Two loudspeakers, separated by a distance of d_1=2.27 m, are in phase. Assume the amplitudes of...

Question:

Two loudspeakers, separated by a distance of {eq}d_1=2.27 m {/eq}, are in phase. Assume the amplitudes of the sound from the speakers are approximately the same position of a listener, who is {eq}d_2 = 4.35 m {/eq} directly in front of one of the speakers. Consider the audible range for normal hearing, {eq}20 Hz to 20kHz {/eq}.

a. What is the lowest frequency that gives the minimum signal at the listener's ear?

b. What is the lowest frequency that gives the maximum signal at the listener's ear?

(Take the speed of the sound to be {eq}343 \frac{m}{s} {/eq}).

Sound Interference

Interference is defined as the superimposition of the wavefront which creates the bright and dark fringes. When the extra distance traveled path by the wave is equal to the integral multiple of wavelength then the constructive interference takes place.

Answer and Explanation:

Now, the path of the wave 1 {eq}L_{1} = d_{2} \\ L_{1} =4.35 \ m {/eq}

Now, the path of the wave 2

{eq}L_{2} = \sqrt{d_{1}^{2} + d_{2}^{2}} \\ L_{2} = \sqrt{2.27^{2} + 4.35^{2}} \\ L_{2} = 4.91 \ m {/eq}

(a)

For the destructive interference

{eq}L_{2} - L_{1} = \dfrac{\lambda}{2} \\ 4.91 - 4.35 = \dfrac{\lambda}{2} \\ \lambda = 1.12 \ m {/eq}

Now, the frequency of the signal

{eq}f = \dfrac{c}{\lambda} \\ f = \dfrac{343}{1.12} = 306.25 \ Hz {/eq}

(b)

Now, for the constructive interference

{eq}L_{2} - L_{1} = \lambda \\ 4.91 - 4.35 = \lambda \\ \lambda = 0.56\ m {/eq}

Now, the frequency of the signal

{eq}f = \dfrac{c}{\lambda} \\ f = \dfrac{343}{0.56} = 612.5 \ 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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