The given problem involves understanding the change in the fundamental frequency of a resonating pipe when it is partially filled with water. Let's break it down step by step:
- An open pipe resonates at its fundamental frequency \(f_o\). For an open pipe, the fundamental frequency is given by: \(f_o = \frac{v}{2L}\) where \(v\) is the speed of sound in air and \(L\) is the length of the pipe.
- When the pipe is half-filled with water, it effectively transforms into a closed pipe (closed at the water surface) with a length equal to half the original length of the pipe, that is \(\frac{L}{2}\).
- The fundamental frequency of a closed pipe is given by: \(f = \frac{v}{4L_{\text{{closed}}}}\) where \(L_{\text{{closed}}}\) is \(\frac{L}{2}\).
- Substituting \(L_{\text{{closed}}} = \frac{L}{2}\) in the formula for a closed pipe, we get: \(f = \frac{v}{4 \times \frac{L}{2}} = \frac{v}{2L}\)
- Notice that this result is equal to the original fundamental frequency of the open pipe: \(f = f_o\).
Thus, when an open pipe is half-filled with water, its fundamental frequency remains \(f_o\), which is the same as the frequency of the open pipe. Therefore, the correct answer is \(f_o\).