Question:
A pipe of length \(85\ \mathrm{cm}\) is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below \(1250\ \mathrm{Hz}\). The velocity of sound in air is \(340\ \mathrm{m/s}\).
A pipe of length \(85\ \mathrm{cm}\) is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below \(1250\ \mathrm{Hz}\). The velocity of sound in air is \(340\ \mathrm{m/s}\).
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A closed pipe supports only odd harmonics \(f=(2k+1)\frac{v}{4L}\); count those below 1250 Hz.
Updated On: Nov 11, 2026
- 12
- 8
- 6
- 4
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The Correct Option is C
Solution and Explanation
Fundamental of a closed pipe: \(f_1=\frac{v}{4L}=\frac{340}{4\times0.85}=100\ \mathrm{Hz}\).
Only odd harmonics occur: \(f=100,\,300,\,500,\,700,\,900,\,1100,\,1300,\dots\ \mathrm{Hz}\).
Those below 1250 Hz: 100, 300, 500, 700, 900, 1100, i.e. 6 frequencies (1300 Hz exceeds the limit).
Answer: (3) 6
Only odd harmonics occur: \(f=100,\,300,\,500,\,700,\,900,\,1100,\,1300,\dots\ \mathrm{Hz}\).
Those below 1250 Hz: 100, 300, 500, 700, 900, 1100, i.e. 6 frequencies (1300 Hz exceeds the limit).
Answer: (3) 6
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