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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Important Questions from Oscillations and Waves
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