A tube, closed at one end and containing air, produces, when excited, the fundamental note of frequency `512 Hz`. If the tube is open at both ands the fundamental frequency that can be excited is (in Hz)

Explanation

To solve the problem, we need to understand the relationship between the frequency of sound waves in tubes that are closed at one end and those that are open at both ends.

### Step-by-Step Solution:

1. **Identify the Fundamental Frequency for Closed Tube:**
The fundamental frequency \( f \) of a tube closed at one end 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 tube.

2. **Given Information:**
We know that for the closed tube, the fundamental frequency \( f \) is 512 Hz. Therefore, we can write:
\[ 512 = \frac{V}{4L} \quad \text{(1)} \]

3. **Fundamental Frequency for Open Tube:**
The fundamental frequency \( f_0 \) of a tube open at both ends is given by:
\[ f_0 = \frac{V}{2L} \]

4. **Relate the Two Frequencies:**
From equation (1), we can express \( V \) in terms of \( L \) and the frequency:
\[ V = 512 \times 4L = 2048L \quad \text{(2)} \]

5. **Substituting into the Open Tube Formula:**
Now, substitute equation (2) into the formula for the open tube:
\[ f_0 = \frac{2048L}{2L} \]
This simplifies to:
\[ f_0 = \frac{2048}{2} = 1024 \text{ Hz} \]

6. **Final Answer:**
Therefore, the fundamental frequency that can be excited in the tube open at both ends is:
\[ \boxed{1024 \text{ Hz}} \]