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. `2m//s`
B. `0.5m//s`
C. `1m//s`
D. `3m//s`

Explanation

To find the velocity of the wave, we will analyze the given displacement functions at two different times and determine the distance traveled by the wave in that time interval.

### Step-by-Step Solution:

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

2. **Determine the Maximum Displacement**:
- For the first function at \( t = 0 \):
- The maximum value of \( y \) occurs when \( x = 0 \):
\[ y_{\text{max}} = \frac{1}{2 + 3(0)^2} = \frac{1}{2} \]

- For the second function at \( t = 2 \):
- The maximum value of \( y \) occurs when \( x = 2 \):
\[ y_{\text{max}} = \frac{1}{2} + 3(0)^2 = \frac{1}{2} \]

3. **Analyze the Wave Propagation**:
- The wave shape does not change as it propagates. Thus, the maximum displacement remains the same at both times.

4. **Determine the Distance Traveled**:
- At \( t = 0 \), the wave is at position \( x = 0 \).
- At \( t = 2 \), the wave has moved to position \( x = 2 \).
- Therefore, the distance traveled by the wave is:
\[ \Delta x = 2 - 0 = 2 \text{ meters} \]

5. **Calculate the Velocity**:
- The velocity \( v \) of the wave is given by the formula:
\[ v = \frac{\Delta x}{\Delta t} \]
- Here, \( \Delta t = 2 \) seconds, so:
\[ v = \frac{2 \text{ meters}}{2 \text{ seconds}} = 1 \text{ m/s} \]

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

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