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: The fundamental frequency of a closed organ pipe is calculated using the formula \( f_{\text{closed}} = \frac{v}{4L_{\text{closed}}} \). Given \( L_{\text{closed}} = 20 \, \mathrm{cm} = 0.2 \, \mathrm{m} \), the frequency is \( f_{\text{closed}} = \frac{v}{4 \cdot 0.2} = \frac{v}{0.8} \).

Step 2: The second overtone (third harmonic) of an open organ pipe is given by \( f_{\text{open, overtone}} = \frac{3v}{2L_{\text{open}}} \).

Step 3: Equating the fundamental frequency of the closed pipe to the second overtone of the open pipe yields \( \frac{v}{0.8} = \frac{3v}{2L_{\text{open}}} \). After canceling \( v \), the equation simplifies to \( \frac{1}{0.8} = \frac{3}{2L_{\text{open}}} \). Solving for \( L_{\text{open}} \), we find \( L_{\text{open}} = \frac{3 \cdot 0.8}{2} = 1.2 \, \mathrm{m} \). Therefore, the length of the open organ pipe is \( \mathbf{1.2 \, \mathrm{m}} \).