A closed organ pipe is vibrating in first overtone and is in resonance with another open organ pipe vibrating in third harmonic. The ratio of lengths of the pipes respectively is

Options

A. `1 : 2`
B. `4 : 1`
C. `8 : 3`
D. `3 : 8`

Explanation

To solve the problem, we need to find the ratio of the lengths of a closed organ pipe vibrating in its first overtone and an open organ pipe vibrating in its third harmonic.

### Step-by-Step Solution:

1. **Understanding the Closed Organ Pipe:**
- A closed organ pipe has one end closed and one end open. The fundamental frequency (first harmonic) of a closed pipe is given by:
\[ f_1 = \frac{v}{4L_c} \]
where \( L_c \) is the length of the closed pipe and \( v \) is the speed of sound in air.
- The first overtone (which is the second harmonic for a closed pipe) is given by:
\[ f_2 = \frac{3v}{4L_c} \]

2. **Understanding the Open Organ Pipe:**
- An open organ pipe has both ends open. The fundamental frequency (first harmonic) of an open pipe is given by:
\[ f_1 = \frac{v}{2L_o} \]
where \( L_o \) is the length of the open pipe.
- The third harmonic (which is the second overtone) is given by:
\[ f_3 = \frac{3v}{2L_o} \]

3. **Setting Up the Resonance Condition:**
- According to the problem, the closed organ pipe in the first overtone is in resonance with the open organ pipe in the third harmonic:
\[ f_2 = f_3 \]
- Substituting the expressions for \( f_2 \) and \( f_3 \):
\[ \frac{3v}{4L_c} = \frac{3v}{2L_o} \]

4. **Cancelling Common Terms:**
- We can cancel \( 3v \) from both sides:
\[ \frac{1}{4L_c} = \frac{1}{2L_o} \]

5. **Cross-Multiplying to Find the Ratio:**
- Cross-multiplying gives:
\[ 2L_c = 4L_o \]
- Rearranging this gives:
\[ \frac{L_c}{L_o} = \frac{4}{2} = 2 \]

6. **Final Ratio:**
- Therefore, the ratio of the lengths of the closed organ pipe to the open organ pipe is:
\[ \frac{L_c}{L_o} = 2:1 \]

### Conclusion:
The ratio of the lengths of the closed organ pipe to the open organ pipe is \( 2:1 \).

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