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Question #42873
The Question
A long, taut string of linear mass density and tension lies along the -axis. A small bead of mass is fixed to the string at the origin (). Two harmonic waves are incident from the negative -direction, given by the superposition , where and are the respective wave numbers. The angular frequencies are and , where is the wave speed. The waves are partially transmitted into the region . The transmitted wave can be written as , where and are the phase constants introduced at the boundary. Assuming the phase constants are chosen in the principal range , what is the magnitude of the relative phase shift between the two transmitted components?
Options
- A.
- B.
0
- C.
- D.
Check Your Answer
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
Phase Constant (Epoch)|Superposition of Non-identical Waves|Medium Dependency
Why This Question Is Tricky
Common Mistakes
This question tests the student's understanding of wave transmission across a reactive boundary, which acts as a dispersive phase shifter.
The lethal trap lies in assuming that the phase constant (epoch) is an absolute, medium-independent property or that transmission always occurs with zero phase shift (as it does between two simple ideal strings).
Students often misapply the superposition principle by carrying over the initial phase constants directly to the transmitted waves, ignoring the boundary condition imposed by Newton's Second Law on the mass .
This boundary condition forces a complex transmission coefficient, resulting in a frequency-dependent phase jump.
Failing to derive this dynamic phase shift leads to the incorrect assumption that the relative phase difference remains zero.
Quick Facts About This Question
Question Information
| Subject | Physics |
| Chapter | Waves - Class 11 Part 2 |
| Concept | Phase Constant (Epoch) |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |