Question

For sound waves, if the number of nodes for the 5th harmonic of an open-ended pipe is \(n\) and that for the 9th harmonic of the same pipe with one of its ends closed is \(m\), the ratio \(n/m\) is:

• A. \(\dfrac{3}{5}\)
• B. \(\dfrac{9}{5}\)
• C. \(\dfrac{5}{9}\)
• D. \(1\)

Hint:

For organ-pipe questions, first draw the standing-wave pattern. In an open pipe, both ends are antinodes, whereas in a closed pipe one end is always a node.

Answer 1

The Correct Option is C

Step 1: Recall the open pipe pattern.
An open-ended pipe has antinodes at both ends. For its $p$-th harmonic, the number of nodes equals the harmonic number $p$.
Step 2: Count nodes for the open pipe.
For the 5th harmonic of the open pipe, the number of nodes is $n = 5$.
Step 3: Recall the closed pipe pattern.
A pipe closed at one end has a node at the closed end and an antinode at the open end, and supports only odd harmonics.
Step 4: Count nodes for the closed pipe.
For the 9th harmonic of the closed pipe, the standing-wave pattern contains $m = 9$ nodes.
Step 5: Form the ratio.
$\dfrac{n}{m} = \dfrac{5}{9}$.
Step 6: Select the answer.
This is option C.
\[ \boxed{ \frac{n}{m} = \frac{5}{9} } \]