Third overtone of a closed organ pipe is in unison with fourth harmonic of an open organ pipe . Find the ratio of the lengths of the pipes.

Explanation

To solve the problem of finding the ratio of the lengths of a closed organ pipe and an open organ pipe when their frequencies are in unison, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Frequencies**:
- For a closed organ pipe, the frequencies are given by:
\[ f_n = \frac{(2n-1)V}{4L_1} \]
where \( n \) is the harmonic number (1, 2, 3,...), \( V \) is the speed of sound, and \( L_1 \) is the length of the closed pipe. The frequencies for the first few harmonics are:
- Fundamental (1st harmonic): \( f_1 = \frac{V}{4L_1} \)
- 1st overtone (3rd harmonic): \( f_2 = \frac{3V}{4L_1} \)
- 2nd overtone (5th harmonic): \( f_3 = \frac{5V}{4L_1} \)
- 3rd overtone (7th harmonic): \( f_4 = \frac{7V}{4L_1} \)

- For an open organ pipe, the frequencies are given by:
\[ f_n = \frac{nV}{2L_2} \]
where \( L_2 \) is the length of the open pipe. The frequencies for the first few harmonics are:
- Fundamental (1st harmonic): \( f_1 = \frac{V}{2L_2} \)
- 2nd harmonic: \( f_2 = \frac{2V}{2L_2} \)
- 3rd harmonic: \( f_3 = \frac{3V}{2L_2} \)
- 4th harmonic: \( f_4 = \frac{4V}{2L_2} \)

2. **Identifying the Given Frequencies**:
- The problem states that the third overtone of the closed organ pipe (which corresponds to \( f_4 = \frac{7V}{4L_1} \)) is in unison with the fourth harmonic of the open organ pipe (which corresponds to \( f_4 = \frac{4V}{2L_2} \)).

3. **Setting Up the Equation**:
- We equate the two frequencies:
\[ \frac{7V}{4L_1} = \frac{4V}{2L_2} \]

4. **Simplifying the Equation**:
- Cancel \( V \) from both sides:
\[ \frac{7}{4L_1} = \frac{4}{2L_2} \]
- Simplifying the right side gives:
\[ \frac{7}{4L_1} = \frac{2}{L_2} \]

5. **Cross-Multiplying**:
- Cross-multiply to find the relationship between \( L_1 \) and \( L_2 \):
\[ 7L_2 = 8L_1 \]

6. **Finding the Ratio of Lengths**:
- Rearranging gives:
\[ \frac{L_1}{L_2} = \frac{7}{8} \]
- Therefore, the ratio of the lengths of the closed organ pipe to the open organ pipe is:
\[ L_1 : L_2 = 7 : 8 \]

### Final Answer:
The ratio of the lengths of the pipes is \( 7 : 8 \).