An observer is located directly between two speakers, located 20m apart. The speakers are in...

Question:

An observer is located directly between two speakers, located 20m apart. The speakers are in phase with each other, and both are emitting a sound with a frequency of 60Hz. How far away from the centre should the observer move to get the first destructive interference?

Wave interference

Wave interference corresponds to the phenomenon that occurs when two waves meet while traveling along with the same medium. The interference of waves causes the medium to take on a shape that results from the net effect of the two individual waves upon the particles of the medium. This interference may be constructive or destructive.

Answer and Explanation:

{eq}\begin{align} \\ \end{align} {/eq}

Let:

{eq}\begin{align} R_1 \, \text{and} \, R_2 \to \end{align} {/eq} be the distance between the observer and the speakers

  • Note that we use the midpoint as a reference since we know that we will have constructive interference. Now, if we move to the left by an amount x, the distance {eq}\begin{align} R_1 \end{align} {/eq} increases by x whereas the distance {eq}\begin{align} R_2 \end{align} {/eq} decreases by x. If {eq}\begin{align} R_1 \end{align} {/eq} increases and{eq}\begin{align} R_2 \end{align} {/eq} decreases, the difference between the two ({eq}\begin{align} R_1 - R_2 \end{align} {/eq}) increases by an amount 2x.
  • Now, if this 2x happens to be equal to {eq}\begin{align} \lambda/2 \end{align} {/eq}, we have met the conditions for destructive interference. Therefore:

$$\begin{align} 2x &= \lambda/2 \\ x &= \lambda/4 \\ \end{align} $$

  • If {eq}\begin{align} x = \lambda/4 \end{align} {/eq}, we will have destructive interference. To put it simply, if one move one quarter of a wavelength away from the midpoint, you will get the first destructive interference.

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
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