Three coherent waves having amplitudes 12mm, 6mm and 4mm arrive at a given point with successive phase difference of `pi/2`. Then, the amplitude of the resultant wave is

Options

A. 7mm
B. 10mm
C. 5mm
D. 4.8mm

Explanation

To find the amplitude of the resultant wave from three coherent waves with given amplitudes and phase differences, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Amplitudes and Phase Differences**:
- The amplitudes of the three waves are:
- \( A_1 = 12 \, \text{mm} \)
- \( A_2 = 6 \, \text{mm} \)
- \( A_3 = 4 \, \text{mm} \)
- The phase differences between the waves are \( \frac{\pi}{2} \) radians (or 90 degrees).

2. **Determine the Effective Amplitude of the First Two Waves**:
- Since the phase difference between \( A_1 \) and \( A_2 \) is \( 90^\circ \), we can represent them as vectors in a two-dimensional plane.
- The amplitude \( A_1 \) can be taken along the x-axis, and \( A_2 \) can be taken along the y-axis.
- Thus, we have:
- \( A_1 = 12 \, \text{mm} \) (along x-axis)
- \( A_2 = 6 \, \text{mm} \) (along y-axis)

3. **Calculate the Resultant Amplitude of \( A_1 \) and \( A_2 \)**:
- The resultant amplitude \( A_{13} \) of \( A_1 \) and \( A_3 \) (considering \( A_3 \) is in the opposite direction of \( A_1 \)) can be calculated as:
\[ A_{13} = A_1 - A_3 = 12 \, \text{mm} - 4 \, \text{mm} = 8 \, \text{mm} \]

4. **Combine \( A_{13} \) and \( A_2 \)**:
- Now, we need to find the resultant amplitude of \( A_{13} \) and \( A_2 \) which are perpendicular to each other.
- We can use the Pythagorean theorem to find the resultant amplitude \( A_R \):
\[ A_R = \sqrt{A_{13}^2 + A_2^2} \]
- Substitute the values:
\[ A_R = \sqrt{(8 \, \text{mm})^2 + (6 \, \text{mm})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \, \text{mm} \]

5. **Final Result**:
- The amplitude of the resultant wave is \( 10 \, \text{mm} \).