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NEET PHYSICSWAVE OPTICSMedium

Question

A string is stretched between fixed points separated by 75.0 cm75.0 \text{ cm}. It is observed to have resonant frequencies of 420 Hz420 \text{ Hz} and 315 Hz315 \text{ Hz}. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:

A

155 Hz155 \text{ Hz}

B

205 Hz205 \text{ Hz}

C

10.5 Hz10.5 \text{ Hz}

D

105 Hz105 \text{ Hz}

Step-by-Step Solution

  1. Identify Resonant Frequencies for a Fixed String: For a string fixed at both ends, the resonant frequencies are integer multiples (harmonics) of the fundamental frequency, given by fn=nf1f_n = n f_1, where n=1,2,3,n = 1, 2, 3, \dots and f1f_1 is the fundamental (lowest) resonant frequency .
  2. Set up the Equations: The problem states there are no other resonant frequencies between 315 Hz315 \text{ Hz} and 420 Hz420 \text{ Hz}, which means they are consecutive harmonics. Let them be the nn-th and (n+1)(n+1)-th harmonics: fn=nf1=315 Hzf_n = n f_1 = 315 \text{ Hz} fn+1=(n+1)f1=420 Hzf_{n+1} = (n+1) f_1 = 420 \text{ Hz}
  3. Calculate the Fundamental Frequency: The difference between any two consecutive harmonic frequencies of a string fixed at both ends is equal to the fundamental frequency: fn+1fn=(n+1)f1nf1=f1f_{n+1} - f_n = (n+1)f_1 - n f_1 = f_1 f1=420 Hz315 Hz=105 Hzf_1 = 420 \text{ Hz} - 315 \text{ Hz} = 105 \text{ Hz} Therefore, the lowest resonant frequency (fundamental frequency) is 105 Hz105 \text{ Hz}.

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 OPTICSstringstretchedbetweenpointsseparated

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