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

Options

A. L/2
B. 4L
C. L
D. 2L

Answer

To solve the problem, we need to find the length of the open organ pipe given that its second overtone has the same frequency as the first overtone of a closed pipe of length \( L \).

### Step-by-Step Solution:

1. **Understanding the Frequencies**:
- The frequency of the second overtone of an open organ pipe can be expressed as:
\[ f_o = \frac{3v}{2L'} \]
where \( L' \) is the length of the open pipe and \( v \) is the speed of sound.

2. **Finding the Frequency of the Closed Pipe**:
- The frequency of the first overtone of a closed organ pipe is given by:
\[ f_c = \frac{3v}{4L} \]
where \( L \) is the length of the closed pipe.

3. **Setting the Frequencies Equal**:
- According to the problem, the second overtone of the open pipe has the same frequency as the first overtone of the closed pipe:
\[ f_o = f_c \]
- Therefore, we can write:
\[ \frac{3v}{2L'} = \frac{3v}{4L} \]

4. **Canceling Common Terms**:
- We can cancel \( 3v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{1}{2L'} = \frac{1}{4L} \]

5. **Cross-Multiplying**:
- Cross-multiplying gives us:
\[ 4L = 2L' \]

6. **Solving for \( L' \)**:
- Dividing both sides by 2:
\[ L' = 2L \]

7. **Conclusion**:
- The length of the open organ pipe \( L' \) is \( 2L \).

### Final Answer:
The length of the open organ pipe will be \( 2L \).