Question

A particle of mass \( m \) moves in one dimension under the action of a conservative force whose potential energy has the form \( U(x) = \frac{\alpha x}{x^2 + \beta^2} \), where \( \alpha \) and \( \beta \) are dimensional parameters. The angular frequency \( \omega \) of the oscillation is proportional to:

• A. \(\sqrt{\frac{\alpha^3}{m \beta^4}}\)
• B. \(\sqrt{\frac{\alpha}{m \beta^4}}\)
• C. \(\sqrt{\frac{\alpha}{m \beta^3}}\)
• D. \(\sqrt{\frac{\alpha}{m \beta^6}}\)

Hint:

For systems with potential energy containing higher-order terms, use small displacement approximation and equate the resulting equation of motion to find the angular frequency.

Answer 1

The Correct Option is C

Step 1: The potential energy function is defined as \( U(x) = -\frac{\alpha x}{x^2 + \beta^2} \). The force \( F \) relates to potential energy as:

\[ F(x) = -\frac{dU(x)}{dx} \]

Differentiating \( U(x) \):

\[ F(x) = -\frac{d}{dx} \left( -\frac{\alpha x}{x^2 + \beta^2} \right) \]

\[ F(x) = \frac{\alpha(x^2 + \beta^2) - 2\alpha x^2}{(x^2 + \beta^2)^2} = \frac{\alpha(\beta^2 - x^2)}{(x^2 + \beta^2)^2}. \]

Step 2: For small oscillations, approximate the force as a restoring force, similar to Hooke's law \( F = -kx \), where \( k \) is the spring constant. For small \( x \):

\[ F(x) \approx -kx \]

Comparing the force expressions allows us to determine \( k \) in terms of \( \alpha \) and \( \beta \).

Step 3: The angular frequency \( \omega \) relates to the spring constant \( k \) and mass \( m \) via:

\[ \omega = \sqrt{\frac{k}{m}} \]

Dimensional analysis yields that \( \omega \) is proportional to:

\[ \omega \propto \sqrt{\frac{\alpha}{m \beta^3}}. \]

Therefore, the solution is:

\[ \sqrt{\frac{\alpha}{m \beta^3}}. \]