Question

Question: Two sound waves, from two different sources with the same frequency,540 Hz,travel in the same direction at330 m/s. The sources are in phase. What is the phase difference of the wave at a point that is 4.40 mfrom one sound andfrom the other?

Short answer

Answer

The phase difference of the wave at a point that is 4.40 m from one sound and 4.00m from the other is 4. 12 rad.

Step-by-step solution

Given

1. Frequency of the wave$f=54\text{0}\text{Hz}$
2. Velocity of the wave 330 m/s
3. Separation distance from one source is$x_1=4.\text{40m}$
4. Separation distance from the other source is$x_2=4.\text{00m}$

Determining the concept

Phase difference is the difference between the phase constants of the two waves. Using the given information, find the phase constant of each wave. Using these phase constant values, find the phase difference.

Formula is as follow:

$∆f=2\pi \left(\frac{x_1}{\lambda }+ft\right)-2\pi \left(\frac{x_2}{\lambda }-ft\right)$

where,$∆\varphi$ is change in phase difference, $\lambda$is wavelength, x is displacement,tis time and f is frequency.

Determining the phase difference

Let, the separation between the point and two sources beand, then by using the formula, find the phase difference,

$$\begin{gathered} ∆f=f_1-f_2 \\ =2\pi \left(\frac{x_1}{\lambda }+ft\right)-2\pi \left(\frac{x_2}{\lambda }-ft\right) \\ =2\pi \left(\frac{x_1-x_2}{\lambda }\right) \end{gathered}$$

For the given values, the above equation becomes-

$$\begin{gathered} ∆f=2\pi \left(\frac{4.40\text{m}-4.00\text{m}}{\frac{v}{f}}\right) \\ =2\pi \left(\frac{4.40\text{m}-4.00\text{m}}{\frac{330m/s}{540\text{Hz}}}\right) \\ =2\pi \left(\frac{0.40\text{m}}{0.62\text{m}}\right) \\ =4.11\text{rad} \end{gathered}$$

Hence, the phase difference of the wave at a point that is 4.40 m from one sound and 4.00m from the other is 4.12 rad .