Question

The fundamental frequency of a closed organ pipe of length \( 20 \, \mathrm{cm} \) is equal to the second overtone of an organ pipe open at both ends. What is the length of the organ pipe open at both ends?

• A. \( 1.0 \, \mathrm{m} \)
• B. \( 1.2 \, \mathrm{m} \)
• C. \( 1.4 \, \mathrm{m} \)
• D. \( 1.6 \, \mathrm{m} \)

Hint:

For organ pipes: Closed pipe fundamental frequency: \( f = \frac{v}{4L} \), Open pipe \( n \)-th harmonic frequency: \( f_n = \frac{nv}{2L} \). Equate the respective frequencies for comparisons.

Answer 1

The Correct Option is B

Step 1: Recall the frequency of a closed organ pipe. 
The fundamental frequency of a closed organ pipe is given by: \[ f_{\text{closed}} = \frac{v}{4L_{\text{closed}}}, \] where: \( v \) is the speed of sound in air, \( L_{\text{closed}} \) is the length of the closed organ pipe. Substitute \( L_{\text{closed}} = 20 \, \mathrm{cm} = 0.2 \, \mathrm{m} \): \[ f_{\text{closed}} = \frac{v}{4 \cdot 0.2} = \frac{v}{0.8}. \] 

Step 2: Recall the frequency of the second overtone of an open pipe. 
For an organ pipe open at both ends, the frequency of the second overtone (third harmonic) is: \[ f_{\text{open, overtone}} = \frac{3v}{2L_{\text{open}}}, \] where \( L_{\text{open}} \) is the length of the open pipe.

 Step 3: Equate the frequencies. 
Given that the fundamental frequency of the closed pipe is equal to the second overtone of the open pipe: \[ \frac{v}{0.8} = \frac{3v}{2L_{\text{open}}}. \] Simplify by canceling \( v \) from both sides: \[ \frac{1}{0.8} = \frac{3}{2L_{\text{open}}}. \] Rearrange to find \( L_{\text{open}} \): \[ L_{\text{open}} = \frac{3 \cdot 0.8}{2} = 1.2 \, \mathrm{m}. \] Thus, the length of the open organ pipe is \( \mathbf{1.2 \, \mathrm{m}} \).