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Question #42880

Difficulty: 8/10 | 🟢 Free

The Question

A uniform string of length $L$ (m) and linear mass density $\mu$ (kg/m) is stretched with a constant tension $T$ (N) along the $x$ -axis from $x=0$ to $x=L$ . The end at $x=0$ is rigidly fixed. The end at $x=L$ is attached to a solid block of mass $M$ (kg) and uniform cross-sectional area $S$ (m $^2$ ). The block floats in equilibrium in a liquid of density $\rho_L$ (kg/m $^3$ ), constrained to move only in the vertical $y$ -direction. The system is set into small transverse vertical oscillations. The restoring force on the block is purely due to buoyancy and the string's tension. It is observed that the fundamental angular frequency $\omega$ (rad/s) of this coupled system is exactly equal to the natural angular frequency of the isolated block oscillating vertically in the liquid. Which of the following relations must hold true? (Assume $g$ is the acceleration due to gravity in m/s $^2$ )

Options

A.
$\rho_L S g = \frac{\pi^2 M T}{4 L^2 \mu}$
B.
$\rho_L S g = \frac{\pi^2 M T}{L^2 \mu}$
C.
$\rho_L S g = \frac{4 \pi^2 M T}{L^2 \mu}$
D.
$\rho_L S g = \frac{M T}{L^2 \mu}$

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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

Energy Distribution in Standing Waves|Potential Energy Reference|Vertical Oscillations

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle dynamic, time-dependent boundary conditions (mobile constraints) in wave mechanics, rather than defaulting to stationary nodes.

The lethal trap lies in assuming that a massive block at the boundary automatically enforces a fixed node ( $qL = \pi$ ), which would lead to Option B.

By applying D'Alembert's principle and coupling the string's energy with the potential well of the buoyant mass, students must derive a transcendental equation for the allowed frequencies.

The specific resonance condition provided perfectly cancels the inertial and potential terms of the boundary mass, forcing the boundary to behave counter-intuitively as a perfect antinode ( $\cos(qL)=0$ ).

Failing to construct the correct differential equation for the boundary force balance is the primary failure point.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Waves - Class 11 Part 2
Concept	Energy Distribution in Standing Waves
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free