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 (denoted as \( L_O \)) 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 Frequencies of the Pipes:**
- For an open pipe, the frequency of the second overtone (which is the third harmonic) is given by:
\[ f_{open} = \frac{3v}{2L_O} \]
- For a closed pipe, the frequency of the first overtone (which is the third harmonic) is given by:
\[ f_{closed} = \frac{3v}{4L_C} \]
where \( L_C \) is the length of the closed pipe.

2. **Set the Frequencies Equal:**
- According to the problem, the frequencies of the second overtone of the open pipe and the first overtone of the closed pipe are equal:
\[ f_{open} = f_{closed} \]
- Thus, we can write:
\[ \frac{3v}{2L_O} = \frac{3v}{4L_C} \]

3. **Cancel Common Terms:**
- We can cancel \( 3v \) from both sides of the equation:
\[ \frac{1}{2L_O} = \frac{1}{4L_C} \]

4. **Cross Multiply:**
- Cross multiplying gives us:
\[ 4L_C = 2L_O \]

5. **Solve for \( L_O \):**
- Rearranging the equation to find \( L_O \):
\[ L_O = 2L_C \]

6. **Substitute the Length of the Closed Pipe:**
- We know that the length of the closed pipe \( L_C = 2 \) meters (as given in the problem):
\[ L_O = 2 \times 2 = 4 \text{ meters} \]

### Final Answer:
The length of the open pipe \( L_O \) is **4 meters**.

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