Question

A loudspeaker at the origin emits a 120 Hz tone on a day when the speed of sound is 340 m/s. The phase difference between two points on the x-axis is 5.5 rad. What is the distance between these two points?

Short answer

The distance between the two points is$2.5m$.

Step-by-step solution

Given information

The frequency of the wave is 120 Hz.

Speed of sound is $340m/s$.

The phase difference between the two points is$∆\varphi =5.5rad$

Wavelength of the wave

The wavelength of the wave is given by $\lambda =\frac{v}{f}$.

Substituting the given values

$$\begin{gathered} \lambda =\frac{340}{120} \\ \lambda =2.83m \end{gathered}$$

The distance between the two points

The phase difference of the waves is given by $∆\varphi =\frac{2\mathrm{\pi }}{\lambda }∆x$.

Substitute given values

$$\begin{gathered} 5.5=\frac{2\mathrm{\pi }}{2.83}∆x \\ ∆x=2.477 \\ ∆x\approx 2.5m \end{gathered}$$

Therefore, the distance between two points is$2.5m$.