The equilibrium position will shift to a point where the resultant force is equal to zero. kxeq=qE⇒xeq=kqE. Energy =21mω2[A2+(kqE)2]=21mω2A2+21kq2E2.
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?
Held on 15 Apr 2018 · Verified 6 Jul 2026.
The total energy of the system is 21mω2A2+21kq2E2
The new equilibrium position is at a distance k2qE from x=0
The new equilibrium position is at a distance 2kqE from x=0
The total energy of the system is 21mω2A2−21kq2E2 .
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Work through every JEE Main Waves & Oscillations PYQ, year by year.