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Question #42893

Difficulty: 8/10 | 🟢 Free

The Question

A simple pendulum consists of a small bob of mass mm and positive charge qq suspended by a massless, inextensible string of length LL. The point of suspension is at OO. The equilibrium position of the bob is at PP, where OP=LOP = L. A fixed point charge +Q+Q is placed at point SS, such that PP is the midpoint of the vertical line segment OSOS (with SS located directly below PP). The pendulum is displaced by a small angle θ\theta from its equilibrium position and released. Assuming mg>Qq2πϵ0L2mg > \frac{Qq}{2\pi\epsilon_0 L^2}, what is the angular frequency ω\omega of the resulting simple harmonic motion?

Options

  • A.

    gL−Qq2πϵ0mL3\sqrt{\frac{g}{L} - \frac{Qq}{2\pi\epsilon_0 m L^3}}

  • B.

    gL−Qq4πϵ0mL3\sqrt{\frac{g}{L} - \frac{Qq}{4\pi\epsilon_0 m L^3}}

  • C.

    gL+Qq4πϵ0mL3\sqrt{\frac{g}{L} + \frac{Qq}{4\pi\epsilon_0 m L^3}}

  • D.

    gL−Qq8πϵ0mL3\sqrt{\frac{g}{L} - \frac{Qq}{8\pi\epsilon_0 m L^3}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Coulomb's Law Vector Form|Gravitational Potential Energy (System)|Simple Pendulum Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of small-angle approximations in coupled force fields.

A common trap is to assume the electrostatic force acts purely vertically with a constant magnitude Fe=Qq4πϵ0L2F_e = \frac{Qq}{4\pi\epsilon_0 L^2} at all small angles, leading to an incorrect effective gravity geff=g−Femg_{eff} = g - \frac{F_e}{m} (Option B).

However, as the bob moves, the distance vector r⃗\vec{r} changes both direction and magnitude.

The tangential restoring force receives equal linear contributions from both the xx-component of the Coulomb force and the projection of its yy-component onto the tangential axis, introducing a critical factor of 2.

Students must either perform a proper Taylor expansion of the potential energy or rigorously linearize the force vector to avoid this trap.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectric Charges and Fields - Class 12
ConceptCoulomb's Law Vector Form
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree