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

In a resonance tube experiment using a tuning fork of frequency 340 Hz340\text{ Hz}, the first two consecutive resonance lengths of the air column closed at one end are 22 cm22\text{ cm} and 72 cm72\text{ cm}. If a pipe open at both ends with a physical length of 28 cm28\text{ cm} is operated in the same environment, what is its fundamental frequency when end corrections at both open ends are taken into account?

  1. 500 Hz500\text{ Hz}Answer
  2. B
    250 Hz250\text{ Hz}
  3. C
    680 Hz680\text{ Hz}
  4. D
    1000 Hz1000\text{ Hz}

Answer

The fundamental frequency of the open pipe is 500 Hz500\text{ Hz}.
The difference between successive resonant lengths in the closed tube gives half a wavelength (L2−L1=0.50 m  ⟹  λ=1.00 mL_2 - L_1 = 0.50\text{ m} \implies \lambda = 1.00\text{ m}). Using the tuning fork frequency 340 Hz340\text{ Hz}, the speed of sound is 340 m/s340\text{ m/s}. The end correction is e=λ/4−L1=0.25 m−0.22 m=0.03 me = \lambda / 4 - L_1 = 0.25\text{ m} - 0.22\text{ m} = 0.03\text{ m}. For a pipe open at both ends, end corrections apply at both openings, making the effective length Leff=0.28 m+2(0.03 m)=0.34 mL_{\text{eff}} = 0.28\text{ m} + 2(0.03\text{ m}) = 0.34\text{ m}. Its fundamental frequency is f0=v/(2Leff)=340/(2×0.34)=500 Hzf_0 = v / (2 L_{\text{eff}}) = 340 / (2 \times 0.34) = 500\text{ Hz}.

Step-by-Step Solution

1
Determine the wavelength and speed of sound from the resonance tube data.
λ=2(L2−L1)=2(0.72 m−0.22 m)=1.00 m\lambda = 2(L_2 - L_1) = 2(0.72\text{ m} - 0.22\text{ m}) = 1.00\text{ m}. Speed of sound v=fλ=340 Hz×1.00 m=340 m/sv = f \lambda = 340\text{ Hz} \times 1.00\text{ m} = 340\text{ m/s}.
The distance between consecutive resonance positions in a closed pipe is equal to half a wavelength.
2
Calculate the end correction ee of the tube.
L1+e=λ4  ⟹  0.22 m+e=0.25 m  ⟹  e=0.03 m=3 cmL_1 + e = \frac{\lambda}{4} \implies 0.22\text{ m} + e = 0.25\text{ m} \implies e = 0.03\text{ m} = 3\text{ cm}.
The first resonance of a pipe closed at one end occurs when the effective length equals one quarter of a wavelength.
3
Calculate the effective length LeffL_{\text{eff}} of the open pipe.
Leff=L+2e=28 cm+2(3 cm)=34 cm=0.34 mL_{\text{eff}} = L + 2e = 28\text{ cm} + 2(3\text{ cm}) = 34\text{ cm} = 0.34\text{ m}.
A pipe open at both ends requires an end-correction term added at each open boundary.
4
Calculate the fundamental frequency of the open pipe.
f0=v2Leff=340 m/s2×0.34 m=500 Hzf_0 = \frac{v}{2 L_{\text{eff}}} = \frac{340\text{ m/s}}{2 \times 0.34\text{ m}} = 500\text{ Hz}.
The fundamental wavelength of a pipe open at both ends is twice its effective length.

Key Concept

Resonance tube end correction and boundary conditions of open vs closed pipes
Estimated Time:2m 0s
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Topics

Resononace, Vibrating Strings, and Air Columns in Pipes

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