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Waves question

2023 · 12 Apr · Shift 1 · Q66
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Waves question

2023 · 12 Apr · Shift 1 · Q66

JEE MainPhysicsWavesNumerical+4 / −1
For a certain organ pipe, the first three resonance frequencies are in the ratio of 1:3:51:3:51:3:5 respectively. If the frequency of fifth harmonic is 405 Hz405 \mathrm{~Hz}405 Hz and the speed of sound in air is 324 ms−1324 \mathrm{~ms}^{-1}324 ms−1 the length of the organ pipe is ‾\underline{\hspace{2cm}}​m\mathrm{m}m.
Numerical answer
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Correct answer: 1

  1. Identify the type of organ pipe

    The first three resonance frequencies are in the ratio 1:3:51:3:51:3:5 This is the characteristic of a closed organ pipe (one end closed), because only odd harmonics are present: f1:f3:f5=1:3:5f_1:f_3:f_5 = 1:3:5f1​:f3​:f5​=1:3:5

  2. Use the frequency of the fifth harmonic

    For a closed pipe, allowed harmonics are: fn=nv4L,n=1,3,5,…f_n = \frac{n v}{4L}, \quad n=1,3,5,\dotsfn​=4Lnv​,n=1,3,5,…

    Given fifth harmonic frequency: f5=405 Hzf_5 = 405\,\text{Hz}f5​=405Hz

    So, 405=5v4L405 = \frac{5v}{4L}405=4L5v​

    Substitute v=324 m/sv = 324\,\text{m/s}v=324m/s: 405=5×3244L405 = \frac{5\times 324}{4L}405=4L5×324​

  3. Solve for LLL

    405=16204L405 = \frac{1620}{4L}405=4L1620​ 405=405L405 = \frac{405}{L}405=L405​

    Therefore, L=1 mL = 1\,\text{m}L=1m

  4. Final answer

    The length of the organ pipe is 1 m\boxed{1\,\text{m}}1m​

  5. Comparison with stored answer

    Stored correct answer = 111

    Our derived answer = 111

    Hence, the answer agrees with the stored correct answer.

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