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, $v=f\lambda$.
$$\begin{gathered} \therefore \lambda =\frac{v}{f} \\ \Rightarrow \lambda =\frac{350}{100}=3.5\mathrm{m} \end{gathered}$$
Distance travelled by the particle:
Δx = (350 × 2.5 × 10⁻³) m

Phase difference is given by:
$$\begin{gathered} \varphi =\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times ∆x \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{we}\mathrm{get}: \\ \varphi =\left(\frac{2\mathrm{\pi }\times 350\times 2.5\times {10}^{-3}}{3.5}\right) \\ \Rightarrow \varphi =\left(\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$
(b) For the second case:
Distance between the two points:
$∆x$ = 10 cm = 0.1 m
$$\begin{gathered} \Rightarrow \varphi =\frac{2\pi }{\lambda }∆x \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation},\mathrm{we}\mathrm{get}: \\ \varphi =\frac{2\mathrm{\pi }\times 0.1}{3.5}=\frac{2\mathrm{\pi }}{35} \end{gathered}$$
The phase difference between the two points is $\frac{2\pi }{35}$.