Motion of Charges in an Electric Field question from JEE Main 2018 (15 Apr, Shift 1)

Answer: (b)

1. The field adds a constant force $qE$, so the new mean position is at $x_0 = \frac{qE}{k}$ (rules out (a) and (d)).
2. Take the field switched on as the body passes $x = 0$ with speed $\omega A$; it is then $x_0$ from the new mean.
3. Energy of the new oscillation: $\frac{1}{2}m\omega^2A^2 + \frac{1}{2}kx_0^2 = \frac{1}{2}m\omega^2A^2 + \frac{1}{2}\frac{q^2E^2}{k}$
4. So the answer is (b).