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Two loudspeakers located along the x-axis as shown in the figure produce sounds of equal...

Question:

Two loudspeakers located along the x-axis as shown in the figure produce sounds of equal frequency. Speaker 1 is at the origin, while the location of speaker 2 be varied by a remote control wielded by the listener. He notices maxima in the sound intensity when speaker 2 is located at x = 0.840 m and 1.12 m, but at no points in between.

What is the frequency of the sound? Assume the speed of sound is 340 m/s.

f = _____ Hz

Frequency of the Sound From Constructive Interference

When two sound waves of same frequency, amplitude and wavelength interfere at a point, if the path difference between the two interfering waves are either zero or an integer multiple of the wavelength, constructive interference or large sound occurs at that point. If the path difference is an odd integer multiple of half wavelength, the destructive interference or minimum sound is observed at that point.

Answer and Explanation:

Given points

  • Speakers are producing same frequency F
  • First speaker is at the origin that is at x = 0 m
  • Sound maxima is observed when the second speaker is at {eq}x_1 = 0.840 \ m {/eq}
  • The next sound maxima is observed when the second speaker is at {eq}x_2 = 1.12 \ m {/eq}
  • Speed of sound in air v = 340 m/s

Let {eq}\lambda {/eq} be the wavelength of the sound.

At the position {eq}x_1 {/eq} we get some order of constructive interference, so we can write {eq}\Delta x_1 = m \lambda {/eq}

Here the path difference {eq}\Delta x_1 = x_1 - x \\ \Delta x_1 = 0.840 - 0.0 \\ \Delta x_1 = 0.840 \ m {/eq}

Again next constructive interference is observed at {eq}x_2 {/eq}.

Then we can write {eq}\Delta x_2 = ( m+1 ) \lambda {/eq}

Here the path difference {eq}\Delta x_2 = 1.12 \ m {/eq}

If we subtract the first path difference from the second path difference we get

{eq}\Delta x_2 - \Delta x_1 = ( m+1 ) \lambda = m \lambda \\ \implies 1.12 - 0.840 = \lambda {/eq}

So wavelength of the sound {eq}\lambda = 0.28 \ \ m {/eq}

So frequency of the sound {eq}F = \dfrac { v } { \lambda } \\ F = \dfrac { 340 } { 0.28 } \\ F = 1214.286 \ \ Hz {/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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