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NEET PHYSICSWAVE OPTICSMedium

Question

Two identical piano wires kept under the same tension TT have a fundamental frequency of 600 Hz600 \text{ Hz}. The fractional increase in the tension of one of the wires which will lead to the occurrence of 6 beats/s6 \text{ beats/s} when both the wires oscillate together would be:

A

0.020.02

B

0.030.03

C

0.040.04

D

0.010.01

Step-by-Step Solution

  1. Identify the relation between frequency and tension: The fundamental frequency ff of a stretched wire is given by f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, which means fTf \propto \sqrt{T} .
  2. Determine the fractional change relation: By taking the natural logarithm on both sides and differentiating, we get the relationship for small changes: Δff=12ΔTT\frac{\Delta f}{f} = \frac{1}{2} \frac{\Delta T}{T}.
  3. Calculate the fractional change in frequency: The beat frequency is the difference between the new and old frequencies, so Δf=6 Hz\Delta f = 6 \text{ Hz} . The original frequency is f=600 Hzf = 600 \text{ Hz}. Therefore, the fractional change in frequency is Δff=6600=0.01\frac{\Delta f}{f} = \frac{6}{600} = 0.01.
  4. Calculate the fractional increase in tension: Rearranging the relation from step 2, the fractional change in tension is ΔTT=2×Δff\frac{\Delta T}{T} = 2 \times \frac{\Delta f}{f}. Substitute the value: ΔTT=2×0.01=0.02\frac{\Delta T}{T} = 2 \times 0.01 = 0.02. The fractional increase in tension is 0.020.02.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSidenticaltensionfundamentalfrequencyfractional

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