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Waves question

2023 · 8 Apr · Shift 1 · Q67
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Waves question

2023 · 8 Apr · Shift 1 · Q67

JEE MainPhysicsWavesNumerical+4 / −1
An organ pipe 40 cm40 \mathrm{~cm}40 cm long is open at both ends. The speed of sound in air is 360 ms−1360 \mathrm{~ms}^{-1}360 ms−1. The frequency of the second harmonic is ‾Hz\underline{\hspace{2cm}}\mathrm{Hz}​Hz.
Numerical answer
View written solutionFree

Correct answer: 900

  1. Identify the type of pipe and harmonic formula

For an organ pipe open at both ends, all harmonics are present, and the frequency of the nnnth harmonic is

fn=nv2Lf_n = \frac{n v}{2L}fn​=2Lnv​

where:

  • v=360 m s−1v = 360\,\text{m s}^{-1}v=360m s−1
  • L=40 cm=0.40 mL = 40\,\text{cm} = 0.40\,\text{m}L=40cm=0.40m
  • For the second harmonic, n=2n=2n=2
  1. Substitute into the formula
f2=2×3602×0.40f_2 = \frac{2 \times 360}{2 \times 0.40}f2​=2×0.402×360​
  1. Simplify
f2=3600.40=900 Hzf_2 = \frac{360}{0.40} = 900\,\text{Hz}f2​=0.40360​=900Hz
  1. Final answer

The frequency of the second harmonic is

900 Hz\boxed{900\,\text{Hz}}900Hz​
  1. Comparison with stored answer

Stored correct answer: 900900900

My derived answer is also 900900900, so they match.

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