Two waves of amplitude `A_0 and xA_0` pass through a region . If ` x gt 1` , the difference in the maximum and minimum resultant amplitude possible is

Options

A. `(x+1)A_0`
B. `(x-1)A_0`
C. `2xA_0`
D. `2A_0`

Answer

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

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

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

4. **Find the Difference Between Maximum and Minimum Amplitudes:**
- The difference \( D \) between the maximum and minimum resultant amplitudes is:
\[ D = A_{\text{max}} - A_{\text{min}} = (x + 1)A_0 - (x - 1)A_0 \]
- Simplifying this gives:
\[ D = (x + 1)A_0 - (x - 1)A_0 = (x + 1 - x + 1)A_0 = 2A_0 \]

5. **Conclusion:**
- The difference in the maximum and minimum resultant amplitude possible is:
\[ D = 2A_0 \]

### Final Answer:
The difference in the maximum and minimum resultant amplitude possible is \( 2A_0 \).