Question

A particle of mass $m$ is moving along the $x$-axis under the potential $V(x)=\frac{kx^2}{2}+\frac{\lambda}{x}$, where $k$ and $\lambda$ are positive constants. The particle is slightly displaced from its equilibrium position. The particle oscillates with the angular frequency $\omega$ given by

• A. $3\frac{k}{m}$
• B. $3\frac{m}{k}$
• C. $\sqrt{\frac{k}{m}}$
• D. $\sqrt{3\frac{m}{k}}$
• E. $\sqrt{3\frac{k}{m}}$

Hint:

For potential energy problems, the "restoring force constant" is always the second derivative of the potential evaluated at the stable equilibrium point.

Answer 1

Answer: (E)

Concept:
For small oscillations about an equilibrium position $x_0$, the angular frequency is: \[ \omega = \sqrt{\frac{k_{eff}}{m}}, \text{ where } k_{eff} = \left. \frac{d^2V}{dx^2} \right|_{x=x_0} \] Equilibrium occurs where the force is zero: $F = -\frac{dV}{dx} = 0$.

Step 1: Find the equilibrium position $x_0$.
\[ V(x) = \frac{1}{2}kx^2 + \lambda x^{-1} \] \[ \frac{dV}{dx} = kx - \lambda x^{-2} = 0 \Rightarrow kx = \frac{\lambda}{x^2} \Rightarrow x^3 = \frac{\lambda}{k} \Rightarrow x_0 = \left(\frac{\lambda}{k}\right)^{1/3} \]

Step 2: Calculate the effective spring constant $k_{eff}$.
\[ \frac{d^2V}{dx^2} = k + 2\lambda x^{-3} \] At $x = x_0$: \[ k_{eff} = k + \frac{2\lambda}{x_0^3} = k + \frac{2\lambda}{(\lambda/k)} = k + 2k = 3k \]

Step 3: Determine the angular frequency $\omega$.
\[ \omega = \sqrt{\frac{k_{eff}}{m}} = \sqrt{\frac{3k}{m}} \]