Question

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.

Answer 1

your question is incomplete. A complete question is ----> 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?

(1) The total energy of the system is $\frac{1}{2}m\omega^2A^2+\frac{1}{2}\frac{q^2E^2}{k}$.

(2) The new equilibrium position is at a distance $\frac{2qE}{k}$ from x = 0.

(3) The new equilibrium position is at a distance $\frac{qE}{2k}$ from x = 0.

(4) The total energy of the system is $\frac{1}{2}m\omega^2A^2-\frac{1}{2}\frac{q^2E^2}{k}$.

solution : at equilibrium,

spring force = electrostatic force

or, kx = qE

or, x = qE/k,

now total energy = $\frac{1}{2}k(A^2+x^2)$

= $\frac{1}{2}k\left[A^2+\left(\frac{qE}{k}\right)^2\right]</p><p>= [tex]\frac{1}{2}k\left[A^2+\frac{q^2E^2}{k^2}\right]</p><p>= [tex]\frac{1}{2}m\omega^2A^2+\frac{1}{2}\frac{q^2E^2}{k}$

hence, it is clear that option (1) is correct