The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. The length of the open pipe is

Options

A. ` 8 m `
B. ` 4 m `
C. `2 m`
D. ` 1 m `

Explanation

To solve the problem, we need to find the length of the open pipe given that the second overtone of the open pipe has the same frequency as the first overtone of a closed pipe that is 2 meters long.

### Step-by-Step Solution:

1. **Understand the Harmonics:**
- For an open pipe, the harmonics are given by the formula:
- \( f_n = n \frac{V}{2L} \) where \( n \) is the harmonic number, \( V \) is the speed of sound, and \( L \) is the length of the pipe.
- The second overtone corresponds to the third harmonic (\( n = 3 \)).
- Thus, the frequency of the second overtone of the open pipe is:
\[ f_{open} = 3 \frac{V}{2L_{open}} \]

2. **Closed Pipe Frequencies:**
- For a closed pipe, the harmonics are given by the formula:
- \( f_n = n \frac{V}{4L} \) where \( n \) is the harmonic number.
- The first overtone corresponds to the second harmonic (\( n = 2 \)).
- Thus, the frequency of the first overtone of the closed pipe is:
\[ f_{closed} = 2 \frac{V}{4L_{closed}} = \frac{V}{2L_{closed}} \]

3. **Set the Frequencies Equal:**
- According to the problem, the frequency of the second overtone of the open pipe is equal to the frequency of the first overtone of the closed pipe:
\[ 3 \frac{V}{2L_{open}} = \frac{V}{2L_{closed}} \]

4. **Cancel Out Common Terms:**
- We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ 3 \cdot \frac{1}{2L_{open}} = \frac{1}{2L_{closed}} \]

5. **Cross-Multiply to Solve for Lengths:**
- Cross-multiplying gives:
\[ 3 \cdot 2L_{closed} = 2L_{open} \]
- Simplifying this:
\[ 3L_{closed} = L_{open} \]

6. **Substitute the Length of the Closed Pipe:**
- We know that \( L_{closed} = 2 \) meters (as given in the problem).
- Substituting this value:
\[ L_{open} = 3 \cdot 2 = 6 \text{ meters} \]

### Final Answer:
The length of the open pipe is **6 meters**.