`y_(1)=Asin(omegat-kx)&y_(2)=Asin(omegat-kx+delta)`: these two equations are representing two waves. Then the amplitude of the resulting wave is

Options

A. `2Acosdelta`
B. `2Acos(delta//2)`
C. `Acosdeltasindelta`
D. none of these

Answer

To find the amplitude of the resulting wave from the two given wave equations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Waves**:
The two wave equations are:
\[ y_1 = A \sin(\omega t - kx) \]
\[ y_2 = A \sin(\omega t - kx + \delta) \]

2. **Superposition of Waves**:
When two waves interfere, the resultant wave \( y \) can be expressed as:
\[ y = y_1 + y_2 \]
Therefore, we have:
\[ y = A \sin(\omega t - kx) + A \sin(\omega t - kx + \delta) \]

3. **Using the Phasor Representation**:
We can represent the two waves as phasors. The first wave \( y_1 \) can be represented as a vector of length \( A \) at an angle of \( 0 \) degrees (or radians), and the second wave \( y_2 \) as a vector of length \( A \) at an angle \( \delta \).

4. **Resultant Amplitude Calculation**:
The resultant amplitude \( A_R \) can be calculated using the formula:
\[ A_R = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos(\delta)} \]
This simplifies to:
\[ A_R = \sqrt{2A^2 + 2A^2 \cos(\delta)} \]

5. **Factor Out Common Terms**:
We can factor out \( 2A^2 \):
\[ A_R = \sqrt{2A^2(1 + \cos(\delta))} \]

6. **Use the Cosine Identity**:
We can use the identity \( 1 + \cos(\delta) = 2 \cos^2(\delta/2) \):
\[ A_R = \sqrt{2A^2 \cdot 2 \cos^2(\delta/2)} \]

7. **Final Result for Amplitude**:
This simplifies to:
\[ A_R = 2A \cos(\delta/2) \]

Thus, the amplitude of the resulting wave is:
\[ A_R = 2A \cos\left(\frac{\delta}{2}\right) \]