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Consider a particle of mass m that is subjected to a conservative force described by a potential energy function V(x)=\frac{C x}{x^2+a^2}, where \mathrm{C} and a are positive constants. The position or positions of the stable equilibrium is or are given as :

Option: 1

x=+a \, \, only


Option: 2

\mathrm{x}=- a \, \, only


Option: 3

x=-\frac{a}{2} \, \, and \, \, +\frac{a}{2}


Option: 4

\mathrm{x}=-\mathrm{a} \, \, and \, \, +\mathrm{a}


Answers (1)

best_answer

In the equilibrium position of the particle, the net force acting on it is zero, which can be determined by calculating the derivative of the potential energy function with respect to position. To find the equilibrium positions, we set the derivative of the potential energy function equal to zero:

\frac{d V}{d x}=\frac{C\left(a^2-x^2\right)}{\left(x^2+a^2\right)^2}=0

Solving this equation, we find two equilibrium positions:

\begin{aligned} & x_1=a \\ & x_2=-a \end{aligned}

To determine the nature of these equilibrium positions, we examine the second derivative of the potential energy function:

\frac{d^2 V}{d x^2}=\frac{2 C x\left(x^2-3 a^2\right)}{\left(x^2+a^2\right)^3}

Evaluating the second derivative at x_1 and  x_2, we find: 

\begin{aligned} & \left.\frac{d^2 V}{d x^2}\right|_{x_1}<0 \\ & \left.\frac{d^2 V}{d x^2}\right|_{x_2}>0 \end{aligned}

This indicates that at \mathrm{x}=\mathrm{a}, there is a maxima and at \mathrm{x}=-\mathrm{a}, there is a minima. Therefore, x_1 represents a position of unstable equilibrium, while x_2 represents a position of stable equilibrium.

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