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 find the phase difference between two waves of equal amplitude that superpose to give a resultant wave with the same amplitude as either of the individual waves, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Let the amplitude of both waves be \( a_1 = a \) and \( a_2 = a \).
- The resultant amplitude \( R \) is also given to be \( R = a \).

2. **Use the Formula for Resultant Amplitude:**
- The formula for the resultant amplitude \( R \) when two waves of amplitudes \( a_1 \) and \( a_2 \) interfere is:
\[ R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi} \]
- Substituting the values, we have:
\[ R = \sqrt{a^2 + a^2 + 2a \cdot a \cos \phi} \]
- This simplifies to:
\[ R = \sqrt{2a^2 + 2a^2 \cos \phi} \]

3. **Set Up the Equation:**
- Since \( R = a \), we can set up the equation:
\[ a = \sqrt{2a^2 + 2a^2 \cos \phi} \]

4. **Square Both Sides:**
- Squaring both sides to eliminate the square root gives:
\[ a^2 = 2a^2 + 2a^2 \cos \phi \]

5. **Rearrange the Equation:**
- Rearranging the equation leads to:
\[ a^2 - 2a^2 = 2a^2 \cos \phi \]
- This simplifies to:
\[ -a^2 = 2a^2 \cos \phi \]

6. **Solve for \( \cos \phi \):**
- Dividing both sides by \( 2a^2 \) gives:
\[ \cos \phi = -\frac{1}{2} \]

7. **Find the Phase Difference \( \phi \):**
- The angle whose cosine is \( -\frac{1}{2} \) is:
\[ \phi = \cos^{-1}\left(-\frac{1}{2}\right) \]
- This corresponds to:
\[ \phi = 120^\circ \quad \text{or} \quad \phi = \frac{2\pi}{3} \text{ radians} \]

8. **Conclusion:**
- Therefore, the phase difference between the two waves is \( \phi = \frac{2\pi}{3} \text{ radians} \).