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, 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?

We have a body of mass $$M$$ carrying charge $$q$$ which is attached to a spring of spring constant $$k$$ and is free to move along the $$x$$-axis. In the absence of any external field the restoring force of the spring alone produces simple harmonic motion (SHM). For SHM the standard formula for angular frequency is

$$\omega=\sqrt{\dfrac{k}{M}}$$

and, if the amplitude (maximum displacement from the equilibrium position) is $$A$$, the total mechanical energy is

$$E_{\text{old}}=\dfrac12\,M\omega^{2}A^{2}=\dfrac12\,kA^{2}.$$

Now an electric field $$E$$ is applied along the $$+x$$ direction. A charge in a uniform electric field experiences a constant force whose magnitude is given by the well-known relation

$$F_{e}=qE.$$

Because the field is along $$+x$$, the electric force on the charge also points along $$+x$$. The spring at a general position $$x$$ exerts the force $$F_{s}=-kx$$ (Hooke’s law). Hence, after the field is switched on, the net force on the mass becomes

$$F_{\text{net}}=-kx+qE.$$

The (new) equilibrium position $$x_{0}$$ is obtained by setting this net force to zero:

$$-kx_{0}+qE=0\;\;\Longrightarrow\;\;x_{0}=\dfrac{qE}{k}.$$

Because $$x_{0}\neq\dfrac{2qE}{k}$$ and also $$x_{0}\neq\dfrac{qE}{2k},$$ both option B and option C are incorrect.

Although the equilibrium point has shifted, the curvature of the potential, and hence the effective spring constant, has not changed; therefore the angular frequency is still $$\omega=\sqrt{k/M}$$ and the motion about the new equilibrium continues to be SHM.

To compute the total energy in the presence of the field we first write the complete potential energy. The potential energy of the spring is the familiar

$$U_{s}=\dfrac12\,k x^{2},$$

while the potential energy of a charge in a constant electric field is, from electrostatics,

$$U_{e}=-qEx.$$

Hence the combined potential energy is

$$U(x)=\dfrac12\,k x^{2}-qEx.$$

It is often convenient to rewrite this expression by “completing the square.” We have

$$\begin{aligned} U(x) &=\dfrac12\,k\!\left[x^{2}-2\left(\dfrac{qE}{k}\right)x\right] \\[2mm] &=\dfrac12\,k\!\left[(x-\dfrac{qE}{k})^{2}-\left(\dfrac{qE}{k}\right)^{2}\right] \\[2mm] &=\dfrac12\,k\,(x-x_{0})^{2}-\dfrac12\,\dfrac{q^{2}E^{2}}{k}, \end{aligned}$$

where $$x_{0}=qE/k$$ as obtained earlier. Notice that the last term is a constant: it does not depend on the configuration of the system and therefore can be changed by adding any convenient reference value without affecting the dynamics. If we add the constant $$\dfrac12\,\dfrac{q^{2}E^{2}}{k}$$ to every value of the potential (an operation that merely shifts the zero of energy), the potential energy becomes

$$U^{\prime}(x)=\dfrac12\,k\,(x-x_{0})^{2}+\dfrac12\,\dfrac{q^{2}E^{2}}{k}.$$

Using this shifted reference the mechanical energy (kinetic plus potential) is

$$E=\dfrac12\,M v^{2}+\dfrac12\,k\,(x-x_{0})^{2}+\dfrac12\,\dfrac{q^{2}E^{2}}{k}.$$

During an oscillation the quantity $$|x-x_{0}|$$ varies between $$0$$ and the amplitude $$A$$ (now understood to be measured from the new equilibrium). At either extreme point the velocity becomes zero, so the maximum (and, for SHM, constant) energy stored in the system is

$$\begin{aligned} E_{\text{total}} &=\dfrac12\,kA^{2}+\dfrac12\,\dfrac{q^{2}E^{2}}{k} \\[2mm] &=\dfrac12\,M\omega^{2}A^{2}+\dfrac12\,\dfrac{q^{2}E^{2}}{k}. \end{aligned}$$

This is exactly the expression quoted in option A. Option D differs only in the overall sign of the constant term; that sign depends on the arbitrary choice of the zero of potential energy, and the statement in option A is the one consistent with the positive shift we have just introduced. Therefore option D is not taken as the correct alternative in the language of the question.

Thus, out of the four given statements, only option A is correct.

Hence, the correct answer is Option A.