Phase difference between two waves having same frequency `(v)` and same amplitude `(A)` is `2pi//3`. If these waves superimpose each other, then resultant amplitude will be–

Options

A. `2A`
B. `0`
C. `A`
D. `A^(2)`

Answer

To find the resultant amplitude when two waves superimpose with a phase difference of \( \frac{2\pi}{3} \), we can use the formula for the resultant amplitude of two waves. Here are the steps to solve the problem:

### Step 1: Identify the parameters
We have two waves with:
- Amplitude \( A_1 = A \)
- Amplitude \( A_2 = A \)
- Phase difference \( \Psi = \frac{2\pi}{3} \)

### Step 2: Use the formula for resultant amplitude
The formula for the resultant amplitude \( A_r \) when two waves superimpose is given by:
\[ A_r = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\Psi)} \]

### Step 3: Substitute the values into the formula
Since \( A_1 = A \) and \( A_2 = A \), we can substitute these values into the formula:
\[ A_r = \sqrt{A^2 + A^2 + 2A \cdot A \cdot \cos\left(\frac{2\pi}{3}\right)} \]

### Step 4: Simplify the expression
This simplifies to:
\[ A_r = \sqrt{2A^2 + 2A^2 \cos\left(\frac{2\pi}{3}\right)} \]

### Step 5: Calculate \( \cos\left(\frac{2\pi}{3}\right) \)
We know that:
\[ \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]
Substituting this value gives:
\[ A_r = \sqrt{2A^2 + 2A^2 \left(-\frac{1}{2}\right)} \]

### Step 6: Further simplify
This becomes:
\[ A_r = \sqrt{2A^2 - A^2} = \sqrt{A^2} = A \]

### Conclusion
Thus, the resultant amplitude when the two waves superimpose is:
\[ \boxed{A} \]

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