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Waves question

2006 · Shift 0 · Q113
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Waves question

2006 · Shift 0 · Q113

JEE MainPhysicsWavesMCQ+4 / −1
A string is stretched between fixed points separated by 75.0cm.75.0cm.75.0cm. It is observed to have resonant frequencies of 420Hz420Hz420Hz and 315Hz315Hz315Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
  1. A
    105Hz105Hz105Hz
  2. B
    1.05Hz1.05Hz1.05Hz
  3. C
    1050Hz1050Hz1050Hz
  4. D
    10.5Hz10.5Hz10.5Hz
View written solutionFree

Correct answer: A

  1. Resonant frequencies of a string fixed at both ends

For a string fixed at both ends, allowed resonant frequencies are:

fn=nf1f_n = n f_1 fn​=nf1​

where:

  • n=1,2,3,…n = 1,2,3,\dotsn=1,2,3,…
  • f1f_1f1​ is the fundamental (lowest resonant frequency).

So all resonant frequencies are equally spaced by f1f_1f1​.


  1. Use the given frequencies

Two resonant frequencies are given:

420 Hz,315 Hz420\,\text{Hz}, \quad 315\,\text{Hz}420Hz,315Hz

And it is stated that there are no other resonant frequencies between them.

This means these two frequencies must be consecutive harmonics.

Hence their difference equals the fundamental frequency:

f1=420−315=105 Hzf_1 = 420 - 315 = 105\,\text{Hz}f1​=420−315=105Hz
  1. Check

If f1=105 Hzf_1 = 105\,\text{Hz}f1​=105Hz, then:

315=3×105315 = 3 \times 105315=3×105 420=4×105420 = 4 \times 105420=4×105

So these are indeed consecutive harmonics (3rd3^{\text{rd}}3rd and 4th4^{\text{th}}4th), with no resonance in between.


  1. Final answer

The lowest resonant frequency is:

105 Hz\boxed{105\,\text{Hz}}105Hz​

So the correct option is A.


  1. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

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