A wave is propagating along x-axis. The displacement of particles of the medium in z-direction at t=0 is given by : `z= exp[-(x+2)^(2)]` where 'x' is in meter. At t=1s, the same wave disturbance is given by `z=exp[-(x-2)^(2)]`. Then the wave propagation velocity is :-

Options

A. 4 m/s in + x direction
B. 4 m/s in - x direction
C. 2 m/s in + x direction
D. 2 m/s in - x direction

Explanation

To determine the wave propagation velocity given the displacement equations at two different times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Displacement Equations**:
- At \( t = 0 \): The displacement is given by
\[ z = e^{-(x + 2)^2} \]
- At \( t = 1 \): The displacement is given by
\[ z = e^{-(x - 2)^2} \]

2. **Formulate the General Wave Equation**:
- The general form of a wave traveling in the positive x-direction can be represented as:
\[ z = e^{-(x - vt)^2} \]
- For the negative x-direction, it would be:
\[ z = e^{-(x + vt)^2} \]

3. **Set Up the Equations**:
- For \( t = 0 \):
\[ z = e^{-(x + 2)^2} \implies z = e^{-(x - (-2))^2} \]
- For \( t = 1 \):
\[ z = e^{-(x - 2)^2} \implies z = e^{-(x - 2)^2} \]

4. **Relate the Two Displacement Equations**:
- At \( t = 1 \), we can relate the two equations:
\[ e^{-(x + 2 - v)^2} = e^{-(x - 2)^2} \]
- This implies:
\[ -(x + 2 - v)^2 = -(x - 2)^2 \]

5. **Expand and Simplify**:
- Expanding both sides gives:
\[ (x + 2 - v)^2 = (x - 2)^2 \]
- Expanding both sides:
\[ (x^2 + 4 + v^2 - 4x - 4v + 4x) = (x^2 - 4 + 4) \]
- This simplifies to:
\[ v^2 - 4v + 4 = 0 \]

6. **Solve for v**:
- Factoring or using the quadratic formula:
\[ v^2 - 4v = 0 \implies v(v - 4) = 0 \]
- This gives us two solutions: \( v = 0 \) or \( v = 4 \).

7. **Determine the Valid Solution**:
- Since a wave must propagate, we discard \( v = 0 \).
- Thus, the wave propagation velocity is:
\[ v = 4 \text{ m/s} \]

### Final Answer:
The wave propagation velocity is \( \mathbf{4 \, m/s} \).