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Oscillations

JEE Main level, single correct, medium

96 of 194

Question 96 of 194

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?

Show answer and solution

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

The equilibrium position will shift to a point where the resultant force is equal to zero.

k{x}_{eq}=qE\Rightarrow {x}_{eq}=\dfrac{qE}{k}

. Energy

=\dfrac{1}{2}m{\omega }^{2}\left[{A}^{2}+{\left(\dfrac{qE}{k}\right)}^{2}\right]=\dfrac{1}{2}m{\omega }^{2}{A}^{2}+\dfrac{1}{2}\dfrac{{q}^{2}{E}^{2}}{k}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Oscillations (Energy of Simple Harmonic Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.