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Question #42643

Difficulty: 8/10 | 🟢 Free

The Question

Two transverse waves propagate on a taut string of linear mass density μ=0.01 kg/m\mu = 0.01 \text{ kg/m} under a tension T=100 NT = 100 \text{ N}. The individual wave displacements are given by y1(x,t)=Asin⁡(kx−ωt)y_1(x,t) = A \sin(kx - \omega t) and y2(x,t)=Asin⁡(kx+ωt+ϕ)y_2(x,t) = A \sin(kx + \omega t + \phi). The resultant displacement of the string is y=y1+y2y = y_1 + y_2. At x=0x = 0 and t=0t = 0, the instantaneous spatial slope of the string is ∂y∂x=0.03\frac{\partial y}{\partial x} = 0.03, and the phase constant is ϕ=π3\phi = \frac{\pi}{3}. What is the magnitude of the transverse particle velocity of the string at x=0x = 0 and t=0t = 0?

Options

  • A.

    1 m/s1 \text{ m/s}

  • B.

    2 m/s2 \text{ m/s}

  • C.

    3 m/s3 \text{ m/s}

  • D.

    0 m/s0 \text{ m/s}

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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

Particle Velocity vs. Wave Velocity|Phase Constant Determination|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the limits of the traveling wave relation vp=−v∂y∂xv_p = -v \frac{\partial y}{\partial x}.

Students often incorrectly apply this formula to the resultant profile of superposed waves traveling in opposite directions, which would yield the trap answer of 3 m/s3 \text{ m/s}.

By introducing a phase constant ϕ\phi, the problem forces the student to manually apply the principle of superposition to the partial derivatives (particle velocity and spatial slope) rather than blindly multiplying the resultant slope by the wave speed.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterWaves - Class 11 Part 2
ConceptParticle Velocity vs. Wave Velocity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree