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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?
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: $\frac{{2qE}}{k}$ from $x = 0$
The new equilibrium position is at a distance: $\frac{{qE}}{{2k}}$ from $x = 0$
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 shifto point where resultant force $=0$
$k{x_{eq}} = {\text{qE}} \Rightarrow {{\text{x}}_{{\text{eq}}}} = \frac{{{\text{qE}}}}{{\text{k}}}$
Total energy $=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{A}^{2}+\frac{1}{2} \mathrm{k} \mathrm{x}_{\mathrm{eq}}^{2}$
Total energy $=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{A}^{2}+\frac{1}{2} \frac{\mathrm{q}^{2} \mathrm{E}^{2}}{\mathrm{k}}$
