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Difficulty: MediumResononace, Vibrating Strings, and Air Columns in Pipes

A pipe of length 0.80 m0.80\text{ m} is closed at one end and open at the other. If the speed of sound in air is 320 m/s320\text{ m/s}, what is the frequency of its first overtone?

  1. A
    100 Hz100\text{ Hz}
  2. B
    200 Hz200\text{ Hz}
  3. 300 Hz300\text{ Hz}Answer
  4. D
    400 Hz400\text{ Hz}

Answer

The frequency of the first overtone is 300 Hz300\text{ Hz}.
For an air column closed at one end, standing wave resonance occurs only at odd harmonic frequencies given by fn=nv4Lf_n = \frac{n v}{4L} for n=1,3,5,…n = 1, 3, 5, \dots. The fundamental frequency (n=1n = 1) is f1=3204×0.80=100 Hzf_1 = \frac{320}{4 \times 0.80} = 100\text{ Hz}. The first overtone is the very next resonant mode, which corresponds to the third harmonic (n=3n = 3), giving f3=3×100 Hz=300 Hzf_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}.

Step-by-Step Solution

1
Calculate the fundamental frequency of the closed pipe
f1=v4L=3204×0.80=100 Hzf_1 = \frac{v}{4L} = \frac{320}{4 \times 0.80} = 100\text{ Hz}
For a pipe closed at one end, the fundamental wavelength is λ1=4L\lambda_1 = 4L.
2
Determine the harmonic number for the first overtone
First overtone = 3rd harmonic (f3=3f1f_3 = 3 f_1)
A pipe closed at one end produces only odd harmonics (fn=nf1f_n = n f_1 where n=1,3,5,…n = 1, 3, 5, \dots).
3
Compute the first overtone frequency
f3=3×100 Hz=300 Hzf_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}
Multiplying the fundamental frequency by 3 yields the first overtone frequency.

Key Concept

Resonance and Harmonics in Closed Air Columns
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Topics

Resononace, Vibrating Strings, and Air Columns in Pipes

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