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 find the equation of the other wave that superimposes with the wave represented by \( y = a \sin(kx - \omega t) \) to form a stationary wave with a node at \( x = 0 \), we can follow these steps:

### Step 1: Understanding the Condition for a Node
A node in a stationary wave is a point where the amplitude is always zero. For the wave to have a node at \( x = 0 \), the resultant wave function must satisfy \( y(0, t) = 0 \) for all times \( t \).

### Step 2: Write the Expression for the Resultant Wave
The resultant wave \( y \) can be expressed as the sum of the two waves:
\[ y = y_1 + y_2 \]
where \( y_1 = a \sin(kx - \omega t) \) is the first wave and \( y_2 \) is the wave we need to find.

### Step 3: Condition at \( x = 0 \)
Substituting \( x = 0 \) into the equation:
\[ y(0, t) = a \sin(0 - \omega t) + y_2(0, t) = a \sin(-\omega t) + y_2(0, t) \]
Since \( \sin(-\omega t) = -\sin(\omega t) \), we have:
\[ y(0, t) = -a \sin(\omega t) + y_2(0, t) \]
For this to equal zero for all \( t \):
\[ y_2(0, t) = a \sin(\omega t) \]

### Step 4: Form of the Second Wave
To ensure that \( y_2 \) is a wave that satisfies the condition of being inverted with respect to the first wave, we can write:
\[ y_2 = -a \sin(kx - \omega t) \]
This wave is inverted because it has a negative sign in front of the amplitude.

### Step 5: Final Form of the Second Wave
However, we can also express the second wave in terms of a positive sine function:
\[ y_2 = a \sin(kx + \omega t) \]
This form also satisfies the condition of being inverted relative to the first wave, as it will still create nodes at the same points.

### Conclusion
Thus, the equation of the other wave is:
\[ y_2 = a \sin(kx + \omega t) \]