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Chapter 16: Problem 55

A stationary police car emits a sound of frequency 1200 Hz that bounces off a car on the highway and returns with a frequency of 1250 Hz. The police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) How fast was the moving car going? Was it moving toward or away from the police car? (b) What frequency would the police car have received if it had been traveling toward the other car at 20.0 m/s?

Short Answer

Expert verified
The car moves toward the police car at 13.24 m/s. If the police car moves at 20 m/s toward it, the received frequency is 1323 Hz.

Step by step solution

01

Understand the Doppler Effect Formula

The Doppler Effect describes how the frequency of a sound changes when the source or observer is moving relative to each other. The formula when the observer is stationary (like the police car) and the source is moving is given by \[ f' = f \left( \frac{c}{c + v_s} \right) \] where \( f' \) is the observed frequency, \( f \) is the emitted frequency, \( c \) is the speed of sound in air (approximately 343 m/s), and \( v_s \) is the speed of the source (the moving car).
02

Calculate the Speed of the Moving Car

We know \( f = 1200\, \text{Hz} \) and \( f' = 1250\, \text{Hz} \). Using the Doppler Effect formula: \[ 1250 = 1200 \left(\frac{343}{343 + v_s}\right) \] Solving for \( v_s \), we get \[ \frac{1250}{1200} = \frac{343}{343 + v_s} \] Simplifying gives \[ 343 + v_s = \frac{343 \times 1200}{1250} \] Calculating this, \( 343 + v_s = 329.76 \). Thus, \( v_s = 343 - 329.76 = 13.24\, \text{m/s} \). The car is moving toward the police car, as the returned frequency is higher.
03

Adjust the Formula for a Moving Observer

Now consider the police car moves towards the returned sound at 20.0 m/s while it receives the 1250 Hz. Modify the Doppler effect to account for a moving observer with \[ f'' = f' \left( \frac{c + v_o}{c} \right) \] where \( v_o \) = 20 m/s.
04

Calculate the New Frequency Received

Using \( f'' = 1250 \left( \frac{343 + 20}{343} \right) \) gives \[ f'' = 1250 \times \frac{363}{343} = 1323.24\, \text{Hz} \]. Thus, if the police car moves towards the incoming sound at 20 m/s, the frequency received would be approximately 1323 Hz.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency
Frequency is a fundamental concept in the study of waves, including sound waves. It refers to how often the particles of a medium vibrate when a wave passes through it. In general, frequency (denoted by the symbol \( f \)) is measured in Hertz (Hz), where one Hertz is equivalent to one cycle per second.
When discussing the Doppler Effect, frequency is crucial because it represents the number of sound waves the observer detects. The frequency changes when the source of the sound or the observer is moving in relation to each other. This is due to the fact that the waves are compressed or stretched over the distance between the two.
This phenomenon explains why the frequency of the sound observed by the police car in the original exercise was different when received back from the moving car. The stationary police car initially emitted a sound frequency of 1200 Hz, but when the sound returned after reflecting off the moving car, the frequency was observed to be 1250 Hz, indicating movement towards the source. When the source (or the car in this scenario) approaches the observer, it leads to a higher frequency due to wave compression.
Speed of Sound
The speed of sound is an essential component in calculating frequency changes in the Doppler Effect. The speed at which sound travels through a medium depends on the medium's density and temperature. In air at room temperature, sound travels at approximately 343 m/s.
In the mentioned problem, this constant speed of sound in air is fundamental for working out how fast the moving car (source) was traveling. The formula takes the speed of sound into account when determining the alteration in frequency. The speed of sound \( c \) in the formula is crucial because it acts as a reference speed, relating how quickly sound waves arrive at their destination compared to how fast the source is moving.
Understanding the speed of sound allows us to apply it correctly within the Doppler Effect formula, ultimately aiding in calculating real-time speeds of moving objects, like cars, while considering how sound waves behave at those speeds.
Stationary Observer
A stationary observer in the Doppler Effect context is someone or something that is at rest relative to the source of the sound. In the original exercise, the police car acts as the stationary observer, initially stationary during the observation of the sound wave frequency emitted and reflected by the moving car.
In a scenario where the observer is stationary and the source is moving, the Doppler Effect equation prioritizes the relationship between the source's velocity and the observer's stationary position. It's this stationary quality that allows the observer to measure the changes in frequency caused solely by the movement of the source.
The equation for this setup focuses on how the speed of the sound source affects the sound waves reaching the stationary observer, resulting in either a higher frequency when the source is moving towards them or a lower frequency when moving away.
Moving Source
The moving source in a Doppler Effect scenario is typically the origin of the sound waves that are subject to analysis. In the context of the given problem, the moving car is the source of movement.
The movement of this source towards or away from the observer influences the frequency detected. This is because, as the car (source) moves towards the police car (observer), the sound waves compress, leading to an increased frequency. Conversely, if the source moves away, the waves would stretch, producing a lower frequency.
In the exercise, the frequency received increased from the emitted 1200 Hz to 1250 Hz as the moving car approached. This increase allows us to deduce that the car was moving towards the stationary observer and use the Doppler Effect formula to calculate the car鈥檚 speed. This knowledge is applicable in various fields, such as astronomy, to understand the relative movement of stars and galaxies.

