A wave pulse in a string is described by the equation `y_1=5/((3x-4t)^2+2)` and another wave pulse in the same string is described by .. The values of `y_2=(-5)/((3x+4t-6)^2+2)` and x are in metres and t is in seconds. Which of the following statements is correct ?

Options

A. `y_1` travels along -x - direction and `y_2` along + x - direction
B. Both `y_1 and y_2` travel along +x - direction
C. At `x = 1m, y_1 and y_2` always cancel
D. At time t = 1s, `y_1 and y_2` exactly cancel everywhere

Answer

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

1. \( y_1 = \frac{5}{(3x - 4t)^2 + 2} \)
2. \( y_2 = \frac{-5}{(3x + 4t - 6)^2 + 2} \)

We want to determine the behavior of these wave pulses and identify which statements about them are correct.

### Step 1: Analyze the wave equations

**Wave 1: \( y_1 \)**

The equation \( y_1 = \frac{5}{(3x - 4t)^2 + 2} \) indicates that this wave pulse is dependent on the term \( (3x - 4t) \).

- The term \( 3x - 4t \) suggests that as time \( t \) increases, the wave pulse moves in the positive x-direction because the coefficient of \( t \) is negative.

**Wave 2: \( y_2 \)**

The equation \( y_2 = \frac{-5}{(3x + 4t - 6)^2 + 2} \) indicates that this wave pulse is dependent on the term \( (3x + 4t - 6) \).

- The term \( 3x + 4t - 6 \) suggests that as time \( t \) increases, this wave pulse moves in the negative x-direction because the coefficient of \( t \) is positive.

### Step 2: Determine the cancellation of the waves

To find out where these two wave pulses cancel each other, we need to set \( y_1 \) equal to \( y_2 \):

\[ \frac{5}{(3x - 4t)^2 + 2} = \frac{-5}{(3x + 4t - 6)^2 + 2} \]

Cross-multiplying gives:

\[ 5 \cdot ((3x + 4t - 6)^2 + 2) = -5 \cdot ((3x - 4t)^2 + 2) \]

This implies:

\[ (3x + 4t - 6)^2 + 2 = -(3x - 4t)^2 - 2 \]

Since both sides cannot be equal (one side is positive and the other is negative), we conclude that they do not cancel each other everywhere.

### Step 3: Evaluate at specific points

**At \( x = 1 \)**:

1. For \( y_1 \):
\[ y_1 = \frac{5}{(3(1) - 4t)^2 + 2} = \frac{5}{(3 - 4t)^2 + 2} \]

2. For \( y_2 \):
\[ y_2 = \frac{-5}{(3(1) + 4t - 6)^2 + 2} = \frac{-5}{(4t - 3)^2 + 2} \]

Setting \( y_1 = y_2 \) at \( x = 1 \) gives us a condition for cancellation.

**At \( t = 1 \)**:

1. For \( y_1 \):
\[ y_1 = \frac{5}{(3(1) - 4(1))^2 + 2} = \frac{5}{(3 - 4)^2 + 2} = \frac{5}{1 + 2} = \frac{5}{3} \]

2. For \( y_2 \):
\[ y_2 = \frac{-5}{(3(1) + 4(1) - 6)^2 + 2} = \frac{-5}{(3 + 4 - 6)^2 + 2} = \frac{-5}{1 + 2} = \frac{-5}{3} \]

At \( t = 1 \), \( y_1 \) and \( y_2 \) do not cancel each other out everywhere.

### Conclusion

From the analysis, we conclude that:

- \( y_1 \) travels in the positive x-direction.
- \( y_2 \) travels in the negative x-direction.
- They cancel each other at specific points but not everywhere.

### Correct Statement

The correct statement is that at \( x = 1 \), \( y_1 \) and \( y_2 \) cancel each other out, but they do not cancel everywhere at \( t = 1 \).

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