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 find the amplitude of the resulting wave when three waves with different amplitudes and phase differences are superimposed, we can follow these steps:

### Step 1: Identify the amplitudes and phase differences
We have three waves with the following amplitudes:
- Wave 1: \( A_1 = 7 \, \text{mm} \)
- Wave 2: \( A_2 = 3 \, \text{mm} \)
- Wave 3: \( A_3 = 11 \, \text{mm} \)

The phase differences between successive waves are given as \( \frac{\pi}{2} \).

### Step 2: Represent the waves as vectors
Since the phase difference is \( \frac{\pi}{2} \), we can represent these waves as vectors in a 2D plane:
- The first wave (7 mm) can be represented along the x-axis.
- The second wave (3 mm) can be represented along the y-axis.
- The third wave (11 mm) will be at an angle of \( \frac{\pi}{2} \) from the second wave, which means it will also be perpendicular to the first two waves.

### Step 3: Calculate the resultant of the first two waves
To find the resultant of the first two waves, we can use the Pythagorean theorem since they are perpendicular:
\[ R_{12} = \sqrt{A_1^2 + A_2^2} = \sqrt{(7 \, \text{mm})^2 + (3 \, \text{mm})^2} \]
Calculating this gives:
\[ R_{12} = \sqrt{49 + 9} = \sqrt{58} \, \text{mm} \]

### Step 4: Calculate the resultant with the third wave
Now, we need to find the resultant of \( R_{12} \) and the third wave \( A_3 \):
\[ R = \sqrt{R_{12}^2 + A_3^2} = \sqrt{(\sqrt{58})^2 + (11 \, \text{mm})^2} \]
Calculating this gives:
\[ R = \sqrt{58 + 121} = \sqrt{179} \, \text{mm} \]

### Step 5: Final calculation
To find the numerical value:
\[ R \approx 13.38 \, \text{mm} \]

### Conclusion
The amplitude of the resulting wave is approximately \( 13.38 \, \text{mm} \).