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Question #43545
The Question
A rigid, thin rod of length is suspended from a frictionless pivot at one of its ends (). The linear mass density of the rod varies along its length as , where is a positive constant and is the distance from the pivot. If the rod is displaced by a small angle from the vertical and released, what is the time period of its resulting simple harmonic motion? (Assume acceleration due to gravity is and air resistance is negligible).
Options
- A.
- B.
- C.
- D.
Check Your Answer
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
SHM Definition via Differential Equation|Physical Pendulum Inertia
Why This Question Is Tricky
Common Mistakes
This question tests the rigorous derivation of the SHM differential equation for a physical pendulum, specifically targeting the trap of misapplying standard uniform-body formulas.
Students often memorize the physical pendulum formula but fail to correctly evaluate (the distance to the center of mass) for a non-uniform density distribution.
A common mistake is calculating the correct moment of inertia via integration but incorrectly assuming the center of mass remains at (yielding Option A), or calculating the correct COM but using the standard uniform rod inertia (yielding Option D).
By introducing a power-law density distribution, the problem forces students to independently integrate both the restoring torque's lever arm and the rotational inertia, filtering out formula-memorizers.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Oscillations - Class 11 Part 2 |
| Concept | SHM Definition via Differential Equation |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |