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. 2b / a
B. b / a
C. b / 2a
D. a / 2b
Answer: Option C
Solution (By JKSSB Mock Tests)
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.
Explanation:
Area under v-t graph gives displacement (distance if velocity doesn't change sign). Slope of x-t gives velocity. Slope of v-t gives acceleration. Area a-t gives change in velocity.
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