Question

A sound wave of frequency $100Hz$ is travelling in air. The speed of sound in air is $350m{s}^{-1}$.

(a) By how much is the phase changed at a given point in $2.5ms$?

(b) What is the phase difference at a given instant between two points separated by a distance of $10.0cm$ along the direction of propagation?

Answer 1

Step 1: Given data

Frequency, $f=100Hz$

Speed of sound in air $v=350m{s}^{-1}$

Phase difference $=\phi$

The distance travelled by the particle $=∆x$

Wavelength $=\lambda$

Time ,$t=2.5ms$

Step 2: (a) Find phase change

The velocity is given by

$$\begin{gathered} v=f\lambda \\ \lambda =v/f \\ \lambda =350/100 \\ \lambda =3.5m \end{gathered}$$

In $2.5ms$, the distance travelled by the particle,

$\Delta x=v\times t$

$\Delta x=(350x2.5x{10}^{-3})m$

Now, phase difference

$$\begin{gathered} \phi =(2\pi /\lambda )\Delta x \\ \phi =[2\pi x350x2.5\times {10}^{-3}]/[3.5] \\ \phi =\pi /2 \end{gathered}$$

Hence, the phase changed at a given point is $\phi =\pi /2$

Step 3: (b) Phase difference between two point

Distance between two points travelling along the direction of propagation is

$$\begin{gathered} \Delta x=10cm \\ \Delta x=0.1m \end{gathered}$$

The phase difference is calculated by,

$\phi =(2\pi /\lambda )\Delta x$

On substituting the values,

$$\begin{gathered} \phi =2\pi (0.1)/3.5 \\ \phi =2\pi /35 \end{gathered}$$

Hence, Phase difference between two point is $\phi =2\pi /35$.