Three waves of amplitudes 7 mm, 3 mm and 11 mm with a successive phase difference of `(pi)/(2)` are super - imposed. The amplitude of the resulting wave will be

Options

A. 5 mm
B. 12 mm
C. 8 mm
D. 6 mm

Answer

To solve the problem of finding the amplitude of the resulting wave from the superposition of three waves with given amplitudes and phase differences, we can follow these steps:

### Step 1: Identify the Amplitudes and Phase Differences
We have three waves with the following amplitudes:
- \( A_1 = 7 \, \text{mm} \)
- \( A_2 = 3 \, \text{mm} \)
- \( A_3 = 11 \, \text{mm} \)

The phase differences between the waves are:
- The phase difference between \( A_1 \) and \( A_2 \) is \( \frac{\pi}{2} \).
- The phase difference between \( A_2 \) and \( A_3 \) is \( \frac{\pi}{2} \).
- The phase difference between \( A_1 \) and \( A_3 \) is \( \pi \) (completely out of phase).

### Step 2: Calculate the Resultant Amplitude of \( A_1 \) and \( A_3 \)
Since \( A_1 \) and \( A_3 \) are out of phase, we can calculate their resultant amplitude \( A_{13} \) as:
\[ A_{13} = |A_3 - A_1| = |11 \, \text{mm} - 7 \, \text{mm}| = 4 \, \text{mm} \]

### Step 3: Combine \( A_{13} \) with \( A_2 \)
Now, we need to combine the resultant amplitude \( A_{13} \) with \( A_2 \) (which has a phase difference of \( \frac{\pi}{2} \)). The formula for the resultant amplitude \( A \) when two waves are at a phase difference is given by:
\[ A = \sqrt{A_{13}^2 + A_2^2 + 2 A_{13} A_2 \cos(\phi)} \]
where \( \phi \) is the phase difference. Since \( \phi = \frac{\pi}{2} \), \( \cos(\frac{\pi}{2}) = 0 \). Thus, the formula simplifies to:
\[ A = \sqrt{A_{13}^2 + A_2^2} \]

### Step 4: Substitute the Values
Substituting the values we have:
\[ A = \sqrt{(4 \, \text{mm})^2 + (3 \, \text{mm})^2} \]
\[ A = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{mm} \]

### Final Answer
The amplitude of the resulting wave is \( 5 \, \text{mm} \).

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