A plane progressive wave of frequency 25 Hz, amplitude `2.5 xx 10^(-5) m` and initial phase zero moves along the negative x-direction with a velocity of 300 m/s. A and B are two points 6 m apart on the line of propagation of the wave. At any instant the phase different between A and B is `phi`. The maximum difference in the displacements of particle at A and B is `Delta`

Explanation

To solve the problem step by step, we will follow these key points:

### Step 1: Identify the given values
- Frequency (f) = 25 Hz
- Amplitude (A) = \(2.5 \times 10^{-5}\) m
- Velocity (v) = 300 m/s
- Distance between points A and B (d) = 6 m

### Step 2: Calculate the wavelength (λ)
Using the formula for wave speed:
\[ v = f \cdot \lambda \]
We can rearrange this to find the wavelength:
\[ \lambda = \frac{v}{f} \]
Substituting the values:
\[ \lambda = \frac{300 \, \text{m/s}}{25 \, \text{Hz}} = 12 \, \text{m} \]

### Step 3: Calculate the phase difference (φ) between points A and B
The phase difference can be calculated using the formula:
\[ \phi = \frac{2\pi}{\lambda} \cdot d \]
Substituting the values:
\[ \phi = \frac{2\pi}{12} \cdot 6 = \pi \, \text{radians} \]

### Step 4: Calculate the maximum difference in displacements (Δ)
The maximum difference in displacement between two points in a wave is given by:
\[ \Delta = 2A \cdot \sin\left(\frac{\phi}{2}\right) \]
Substituting the values:
\[ \Delta = 2 \cdot (2.5 \times 10^{-5}) \cdot \sin\left(\frac{\pi}{2}\right) \]
Since \(\sin\left(\frac{\pi}{2}\right) = 1\):
\[ \Delta = 2 \cdot (2.5 \times 10^{-5}) = 5 \times 10^{-5} \, \text{m} \]

### Final Answer
The maximum difference in the displacements of particles at A and B is \(5 \times 10^{-5}\) m.

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