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?

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Verified Solution

Answer & Solution

Correct Answer

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

Detailed explanation

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 \(=\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{A}^{2}+\frac{1}{2} \frac{\mathrm{q}^{2} \mathrm{E}^{2}}{\mathrm{k}}\)

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