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The drawing shows a loudspeaker A and point C, where a listener is positioned. A second...

Question:

The loudspeaker {eq}A {/eq} and point {eq}C {/eq}, where a listener is positioned(1 m) at 60°. A second loudspeaker {eq}B {/eq} is located somewhere to the right of loudspeaker {eq}A {/eq}. Both speakers vibrate in phase and are playing a {eq}64.6-Hz {/eq} tone. The speed of sound is {eq}343 \frac{m}{s} {/eq}. What is the closest to speaker {eq}A {/eq} that speaker {eq}B {/eq} can be located, so that the listener hears no sound?

Destructive Interference

The sound waves cancel out when the path difference between them is integral odd multiples of wavelength and waves amplify when the difference is even multiples.

Answer and Explanation:

The schematic diagram is shown below (in accordance with the problem),

Schematic Diagram (Ms Word)

Interference at point C must be destructive in order to hear no sound.

Mathematically,

{eq}d_2 -d_1 = \dfrac{n \lambda}{2}, \;\; n = 1, 3, 5... {/eq}

We know that {eq}v_{(wave)} = \lambda f\\ where,\\ v_{(wave)} = velocity\\ \lambda = wavelength\\ f = frequency {/eq}

Plugin the given values we get,

{eq}v_{(wave)} = \lambda f\\ \lambda = \dfrac{343}{64.6}\\ \lambda = 9.91\;m {/eq}

For minimum distance, n = 1 so,

{eq}d_2 = \dfrac{n \lambda}{2}+d_1\\ d_2 = \dfrac{9.91}{2}+1 = 5.95\;m {/eq}

Therefore,

{eq}x_1 = (1) \cos60° = 0.500\; m\\ y = (1) \sin60° = 0.866\;m {/eq}

Thus,

{eq}x^2_2 +y^2 = d^2_2\\ d_2 = \sqrt{(5.95)^2+(0.866)^2}\\ d_2 = \sqrt{35.40 - 0.749}\\ d_2 = 5.88\;m {/eq}

So the minimum distance will be,

{eq}x_1 +x_2 = 0.500+5.88 = 6.38\;m {/eq}


Learn more about this topic:

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Constructive and Destructive Interference

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16
110K

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