Question

A particle of mass $m$ moves in the positive $x$ direction under the potential $V(x)=\frac{1}{2}kx^{2}+\frac{\lambda}{2x^{2}}$. If the particle is slightly displaced from equilibrium, the angular frequency $\omega$ is ............... (Give answer in units of $\sqrt{k/m}$ as an integer.)
 

Hint:

For small oscillations, always evaluate the second derivative of potential at equilibrium.

Answer 1

Correct Answer: 2

Step 1: Find the equilibrium position.
Equilibrium occurs where $V'(x)=kx - \frac{\lambda}{x^{3}} = 0$. Solving gives $x_{0}^{4} = \frac{\lambda}{k}$.

Step 2: Use the formula for small oscillations.
For small oscillations, $\omega^{2} = \frac{1}{m}V''(x_0)$. Differentiate:
$V''(x)=k + \frac{3\lambda}{x^{4}}$. Evaluate at $x_0$ using $x_0^{4}=\lambda/k$:
$V''(x_0)=k + 3k = 4k$.

Step 3: Final expression.
$\omega = \sqrt{\frac{4k}{m}} = 2\sqrt{\frac{k}{m}}$.
Thus, in units of $\sqrt{k/m}$, the answer is 2.