Q 11-14-001NEETNEET 2026Top questionMedium
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 $\dfrac{n}{m}$ is :
Answer: (C) $1$
Open pipe (antinodes at both ends): the $p^{\text{th}}$ harmonic has length $L = p\dfrac{\lambda}{2}$ and contains $p$ nodes. For the $5^{\text{th}}$ harmonic, $n = 5$.
Closed pipe (node at the closed end, antinode at the open end): the $(2k-1)^{\text{th}}$ harmonic has $L = (2k-1)\dfrac{\lambda}{4}$ and contains $k$ nodes (counting the closed end). For the $9^{\text{th}}$ harmonic, $2k - 1 = 9$, so $k = 5$ and $m = 5$.
$$\frac{n}{m} = \frac{5}{5} = 1$$
Solution by Sreeraj P, M.Sc Physics