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JEE Main 201815 April
Exam
JEE Main
Year
2018
Date
15 April
Subject
Physics
Chapter
Oscillations

JEE Main 2018 · 15 April · Physics · Oscillations

A body of mass M and charge q is connected to a spring of spring...

A body of mass MM and charge qq is connected to a spring of spring constant kk. It is oscillating along xx-direction about its equilibrium position in the horizontal plane, taken to be at x=0x = 0, with an amplitude AA. An electric field EE is applied along the xx-direction. Which of the following statements is correct?

Options

  1. AThe total energy of the system is 12mω2A2+12q2E2k\dfrac{1}{2}m{\omega}^{2}{A}^{2} + \dfrac{1}{2}\dfrac{{q}^{2}{E}^{2}}{k}
  2. BThe new equilibrium position is at a distance 2qEk\dfrac{2qE}{k} from x=0x = 0
  3. CThe new equilibrium position is at a distance qE2k\dfrac{qE}{2k} from x=0x = 0
  4. DThe total energy of the system is 12mω2A2−12q2E2k\dfrac{1}{2}m{\omega}^{2}{A}^{2} - \dfrac{1}{2}\dfrac{{q}^{2}{E}^{2}}{k} .

Correct answer

A. The total energy of the system is 12mω2A2+12q2E2k\dfrac{1}{2}m{\omega}^{2}{A}^{2} + \dfrac{1}{2}\dfrac{{q}^{2}{E}^{2}}{k}

Step-by-step solution

The equilibrium position will shift to a point where the resultant force is equal to zero. kxeq=qE⇒xeq=qEkk{x}_{eq} = qE ⇒ {x}_{eq} = \dfrac{qE}{k}. Energy =12mω2[A2+(qEk)2]=12mω2A2+12q2E2k= \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}.
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