Resting on a smooth horizontal table, a block of mass m with a charge q is connected to a wall by an unstressed spring having a spring constant k , as illustrated in the figure. Subsequently, a horizontal electric field E is switched on parallel to the spring. Determine the amplitude of the resulting simple harmonic motion of the block.

Step-by-step solution

When the horizontal electric field E is switched on, a constant electric force F = qE acts on the block. This constant force shifts the equilibrium (mean) position of the spring-mass system. Let the new equilibrium position be at a distance x₀ from the initial unstretched position. At equilibrium, the spring force balances the electric force: k x₀ = qE x₀ = qE k Initially, the block is at rest at the unstretched position ( x = 0 ). Since its velocity is zero at this point, it acts as an extreme position for the res