A body of mass $M$ and charge $q$ is connected to a spring of spring constant $k$. It is oscillating along…
A body of mass $M$ and charge $q$ is connected to a spring of spring constant $k$. It is oscillating along $\mathrm{x}$-direction about its equilibrium position, taken to be at $x=0$, with an amplitude $A$. An electric field $E$ is applied along the $\mathrm{x}$-direction. Which of the following statements is correct?
The total energy of the system is
$$
\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}
$$
The new equilibrium position is at a distance: $\frac{2 q E}{k}$ from $x=0$
The new equilibrium position is at a distance: $\frac{q E}{2 k}$ from $\mathrm{x}=0$
The total energy of the system is $\frac{1}{2} m \omega^2 A^2-\frac{1}{2} \frac{q^2 E^2}{k}$
Solution
Equilibrium position will shift to point where resultant force $=0$
$
\mathrm{kx}_{\text {eq }}=\mathrm{qE} \Rightarrow \mathrm{x}_{\text {eq }}=\frac{\mathrm{qE}}{\mathrm{k}}
$
Total energy $=\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2+\frac{1}{2} \mathrm{kx}_{\text {eq }}^2$
Total energy $=\frac{1}{2} m \omega^2 A^2+\frac{1}{2} \frac{q^2 E^2}{k}$