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

A. There is no position at which resultant displacement will be zero at all times
B. There is no time at which resultant displacement will be zero everywhere.
C. Both waves travel in same directions.
D. Both waves travel in oppoisite directions.

Options

A. There is no position at which resultant displacement will be zero at all times
B. There is no time at which resultant displacement will be zero everywhere.
C. Both waves travel in same directions.
D. Both waves travel in oppoisite directions.

Explanation

To solve the problem, we need to analyze the two wave functions given:

1. \( y_1 = \frac{1}{2 + (2x - 3t)^2} \)
2. \( y_2 = -\frac{1}{2 + (2x + 3t - 6)^2} \)

We will evaluate the options provided based on the behavior of these wave functions.

### Step 1: Determine the condition for resultant displacement to be zero

The resultant displacement \( y' \) is given by:

\[ y' = y_1 + y_2 \]

For the resultant displacement to be zero at a particular position \( x \) and time \( t \), we need:

\[ y_1 + y_2 = 0 \]

This means:

\[ y_1 = -y_2 \]

### Step 2: Set up the equation

Substituting the expressions for \( y_1 \) and \( y_2 \):

\[ \frac{1}{2 + (2x - 3t)^2} = -\left(-\frac{1}{2 + (2x + 3t - 6)^2}\right) \]

This simplifies to:

\[ \frac{1}{2 + (2x - 3t)^2} = \frac{1}{2 + (2x + 3t - 6)^2} \]

### Step 3: Cross-multiply and simplify

Cross-multiplying gives:

\[ 2 + (2x + 3t - 6)^2 = 2 + (2x - 3t)^2 \]

This simplifies to:

\[ (2x + 3t - 6)^2 = (2x - 3t)^2 \]

### Step 4: Expand both sides

Expanding both sides:

\[ (2x + 3t - 6)(2x + 3t - 6) = (2x - 3t)(2x - 3t) \]

This leads to:

\[ (2x + 3t - 6)^2 - (2x - 3t)^2 = 0 \]

### Step 5: Factor the expression

Using the difference of squares:

\[ [(2x + 3t - 6) - (2x - 3t)][(2x + 3t - 6) + (2x - 3t)] = 0 \]

This gives us two equations to solve:

1. \( 6t - 6 = 0 \) → \( t = 1 \)
2. \( 4x - 6 = 0 \) → \( x = \frac{3}{2} \)

### Step 6: Analyze the results

The first equation shows that at \( t = 1 \), the displacement can be zero at \( x = \frac{3}{2} \). This indicates that there is a position where the resultant displacement can be zero at specific times.

### Step 7: Determine the direction of wave propagation

To determine the direction of wave propagation, we look at the arguments of the wave functions:

- \( y_1 \) has a term \( (2x - 3t) \) which indicates it is moving in the positive x-direction.
- \( y_2 \) has a term \( (2x + 3t - 6) \) which can be rewritten as \( (2x - 3t + 6) \), indicating it is moving in the negative x-direction.

### Conclusion

Based on the analysis:

- **A**: False, as there is a position where the resultant displacement can be zero at specific times.
- **B**: False, as there is a time when the resultant displacement can be zero at a specific position.
- **C**: False, as the waves travel in opposite directions.
- **D**: True, both waves travel in opposite directions.

### Final Answer

The correct option is **D**: Both waves travel in opposite directions.

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