The resultant amplitude, when two waves of two waves of same frequency but with amplitudes `a_(1)` and `a_(2)` superimpose at phase difference of `pi//2` will be :-

Options

A. `a_(1) + a_(2)`
B. `a_(1) - a_(2)`
C. `sqrt(a_(1)^(2) + a_(2)^(2))`
D. `a_(1)^(2) + a_(2)^(2)`

Explanation

To find the resultant amplitude when two waves of the same frequency but with amplitudes \( a_1 \) and \( a_2 \) superimpose at a phase difference of \( \frac{\pi}{2} \), we can follow these steps:

### Step 1: Write the equations of the two waves
The equations of the two waves can be expressed as:
- \( y_1 = a_1 \sin(\omega t) \)
- \( y_2 = a_2 \sin(\omega t + \frac{\pi}{2}) \)

### Step 2: Convert the second wave using trigonometric identities
Using the trigonometric identity for sine, we can rewrite \( y_2 \):
- \( y_2 = a_2 \sin(\omega t + \frac{\pi}{2}) = a_2 \cos(\omega t) \)

### Step 3: Write the resultant wave equation
The resultant wave \( y \) is the sum of the two waves:
- \( y = y_1 + y_2 = a_1 \sin(\omega t) + a_2 \cos(\omega t) \)

### Step 4: Use the Pythagorean theorem to find the resultant amplitude
To find the resultant amplitude \( A \), we can treat the components \( a_1 \) and \( a_2 \) as the legs of a right triangle:
- The resultant amplitude \( A \) can be calculated using the formula:
\[ A = \sqrt{a_1^2 + a_2^2} \]

### Step 5: Conclusion
Thus, the resultant amplitude when two waves of the same frequency and a phase difference of \( \frac{\pi}{2} \) superimpose is:
\[ A = \sqrt{a_1^2 + a_2^2} \]