Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources ?

Explanation

To solve the problem, we need to determine the phase difference between two sound waves arriving at a point from two sources. We will analyze two cases: one where the point is on the line joining the sources and the other where the point is on the perpendicular bisector of the line joining the sources.

### Given Data:
- Separation between the two sources (S1 and S2): \(d = 10 \, \text{cm}\)
- Wavelength of the sound waves: \(\lambda = 5.0 \, \text{cm}\)
- Distance from one source to the point (P): \(D = 20 \, \text{cm}\)

### (a) Phase Difference on the Line Joining the Sources

1. **Calculate the distance from S2 to the point P:**
- The distance from S1 to P is given as \(D = 20 \, \text{cm}\).
- The distance from S2 to P can be calculated as:
\[ D_{S2} = D + d = 20 \, \text{cm} + 10 \, \text{cm} = 30 \, \text{cm} \]

2. **Calculate the path difference:**
- The path difference (\(\Delta x\)) is the difference in distance from the two sources to the point P:
\[ \Delta x = D_{S2} - D_{S1} = 30 \, \text{cm} - 20 \, \text{cm} = 10 \, \text{cm} \]

3. **Calculate the phase difference:**
- The phase difference (\(\Delta \phi\)) is given by the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
- Substituting the values:
\[ \Delta \phi = \frac{2\pi}{5 \, \text{cm}} \times 10 \, \text{cm} = 4\pi \, \text{radians} \]

### (b) Phase Difference on the Perpendicular Bisector

1. **Identify distances from both sources to point P:**
- On the perpendicular bisector, the distances from S1 and S2 to point P are equal. Let this distance be \(d\).
- Therefore, \(D_{S1} = D_{S2} = d\).

2. **Calculate the path difference:**
- Since both distances are equal, the path difference (\(\Delta x\)) is:
\[ \Delta x = D_{S2} - D_{S1} = d - d = 0 \]

3. **Calculate the phase difference:**
- The phase difference is:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{5 \, \text{cm}} \times 0 = 0 \, \text{radians} \]

### Summary of Results:
- (a) Phase difference on the line joining the sources: \(4\pi \, \text{radians}\)
- (b) Phase difference on the perpendicular bisector: \(0 \, \text{radians}\)