Two waves of amplitudes `A_(0)` & `xA_(0)` pass through a region. If `x gt 1`, the difference in the maximum and minimum resultant amplitude possible it

Options

A. `(x+1)A_(0)`
B. `(x-1)A_(0)`
C. `2xA_(0)`
D. `2A_(0)`

Explanation

To solve the problem, we need to find the difference between the maximum and minimum resultant amplitudes of two waves with given amplitudes \( A_0 \) and \( xA_0 \), where \( x > 1 \).

### Step-by-Step Solution:

1. **Identify the Amplitudes**:
- Let the amplitude of the first wave be \( A_1 = A_0 \).
- Let the amplitude of the second wave be \( A_2 = xA_0 \).

2. **Calculate Maximum Resultant Amplitude**:
- The maximum resultant amplitude occurs when the two waves are in phase. This is given by:
\[ A_{\text{max}} = A_1 + A_2 = A_0 + xA_0 = (1 + x)A_0 \]

3. **Calculate Minimum Resultant Amplitude**:
- The minimum resultant amplitude occurs when the two waves are out of phase. This is given by:
\[ A_{\text{min}} = |A_1 - A_2| = |A_0 - xA_0| \]
- Since \( x > 1 \), we have:
\[ A_{\text{min}} = xA_0 - A_0 = (x - 1)A_0 \]

4. **Find the Difference Between Maximum and Minimum Amplitude**:
- Now, we can find the difference between the maximum and minimum resultant amplitudes:
\[ \Delta A = A_{\text{max}} - A_{\text{min}} = (1 + x)A_0 - (x - 1)A_0 \]
- Simplifying this gives:
\[ \Delta A = (1 + x - x + 1)A_0 = 2A_0 \]

5. **Final Result**:
- Therefore, the difference in the maximum and minimum resultant amplitude possible is:
\[ \Delta A = 2A_0 \]