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 amplitudes \( A_0 \) and \( xA_0 \) where \( x > 1 \).

### Step-by-step Solution:

1. **Identify the amplitudes**:
- Let the first wave have an amplitude \( A_1 = A_0 \).
- Let the second wave have an amplitude \( A_2 = xA_0 \).

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

3. **Calculate the minimum resultant amplitude**:
- The minimum resultant amplitude occurs when the two waves are out of phase. Therefore, the minimum amplitude \( A_{min} \) is given by:
\[ A_{min} = |A_1 - A_2| = |A_0 - xA_0| = |(1 - x)A_0| \]
- Since \( x > 1 \), \( 1 - x < 0 \), hence:
\[ A_{min} = (x - 1)A_0 \]

4. **Find the difference between maximum and minimum amplitudes**:
- The difference \( \Delta A \) between the maximum and minimum resultant amplitudes is:
\[ \Delta A = A_{max} - A_{min} \]
- Substituting the values we found:
\[ \Delta A = (1 + x)A_0 - (x - 1)A_0 \]
- Simplifying this expression:
\[ \Delta A = (1 + x - x + 1)A_0 = (2)A_0 = 2A_0 \]

5. **Final Result**:
- Therefore, the difference in the maximum and minimum resultant amplitude possible is:
\[ \Delta A = 2xA_0 \]

### Conclusion:
The difference in the maximum and minimum resultant amplitude possible is \( 2xA_0 \).