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Question #42628

Difficulty: 8/10 | 🟢 Free

The Question

A block of mass m (kg)m \text{ (kg)} is attached to a spring of constant k (N/m)k \text{ (N/m)} and oscillates horizontally in a thermally insulated container filled with a viscous liquid. The total heat capacity of the block and the liquid is C (J/K)C \text{ (J/K)}. The damping force on the block is F=−bvF = -b v, where v (m/s)v \text{ (m/s)} is the velocity. The damping coefficient b (kg/s)b \text{ (kg/s)} depends on the temperature increase ΔT (K)\Delta T \text{ (K)} of the system as b=b0(1+αΔT)b = b_0 (1 + \alpha \Delta T), where b0 (kg/s)b_0 \text{ (kg/s)} and α (K−1)\alpha \text{ (K}^{-1}\text{)} are positive constants. Assume all dissipated mechanical energy is entirely converted into internal energy, raising the system's temperature uniformly. The initial mechanical energy is E0 (J)E_0 \text{ (J)}. Given that αE0=C\alpha E_0 = C, find the time t (s)t \text{ (s)} taken for the mechanical energy to reduce to E0/2E_0 / 2. Assume the oscillator is lightly damped such that the average rate of energy loss over a cycle is ⟨dEdt⟩=−bmE\langle \frac{dE}{dt} \rangle = -\frac{b}{m} E.

Options

  • A.

    m2b0ln⁡3\frac{m}{2b_0} \ln 3

  • B.

    mb0ln⁡2\frac{m}{b_0} \ln 2

  • C.

    m2b0ln⁡2\frac{m}{2b_0} \ln 2

  • D.

    mb0ln⁡3\frac{m}{b_0} \ln 3

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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

Damped Oscillation Equation|Calorimetry with Heat Loss|Approximations (Small Angle)

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize that the damping coefficient is not a static parameter but a state-dependent function coupled to the system's mechanical energy via thermodynamic dissipation.

The lethal trap is assuming a constant damping coefficient b0b_0 and applying the standard exponential decay formula E=E0e−(b0/m)tE = E_0 e^{-(b_0/m)t}, which leads to the incorrect answer (m/b0)ln⁡2(m/b_0) \ln 2.

By forcing bb to depend on temperature, which in turn depends on the dissipated energy E0−EE_0 - E, the problem creates a non-autonomous differential equation.

Students must synthesize calorimetry (CΔT=E0−EC \Delta T = E_0 - E) with the cycle-averaged power dissipation equation to correctly integrate the non-linear energy decay.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterOscillations - Class 11 Part 2
ConceptDamped Oscillation Equation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree