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Two point sources are in phase and emit a 482 Hz tone. What is the phase difference at a point...

Question:

Two point sources are in phase and emit a 482 Hz tone. What is the phase difference at a point that is 3.0 m from one source and 8.0 m from the other? (Take the air temperature to be 20{eq}^{\circ} {/eq}C.) ( )rad

Phase Difference at a Point Due to Two Sound Sources

Sources are said to be in phase if they produce sounds of same phase. When the sound waves travel in the air or in some medium, each wavelength corresponds to a phase difference of {eq}2 \pi {/eq}. So if we can calculate the path difference in terms of wavelength of the sound phase difference at a point can be calculated. If we know the frequency F of the sound and its velocity v in a medium wavelength of the sound can be calculated from the relation {eq}v = F \lambda {/eq}. Here {eq}\lambda {/eq} is the wavelength of the wave.

Answer and Explanation:

Given points

  • Frequency of the sound produced by both the sources F = 482 Hz
  • Distance from the first source to the point {eq}d_1 = 3 \ \ m {/eq}
  • Distance from the second source to the point {eq}d_2 = 8 \ \ m {/eq}
  • Speed of sound at 20 degree Celsius {eq}v = 343 \ \ m/s {/eq}

Wavelength of the sound in air {eq}\lambda = \dfrac { v } { F } \\ \lambda = \dfrac { 343 } { 482 } \\ \lambda = 7.116 \times 10^{-2 } \ \ m {/eq}

Path difference between the sound reaching at the point {eq}\Delta d = d_2 - d_1 \\ \Delta d = 8 - 3 \\ \Delta d = 5 m {/eq}

Number of wavelengths can exist in the given path difference {eq}n = \dfrac { \Delta d } { \lambda } \\ n = \dfrac { 5 } { 7.116 \times 10^{-1 } } \\ n = 7.0264 {/eq}

Each wavelength corresponds to {eq}2 \pi {/eq} phase difference

So the total phase difference {eq}\Delta \Phi = n \times 2 \pi \\ \Delta \Phi = 7.0264 \times 2 \pi \\ \Delta \Phi = 14.0528 \pi \ \ rad {/eq}


Learn more about this topic:

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Interference Patterns of Sound Waves

from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5
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