Question

The harmonic mode which resonates with a closed pipe of length 22 cm, when excited by a 1875 Hz source and the number of nodes present in it respectively are (velocity of sound in air = 330 m/s)

• A. 1st, 1
• B. 3rd, 1
• C. 3rd, 2
• D. 5th, 4
• E. 5th, 3

Hint:

Closed pipe → only odd harmonics allowed and nodes = $\frac{n+1}{2}$.

Answer 1

Answer: (E)

Concept: A closed pipe (one end closed, one end open) supports standing waves with the following properties:
• Closed end → displacement node
• Open end → displacement antinode
• Only odd harmonics are allowed: $n = 1, 3, 5, 7, \dots$ The frequency of the $n^{\text{th}}$ harmonic is: \[ f_n = \frac{n v}{4L}, \quad n = 1,3,5,... \]

Step 1: Convert given length: \[ L = 22 \text{ cm} = 0.22 \text{ m} \]

Step 2: Substitute values: \[ f = 1875 \, \text{Hz}, \quad v = 330 \, \text{m/s} \] \[ 1875 = \frac{n \cdot 330}{4 \times 0.22} \]

Step 3: Simplify denominator: \[ 4 \times 0.22 = 0.88 \] \[ 1875 = \frac{330n}{0.88} \]

Step 4: Solve for $n$: \[ n = \frac{1875 \times 0.88}{330} \] \[ n = \frac{1650}{330} = 5 \]

Step 5: Interpretation:
• $n = 5$ → 5th harmonic
• Valid because closed pipe allows only odd harmonics

Step 6: Number of nodes: In a closed pipe:
• Number of nodes = $\frac{n+1}{2}$ \[ \text{Nodes} = \frac{5+1}{2} = 3 \]

Step 7: Physical understanding:
• Node at closed end
• Alternating nodes and antinodes along length
• Total 3 nodes appear in 5th harmonic Final Answer: \[ \text{5th harmonic, 3 nodes} \]