Question

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60 cm, the length of the closed pipe will be :

• A. 60 cm
• B. 45 cm
• C. 30 cm
• D. 15 cm

Answer 1

The Correct Option is D

To solve this problem, we need to understand the relation between the frequencies of closed and open organ pipes.

1. 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 and \(L_{\text{closed}}\) is the length of the closed pipe.
2. The first overtone (or second harmonic) frequency of an open organ pipe is \(f_{\text{open}} = \frac{2v}{2L_{\text{open}}} = \frac{v}{L_{\text{open}}}\) where \(L_{\text{open}}\) is the length of the open pipe. 
3. According to the problem statement, \(f_{\text{closed}} = f_{\text{open}}\).
4. Plugging in the formulas into this equation, we have: \(\frac{v}{4L_{\text{closed}}} = \frac{v}{L_{\text{open}}}\).
5. By cancelling \(v\) from both sides and solving for \(L_{\text{closed}}\), we get: \(4L_{\text{closed}} = L_{\text{open}}\) 
   Which simplifies to: \(L_{\text{closed}} = \frac{L_{\text{open}}}{4}\).
6. Given \(L_{\text{open}} = 60 \text{ cm}\), plug this value into the equation: \(L_{\text{closed}} = \frac{60}{4} = 15 \text{ cm}\).

Therefore, the length of the closed pipe is 15 cm.

Hence, the correct answer is 15 cm.

Answer 2

The relationship for the fundamental frequency of a closed organ pipe (length \( L_1 \)) and the first overtone of an open organ pipe (length \( L_2 \)) is given by:

\[ \frac{\lambda}{4} = L_1 \quad \text{and} \quad 2 \left( \frac{\lambda}{2} \right) = \lambda \]

The velocity of sound is \( v \), thus:

\[ v = f \lambda \]

For the closed pipe:

\[ v = f_1 (4L_1) \]

For the open pipe:

\[ f_2 = \frac{v}{2L_2} \]

Equating the fundamental frequency of the closed pipe to the first overtone of the open pipe:

\[ f_1 = f_2 \] \[ \frac{v}{4L_1} = \frac{v}{2L_2} \] \[ L_2 = 4L_1 \]

Given \( L_2 = 60 \, \text{cm} \):

\[ 60 = 4 \times L_1 \] \[ L_1 = 15 \, \text{cm} \]