A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is `350 ms^-1` (a) By how much is the phase changed at a given point in 2.5 ms ? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation ?

Explanation

To solve the problem step by step, we will address both parts (a) and (b) of the question.

### Given Data:
- Frequency of sound wave, \( f = 100 \, \text{Hz} \)
- Speed of sound in air, \( v = 350 \, \text{ms}^{-1} \)
- Time interval for part (a), \( t = 2.5 \, \text{ms} = 2.5 \times 10^{-3} \, \text{s} \)
- Distance for part (b), \( d = 10.0 \, \text{cm} = 0.1 \, \text{m} \)

### (a) Phase Change at a Given Point in 2.5 ms

1. **Calculate the wavelength (\( \lambda \))**:
\[ \lambda = \frac{v}{f} = \frac{350 \, \text{ms}^{-1}}{100 \, \text{Hz}} = 3.5 \, \text{m} \]

2. **Calculate the distance travelled by the wave in 2.5 ms**:
\[ \Delta x = v \times t = 350 \, \text{ms}^{-1} \times 2.5 \times 10^{-3} \, \text{s} = 0.875 \, \text{m} \]

3. **Calculate the phase change (\( \phi \))**:
\[ \phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{3.5} \times 0.875 \]
\[ \phi = \frac{2\pi \times 0.875}{3.5} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{radians} \]

### (b) Phase Difference Between Two Points Separated by 10.0 cm

1. **Convert distance to meters**:
\[ d = 10.0 \, \text{cm} = 0.1 \, \text{m} \]

2. **Calculate the phase difference (\( \Delta \phi \))**:
\[ \Delta \phi = \frac{2\pi}{\lambda} \times d = \frac{2\pi}{3.5} \times 0.1 \]
\[ \Delta \phi = \frac{2\pi \times 0.1}{3.5} = \frac{0.2\pi}{3.5} = \frac{2\pi}{35} \, \text{radians} \]

### Final Answers:
- (a) The phase change at a given point in 2.5 ms is \( \frac{\pi}{2} \, \text{radians} \).
- (b) The phase difference at a given instant between two points separated by a distance of 10.0 cm is \( \frac{2\pi}{35} \, \text{radians} \).