The phase difference between two waves reaching a point is `(pi)/(2)`. What is the resultant amplitude. If the individual amplitude are 3 mm and 4mm ?

Options

A. 1mm
B. 2mm
C. 3mm
D. 5mm

Answer

To find the resultant amplitude of two waves with given individual amplitudes and a phase difference, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Amplitude of wave A, \( A = 3 \, \text{mm} \)
- Amplitude of wave B, \( B = 4 \, \text{mm} \)
- Phase difference, \( \theta = \frac{\pi}{2} \)

2. **Understand the Phase Difference**:
- A phase difference of \( \frac{\pi}{2} \) radians means that the two waves are at right angles to each other. This can be visualized as two vectors, one representing each wave.

3. **Use the Resultant Amplitude Formula**:
- The formula for the resultant amplitude \( R \) when two waves are at a phase difference \( \theta \) is given by:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]
- Since \( \theta = \frac{\pi}{2} \), we know that \( \cos\left(\frac{\pi}{2}\right) = 0 \). Therefore, the formula simplifies to:
\[ R = \sqrt{A^2 + B^2} \]

4. **Substitute the Values**:
- Now substitute the values of \( A \) and \( B \):
\[ R = \sqrt{(3 \, \text{mm})^2 + (4 \, \text{mm})^2} \]

5. **Calculate the Squares**:
- Calculate \( 3^2 \) and \( 4^2 \):
\[ R = \sqrt{9 \, \text{mm}^2 + 16 \, \text{mm}^2} \]

6. **Add the Squares**:
- Add the results:
\[ R = \sqrt{25 \, \text{mm}^2} \]

7. **Take the Square Root**:
- Finally, take the square root:
\[ R = 5 \, \text{mm} \]

### Final Result:
The resultant amplitude of the two waves is \( 5 \, \text{mm} \).

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