The fundamental frequency of a closed organ pipe is 256 Hz. If the pipe is cut in two halves, then the difference in fundamental notes produced by two pipes is `x xx 128` Hz. Find the value of x.
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Explanation

To solve the problem, we need to find the value of \( x \) given that the fundamental frequency of a closed organ pipe is 256 Hz and that the pipe is cut in half. We will calculate the fundamental frequencies of both halves and find their difference.

### Step-by-Step Solution:

1. **Understanding the Fundamental Frequency of a Closed Organ Pipe:**
The fundamental frequency \( f \) of a closed organ pipe is given by the formula:
\[ f = \frac{v}{4L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the pipe.

2. **Given Data:**
We know the fundamental frequency \( f = 256 \) Hz.

3. **Finding the Speed of Sound:**
Rearranging the formula to find \( v \):
\[ v = 4Lf \]
Substituting \( f = 256 \) Hz:
\[ v = 4L \times 256 = 1024L \]

4. **Cutting the Pipe in Half:**
When the pipe is cut in half, the length of each half becomes \( \frac{L}{2} \).

5. **Calculating the Frequencies of the Halved Pipes:**
- For the first half (which behaves as an open organ pipe):
\[ f_{\text{open}} = \frac{v}{2L} = \frac{1024L}{2L} = 512 \text{ Hz} \]
- For the second half (which behaves as a closed organ pipe):
\[ f_{\text{closed}} = \frac{v}{4 \cdot \frac{L}{2}} = \frac{1024L}{2L} = 512 \text{ Hz} \]

6. **Finding the Difference in Frequencies:**
The difference in the fundamental frequencies of the two pipes is:
\[ \Delta f = f_{\text{open}} - f_{\text{closed}} = 512 \text{ Hz} - 256 \text{ Hz} = 256 \text{ Hz} \]

7. **Setting Up the Equation:**
According to the problem, this difference can also be expressed as:
\[ \Delta f = x \times 128 \]
Substituting the value we found:
\[ 256 = x \times 128 \]

8. **Solving for \( x \):**
Dividing both sides by 128:
\[ x = \frac{256}{128} = 2 \]

### Final Answer:
The value of \( x \) is \( 2 \).