A pair of monochromatic waves of amplitude A and 2 A are travelling in same direction. Determine the amplitude of the resultant wave if both the waves are superimposed and phase difference of `45^@`.

Answer

To find the amplitude of the resultant wave when two monochromatic waves with different amplitudes are superimposed, we can use the formula for the resultant amplitude when two waves interfere:

\[ R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi} \]

Where:
- \( R \) is the resultant amplitude,
- \( A_1 \) and \( A_2 \) are the amplitudes of the two waves,
- \( \phi \) is the phase difference between the two waves.

### Step 1: Identify the given values
- Amplitude of the first wave, \( A_1 = A \)
- Amplitude of the second wave, \( A_2 = 2A \)
- Phase difference, \( \phi = 45^\circ \)

### Step 2: Convert the phase difference to cosine
We know that:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \approx 0.7071 \]

### Step 3: Substitute the values into the formula
Now, substituting the values into the resultant amplitude formula:
\[ R = \sqrt{A^2 + (2A)^2 + 2 \cdot A \cdot 2A \cdot \cos(45^\circ)} \]

### Step 4: Calculate each term
- \( A^2 = A^2 \)
- \( (2A)^2 = 4A^2 \)
- \( 2 \cdot A \cdot 2A \cdot \cos(45^\circ) = 4A^2 \cdot \frac{1}{\sqrt{2}} = \frac{4A^2}{\sqrt{2}} \)

### Step 5: Combine the terms
Now we can combine the terms:
\[ R = \sqrt{A^2 + 4A^2 + \frac{4A^2}{\sqrt{2}}} \]
\[ = \sqrt{5A^2 + \frac{4A^2}{\sqrt{2}}} \]

### Step 6: Factor out \( A^2 \)
\[ R = A \sqrt{5 + \frac{4}{\sqrt{2}}} \]

### Step 7: Simplify the expression
To simplify \( \frac{4}{\sqrt{2}} \):
\[ \frac{4}{\sqrt{2}} = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \]
Thus,
\[ R = A \sqrt{5 + 2\sqrt{2}} \]

### Step 8: Calculate the numerical value
To find a numerical approximation, we can calculate:
\[ 5 + 2\sqrt{2} \approx 5 + 2 \cdot 1.414 \approx 5 + 2.828 \approx 7.828 \]
So,
\[ R \approx A \sqrt{7.828} \approx A \cdot 2.7988 \]

### Final Result
The amplitude of the resultant wave is approximately:
\[ R \approx 2.7988A \]