An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by `100 Hz` then the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

Options

A. `200 Hz`
B. `300 Hz`
C. `240 Hz`
D. `480 Hz`

Explanation

To solve the problem, we need to find the fundamental frequency of the open pipe given that the frequency of the third harmonic of the closed pipe is 100 Hz higher than the fundamental frequency of the open pipe.

### Step-by-Step Solution:

1. **Understanding the Frequencies**:
- Let the fundamental frequency of the open pipe be \( f \).
- The frequency of the third harmonic of the closed pipe will be denoted as \( f' \).

2. **Relationship Between Frequencies**:
- According to the problem, we have:
\[ f' = f + 100 \text{ Hz} \]

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

4. **Frequency of the Closed Pipe**:
- The third harmonic frequency of a closed pipe is given by:
\[ f' = \frac{3v}{4L} \]

5. **Setting Up the Equation**:
- From the relationship established in step 2, we can substitute \( f' \):
\[ \frac{3v}{4L} = \frac{v}{2L} + 100 \]

6. **Solving for \( v \)**:
- To eliminate \( L \), we can multiply through by \( 4L \):
\[ 3v = 2v + 400L \]
- Rearranging gives:
\[ 3v - 2v = 400L \implies v = 400L \]

7. **Substituting Back to Find \( f \)**:
- Now substitute \( v \) back into the equation for the fundamental frequency of the open pipe:
\[ f = \frac{v}{2L} = \frac{400L}{2L} = 200 \text{ Hz} \]

### Conclusion:
The fundamental frequency of the open pipe is \( \boxed{200 \text{ Hz}} \).