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Question #43116

Difficulty: 8/10 | 🟢 Free

The Question

A small bob of mass $m$ and charge $q$ is suspended by a massless string of length $L$ in a uniform downward gravitational field $g$ . The bob is subjected to a weak viscous drag force $\vec{F}_d = -b\vec{v}$ , where $\vec{v}$ is its velocity and $b$ is a positive constant. The bob is displaced by a small angle $\theta_0 \ll 1 \text{ rad}$ and released from rest. As the bob oscillates, it emits electromagnetic radiation due to its acceleration. Assuming non-relativistic motion and that the instantaneous power radiated is given by the Larmor formula $P = \frac{q^2 a^2}{6\pi\epsilon_0 c^3}$ (where $a$ is the acceleration, $c$ is the speed of light, and $\epsilon_0$ is the vacuum permittivity), find the time $t$ after which the amplitude of oscillation reduces to $\theta_0 / e$ . (Assume the energy loss per cycle is very small compared to the total energy).

Options

A.
$\frac{12\pi\epsilon_0 m c^3 L}{6\pi\epsilon_0 b c^3 L + q^2 g}$
B.
$\frac{6\pi\epsilon_0 m c^3 L}{6\pi\epsilon_0 b c^3 L + q^2 g}$
C.
$\frac{12\pi\epsilon_0 m c^3 L}{6\pi\epsilon_0 b c^3 L - q^2 g}$
D.
$\frac{12\pi\epsilon_0 m c^3 L^2}{6\pi\epsilon_0 b c^3 L^2 + q^2 g}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Source of EM Waves|Simple Pendulum Small Angle Approximation|Calculus (Differential Equations)

Why This Question Is Tricky

Common Mistakes

The primary trap tests the student's ability to couple non-conservative mechanical forces with radiation reaction forces by averaging power over an oscillation cycle.

A common misconception is equating the energy decay time constant directly to the amplitude decay time constant, leading to a missing factor of 2 (Option B).

Furthermore, students often misapply the Larmor formula by using the constant gravitational acceleration $g$ instead of the time-varying tangential acceleration $a = -\omega^2 x$ , or they fail to recognize that $\langle a^2 \rangle = \omega^2 \langle v^2 \rangle$ for a harmonic oscillator, which is essential to express the radiation power loss purely in terms of the instantaneous mechanical energy $E$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Waves - Class 12
Concept	Source of EM Waves
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free