Question

A sound wave frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s⁻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?

Answer 1

Given:
Speed of sound in air v = 350 m/s
Frequency of sound wave f = 100 Hz
a) As we know, 

\[v = f\lambda\]

\[\therefore   \lambda = \frac{v}{f}\] 

\[ \Rightarrow \lambda = \frac{350}{100} = 3 . 5  m\]

Distance travelled by the particle:
Δx = (350 × 2.5 × 10⁻³) m

Phase difference is given by:

\[\phi = \frac{2\pi}{\lambda} \times  ∆ x\] 

\[\text { On  substituting  the  values  we  get: }\] 

\[\phi = \left( \frac{2\pi \times 350 \times 2 . 5 \times {10}^{- 3}}{3 . 5} \right)\] 

\[ \Rightarrow   \phi = \left( \frac{\pi}{2} \right)\]

(b) For the second case:
Distance between the two points:

\[∆ x\]= 10 cm = 0.1 m

\[\Rightarrow   \phi = \frac{2\pi}{\lambda} ∆ x\] 

\[\text { On  substituting  the  respective  values  in  the  above  equation,   we  get: }\] \[\phi = \frac{2\pi \times 0 . 1}{3 . 5} = \frac{2\pi}{35}\]

The phase difference between the two points is \[\frac{2\pi}{35}\]