For an open-ended pipe, the $k^{th}$ harmonic corresponds to the $k^{th}$ overtone plus the fundamental. The number of nodes in an open pipe for the $k^{th}$ harmonic is given by $n = k$. Here, for the $5^{th}$ harmonic, $n = 5$.
For a pipe closed at one end, only odd harmonics exist. The $p^{th}$ harmonic (where $p$ is odd) corresponds to the frequency $f_p = p \times f_1$. The number of nodes $m$ for the $p^{th}$ harmonic in a closed pipe is given by $m = \frac{p+1}{2}$. For the $9^{th}$ harmonic, $p = 9$. Thus, $m = \frac{9+1}{2} = 5$.
We have $n = 5$ and $m = 5$. Therefore, the ratio is: $$ \frac{n}{m} = \frac{5}{5} = 1 $$
Final Answer: 1
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