JEE Main 2018PhysicsOscillationsHardMCQ

JEE Main 2018Oscillations Question with Solution

JEE Main 2018 (15 Apr)

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 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?

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Show full solutionCorrect option: A
Correct answer
AThe total energy of the system is 12mω2A2+12q2E2k

Step-by-step explanation

The equilibrium position will shift to a point where the resultant force is equal to zero. kxeq=qE  xeq=qEk. Energy =12mω2A2+qEk2=12mω2A2+12q2E2k.

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About this question

This is a previous-year question from JEE Main 2018, covering the Oscillations chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.