Two waves of equal amplitude when superposed, give a resultant wave having an amplitude equal to that of either wave. The phase difference between the two waves is

Options

A. `(pi)/(3)` radian
B. zero
C. `(pi)/(2)` radian
D. `(2pi)/(3)` radian

Answer

To solve the problem, we need to find the phase difference between two waves of equal amplitude that result in a resultant wave with the same amplitude as either of the individual waves.

### Step-by-Step Solution:

1. **Define the Amplitudes**:
Let the amplitude of both waves be \( A_1 = A \) and \( A_2 = A \).

2. **Resultant Amplitude**:
The resultant amplitude \( A_R \) when two waves superpose can be expressed using the formula:
\[ A_R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi} \]
Given that \( A_R = A \), we can substitute the values:
\[ A = \sqrt{A^2 + A^2 + 2 A \cdot A \cos \phi} \]

3. **Simplify the Equation**:
Squaring both sides gives:
\[ A^2 = A^2 + A^2 + 2A^2 \cos \phi \]
This simplifies to:
\[ A^2 = 2A^2 + 2A^2 \cos \phi \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ A^2 - 2A^2 = 2A^2 \cos \phi \]
Simplifying further leads to:
\[ -A^2 = 2A^2 \cos \phi \]

5. **Solving for Cosine**:
Dividing both sides by \( A^2 \) (assuming \( A \neq 0 \)):
\[ -1 = 2 \cos \phi \]
Thus, we find:
\[ \cos \phi = -\frac{1}{2} \]

6. **Finding the Phase Difference**:
The angle \( \phi \) that corresponds to \( \cos \phi = -\frac{1}{2} \) is:
\[ \phi = 120^\circ \quad \text{or} \quad \phi = \frac{2\pi}{3} \text{ radians} \]

### Final Answer:
The phase difference between the two waves is \( 120^\circ \) or \( \frac{2\pi}{3} \) radians.

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