*The potential energy function of a particle of mass \(m\) is \(V=c x /\left(x^{2}+\right.\) \(\left.a^{2}\right)\), where \(c\) and \(a\) are positive constants. Sketch \(V\) as a function of \(x\). Find the position of stable equilibrium, and the period of small oscillations about it. Given that the particle starts from this point with velocity \(v\), find the ranges of values of \(v\) for which it (a) oscillates, (b) escapes to \(-\infty\), and \((c)\) escapes to \(+\infty\).

To sketch the potential energy function, we should notice its basic characteristics. As x approaches infinity or negative infinity, the denominator dominates, and the function approaches 0. At x=0, the potential energy function has its maximum value c/a. V(x) goes to 0 in both directions on the x-axis. Now we can sketch the function like this: 1. Draw the x and V(x) axis 2. Plot a point at (0, c/a), which represents the maximum value of V(x) 3. Create a curve that is symmetric with respect to the y-axis and approaches 0 as x approaches infinity or negative infinity.

A stable equilibrium position occurs when the first derivative of the potential energy function with respect to x is equal to 0, and the second derivative is positive. Let's find the first and second derivatives: \(V(x) = \frac{cx}{x^2 + a^2}\) \(\frac{dV}{dx} = \frac{c(x^{2}+a^{2}) - 2cx^{2}}{(x^{2}+a^{2})^{2}}\) Now, set \(\frac{dV}{dx} = 0\) to find equilibrium positions: \(0 = \frac{ca^{2} - cx^{2}}{(x^{2}+a^{2})^{2}}\) This simplifies to \(x^{2} = a^{2}\), so the stable equilibrium positions are x = a and x = -a. Now, find the second derivative: \(\frac{d^{2}V}{dx^{2}} = \frac{-2c(3x^{2}a^{2} - a^{4})}{(x^{2}+a^{2})^{3}}\) Let's check if either x = a or x = -a are stable. Plug x = a into the second derivative: \(\frac{d^{2}V}{dx^{2}}|_{x=a} = \frac{-6ca^{2}}{(2a^{2})^{3}} < 0\) This means that x = a is unstable. Now, plug x = -a: \(\frac{d^{2}V}{dx^{2}}|_{x=-a} = \frac{6ca^{2}}{(2a^{2})^{3}} > 0\) This means that x = -a is the stable equilibrium position.

To find the period of small oscillations about the stable equilibrium point, we can use the formula: \(T = 2 \pi \sqrt{\frac{m}{k}}\) where \(T\) is the period, \(m\) is the mass of the particle, and \(k\) is the effective spring constant at the equilibrium point, which can be calculated from the second derivative of the potential energy evaluated at the stable equilibrium position: \(k = \frac{6ca^{2}}{(2a^{2})^{3}}\) \(T = 2 \pi \sqrt{\frac{m}{k}} = 2 \pi \sqrt{\frac{m(2a^{2})^{3}}{6ca^{2}}}\)

We can find these by analyzing the total mechanical energy \(E\) of the system: \(E = K + V = \frac{1}{2}mv^{2} + V(x)\) For oscillation, the total energy must be trapped within the local potential well created by V(x) around the stable equilibrium position x = -a. Thus, the total energy must satisfy the inequality: \(0 \le E = \frac{1}{2}mv^{2} + V(-a) < V(0) = \frac{c}{a}\) Since \(V(-a) = 0\), we get: \(0 \le \frac{1}{2}mv^{2} < \frac{c}{a}\), so the oscillating range of velocities is: \(0 \le v < \sqrt{\frac{2c}{ma}}\) For escape to \(-\infty\), the particle must have enough energy to escape the potential well, meaning that the total energy must be greater than or equal to the energy at x = 0: \(E = \frac{1}{2}mv^{2} + V(-a) \ge V(0) = \frac{c}{a}\) \(v \ge \sqrt{\frac{2c}{ma}}\) For escape to \(+\infty\), the particle must be able to cross the unstable equilibrium point at x = a. The critical energy for this is \(E = V(a)\). So, for escape to \(+\infty\), \(E = \frac{1}{2}mv^{2} + V(-a) \ge V(a) = \frac{-c}{a}\) \(v \ge \sqrt{\frac{-2c}{ma}}\) However, this result implies an imaginary velocity, indicating that there is no range of initial velocities for which the particle escapes to \(+\infty\). In conclusion, the ranges of velocities are: a) Oscillation: \(0 \le v < \sqrt{\frac{2c}{ma}}\) b) Escape to \(-\infty\): \(v \ge \sqrt{\frac{2c}{ma}}\) c) Escape to \(+\infty\): There is no range of initial velocities for escape to \(+\infty\).