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 address this, we must first understand the relationship between the frequencies produced by closed and open organ pipes.

1. The fundamental frequency for a closed organ pipe is determined by the formula: \(f_{\text{closed}} = \frac{v}{4L_{\text{closed}}}\). Here, \(v\) represents the speed of sound in air, and \(L_{\text{closed}}\) is the physical length of the closed pipe.
2. The frequency of the first overtone (which is also the second harmonic) for an open organ pipe is given by: \(f_{\text{open}} = \frac{2v}{2L_{\text{open}}} = \frac{v}{L_{\text{open}}}\). In this equation, \(L_{\text{open}}\) denotes the length of the open pipe.
3. The problem statement specifies that the frequencies are equal: \(f_{\text{closed}} = f_{\text{open}}\).
4. Substituting the respective formulas into this equality yields: \(\frac{v}{4L_{\text{closed}}} = \frac{v}{L_{\text{open}}}\).
5. By eliminating \(v\) from both sides and rearranging to solve for \(L_{\text{closed}}\), we obtain: \(4L_{\text{closed}} = L_{\text{open}}\). This simplifies to: \(L_{\text{closed}} = \frac{L_{\text{open}}}{4}\).
6. Given that \(L_{\text{open}} = 60 \text{ cm}\), we can substitute this value into the derived equation: \(L_{\text{closed}} = \frac{60}{4} = 15 \text{ cm}\).

Consequently, the length of the closed pipe is determined to be 15 cm.

Therefore, the correct measurement is 15 cm.