A wave is represented by the equation `y = a sin(kx - omega t)` is superimposed with another wave to form a stationary wave such that the point x = 0 is a node. Then the equation of other wave is :-

Options

A. `y =a cos(kx - omega t)`
B. `y= a cos(kx + omega t)`
C. `y = -a sin (kx + omega t)`
D. `y = a sin (kx + omega t)`

Explanation

To solve the problem, we need to find the equation of the second wave that, when superimposed with the first wave given by \( y = a \sin(kx - \omega t) \), results in a stationary wave with a node at \( x = 0 \).

### Step-by-Step Solution:

1. **Understand the Condition for a Node**:
At a node, the displacement of the wave is always zero. This means that the resultant wave formed by the superposition of the two waves must equal zero at that point.

2. **Write the Equation for the Resultant Wave**:
The resultant wave \( y_{\text{resultant}} \) can be expressed as:
\[ y_{\text{resultant}} = y + y' = 0 \]
where \( y' \) is the equation of the second wave.

3. **Substituting the First Wave**:
Substitute the first wave equation into the resultant equation:
\[ a \sin(k \cdot 0 - \omega t) + y' = 0 \]
Simplifying this gives:
\[ a \sin(-\omega t) + y' = 0 \]
Since \( \sin(-\theta) = -\sin(\theta) \), we have:
\[ -a \sin(\omega t) + y' = 0 \]
Thus, we can express \( y' \) as:
\[ y' = a \sin(\omega t) \]

4. **Formulate the Equation of the Second Wave**:
The general form of a wave can be expressed as \( y' = A \sin(kx + \omega t) \) for some amplitude \( A \). To satisfy the condition of a node at \( x = 0 \), we can write:
\[ y' = a \sin(kx + \omega t) \]

5. **Final Equation**:
Therefore, the equation of the second wave that superimposes with the first wave to form a stationary wave with a node at \( x = 0 \) is:
\[ y' = a \sin(kx + \omega t) \]

6. **Conclusion**:
The correct option for the equation of the second wave is option D: \( y' = a \sin(kx + \omega t) \).