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For sound waves, if the number of nodes for the $5^{\text{th}}$ harmonic of an open-ended pipe is $n$ and that for the $9^{\text{th}}$ harmonic of the same pipe with one of its ends closed is $m$, the ratio $\frac{n}{m}$ is :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
1

To solve this problem, we need to understand the harmonic concepts of sound waves in open and closed pipes.

Open Pipes: In an open-ended pipe, both ends are open. The number of nodes (points of zero amplitude) for any harmonic is equal to the harmonic number. Therefore, for an open pipe:

  • \(n = \text{harmonic number}\)

Closed Pipes: In a pipe closed at one end, an antinode (point of maximum amplitude) occurs at the open end and a node at the closed end. Only odd harmonics are possible. For the \(k^{\text{th}}\) harmonic, where \(k\) is odd, the number of nodes is:

  • \(m = \frac{k + 1}{2}\)

We are given:

  • The \(5^{\text{th}}\) harmonic of an open-ended pipe, thus \(n = 5\).
  • The \(9^{\text{th}}\) harmonic of the same pipe with one end closed, meaning this is the 5th odd harmonic, so \(m = \frac{9 + 1}{2} = 5\).

Therefore, the ratio \(\frac{n}{m}\) is:

  • \(\frac{n}{m} = \frac{5}{5} = 1\)

Thus, the answer is:

  • 1

This matches the correct answer in the options provided.

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