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

Concept:

• In an open organ pipe, both ends are antinodes.

• In the \(n^{\text{th}}\) harmonic of an open pipe, the number of nodes equals the harmonic number.

• In a closed organ pipe, one end is a node and the other end is an antinode.

• Only odd harmonics are present in a closed pipe.

Step 1: Find the number of nodes in the open pipe.
For the 5th harmonic of an open pipe, \[ n=5 \]

Step 2: Find the number of nodes in the closed pipe.
For the 9th harmonic of a closed pipe, the standing wave pattern contains \[ m=9 \] nodes.

Step 3: Calculate the required ratio.
\[ \frac{n}{m} = \frac{5}{9} \] \[ \boxed{ \frac{n}{m} = \frac{5}{9} } \]

Step 4: Select the correct answer.
\[ \boxed{\text{Option (C)}} \]