A certain dog whistle operates at 23.5 kHz, while another (brand X) operates at an unknown...


A certain dog whistle operates at 23.5 kHz, while another (brand X) operates at an unknown frequency. If neither whistle can be heard by humans when played separately, but a shrill whine of frequency 9200 Hz occurs when they are played simultaneously, estimate the operating frequency of brand X.

Interference of Sound

Sound waves also will interfere each other and interference voice pattern will be produced. The interference voice pattern is the occurrence of alternate maxima and minima of sound. One maxima and minima together is known as a beat. Normally the minima may not be heard. So beats are counted by just counting the maxima. Number of beats happening per second is known as beat frequency. Beat frequency is equal to the magnitude of the difference between the interfering frequencies of the sound.

Answer and Explanation:

Given data

  • Frequency of the first dog whistle {eq}F_1 = 23.5 \times 10^3 Hz {/eq}
  • The beat frequency heard when the first whistle and unknown brand whistle X blown together {eq}F_b = 9200 \ Hz {/eq}

Let {eq}F_2 {/eq} be the frequency of the brand X whistle.

We have the equation for beat frequency {eq}F_b = | F_1 - F_2 | {/eq}

Therefore the operating frequency of the second whistle {eq}F_2 = F_1 \pm F_b \\ \implies F_2 = 23.5 \times 10^3 \pm 9200 {/eq}

Since the second whistle is also a dog whistle, the lower frequency obtaining can be neglected as it comes in the human audible range.

So the frequency of the second whistle {eq}F_2 = 23.5 \times 10^3 + 9200 \\ F_2 = 3.27 \times 10^4 \ Hz {/eq}

Learn more about this topic:

Interference Patterns of Sound Waves

from MTEL Physics (11): Practice & Study Guide

Chapter 16 / Lesson 5

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