Sound waves of frequency `660 H_(Z)` fall normally on perfectly reflecting wall. The distance from the wall at which the air particles have the maximum amplitude of vibration is (speed of sound in air = `330 m//s`)

Explanation

To solve the problem, we need to determine the distance from the wall at which the air particles have the maximum amplitude of vibration when sound waves of frequency 660 Hz fall normally on a perfectly reflecting wall. The speed of sound in air is given as 330 m/s.

### Step-by-Step Solution:

1. **Identify the Properties of the Wave:**
- Frequency (f) = 660 Hz
- Speed of sound (v) = 330 m/s

2. **Calculate the Wavelength (λ):**
- We use the formula that relates speed, frequency, and wavelength:
\[ v = f \cdot \lambda \]
- Rearranging the formula to find the wavelength:
\[ \lambda = \frac{v}{f} \]
- Substituting the known values:
\[ \lambda = \frac{330 \, \text{m/s}}{660 \, \text{Hz}} = 0.5 \, \text{m} \]

3. **Identify the Node and Antinode:**
- At the wall (reflecting surface), there is a node (point of no displacement).
- The first antinode (point of maximum displacement) occurs at a distance of \( \frac{\lambda}{4} \) from the wall.

4. **Calculate the Distance to the First Antinode:**
- We calculate the distance from the wall to the first antinode:
\[ \text{Distance to antinode} = \frac{\lambda}{4} = \frac{0.5 \, \text{m}}{4} = 0.125 \, \text{m} \]
- Converting to a fraction:
\[ 0.125 \, \text{m} = \frac{1}{8} \, \text{m} \]

5. **Final Answer:**
- The distance from the wall at which the air particles have the maximum amplitude of vibration is \( \frac{1}{8} \, \text{m} \).

### Summary:
The distance from the wall at which the air particles have the maximum amplitude of vibration is \( \frac{1}{8} \, \text{m} \).

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