A particle of mass \(m\) moves (in the region \(x>0\) ) under a force \(F=\) \(-k x+c / x\), where \(k\) and \(c\) are positive constants. Find the corresponding potential energy function. Determine the position of equilibrium, and the frequency of small oscillations about it.

Recall that the force F is the negative derivative of the potential energy function U(x). Therefore, we have: \(F=-\frac{dU(x)}{dx}\) Given that \(F=-k x+c/x\), we can integrate the force function with respect to x to find the potential energy function: \(U(x)=-\int F dx = -\int (-k x+\frac{c}{x}) dx\) Now, we can integrate: \(U(x) = -(-k\frac{x^2}{2} + c\ln|x|) + A\) Since the potential energy is defined up to an additive constant A, we can set our potential energy function as: \(U(x) = k\frac{x^2}{2} - c\ln|x|\)

The position of equilibrium is the point at which the force F(x) equals zero. We need to find the value of x that satisfies this condition: \(0 = -k x+c/x\) Now, we can solve for x: \(k x^2 = c\) \(x^2 = \frac{c}{k}\) \(x_\text{eq} = \sqrt{\frac{c}{k}}\) So, the position of equilibrium is \(x_\text{eq} = \sqrt{\frac{c}{k}}\).

To find the frequency of small oscillations, we can use the formula: \(\omega = \sqrt{\frac{k_\text{eff}}{m}}\) Here, \(k_\text{eff}\) is the effective spring constant, which is the second derivative of the potential energy function U(x) evaluated at the equilibrium position. Let's find this value: \(\frac{d^2U(x)}{dx^2} = k - \frac{c}{x^3}\) Now, we need to evaluate this expression at the equilibrium position, \(x_\text{eq}\): \(k_\text{eff} = k - \frac{c}{(\sqrt{\frac{c}{k}})^3} = k - k = 0\) Since the effective spring constant is equal to zero, the frequency of small oscillations is also zero: \(\omega = \sqrt{\frac{k_\text{eff}}{m}} = \sqrt{\frac{0}{m}} = 0\) The frequency of small oscillations about the equilibrium position is 0.