Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase....

Question:

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from the speaker.

Take the speed of sound in air to be 344 m/s. What is the closest you can be to speaker B and be at a point of destructive interference?

Destructive Interference:

When the waves are out of phase during superposition at a point, then the interference is known to be destructive. The destructive interference takes place when the path difference between the wave must be equal to the half of wavelength.

Answer and Explanation:

Given Data:

  • Frequency of the wave {eq}\rm (f) = 172 \ Hz {/eq}
  • Speed of the sound {eq}\rm (V) = 344 \ m/s {/eq}

Now, the wavelength of the wave would be

{eq}\rm \lambda = \dfrac{V}{f} \\ \lambda = \dfrac{344}{172} \\ \lambda = 2 \ m {/eq}

Since, the distance from the Speaker A is 8 m which is the integral multiple of the wavelength, therefore, we can say if at this point, the distanc of the speaker B must be half of the wavelength.

hence the closest distance of the speaker B would be

{eq}\rm d = \dfrac{\lambda}{2} \\ d = \dfrac{2}{2} \\ d = 1\ 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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