Question:
What should be the length of a closed pipe to produce resonance with sound wave of wavelength 62 cm, in fundamental mode? (Neglect end correction)
What should be the length of a closed pipe to produce resonance with sound wave of wavelength 62 cm, in fundamental mode? (Neglect end correction)
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Closed pipes support only odd harmonics; fundamental length is one-fourth of wavelength.
Updated On: Feb 18, 2026
- \(31 \, \text{cm}\)
- \(15.5 \, \text{cm}\)
- \(20.6 \, \text{cm}\)
- \(46.5 \, \text{cm}\)
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The Correct Option is B
Solution and Explanation
Step 1: Condition for fundamental mode in closed pipe.
For a closed pipe, the fundamental mode occurs when \[ L = \frac{\lambda}{4}. \]
Step 2: Substituting wavelength.
Given \( \lambda = 62 \, \text{cm} \), \[ L = \frac{62}{4} = 15.5 \, \text{cm}. \]
Step 3: Conclusion.
The required length of the closed pipe is \(15.5 \, \text{cm}\).
For a closed pipe, the fundamental mode occurs when \[ L = \frac{\lambda}{4}. \]
Step 2: Substituting wavelength.
Given \( \lambda = 62 \, \text{cm} \), \[ L = \frac{62}{4} = 15.5 \, \text{cm}. \]
Step 3: Conclusion.
The required length of the closed pipe is \(15.5 \, \text{cm}\).
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