RE-NEET 2026 Physics Harmonics in Pipes MCQ Question

In an open-ended (open-open) pipe every harmonic is allowed, and the $p$th harmonic carries $p$ nodes inside the pipe. The 5th harmonic therefore has

$n = 5$

With one end closed the pipe supports only odd harmonics, and the $p$th harmonic (p odd) carries $\frac{p+1}{2}$ nodes. For the 9th harmonic,

$m = \frac{9+1}{2} = 5$

So $\frac{n}{m} = \frac{5}{5} = 1$.

The other options come from counting antinodes instead of nodes, or from applying the open-pipe rule to the closed pipe, which would wrongly allow its even harmonics.