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Question #42853

Difficulty: 8/10 | 🟢 Free

The Question

A massless disk of radius R=0.5 mR = 0.5 \text{ m} is pivoted at its center and can rotate freely in a horizontal plane. Two identical long strings, each of linear mass density μ=0.1 kg/m\mu = 0.1 \text{ kg/m} and under tension T=40 NT = 40 \text{ N}, are attached to a pin on the rim of the disk. String 1 extends along the negative x-axis (x<0x < 0) and String 2 extends along the positive x-axis (x>0x > 0). The transverse displacement of the strings at x=0x=0 is constrained to y(0,t)=Rθ(t)y(0,t) = R\theta(t), where θ(t)\theta(t) is the angular displacement of the disk. The disk is subjected to a viscous rotational damping torque τd=−bω\tau_d = -b \omega, where b=1.0 N m sb = 1.0 \text{ N m s} and ω=θ˙\omega = \dot{\theta}. Two rectangular wave pulses are sent towards the disk from opposite sides. The incident pulses are characterized by their constant transverse velocities vy1=∂yi1∂t=0.4 m/sv_{y1} = \frac{\partial y_{i1}}{\partial t} = 0.4 \text{ m/s} (for Pulse 1 on String 1) and vy2=∂yi2∂t=0.2 m/sv_{y2} = \frac{\partial y_{i2}}{\partial t} = 0.2 \text{ m/s} (for Pulse 2 on String 2). Both pulses last for a duration t0=0.5 st_0 = 0.5 \text{ s} and are timed such that they arrive at the disk simultaneously, overlapping completely. Calculate the total energy dissipated by the damping torque during this interaction.

Options

  • A.

    0.18 J0.18 \text{ J}

  • B.

    0.10 J0.10 \text{ J}

  • C.

    0.20 J0.20 \text{ J}

  • D.

    0.02 J0.02 \text{ J}

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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

Superposition of Non-identical Waves|Work-Energy Theorem for Rotation

Why This Question Is Tricky

Common Mistakes

Students incorrectly apply the Superposition Principle directly to energy, assuming the total energy dissipated is the linear sum of the energies dissipated by each pulse individually (Ed1+Ed2=0.10 JE_{d1} + E_{d2} = 0.10 \text{ J}).

This misapplies the Work-Energy Theorem for non-conservative forces.

While wave displacements and velocities superpose linearly, the power dissipated by a viscous damping force depends on the square of the superposed velocity (P∝v2P \propto v^2).

The cross-term in the squared velocity means the non-conservative work done during simultaneous overlap is significantly different from the sum of individual works.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWaves - Class 11 Part 2
ConceptSuperposition of Non-identical Waves
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree