Question:

A pipe with 30 cm Length is open at both ends. Which harmonic mode of the pipe resonates a 1.65 kHz source? (Velocity of sound in air = 330 m/s)

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Remember that for pipes open at both ends, only integer harmonics (1st, 2nd, 3rd, etc.) are possible. The fundamental frequency corresponds to \(n=1\).
Updated On: Mar 22, 2025
  • \(2\)
  • \(3\)
  • \(3.5\)
  • \(2.5\)
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The Correct Option is B

Solution and Explanation

For a pipe open at both ends, the resonant frequencies are given by: \[ f_n = n \frac{v}{2L} \] where \(n\) is a positive integer (harmonic number), \(v\) is the velocity of sound, and \(L\) is the length of the pipe. Given: \[ L = 0.3 \, {m}, \quad v = 330 \, {m/s}, \quad f = 1.65 \, {kHz} = 1650 \, {Hz} \] Solving for \(n\): \[ 1650 = n \frac{330}{2 \times 0.3} \quad \Rightarrow \quad 1650 = n \times 550 \quad \Rightarrow \quad n = \frac{1650}{550} = 3 \] Thus, the third harmonic mode resonates with a 1.65 kHz source.
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