The displacement function of a wave travelling along positive x-direction is `y =(1)/(2 + 3x^(2))at `t=0` and by `y = (1)/(2) + 3(x - 2)^(2))` at `t = 2 s`, where `y` and `x` are in metre. The velocity of the wave is

Options

A. (a) `2m//s`
B. (b) `0.5m//s`
C. (c) `1m//s`
D. (d) `3m//s`

Explanation

To find the velocity of the wave given the displacement functions at two different times, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Displacement Functions**:
At \( t = 0 \):
\[ y = \frac{1}{2 + 3x^2} \]
At \( t = 2 \):
\[ y = \frac{1}{2} + 3(x - 2)^2 \]

2. **Determine the Maximum Displacement**:
For the wave at \( t = 0 \), the maximum displacement occurs at \( x = 0 \):
\[ y(0) = \frac{1}{2 + 3(0)^2} = \frac{1}{2} \]

3. **Set the Maximum Displacement Equal for Both Times**:
Since the wave travels, we can equate the maximum displacement at \( t = 0 \) to the maximum displacement at \( t = 2 \):
\[ \frac{1}{2} = \frac{1}{2} + 3(x - 2)^2 \]

4. **Solve for \( x \)**:
Rearranging the equation:
\[ 3(x - 2)^2 = 0 \]
This implies:
\[ (x - 2)^2 = 0 \implies x - 2 = 0 \implies x = 2 \]

5. **Calculate the Distance Traveled by the Wave**:
The wave has moved from \( x = 0 \) to \( x = 2 \) in \( 2 \) seconds.

6. **Calculate the Velocity of the Wave**:
The velocity \( v \) can be calculated using the formula:
\[ v = \frac{\text{Distance}}{\text{Time}} = \frac{2 \text{ m}}{2 \text{ s}} = 1 \text{ m/s} \]

### Final Answer:
The velocity of the wave is \( 1 \text{ m/s} \).

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