \(x = \pm a\) and one local maximum at \(x = 0\). The potential energy approaches \(+\infty\) as \(x\) goes to \(+\infty\) or \(-\infty\).

(a) For \(E > 0\), the total energy is higher than the energy at the maximum of the potential energy curve, meaning the particle is not trapped in the potential well. (b) If \(E < -\frac{1}{2}ca^4\), the total energy is lower than the energy at both minimum points of the potential energy curve, meaning the particle is trapped in the potential well. (c) For \(-\frac{1}{2}ca^4 < E < 0\), the total energy is in between the minimum and local maximum of the potential, indicating an oscillatory motion case.

(a) For \(E > 0\), the particle is not bound by the potential energy, so it can freely move in any direction. Since the total energy is positive, there will be both potential and kinetic energy for the particle, and its motion will not be confined to the potential well. (b) For the case \(E < -\frac{1}{2}ca^4\), the particle is trapped between the potential well formed by the minima at \(x = \pm a\). As a consequence, the particle's motion will be oscillatory between the minima. (c) In the case \(-\frac{1}{2}ca^4 < E < 0\), the particle's energy is between the local minimum and local maximum of the potential energy. The particle will undergo oscillatory motion, oscillating between the minimum and the local maximum at \(x=0\). It won't be able to reach the other side of the potential barrier as its energy is not high enough.