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

Question

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

Accepted Answer

D. 1. Recall that in an **open pipe** (open at both ends), every harmonic — 1st, 2nd, 3rd, and so on — is allowed, and the number of nodes formed always equals the harmonic number itself.

Given the pipe supports the 5th harmonic,
$$
n = 5
$$

2. Recall that in a **pipe closed at one end**, only **odd harmonics** (1st, 3rd, 5th, 7th, 9th, ...) can exist, and each successive odd harmonic adds one more node than the previous one.

Counting nodes for successive odd harmonics: 1st harmonic → 1 node, 3rd harmonic → 2 nodes, 5th harmonic → 3 nodes, 7th harmonic → 4 nodes, 9th harmonic → 5 nodes.

Therefore, for the 9th harmonic,
$$
m = 5
$$

3. Since both $n$ and $m$ are now known, find their ratio.

Hence,
$$
\frac{n}{m} = \frac{5}{5} = 1
$$

Hence, the answer is **D**.

For sound waves, if the number of nodes for the 5th5^{\text{th}}5th harmonic of an open-ended pipe is nnn and that for the 9th9^{\text{th}}9th harmonic of the same pipe with one of its ends closed is mmm, the ratio nm\frac{n}{m}mn​ is

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