A wave represented by the equation `y=acos(kx-omegat)` is superposed with another wave to form stationary wave such that the point x=0 is a node. The equation for the other wave is:

Options

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

Answer

To find the equation of the second wave that superposes with the first wave to form a stationary wave with a node at \( x = 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the First Wave**:
The first wave is given by the equation:
\[ y = A \cos(kx - \omega t) \]
This wave propagates in the positive x-direction.

2. **Characteristics of Stationary Waves**:
A stationary wave is formed by the superposition of two waves traveling in opposite directions. The points where the amplitude is always zero are called nodes.

3. **Direction of the Second Wave**:
Since the first wave is traveling in the positive x-direction, the second wave must travel in the negative x-direction. Therefore, its general form will be:
\[ y' = A \cos(kx + \omega t) \]

4. **Phase Difference Requirement**:
For the two waves to create nodes, they must have a phase difference of \( \pi \). This means that the second wave can be expressed as:
\[ y' = A \cos(kx + \omega t + \pi) \]
Since \( \cos(\theta + \pi) = -\cos(\theta) \), we can rewrite this as:
\[ y' = -A \cos(kx + \omega t) \]

5. **Combining the Waves**:
The total wave \( y_{net} \) is the sum of the two waves:
\[ y_{net} = y + y' = A \cos(kx - \omega t) - A \cos(kx + \omega t) \]

6. **Finding the Condition at the Node**:
For a node at \( x = 0 \), the amplitude must be zero. Substitute \( x = 0 \):
\[ y_{net} = A \cos(0 - \omega t) - A \cos(0 + \omega t) = A \cos(\omega t) - A \cos(\omega t) = 0 \]
This confirms that the point \( x = 0 \) is indeed a node.

7. **Final Equation of the Second Wave**:
Therefore, the equation for the second wave is:
\[ y' = -A \cos(kx + \omega t) \]