`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 formed by the superposition of the two waves given by the equations \( y_1 = A \sin(\omega t - kx) \) and \( y_2 = A \sin(\omega t - kx + \delta) \), we can follow these steps:

### Step 1: Identify the phases of the waves
The phase of the first wave \( y_1 \) is:
\[ \phi_1 = \omega t - kx \]
The phase of the second wave \( y_2 \) is:
\[ \phi_2 = \omega t - kx + \delta \]

### Step 2: Determine the phase difference
The phase difference \( \Delta \phi \) between the two waves is:
\[ \Delta \phi = \phi_2 - \phi_1 = (\omega t - kx + \delta) - (\omega t - kx) = \delta \]

### Step 3: Use the formula for the resultant amplitude
When two waves of the same amplitude \( A \) interfere, the resultant amplitude \( A_r \) can be calculated using the formula:
\[ A_r = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos(\Delta \phi)} \]
Substituting \( A_1 = A \) and \( A_2 = A \):
\[ A_r = \sqrt{A^2 + A^2 + 2A^2 \cos(\delta)} \]

### Step 4: Simplify the expression
This simplifies to:
\[ A_r = \sqrt{2A^2(1 + \cos(\delta))} \]
Factoring out \( A^2 \):
\[ A_r = A \sqrt{2(1 + \cos(\delta))} \]

### Step 5: Use the trigonometric identity
Using the identity \( 1 + \cos(\delta) = 2 \cos^2\left(\frac{\delta}{2}\right) \):
\[ A_r = A \sqrt{2 \cdot 2 \cos^2\left(\frac{\delta}{2}\right)} = A \cdot 2 \cos\left(\frac{\delta}{2}\right) \]

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