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Two speakers, separated by a distance x = 5.37 m are driven in phase by the same amplifier. A...

Question:

Two speakers, separated by a distance x = 5.37 m are driven in phase by the same amplifier. A listener is located at point A, a distance L = 2.62 m directly in front of one of the speakers. 1) What is the lowest frequency for which there is a maximum signal at A due to constructive interference? The speed of sound can be taken as 3.43  102 ms. 2) What is the second-lowest frequency for which there is a maximum signal? 3) What is the lowest frequency for which there is a minimum signal due to destructive interference? 4) What is the second-lowest frequency for which there is a minimum signal?

Sound Wave Interference

Two sound waves with the same frequency passing through space interfere with each other. These waves create a maximum signal at A (constructive interference) if they are in phase. These waves create a minimum signal at A ( destructive interference) if they are in the opposite phase.

Answer and Explanation:

Sound waves with the same frequency from two spaced sources interfere.

Let L is a distance from the listener to Source1 (2.62 m), x is a distance between two sources (5.37 m), and L2 is a distance from the listener to Source2.

Assuming that two sources and listener are three vertices of a right triangle, L2 can be calculated based on Pythagorean theorem as:

{eq}L2 =\sqrt{L^2+x^2}= 5.98 m {/eq}

The listener receives a maximum signal when constructive interference occurs, which means L2 - L equals multiple wavelengths (zero or even number of one-half wavelength).

The listener receives a minimum signal when destructive interference occurs, which means L2 - L equals an odd number of one-half wavelength.

Frequency equals the speed of sound over wavelength:

{eq}F = V /\lambda {/eq} (a)

, where F is a frequency, V is a speed (343 m/s), and {eq}\lambda {/eq}

is a wavelength.

I. Constructive interference (answers to questions 1 and 2):

{eq}L2 - L = 2n* \lambda/2 {/eq}

, where n is natural number (1, 2, 3, ...)

then

{eq}\lambda=(L2 - L)/n {/eq}

Let's plug-in this formula for above frequency calculation (a):

{eq}F = V * n / (L2 - L) {/eq}

1) The lowest frequency will be if n =1:

{eq}F = V / (L2 - L) = 343 /(5.98 - 2.62) = 102 Hz {/eq}

2) The second lowest frequency will be if n =2:

{eq}F = V * 2/ (L2 - L) = 343 * 2 /(5.98 - 2.62) = 204 Hz {/eq}

II. Destructive interference (answers to questions 3 and 4):

{eq}L2 - L = (2n - 1)* \lambda /2 {/eq}

, where n is a natural number (1, 2, 3, ...)

then

{eq}\lambda = 2 * (L2 - L) / (2n - 1) {/eq}

Let's plug-in this formula for above frequency calculation (a):

{eq}F = V /2 * (2n - 1) / (L2 - L) {/eq}

3) The lowest frequency will be if n =1:

{eq}F = 1/2 * V / (L2 - L) = 0.5 * 343 /(5.98 - 2.62) = 51 Hz {/eq}

4) The second lowest frequency will be if n =2:

{eq}F = 3/2 * V / (L2 - L) = 1.5 * 343 /(5.98 - 2.62) = 153 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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