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PhysicsSHMEnergyHard2 minPYQ_2018

PhysicsHardsingle choice

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?

Options:

Answer:

Solution:

The equilibrium position will shift to a point where the resultant force is equal to zero. $k x_{e q} = q E \Rightarrow x_{e q} = \frac{q E}{k}$ . Energy $= \frac{1}{2} m ω^{2} [A^{2} + {(\frac{q E}{k})}^{2}] = \frac{1}{2} m ω^{2} A^{2} + \frac{1}{2} \frac{q^{2} E^{2}}{k}$ .

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: PYQ_2018

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