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?
Answer: (A) The total energy of the system is $\dfrac12 M\omega^2A^2 + \dfrac12\dfrac{q^2E^2}{k}$
**New equilibrium.** The constant force $qE$ shifts the equilibrium to where $kx_0 = qE$:
$$x_0 = \frac{qE}{k}$$
so options (2) and (3) are wrong. The angular frequency $\omega = \sqrt{k/M}$ is unchanged.
**Energy.** Take the field to be switched on as the body passes $x = 0$ with its maximum speed $v_0 = \omega A$. Measured from the new equilibrium, the body is then at displacement $x_0$ with speed $\omega A$, so its new amplitude $A'$ satisfies
$$\tfrac12 kA'^2 = \tfrac12 M\omega^2A^2 + \tfrac12 kx_0^2$$
The energy of oscillation about the new equilibrium is therefore
$$\frac12 M\omega^2A^2 + \frac12 k\left(\frac{qE}{k}\right)^2 = \frac12 M\omega^2A^2 + \frac12\frac{q^2E^2}{k}$$
This is option (1).
Solution by Sreeraj P, M.Sc Physics