Question

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe $L$ metre long. The length of the open pipe will be

• A. $L$
• B. $2L$
• C. $\frac{L}{2}$
• D. $4L$

Answer 1

The Correct Option is B

For an open organ pipe, the harmonics produced are integer multiples of the fundamental frequency. For a closed organ pipe, only odd harmonics are produced.

Let's analyze the situation step by step:

Step 1: Frequency Formulas

The frequency of the nth harmonic for an open pipe is given by:

\(f_n = n \times \frac{v}{2L_{\text{open}}}\)

For the second overtone of the open pipe, this corresponds to the third harmonic:

\(f_3 = 3 \times \frac{v}{2L_{\text{open}}}\)

The frequency of the nth harmonic for a closed pipe is given by:

\(f_n = n \times \frac{v}{4L_{\text{closed}}}\)

For the first overtone of the closed pipe, this corresponds to the third harmonic:

\(f_1 = \frac{v}{4L_{\text{closed}}}\)

Step 2: Setting Frequencies Equal

Since both frequencies are equal, we equate them:

\(3 \times \frac{v}{2L_{\text{open}}} = \frac{v}{4L_{\text{closed}}}\)

Step 3: Simplifying the Equation

Canceling out \(v\) from both sides:

\(3 \times \frac{1}{2L_{\text{open}}} = \frac{1}{4L_{\text{closed}}}\)

Cross-multiplying gives:

\(12L_{\text{closed}} = 2L_{\text{open}}\)

Dividing both sides by 2:

\(6L_{\text{closed}} = L_{\text{open}}\)

Conclusion:

The length of the open pipe is \(L_{\text{open}} = 6L_{\text{closed}}\).

Therefore, the correct answer is 2L.