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What beat frequencies are possible with tuning forks of frequencies 258, 263, and 267 Hz?

Question:

What beat frequencies are possible with tuning forks of frequencies 258, 263, and 267 Hz?

Beats due to Interference of Sound:

A beat is the result of the interference of sound, this is a combination of one high-intensity sound and the following low-intensity sound. This can be used to determine the difference in the frequencies of the sound waves since sound is a wave that travels in a medium as longitudinal waves. Just like any other coherent beams of waves, sound waves also can interfere with each other given that they exist in the same medium. If their amplitudes add, the interference is said to be constructive interference, and destructive interference if they are "out of phase" and subtract.

Answer and Explanation:

Given:

  • Frequency 1 ({eq}F_1 {/eq}) = 258 Hz
  • Frequency 1 ({eq}F_2 {/eq}) = 263 Hz
  • Frequency 1 ({eq}F_3 {/eq}) = 267 Hz


The given beat characteristic is just the frequency of the beat produced by the interfering sound waves. Beats are described as the difference in the frequency of the waves. We can describe the beat ({eq}b {/eq}) as the difference in frequencies for a pair of waves.

In this case, we can have;

{eq}b_1 = F_3 - F_1\\ b_2 = F_3 - F_2\\ b_3 = F_2 - F_1 {/eq}

The equation already considers that the second frequency has a lower frequency.


Solving for the beat frequencies, we have;

{eq}\begin{align} b_1 &= (267) - (258)\\[0.2cm] b_1 &= \color{red}{9\ \rm Hz}\\ \\ b_2 &= (267) - (263)\\[0.2cm] b_2 &= \color{red}{4\ \rm Hz}\\ \\ b_3 &= (263) - (258)\\[0.2cm] b_3 &= \color{red}{5\ \rm Hz} \end{align} {/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
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