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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

When two pure tones with similar frequencies combine to produce beats, the result is a train of wave packets. That is, the sinusoidal waves are partially localized into packets. Suppose two sinusoidal waves of equal amplitude A, traveling in the same direction, have wave numbers \(\kappa\) and \(\kappa+\Delta \kappa\) and angular frequencies \(\omega\) and \(\omega+\Delta \omega\), respectively. Let \(\Delta x\) be the length of a wave packet, that is, the distance between two nodes of the envelope of the combined sine functions. What is the value of the product \(\Delta x \Delta \kappa ?\)

At a distance of \(20.0 \mathrm{~m}\) from a sound source, the intensity of the sound is \(60.0 \mathrm{~dB}\). What is the intensity (in \(\mathrm{dB}\) ) at a point \(2.00 \mathrm{~m}\) from the source? Assume that the sound radiates equally in all directions from the source.

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).

A meteorite hits the surface of the ocean at a speed of \(8800 \mathrm{~m} / \mathrm{s}\). What are the shock wave angles it produces (a) in the air just before hitting the ocean surface, and (b) in the ocean just after entering? Assume the speed of sound in air and in water is \(343 \mathrm{~m} / \mathrm{s}\) and \(1560 \mathrm{~m} / \mathrm{s}\), respectively.

The Moon has no atmosphere. Is it possible to generate sound waves on the Moon?
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