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Question #43855

Difficulty: 8/10 | 🟢 Free

The Question

A uniform rigid rod of mass $m$ and length $L$ is suspended from a frictionless hinge attached to the ceiling of a train compartment. The train moves along a straight horizontal track with a constant acceleration $a = \sqrt{3}g$ , where $g$ is the acceleration due to gravity. The rod is allowed to reach its stable equilibrium position relative to the train. If the rod is given a small angular displacement in the plane of acceleration, what is the time period $T$ of its small oscillations?

Options

A.
$2\pi \sqrt{\frac{L}{2g}}$
B.
$2\pi \sqrt{\frac{2L}{3g}}$
C.
$2\pi \sqrt{\frac{L}{3g}}$
D.
$2\pi \sqrt{\frac{L}{6g}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Physical Pendulum Inertia|Frame of Reference Dependency|Pseudo-Force

Why This Question Is Tricky

Common Mistakes

This question tests the application of non-inertial frame mechanics to an extended rigid body.

Students often fall into the trap of treating the physical pendulum as a simple point-mass pendulum, incorrectly applying the shortcut formula $T = 2\pi\sqrt{L/g_{\text{eff}}}$ .

Another lethal collision occurs when students correctly identify $g_{\text{eff}}$ but fail to realize that the fictitious force and gravity both act at the center of mass ( $L/2$ ), not at the end of the rod, leading to an incorrect restoring torque calculation.

By forcing the torque balance about the pivot, it differentiates rote memorization from a rigorous understanding of rotational dynamics.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Oscillations - Class 11 Part 2
Concept	Physical Pendulum Inertia
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free