Physics Q4 – NEET-UG 2026

For sound waves, if the number of nodes for the

5^{t h}

harmonic of an open-ended pipe is

n

and that for the

9^{t h}

harmonic of the same pipe with one of its ends closed is

m

, the ratio

\frac{n}{m}

is :

Clue 1 of 3: Understanding Harmonics and Nodes

Recall the definitions of harmonics and the number of nodes for standing waves in open and closed pipes.

🧠 Full Solution Path

Step 1: Determine nodes for an open-ended pipe✦ Active

For an open-ended pipe, the $k^{t h}$ harmonic has $k$ nodes. Given that the pipe is open-ended and it's the $5^{t h}$ harmonic, the number of nodes $n$ is:

n = k = 5

Step 2: Determine nodes for a closed-ended pipe○ Expand

For a closed-ended pipe, only odd harmonics are present. The $k^{t h}$ harmonic (where $k$ is odd) has $\frac{k + 1}{2}$ nodes. Given that the pipe is closed at one end and it's the $9^{t h}$ harmonic, the number of nodes $m$ is:

m = \frac{k + 1}{2} = \frac{9 + 1}{2} = \frac{10}{2} = 5

💡 Teacher's Secret Hint

Remember that for a closed pipe, the harmonic number $k$ refers to the odd multiples of the fundamental frequency.

Step 3: Calculate the ratio n/m○ Expand

Now, calculate the ratio $\frac{n}{m}$ :

\frac{n}{m} = \frac{5}{5} = 1

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Physics Question 4 – NEET-UG 2026

Clue 1 of 3: Understanding Harmonics and Nodes

🧠 Full Solution Path