Question

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)}

• A. $40, 80, 120, 160$
• B. $40, 80, 160, 240$
• C. $40, 120, 200, 280$
• D. $80, 160, 240, 320$

Hint:

Open pipe: all harmonics. Closed pipe: only odd harmonics.

Answer 1

The Correct Option is C

Concept:
For an open organ pipe: \[ f_1=\frac{v}{2L} \] For a closed organ pipe: \[ f_1'=\frac{v}{4L} \] A closed organ pipe produces only odd harmonics: \[ f_1',\ 3f_1',\ 5f_1',\ 7f_1',\dots \] ip

Step 1: Find the fundamental frequency after one end is closed.
Given open pipe fundamental frequency: \[ 80\text{ Hz}=\frac{v}{2L} \] So for the same length: \[ f_1'=\frac{v}{4L}=\frac{80}{2}=40\text{ Hz} \] ip

Step 2: Write the allowed frequencies for closed pipe.
Only odd harmonics occur: \[ 40,\ 3\times 40,\ 5\times 40,\ 7\times 40 \] \[ 40,\ 120,\ 200,\ 280 \] ip Hence, the correct answer is:
\[ \boxed{(C)\ 40,120,200,280} \]