Question:medium

An organ pipe has fundamental frequency $80\text{ Hz}$ . If its one end is closed, the frequencies produced will be (in $\text{Hz}$ ) (Neglect end correction)}

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Open pipe: all harmonics. Closed pipe: only odd harmonics.
Updated On: May 14, 2026
  • $40, 80, 120, 160$
  • $40, 80, 160, 240$
  • $40, 120, 200, 280$
  • $80, 160, 240, 320$
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
An "organ pipe" without qualification usually refers to an open-open pipe. Closing one end changes the boundary conditions and the available resonant frequencies.
Step 2: Key Formula or Approach:
1. Fundamental frequency of open pipe: $f_{open} = \frac{v}{2L}$
2. Fundamental frequency of closed pipe: $f_{closed} = \frac{v}{4L} = \frac{1}{2} f_{open}$
3. Harmonics of a closed pipe are only odd multiples: $f, 3f, 5f, \dots$
Step 3: Detailed Explanation:
Given $f_{open} = 80\text{ Hz}$.
1. When one end is closed, the new fundamental frequency becomes:
\[ f_{closed} = \frac{80}{2} = 40\text{ Hz} \]
2. A closed-at-one-end pipe only supports odd harmonics. The sequence of frequencies will be:
- $1^{st}$ Harmonic: $40\text{ Hz}$
- $3^{rd}$ Harmonic: $3 \times 40 = 120\text{ Hz}$
- $5^{th}$ Harmonic: $5 \times 40 = 200\text{ Hz}$
- $7^{th}$ Harmonic: $7 \times 40 = 280\text{ Hz}$
Step 4: Final Answer:
The frequencies are $40, 120, 200, 280\text{ Hz}$.
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