Question:

The successive frequencies produced by an organ pipe are 330 Hz, 440 Hz and 550 Hz. If the speed of sound in air is 330 m/s then the length of the organ pipe is

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For organ pipes: - Open at both ends: $f_n = n \frac{v}{2L}$, spacing between successive harmonics = $\frac{v}{2L}$. - Closed at one end: $f_n = (2n-1)\frac{v}{4L}$. Check the spacing between frequencies carefully to determine whether the pipe is open or closed.
Updated On: Oct 27, 2025
  • 3 m
  • 1.5 m
  • 2 m
  • 0.75 m
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The Correct Option is B

Solution and Explanation

1. Given frequencies: 330, 440, 550 Hz → common difference $f_2 - f_1 = 110$ Hz.
2. This is the fundamental frequency spacing: $\Delta f = v / 2L$ for a pipe open at both ends.
3. Using $\Delta f = 110~\text{Hz}, v = 330~\text{m/s} \implies L = \frac{v}{2 \Delta f} = \frac{330}{2 \cdot 110} = 1.5~\text{m}$.
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