The amplitude of a wave disturbance propagating along positive X-axis is given by `=1/(1+x^(2))` at t=0 and `y=1/[1+(x-2)^(2)]` at t=4 s where x and y are in metre. The shape of wave diturbance does not change with time. The velocity of the wave is

Options

A. `0.5 m//s`
B. `1 m//s`
C. `2 cm/s`
D. `4 m//s`

Explanation

To solve the problem, we need to determine the velocity of the wave disturbance based on the given amplitude equations at two different times.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
At time \( t = 0 \), the amplitude of the wave is given by:
\[ y = \frac{1}{1 + x^2} \]
This function reaches its maximum when \( x = 0 \), where \( y = 1 \).

2. **Identify the Conditions at \( t = 4 \) seconds:**
At time \( t = 4 \) seconds, the amplitude is given by:
\[ y = \frac{1}{1 + (x - 2)^2} \]
This function also reaches its maximum when \( x = 2 \), where \( y = 1 \).

3. **Determine the Maximum Points:**
- At \( t = 0 \), the maximum point (peak) of the wave is at \( x = 0 \).
- At \( t = 4 \), the maximum point (peak) of the wave is at \( x = 2 \).

4. **Calculate the Distance Traveled by the Wave:**
The distance traveled by the peak of the wave from \( t = 0 \) to \( t = 4 \) seconds is:
\[ \text{Distance} = 2 - 0 = 2 \text{ meters} \]

5. **Calculate the Velocity of the Wave:**
The velocity \( v \) of the wave can be calculated using the formula:
\[ v = \frac{\text{Distance}}{\text{Time}} = \frac{2 \text{ meters}}{4 \text{ seconds}} = 0.5 \text{ m/s} \]

### Final Answer:
The velocity of the wave is \( 0.5 \, \text{m/s} \).

---