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 find the velocity of the wave disturbance, we can follow these steps:

### Step 1: Understand the given wave equations
We are given two wave equations:
1. At \( t = 0 \):
\[ y = \frac{1}{1 + x^2} \]
2. At \( t = 4 \):
\[ y = \frac{1}{1 + (x - 2)^2} \]

### Step 2: Recognize the form of wave propagation
The problem states that the shape of the wave does not change with time. This means that the wave can be described by a function of the form:
\[ y = f(x - vt) \]
where \( v \) is the wave velocity.

### Step 3: Substitute \( t = 0 \) into the wave equation
Substituting \( t = 0 \) into the general wave function gives:
\[ y = f(x) \]
This matches with the first equation:
\[ f(x) = \frac{1}{1 + x^2} \]

### Step 4: Substitute \( t = 4 \) into the wave equation
Now, substituting \( t = 4 \) into the general wave function gives:
\[ y = f(x - 4v) \]
This should match with the second equation:
\[ f(x - 4v) = \frac{1}{1 + (x - 2)^2} \]

### Step 5: Equate the two functions
From the above, we have:
\[ \frac{1}{1 + (x - 4v)^2} = \frac{1}{1 + (x - 2)^2} \]
This implies:
\[ 1 + (x - 4v)^2 = 1 + (x - 2)^2 \]
By simplifying, we get:
\[ (x - 4v)^2 = (x - 2)^2 \]

### Step 6: Expand both sides
Expanding both sides:
\[ x^2 - 8vx + 16v^2 = x^2 - 4x + 4 \]

### Step 7: Simplify the equation
Cancelling \( x^2 \) from both sides gives:
\[ -8vx + 16v^2 = -4x + 4 \]
Rearranging this gives:
\[ (8v - 4)x = 16v^2 - 4 \]

### Step 8: Solve for \( v \)
To find \( v \), we can set the coefficients of \( x \) equal:
\[ 8v - 4 = 0 \implies 8v = 4 \implies v = \frac{4}{8} = 0.5 \text{ m/s} \]

### Conclusion
The velocity of the wave is:
\[ v = 0.5 \text{ m/s} \]