A transverse wave travelling along the positive x-axis, given by `y=A sin (kx - omega t)` is superposed with another wave travelling along the negative x-axis given by `y=A sin (kx + omega t)`. The point `x=0` is

Options

A. a node
B. an antinode
C. neither a node nor an antinode
D. a node or antinode depending on `t`

Explanation

To solve the problem, we need to analyze the superposition of two waves traveling in opposite directions along the x-axis.

### Step-by-Step Solution:

1. **Identify the Waves**:
We have two waves:
- Wave 1 traveling in the positive x-direction:
\[ y_1 = A \sin(kx - \omega t) \]
- Wave 2 traveling in the negative x-direction:
\[ y_2 = A \sin(kx + \omega t) \]

2. **Evaluate at \( x = 0 \)**:
We need to find the resultant wave at the point \( x = 0 \).
- For Wave 1:
\[ y_1(0, t) = A \sin(0 - \omega t) = A \sin(-\omega t) = -A \sin(\omega t) \]
- For Wave 2:
\[ y_2(0, t) = A \sin(0 + \omega t) = A \sin(\omega t) \]

3. **Superpose the Waves**:
Now, we add the two waves together at \( x = 0 \):
\[ y(0, t) = y_1(0, t) + y_2(0, t) = -A \sin(\omega t) + A \sin(\omega t) \]
This simplifies to:
\[ y(0, t) = 0 \]

4. **Interpret the Result**:
The resultant wave at \( x = 0 \) is zero, which indicates that there is complete destructive interference at this point.

5. **Determine the Nature of the Point**:
Since the amplitude is zero at \( x = 0 \), this point is a **node** (where the amplitude is minimum).

### Conclusion:
Thus, the point \( x = 0 \) is a node.