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NEET PHYSICSWAVEMedium

Question

Each of the two strings of length 51.6 cm51.6 \text{ cm} and 49.1 cm49.1 \text{ cm} are tensioned separately by 20 N20 \text{ N} force. Mass per unit length of both the strings is same and equal to 1 g m11 \text{ g m}^{-1}. When both the strings vibrate simultaneously the number of beats is:

A

5

B

7

C

8

D

3

Step-by-Step Solution

  1. Identify the formula for fundamental frequency: The frequency of a vibrating string is given by f=12LTμf = \frac{1}{2L}\sqrt{\frac{T}{\mu}}, where LL is the length, TT is the tension, and μ\mu is the mass per unit length .
  2. Calculate the wave speed (vv): Given tension T=20 NT = 20 \text{ N} and linear mass density μ=1 g m1=103 kg m1\mu = 1 \text{ g m}^{-1} = 10^{-3} \text{ kg m}^{-1}. v=Tμ=20103=20000=1002141.42 m/sv = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = \sqrt{20000} = 100\sqrt{2} \approx 141.42 \text{ m/s}
  3. Calculate the frequencies of both strings:
  • For the first string (L1=51.6 cm=0.516 mL_1 = 51.6 \text{ cm} = 0.516 \text{ m}): f1=v2L1=141.422×0.516=141.421.032137.03 Hzf_1 = \frac{v}{2L_1} = \frac{141.42}{2 \times 0.516} = \frac{141.42}{1.032} \approx 137.03 \text{ Hz}
  • For the second string (L2=49.1 cm=0.491 mL_2 = 49.1 \text{ cm} = 0.491 \text{ m}): f2=v2L2=141.422×0.491=141.420.982144.01 Hzf_2 = \frac{v}{2L_2} = \frac{141.42}{2 \times 0.491} = \frac{141.42}{0.982} \approx 144.01 \text{ Hz}
  1. Calculate the beat frequency: The beat frequency is the difference between the two frequencies . fbeat=f2f1=144.01137.03=6.987f_{\text{beat}} = |f_2 - f_1| = |144.01 - 137.03| = 6.98 \approx 7

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. 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.

PHYSICSWAVEstringslengthtensionedseparatelylength

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