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Two loudspeakers in a 20 degree C room emit 629 Hz sound waves along the x-axis. (a) If the...

Question:

Two loudspeakers in a {eq}20^\circ C {/eq} room emit {eq}629 \ Hz {/eq} sound waves along the {eq}x {/eq} -axis.

(a) If the speakers are in phase, what is the smallest distance between the speakers for which the interference of the sound waves is destructive?

(b) If the speakers are out of phase, what is the smallest distance between the speakers for which the interference of the sound waves is constructive?

Interference of Sound waves

Sound is a longitudinal wave and just like a transverse wave two sound waves will interfere each other and we will get constructive and destructive interference of sound. When constructive interference happens the sound intensity will be maximum and at destructive interference sound intensity can be zero or minimum. The condition for constructive and destructive interference remains the same as that for light. Constructive interference will happen when the path difference between the interfering waves is an integer multiple of wave length of the sound and destructive interference of sound will happen when the path difference is an odd multiple of half wavelength of the sound.

Answer and Explanation:

Given data

  • Temperature of the room {eq}t = 20^o \ C {/eq}
  • Frequency of the emitted sound {eq}F = 629 \ Hz {/eq}
  • Both the speakers emit sound along the x axis .

Speed of sound in the room {eq}v = 331.4 + 0.6 \times t \\ v = 331.4 + 0.6 \times 20 \\ v = 343.4 \ m/s {/eq}

Wavelength of the emitted sound {eq}\lambda = \dfrac { v } { F } \\ \lambda = \dfrac { 343.4 } { 629 } \\ \lambda = 0.546 \ m {/eq}

Part a )

Speakers are in phase.

The condition for destructive interference is path difference of the interfering sound must be an odd integer multiple of half wavelength.

The distance between the speakers is the path difference of the sound.

Let {eq}L_1 {/eq} be the distance of separation between the speakers

Then if {eq}L_1 = \dfrac { \lambda } { 2 } {/eq} we get destructive interference

therefore required separation between the speakers {eq}L_1 = \dfrac { \lambda } { 2 } \\ L_1 = 0.273 \ m {/eq}

Part b)

The speakers are out of phase or having a phase difference of {eq}\pi {/eq} or having an equivalent path difference {eq}\dfrac { \lambda } { 2 } {/eq}

For constructive interference the path difference must be an integer multiple of wave length.

Due to phase difference we have an equivalent path difference of half wavelength. So in addition to this if there is half the wavelength distance between the speakers, we get constructive interference

Then the minimum distance of separation between the speakers {eq}L = \dfrac { \lambda } { 2 } \\ L = 0.273 \ m {/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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