The first overtone of an open pipe has frequency n. The first ovetone of a closed pipe of the same length will have frequency

Options

A. `n//2`
B. 2n
C. `3n//4`
D. `4n//3`

Answer

To solve the problem, we need to find the frequency of the first overtone of a closed pipe given that the first overtone of an open pipe has a frequency \( n \).

### Step-by-Step Solution:

1. **Understand the Frequencies of Open and Closed Pipes:**
- The first overtone of an open pipe is also known as the second harmonic. Its frequency can be expressed as:
\[ f_2 = 2f_1 \]
where \( f_1 \) is the fundamental frequency of the open pipe.

2. **Fundamental Frequency of Open Pipe:**
- The fundamental frequency \( f_1 \) for an open pipe of length \( L \) is given by:
\[ f_1 = \frac{v}{2L} \]
where \( v \) is the speed of sound in air.

3. **Relate Given Frequency \( n \) to Fundamental Frequency:**
- According to the question, the first overtone of the open pipe has a frequency \( n \):
\[ f_2 = n \]
- Since \( f_2 = 2f_1 \), we can write:
\[ 2f_1 = n \implies f_1 = \frac{n}{2} \]

4. **Determine the Fundamental Frequency of Closed Pipe:**
- The fundamental frequency \( f_1' \) for a closed pipe of the same length \( L \) is given by:
\[ f_1' = \frac{v}{4L} \]

5. **Calculate the First Overtone of Closed Pipe:**
- The first overtone of a closed pipe is also known as the third harmonic, which is given by:
\[ f_3' = 3f_1' \]
- Substituting the expression for \( f_1' \):
\[ f_3' = 3 \left( \frac{v}{4L} \right) = \frac{3v}{4L} \]

6. **Relate \( v/L \) to \( n \):**
- From the earlier steps, we know:
\[ f_1 = \frac{n}{2} \implies \frac{n}{2} = \frac{v}{2L} \implies \frac{v}{L} = n \]

7. **Substitute \( v/L \) into the Frequency of Closed Pipe:**
- Now we can substitute \( \frac{v}{L} = n \) into the equation for \( f_3' \):
\[ f_3' = \frac{3}{4} \cdot \frac{v}{L} = \frac{3}{4} n \]

8. **Final Result:**
- Therefore, the frequency of the first overtone of the closed pipe is:
\[ f_3' = \frac{3}{4} n \]

### Conclusion:
The first overtone of a closed pipe of the same length will have a frequency of \( \frac{3}{4} n \).

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