Two periodic waves of amplitudes `A_1 and A_2` pass though a region. If `A_1gtA_2`, the difference in the maximum and minimum resultant amplitude possible is

Options

A. `2A_1`
B. `2A_2`
C. `A_1+A_2`
D. `A_1-A_2`

Answer

To solve the problem of finding the difference in the maximum and minimum resultant amplitude of two periodic waves with amplitudes \( A_1 \) and \( A_2 \) (where \( A_1 > A_2 \)), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Resultant Amplitude**:
The resultant amplitude \( A_R \) of two waves can be calculated using the formula:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi} \]
where \( \phi \) is the phase difference between the two waves.

2. **Maximum Amplitude**:
The maximum resultant amplitude occurs when the two waves are in phase, i.e., \( \phi = 0^\circ \). In this case, \( \cos 0^\circ = 1 \):
\[ A_{\text{max}} = A_1 + A_2 \]

3. **Minimum Amplitude**:
The minimum resultant amplitude occurs when the two waves are out of phase, i.e., \( \phi = 180^\circ \). In this case, \( \cos 180^\circ = -1 \):
\[ A_{\text{min}} = |A_1 - A_2| = A_1 - A_2 \quad (\text{since } A_1 > A_2) \]

4. **Calculating the Difference**:
Now, we need to find the difference between the maximum and minimum resultant amplitudes:
\[ \text{Difference} = A_{\text{max}} - A_{\text{min}} \]
Substituting the values we found:
\[ \text{Difference} = (A_1 + A_2) - (A_1 - A_2) \]
Simplifying this gives:
\[ \text{Difference} = A_1 + A_2 - A_1 + A_2 = 2A_2 \]

5. **Final Result**:
Therefore, the difference in the maximum and minimum resultant amplitude possible is:
\[ \boxed{2A_2} \]