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Chapter 16: Problem 16

A police siren contains at least two frequencies, producing the wavering sound (beats). Explain how the siren sound changes as a police car approaches, passes, and moves away from a pedestrian.

Short Answer

Expert verified
Answer: As a police car approaches a pedestrian, the siren's frequency increases, making it sound higher-pitched due to the Doppler effect, and the beat frequency also increases. While the police car passes by the pedestrian, the siren's frequency remains constant momentarily, as does the beat frequency. As the car moves away from the pedestrian, the siren's frequency decreases, making it sound lower-pitched due to the Doppler effect, and the beat frequency decreases as well.

Step by step solution

01

Understanding beat frequency and Doppler effect

First, we should understand the phenomena of beat frequency and the Doppler effect. Beat frequency is the pulsating sound one hears when two similar frequencies combine. In this case, the two frequencies are produced by the police siren. The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In the context of our exercise, the Doppler effect describes how the frequency of the police siren changes as the car approaches and moves away from the pedestrian.
02

Siren as the police car approaches

When the police car is approaching a pedestrian, due to the Doppler effect, the frequency of the siren heard by the pedestrian increases. This means that the police car's siren sounds higher-pitched than when the car is stationary. During this time, the beat frequency will also increase as the difference between the two siren frequencies becomes larger.
03

Siren as the police car passes by

As the police car passes by the pedestrian, the siren sound's frequency remains constant for a very brief moment. At this point, the pedestrian hears the actual siren frequencies without any modification due to the Doppler effect. The beat frequency remains constant in this moment.
04

Siren as the police car moves away

When the police car begins to move away from the pedestrian, the Doppler effect causes the siren's frequency to decrease in the ears of the pedestrian. The siren will now sound lower-pitched than when the car was stationary. The beat frequency will also decrease as the difference between the two siren frequencies becomes smaller. In summary, as the police car approaches, passes, and moves away from a pedestrian, the sound of the siren changes primarily due to the Doppler effect altering the frequency of the siren. The change in beat frequency is also directly related to the Doppler effect as the difference between the two siren frequencies increases and decreases.

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Most popular questions from this chapter

You are creating a sound wave by shaking a paddle in a liquid medium. How can you increase the speed of the resulting sound wave? a) You shake the paddle harder to give the medium more kinetic energy. b) You vibrate the paddle more rapidly to increase the frequency of the wave. c) You create a resonance with a faster-moving wave in air. d) All of these will work. e) None of these will work. f) Only (a) and (b) will work. g) Only (a) and (c) will work. h) Only (b) and (c) will work.

A sound meter placed \(3 \mathrm{~m}\) from a speaker registers a sound level of \(80 \mathrm{~dB}\). If the volume on the speaker is then turned down so that the power is reduced by a factor of 25 , what will the sound meter read? a) \(3.2 \mathrm{~dB}\) c) \(32 \mathrm{~dB}\) e) \(66 \mathrm{~dB}\) b) \(11 \mathrm{~dB}\) d) \(55 \mathrm{~dB}\)

The sound level in decibels is typically expressed as \(\beta=10 \log \left(I / I_{0}\right),\) but since sound is a pressure wave, the sound level can be expressed in terms of a pressure difference. Intensity depends on the amplitude squared, so the expression is \(\beta=20 \log \left(P / P_{0}\right),\) where \(P_{0}\) is the smallest pressure difference noticeable by the ear: \(P_{0}=2.00 \cdot 10^{-5} \mathrm{~Pa}\). A loud rock concert has a sound level of \(110 . \mathrm{dB}\), find the amplitude of the pressure wave generated by this concert.

A thin aluminum rod of length \(L=2.00 \mathrm{~m}\) is clamped at its center. The speed of sound in aluminum is \(5000 . \mathrm{m} / \mathrm{s}\). Find the lowest resonance frequency for vibrations in this rod.

What has the greatest effect on the speed of sound in air? a) temperature of the air b) frequency of the sound c) wavelength of the sound d) pressure of the atmosphere
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