The separation between a node and the next antinode in a vibrating air column is 25 cm. If the speed of sound in air is `340 m s^-1`, find the frequency of vibration of the air column.

Explanation

To solve the problem, we need to find the frequency of vibration of the air column given the separation between a node and the next antinode and the speed of sound in air.

### Step-by-Step Solution:

1. **Understanding the Relationship Between Node and Antinode**:
- In a vibrating air column, the distance between a node and the next antinode is given by \( \frac{\lambda}{4} \), where \( \lambda \) is the wavelength of the sound wave.

2. **Given Data**:
- Separation between a node and the next antinode = 25 cm = 0.25 m (convert cm to m).
- Speed of sound in air, \( v = 340 \, \text{m/s} \).

3. **Finding the Wavelength**:
- Since the distance between a node and an antinode is \( \frac{\lambda}{4} \), we can express this as:
\[ \frac{\lambda}{4} = 0.25 \, \text{m} \]
- To find \( \lambda \), multiply both sides by 4:
\[ \lambda = 4 \times 0.25 = 1 \, \text{m} \]

4. **Using the Wave Equation**:
- The relationship between speed \( v \), frequency \( f \), and wavelength \( \lambda \) is given by:
\[ v = f \cdot \lambda \]
- Rearranging this equation to solve for frequency \( f \):
\[ f = \frac{v}{\lambda} \]

5. **Substituting the Known Values**:
- Now substitute the values of \( v \) and \( \lambda \):
\[ f = \frac{340 \, \text{m/s}}{1 \, \text{m}} = 340 \, \text{Hz} \]

6. **Final Answer**:
- The frequency of vibration of the air column is \( 340 \, \text{Hz} \).