If the second overtone of an open organ pipe is in unison with third overtone of closed organ pipe, calculate the ratio of length of open organ pipe with closed pipe.

Answer

To solve the problem, we need to find the ratio of the lengths of an open organ pipe (denoted as \( L_O \)) and a closed organ pipe (denoted as \( L_C \)) given that the second overtone of the open pipe is in unison with the third overtone of the closed pipe.

### Step-by-Step Solution:

1. **Identify the Frequencies of the Overtone:**
- The second overtone of an open organ pipe corresponds to the third harmonic. The frequency \( f \) of the nth harmonic for an open pipe is given by:
\[ f_n = \frac{nV}{2L_O} \]
For the second overtone (3rd harmonic), \( n = 3 \):
\[ f_{3} = \frac{3V}{2L_O} \]

2. **Identify the Frequency of the Closed Pipe:**
- The third overtone of a closed organ pipe corresponds to the 7th harmonic. The frequency \( f \) of the nth harmonic for a closed pipe is given by:
\[ f_n = \frac{(2n-1)V}{4L_C} \]
For the third overtone (7th harmonic), \( n = 4 \):
\[ f_{7} = \frac{7V}{4L_C} \]

3. **Set the Frequencies Equal:**
- Since the second overtone of the open pipe is in unison with the third overtone of the closed pipe, we set the frequencies equal to each other:
\[ \frac{3V}{2L_O} = \frac{7V}{4L_C} \]

4. **Cancel \( V \) from Both Sides:**
- We can cancel \( V \) from both sides of the equation:
\[ \frac{3}{2L_O} = \frac{7}{4L_C} \]

5. **Cross Multiply to Solve for the Ratio:**
- Cross multiplying gives us:
\[ 3 \cdot 4L_C = 7 \cdot 2L_O \]
Simplifying this:
\[ 12L_C = 14L_O \]

6. **Rearranging for the Ratio:**
- We can rearrange this to find the ratio of the lengths:
\[ \frac{L_O}{L_C} = \frac{12}{14} = \frac{6}{7} \]

### Final Result:
The ratio of the length of the open organ pipe to the length of the closed organ pipe is:
\[ \frac{L_O}{L_C} = \frac{6}{7} \]