A closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lenghts are in the ratio

Options

A. `1 : 2`
B. `2 : 3`
C. `3 : 4`
D. `4 : 4`

Explanation

To solve the problem, we need to find the ratio of the lengths of a closed organ pipe and an open organ pipe given that their first overtones are identical in frequency.

### Step-by-Step Solution:

1. **Understanding Frequencies of Overtones**:
- For an **open organ pipe**, the frequency of the first overtone (which is the second harmonic) is given by the formula:
\[ f_1 = \frac{2V}{L_1} \]
where \( V \) is the velocity of sound in air and \( L_1 \) is the length of the open pipe.

- For a **closed organ pipe**, the frequency of the first overtone (which is the third harmonic) is given by the formula:
\[ f_2 = \frac{3V}{4L_2} \]
where \( L_2 \) is the length of the closed pipe.

2. **Setting Frequencies Equal**:
- Since the first overtones of both pipes are identical in frequency, we can set the two equations equal to each other:
\[ \frac{2V}{L_1} = \frac{3V}{4L_2} \]

3. **Canceling Common Terms**:
- We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{2}{L_1} = \frac{3}{4L_2} \]

4. **Cross-Multiplying**:
- Cross-multiplying gives us:
\[ 2 \cdot 4L_2 = 3L_1 \]
which simplifies to:
\[ 8L_2 = 3L_1 \]

5. **Finding the Ratio of Lengths**:
- Rearranging the equation to find the ratio of the lengths:
\[ \frac{L_2}{L_1} = \frac{3}{8} \]

6. **Final Ratio**:
- Therefore, the ratio of the lengths of the closed organ pipe to the open organ pipe is:
\[ L_1 : L_2 = 8 : 3 \]

### Conclusion:
The lengths of the closed organ pipe and the open organ pipe are in the ratio \( 8:3 \).

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