Two sound waves having a phase difference of `60^(@)` have path difference of

Options

A. `(lambda)/(6)`
B. `(lambda)/(3)`
C. `2 lambda `
D. `(lambda)/(2)`

Answer

To find the path difference corresponding to a phase difference of \(60^\circ\) (or \(\frac{\pi}{3}\) radians), we can use the relationship between phase difference and path difference.

### Step-by-Step Solution:

1. **Understand the relationship between phase difference and path difference**:
The phase difference \(\phi\) between two waves is related to the path difference \(\Delta x\) by the equation:
\[ \phi = k \cdot \Delta x \]
where \(k\) is the wave number.

2. **Define the wave number \(k\)**:
The wave number \(k\) is defined as:
\[ k = \frac{2\pi}{\lambda} \]
where \(\lambda\) is the wavelength of the sound wave.

3. **Substitute \(k\) into the phase difference equation**:
Replacing \(k\) in the phase difference equation gives:
\[ \phi = \frac{2\pi}{\lambda} \cdot \Delta x \]

4. **Set the phase difference to the given value**:
We know the phase difference \(\phi\) is \(60^\circ\) or \(\frac{\pi}{3}\) radians. So, we can write:
\[ \frac{\pi}{3} = \frac{2\pi}{\lambda} \cdot \Delta x \]

5. **Solve for the path difference \(\Delta x\)**:
Rearranging the equation to solve for \(\Delta x\):
\[ \Delta x = \frac{\lambda}{6} \]
This is derived by multiplying both sides by \(\frac{\lambda}{2\pi}\):
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \frac{\pi}{3} = \frac{\lambda}{6} \]

### Final Answer:
The path difference corresponding to a phase difference of \(60^\circ\) is \(\Delta x = \frac{\lambda}{6}\).

---