JEE Main2018PhysicsOscillationsActual
A body of mass M and charge q is connected to a spring of spring constant k . It is oscillating along x -direction about its equilibrium position in the horizontal plane, taken to be at x = 0 , with an amplitude A . An electric field E is applied along the x -direction. Which of the following statements is correct?
Options
- AThe total energy of the system is 1 2 m ω 2 A 2 + 1 2 q 2 E 2 k
- BThe new equilibrium position is at a distance 2 q E k from x = 0
- CThe new equilibrium position is at a distance q E 2 k from x = 0
- DThe total energy of the system is 1 2 m ω 2 A 2 - 1 2 q 2 E 2 k .
Correct answer
A. The total energy of the system is 1 2 m ω 2 A 2 + 1 2 q 2 E 2 k
Step-by-step solution
The equilibrium position will shift to a point where the resultant force is equal to zero. k x e q = q E     ⇒ x e q = q E k . Energy = 1 2 m ω 2 A 2 + q E k 2 = 1 2 m ω 2 A 2 + 1 2 q 2 E 2 k .