A particle of unit mass subjected to the 1-dimensional potential $$V(x) = \frac{2 \alpha}{x^3} - \frac{3 \beta}{x^2}$$​​
executes small oscillations about its equilibrium position, where α and β are positive constants with appropriate dimensions. The time period of small oscillations is

Correct option is D

Solution:

$$V(x) = \frac{2 \alpha}{x^3} - \frac{3 \beta}{x^2} \quad \text{for equilibrium point} \quad \frac{\partial V}{\partial x} = 0 \Rightarrow -\frac{6 \alpha}{x^4} + \frac{6 \beta}{x^3} = 0 \Rightarrow x_0 = \frac{\alpha}{\beta}$$ 

$$\omega = \sqrt{\frac{\partial^2 V}{\partial x^2} \bigg|_{x = x_0} \over m} \quad \text{where} \quad \frac{\partial^2 V}{\partial x^2} = \frac{24 \alpha}{x^5} - \frac{18 \beta}{x^4} \quad \text{and with } x_0 = \frac{\alpha}{\beta}$$ 

$$\frac{\partial^2 V}{\partial x^2} = \frac{24 \alpha \beta^5}{\alpha^5} - \frac{18 \beta^5}{\alpha^4} = \frac{6 \beta^5}{\alpha^4}$$ 

$$\frac{2 \pi}{T} = \sqrt{\frac{6 \beta^5}{\alpha^4}} \Rightarrow T = \frac{2 \pi \alpha^2}{\sqrt{6 \beta^5}}$$.