In stationary waves, distance between a node and its nearest antinode is 20 cm . The phase difference between two particles having a separation of 60 cm will be

Options

A. Zero
B. `pi//2`
C. `pi`
D. `3 pi //2`

Answer

To solve the problem, we need to find the phase difference between two particles separated by 60 cm in a stationary wave, given that the distance between a node and its nearest antinode is 20 cm.

### Step-by-Step Solution:

1. **Understand the relationship between nodes and antinodes**:
- In a stationary wave, the distance between a node and its nearest antinode is \( \frac{\lambda}{4} \) (where \( \lambda \) is the wavelength).
- Given that this distance is 20 cm, we can express this as:
\[ \frac{\lambda}{4} = 20 \text{ cm} \]

2. **Calculate the wavelength (\( \lambda \))**:
- To find the wavelength, we can rearrange the equation:
\[ \lambda = 4 \times 20 \text{ cm} = 80 \text{ cm} \]

3. **Determine the path difference for the given separation**:
- The path difference (\( \Delta \)) between two points in a wave is related to their separation (\( d \)) as follows:
\[ \Delta = \frac{d}{\lambda} \times \lambda \]
- For a separation of 60 cm:
\[ \Delta = 60 \text{ cm} \]

4. **Relate path difference to phase difference**:
- The phase difference (\( \phi \)) is related to the path difference by the formula:
\[ \phi = \frac{2\pi \Delta}{\lambda} \]

5. **Substitute the values**:
- Now we substitute \( \Delta = 60 \text{ cm} \) and \( \lambda = 80 \text{ cm} \):
\[ \phi = \frac{2\pi \times 60}{80} \]

6. **Simplify the expression**:
- This simplifies to:
\[ \phi = \frac{120\pi}{80} = \frac{3\pi}{2} \]

7. **Final answer**:
- Therefore, the phase difference between the two particles is:
\[ \phi = \frac{3\pi}{2} \]