Standing sound waves are produced in a pipe that is `0.8 m` long, open at one end, and closed at th other. For the fundamental and first two overtone, where along the pipe (measured from the closed end) are
(a) the displacemental antinodes
(b) the pressure antinodes.

Explanation

To solve the problem of standing sound waves in a pipe that is 0.8 m long, open at one end and closed at the other, we will determine the positions of displacement antinodes and pressure antinodes for the fundamental frequency and the first two overtones.

### Step-by-Step Solution:

1. **Understanding the Pipe Configuration**:
- The pipe is closed at one end and open at the other. This means that at the closed end (let's call it point A), there will be a displacement node (minimum displacement) and at the open end (point B), there will be a displacement antinode (maximum displacement).

2. **Finding the Wavelength**:
- For a pipe closed at one end, the fundamental frequency (first harmonic) has a length \( L \) related to the wavelength \( \lambda \) by the formula:
\[ L = \frac{n \lambda}{4} \]
where \( n \) is the harmonic number (1 for fundamental, 2 for first overtone, etc.).
- For the fundamental frequency (n=1):
\[ L = \frac{1 \lambda}{4} \implies \lambda = 4L \]
- For the first overtone (n=3):
\[ L = \frac{3 \lambda}{4} \implies \lambda = \frac{4L}{3} \]

3. **Calculating the Wavelengths**:
- Given \( L = 0.8 \, \text{m} \):
- For the fundamental frequency:
\[ \lambda_1 = 4 \times 0.8 = 3.2 \, \text{m} \]
- For the first overtone:
\[ \lambda_2 = \frac{4 \times 0.8}{3} = \frac{3.2}{3} \approx 1.067 \, \text{m} \]

4. **Finding Displacement Antinodes**:
- The displacement antinodes occur at:
- Fundamental (n=1): \( x = \frac{\lambda_1}{4} \) and \( x = L \)
\[ x_1 = \frac{3.2}{4} = 0.8 \, \text{m} \quad (at \, open \, end) \]
- First overtone (n=3): \( x = \frac{3\lambda_2}{4} \) and \( x = \frac{\lambda_2}{4} \)
\[ x_2 = \frac{3 \times 1.067}{4} \approx 0.800 \, \text{m} \quad (at \, open \, end) \]
\[ x_3 = \frac{1.067}{4} \approx 0.267 \, \text{m} \quad (first \, displacement \, antinode) \]

5. **Finding Pressure Antinodes**:
- Pressure antinodes occur at the displacement nodes:
- For the fundamental frequency:
\[ x = 0 \, \text{m} \quad (closed \, end) \]
- For the first overtone:
\[ x = \frac{\lambda_2}{2} \approx \frac{1.067}{2} \approx 0.5335 \, \text{m} \quad (first \, pressure \, antinode) \]

### Summary of Positions:
- **Displacement Antinodes**:
- Fundamental: \( 0.8 \, \text{m} \)
- First overtone: \( 0.267 \, \text{m} \) and \( 0.8 \, \text{m} \)

- **Pressure Antinodes**:
- Fundamental: \( 0 \, \text{m} \)
- First overtone: \( 0.5335 \, \text{m} \)