Two wave function in a medium along x direction are given by
`y_(1)=1/(2+(2x-3t)^(2))m`, `y_(2)=-1/(2+(2x+3t-6)^(2))m`
Where x is in meters and t is in seconds

Options

A. at `x=(3)/(2)m` the resultant displacement will be zero at all times.
B. at t = 1 s which resultant displacement will be zero everywhere.
C. Both waves travel along the same direction.
D. Both waves travel in the opposite directions.

Answer

To solve the problem involving the two wave functions \( y_1 \) and \( y_2 \), we need to analyze the given equations and determine the resultant displacement and the nature of the waves.

### Step-by-Step Solution:

1. **Identify the Wave Functions:**
The wave functions are given as:
\[ y_1 = \frac{1}{2 + (2x - 3t)^2} \quad \text{(1)} \]
\[ y_2 = -\frac{1}{2 + (2x + 3t - 6)^2} \quad \text{(2)} \]

2. **Determine the Resultant Displacement:**
The resultant displacement \( y \) is the sum of the two wave functions:
\[ y = y_1 + y_2 \]

3. **Set Up the Condition for Zero Resultant Displacement:**
For the resultant displacement to be zero, we need:
\[ y_1 + y_2 = 0 \implies y_1 = -y_2 \]

4. **Equate the Magnitudes:**
From the equations, we can set up the following condition:
\[ \frac{1}{2 + (2x - 3t)^2} = \frac{1}{2 + (2x + 3t - 6)^2} \]

5. **Cross Multiply to Eliminate Fractions:**
Cross-multiplying gives:
\[ 2 + (2x + 3t - 6)^2 = 2 + (2x - 3t)^2 \]

6. **Simplify the Equation:**
This simplifies to:
\[ (2x + 3t - 6)^2 = (2x - 3t)^2 \]

7. **Expand Both Sides:**
Expanding both sides:
\[ (2x + 3t - 6)(2x + 3t - 6) = (2x - 3t)(2x - 3t) \]
This results in:
\[ 4x^2 + 12tx - 24x + 9t^2 - 36t + 36 = 4x^2 - 12tx + 9t^2 \]

8. **Combine Like Terms:**
By simplifying, we can cancel \( 4x^2 \) and \( 9t^2 \) from both sides:
\[ 12tx - 24x - 36t + 36 = -12tx \]

9. **Rearranging the Equation:**
Rearranging gives:
\[ 24tx - 24x - 36t + 36 = 0 \]

10. **Factor the Equation:**
Factoring out common terms:
\[ 12(2tx - 2x - 3t + 3) = 0 \]
This leads to:
\[ 2tx - 2x - 3t + 3 = 0 \]

11. **Solve for x:**
Rearranging gives:
\[ 2x(t - 1) = 3t - 3 \]
Thus:
\[ x = \frac{3(t - 1)}{2(t - 1)} = \frac{3}{2} \quad \text{(for } t \neq 1\text{)} \]

12. **Conclusion:**
At \( x = \frac{3}{2} \), the resultant displacement \( y \) is zero for all times \( t \).