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Question #43543

Difficulty: 8/10 | 🟢 Free

The Question

A smooth, straight horizontal tube is fixed to a vertical axis and rotates with a constant angular velocity $\omega$ . Inside the tube, a light uniform rod of natural length $L_0$ (at temperature $T_0$ ), cross-sectional area $A$ , Young's modulus $Y$ , and coefficient of linear thermal expansion $\alpha$ has one end attached to the axis of rotation. A small block of mass $m$ is attached to the other end of the rod. An ideal spring of force constant $k$ connects the block to the outer closed end of the tube. The temperature of the rod is uniformly increased to $T_0 + \Delta T$ . Assuming the block remains in stable equilibrium, what is the angular frequency $\Omega$ of small radial oscillations of the block about its new equilibrium position? (Assume the tube and spring do not undergo thermal expansion).

Options

A.
$\sqrt{\frac{k}{m} + \frac{YA}{m L_0(1+\alpha\Delta T)} - \omega^2}$
B.
$\sqrt{\frac{k}{m} + \frac{YA}{m L_0}(1+\alpha\Delta T) - \omega^2}$
C.
$\sqrt{\frac{k}{m} + \frac{YA}{m L_0(1+\alpha\Delta T)} + \omega^2}$
D.
$\sqrt{\frac{k}{m} + \frac{YA}{m L_0} - \omega^2}$

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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

Thermal Stress in Constrained Rods|Effective Spring Constant|SHM Definition via Differential Equation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to couple thermal expansion, elasticity, and simple harmonic motion within a non-inertial reference frame.

The primary trap lies in the misapplication of the restoring force constant.

Students often assume that the stiffness of the rod remains $YA/L_0$ or incorrectly add the thermal stress as a direct stiffness multiplier (Option B).

However, the true stiffness is the derivative of the stress with respect to displacement, which depends inversely on the new temperature-adjusted natural length $L_T = L_0(1+\alpha\Delta T)$ .

Furthermore, students must correctly account for the centrifugal softening effect ( $-m\omega^2$ ) in the rotating frame, which reduces the effective restoring force.

Failing to re-evaluate the potential energy's second derivative at the shifted equilibrium leads to incorrect linearizations.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermal Properties of Matter - Class 11 Part 2
Concept	Thermal Stress in Constrained Rods
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free