A particle moves in a conservative force field where its potential energy U is given by U(x) = ax^2 - bx. The particle will be in stable equilibrium at position x equal to: MCQ with Answer and Explanation

A particle moves in a conservative force field where its potential energy U is given by U(x) = ax^2 - bx. The particle will be in stable equilibrium at position x equal to:
A. b / a
B. b / 2a
C. 2b / a
D. a / 2b
Answer: Option B
Solution (By JKExamLibrary)
Equilibrium occurs where the net force is zero, meaning dU/dx = 0. Given U(x) = ax^2 - bx, taking the derivative: dU/dx = 2ax - b = 0. Therefore, x = b / 2a. To confirm it's stable, the second derivative d^2U/dx^2 must be positive. d^2U/dx^2 = 2a. Assuming 'a' is positive, it represents a stable equilibrium point.

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