Resting on a smooth horizontal table, a block of mass m with a...

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 $x_0$ from the initial unstretched position. At equilibrium, the spring force balances the electric force:

$k x_0 = qE \Rightarrow x_0 = \dfrac{qE}{k}$

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 resulting simple harmonic motion.

The amplitude $A$ of the oscillation is the distance between the mean position and the extreme position:

$A = |x_0 - 0| = \dfrac{qE}{k}$

Alternatively, applying the work-energy theorem between the initial position and the position of maximum extension x_\max (where velocity is again zero):

qE x_\max = \dfrac{1}{2} k x_\max^2 \Rightarrow x_\max = \dfrac{2qE}{k}

The amplitude is half of the maximum displacement:

A = \dfrac{x_\max}{2} = \dfrac{qE}{k}