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Question #42856

Difficulty: 8/10 | 🟢 Free

The Question

A transverse wave propagates along a long stretched string of linear mass density μ\mu and tension TT. Due to a viscous surrounding medium, the wave amplitude attenuates exponentially. The transverse displacement of the string is given by: y(x,t)=A0e−γxsin⁡(ωt−kx)y(x,t) = A_0 e^{-\gamma x} \sin(\omega t - kx) where A0A_0, γ\gamma, ω\omega, and kk are positive constants. Which of the following expressions correctly represents the time-averaged total mechanical energy density (energy per unit length) ⟨u⟩\langle u \rangle of the wave at a position xx? Assume all quantities are expressed in coherent SI units.

Options

  • A.

    14A02e−2γx[μω2+T(γ2+k2)]\frac{1}{4} A_0^2 e^{-2\gamma x} \left[ \mu \omega^2 + T(\gamma^2 + k^2) \right]

  • B.

    14A02e−2γx[μω2+Tk2]\frac{1}{4} A_0^2 e^{-2\gamma x} \left[ \mu \omega^2 + T k^2 \right]

  • C.

    12μω2A02e−2γx\frac{1}{2} \mu \omega^2 A_0^2 e^{-2\gamma x}

  • D.

    12A02e−2γx[μω2+T(γ2+k2)]\frac{1}{2} A_0^2 e^{-2\gamma x} \left[ \mu \omega^2 + T(\gamma^2 + k^2) \right]

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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 Density in Transverse Waves|Damped Oscillation Equation|Mechanical Energy Conservation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to derive wave energy from first principles rather than relying on memorized formulas for undamped waves.

The primary trap is assuming the potential energy density depends only on the wave number kk (i.e., ∂y∂x≈−kAcos⁡(ωt−kx)\frac{\partial y}{\partial x} \approx -k A \cos(\omega t - kx)), which leads to the incorrect option B.

Students often forget that spatial attenuation introduces an additional term via the product rule when calculating the spatial derivative ∂y∂x\frac{\partial y}{\partial x}.

This cross-term significantly alters the potential energy density in highly dissipative media, making the standard u=12μω2A2u = \frac{1}{2}\mu\omega^2 A^2 approximation invalid.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWaves - Class 11 Part 2
ConceptEnergy Density in Transverse Waves
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree