An open pipe and a closed pipe are in resonance with each other with their first overtones. The ratio of their lengths are

Options

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

Explanation

To solve the problem of finding the ratio of lengths of an open pipe and a closed pipe that are in resonance with each other at their first overtones, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Pipes**:
- An open pipe has antinodes at both ends, while a closed pipe has a node at one end and an antinode at the other end.

2. **Wavelength in Pipes**:
- For the open pipe, the first overtone corresponds to the second harmonic. The wavelength (λ) for the second harmonic in an open pipe is given by:
\[ \lambda_{open} = \frac{2L_1}{2} = L_1 \]
- For the closed pipe, the first overtone corresponds to the third harmonic. The wavelength for the third harmonic in a closed pipe is given by:
\[ \lambda_{closed} = \frac{4L_2}{3} \]

3. **Frequency Relation**:
- The frequency (f) of a wave is related to its speed (v) and wavelength (λ) by the equation:
\[ f = \frac{v}{\lambda} \]
- For the open pipe (2nd harmonic):
\[ f_{open} = \frac{v}{L_1} \]
- For the closed pipe (3rd harmonic):
\[ f_{closed} = \frac{v}{\frac{4L_2}{3}} = \frac{3v}{4L_2} \]

4. **Setting Frequencies Equal**:
- Since both pipes are in resonance, their frequencies are equal:
\[ \frac{v}{L_1} = \frac{3v}{4L_2} \]

5. **Canceling Common Terms**:
- Canceling the speed of sound (v) from both sides gives:
\[ \frac{1}{L_1} = \frac{3}{4L_2} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ 4L_2 = 3L_1 \]

7. **Finding the Ratio of Lengths**:
- Rearranging this equation gives:
\[ \frac{L_1}{L_2} = \frac{4}{3} \]

### Final Answer:
The ratio of the lengths of the open pipe to the closed pipe is:
\[ \frac{L_1}{L_2} = \frac{4}{3} \]