The stationary wave y = 2a sinkx cos`omega`t in a stretched string is the result of superposition of `y_(1)=a sin(kx-omegat)` and

Options

A. `y_(2)=acos(kx+omegat)`
B. `y_(2)=asin(kx+omegat)`
C. `y_(2)=acos(kx-omegat)`
D. `y_(2)=asin(kx-omegat)`

Explanation

To solve the problem, we need to find the second wave \( y_2 \) that, when superimposed with the first wave \( y_1 = a \sin(kx - \omega t) \), results in the stationary wave given by \( y = 2a \sin(kx) \cos(\omega t) \).

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The resultant wave is given as:
\[ y = 2a \sin(kx) \cos(\omega t) \]
- The first wave is:
\[ y_1 = a \sin(kx - \omega t) \]

2. **Use the Principle of Superposition**:
- According to the principle of superposition, the resultant wave \( y \) can be expressed as:
\[ y = y_1 + y_2 \]
- Therefore, we can express \( y_2 \) as:
\[ y_2 = y - y_1 \]

3. **Substitute the Expressions**:
- Substitute the expressions for \( y \) and \( y_1 \) into the equation for \( y_2 \):
\[ y_2 = 2a \sin(kx) \cos(\omega t) - a \sin(kx - \omega t) \]

4. **Simplify the Expression for \( y_2 \)**:
- We can use the trigonometric identity for sine of a difference:
\[ \sin(kx - \omega t) = \sin(kx) \cos(\omega t) - \cos(kx) \sin(\omega t) \]
- Substitute this identity into the equation:
\[ y_2 = 2a \sin(kx) \cos(\omega t) - a (\sin(kx) \cos(\omega t) - \cos(kx) \sin(\omega t)) \]
- This simplifies to:
\[ y_2 = 2a \sin(kx) \cos(\omega t) - a \sin(kx) \cos(\omega t) + a \cos(kx) \sin(\omega t) \]
- Combine like terms:
\[ y_2 = (2a - a) \sin(kx) \cos(\omega t) + a \cos(kx) \sin(\omega t) \]
- This gives us:
\[ y_2 = a \sin(kx) \cos(\omega t) + a \cos(kx) \sin(\omega t) \]

5. **Use the Sine Addition Formula**:
- The expression can be recognized as:
\[ y_2 = a (\sin(kx + \omega t)) \]
- Therefore, the second wave \( y_2 \) is:
\[ y_2 = a \sin(kx + \omega t) \]

### Final Answer:
The other wave \( y_2 \) is:
\[ y_2 = a \sin(kx + \omega t) \]