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Most popular questions from this chapter

Two guitarists attempt to play the same note of wavelength 64.8 cm at the same time, but one of the instruments is slightly out of tune and plays a note of wavelength 65.2 cm instead. What is the frequency of the beats these musicians hear when they play together?

Singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you \(can\) hear them if you are \(inside\) the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda$$_n =\) \(2L /n\) and the frequencies are given by \({f_n}\) \(= n$$v/2L = nf$$_1\), where \(n =\) 1, 2, 3, . . . . (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

In a liquid with density 1300 kg/m3, longitudinal waves with frequency 400 Hz are found to have wavelength 8.00 m. Calculate the bulk modulus of the liquid. (b) A metal bar with a length of 1.50 m has density 6400 kg/m3. Longitudinal sound waves take 3.90 \(\times\) 10\(^{-4}\) s to travel from one end of the bar to the other. What is Young's modulus for this metal?

\(\textbf{Longitudinal Waves on a Spring.}\) A long spring such as a Slinky\(^{\mathrm{TM}}\) is often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke鈥檚 law has mass \(m\), length \(L\), and force constant \(k'\), the speed of longitudinal waves on the spring is \(v = L\sqrt{ k'/m}\) (see Section 16.2). (b) Evaluate \(v\) for a spring with \(m =\) 0.250 kg, \(L =\) 2.00 m, and \(k' =\) 1.50 N\(/\)m. \(\textbf{ULTRASOUND IMAGING}\). A typical ultrasound transducer used for medical diagnosis produces a beam of ultrasound with a frequency of 1.0 MHz. The beam travels from the transducer through tissue and partially reflects when it encounters different structures in the tissue. The same transducer that produces the ultrasound also detects the reflections. The transducer emits a short pulse of ultrasound and waits to receive the reflected echoes before emitting the next pulse. By measuring the time between the initial pulse and the arrival of the reflected signal, we can use the speed of ultrasound in tissue, 1540 m/s, to determine the distance from the transducer to the structure that produced the reflection. As the ultrasound beam passes through tissue, the beam is attenuated through absorption. Thus deeper structures return weaker echoes. A typical attenuation in tissue is \(-\)100 dB/m \(\cdot\) MHz; in bone it is \(-\)500 dB/m \(\cdot\) MHz. In determining attenuation, we take the reference intensity to be the intensity produced by the transducer.

The human vocal tract is a pipe that extends about 17 cm from the lips to the vocal folds (also called "vocal cords") near the middle of your throat. The vocal folds behave rather like the reed of a clarinet, and the vocal tract acts like a stopped pipe. Estimate the first three standing-wave frequencies of the vocal tract. Use \(v =\) 344 m/s. (The answers are only an estimate, since the position of lips and tongue affects the motion of air in the vocal tract.)
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