Equation for two waves is given as `y_(1)=asin(omegat+phi_(1)), y_(2)=asin(omegat+phi_(2))`.
If ampitude and time period of resultant wave does not change, then calculate `(phi_(1)-phi_(2))`.

Options

A. `pi/3`
B. `2pi/3`
C. `pi/6`
D. `pi/4`

Answer

To solve the problem, we need to analyze the given equations of the two waves and find the phase difference between them under the condition that the amplitude and time period of the resultant wave do not change.

### Step-by-Step Solution:

1. **Write the equations of the two waves**:
\[ y_1 = A \sin(\omega t + \phi_1) \]
\[ y_2 = A \sin(\omega t + \phi_2) \]

2. **Find the resultant wave**:
The resultant wave \( y \) is given by the sum of the two waves:
\[ y = y_1 + y_2 = A \sin(\omega t + \phi_1) + A \sin(\omega t + \phi_2) \]

3. **Factor out the amplitude**:
We can factor out \( A \):
\[ y = A \left( \sin(\omega t + \phi_1) + \sin(\omega t + \phi_2) \right) \]

4. **Apply the sine addition formula**:
Using the formula \( \sin a + \sin b = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \):
\[ y = A \left( 2 \sin\left(\frac{(\omega t + \phi_1) + (\omega t + \phi_2)}{2}\right) \cos\left(\frac{(\omega t + \phi_1) - (\omega t + \phi_2)}{2}\right) \right) \]
Simplifying this gives:
\[ y = 2A \cos\left(\frac{\phi_1 - \phi_2}{2}\right) \sin\left(\omega t + \frac{\phi_1 + \phi_2}{2}\right) \]

5. **Identify the amplitude of the resultant wave**:
The amplitude of the resultant wave is:
\[ A' = 2A \cos\left(\frac{\phi_1 - \phi_2}{2}\right) \]

6. **Set the amplitude condition**:
Since the amplitude of the resultant wave does not change, we have:
\[ A' = A \implies 2A \cos\left(\frac{\phi_1 - \phi_2}{2}\right) = A \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 2 \cos\left(\frac{\phi_1 - \phi_2}{2}\right) = 1 \]

7. **Solve for the phase difference**:
\[ \cos\left(\frac{\phi_1 - \phi_2}{2}\right) = \frac{1}{2} \]
The angles for which cosine equals \( \frac{1}{2} \) are:
\[ \frac{\phi_1 - \phi_2}{2} = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad \frac{\phi_1 - \phi_2}{2} = -\frac{\pi}{3} + 2n\pi \quad (n \in \mathbb{Z}) \]

Therefore:
\[ \phi_1 - \phi_2 = \frac{2\pi}{3} + 4n\pi \quad \text{or} \quad \phi_1 - \phi_2 = -\frac{2\pi}{3} + 4n\pi \]

For the simplest case (where \( n=0 \)):
\[ \phi_1 - \phi_2 = \frac{2\pi}{3} \quad \text{or} \quad \phi_1 - \phi_2 = -\frac{2\pi}{3} \]

### Final Answer:
\[ \phi_1 - \phi_2 = \frac{2\pi}{3} \text{ radians} \quad \text{or} \quad \phi_1 - \phi_2 = -\frac{2\pi}{3} \text{ radians} \]