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(a) A sound source producing \(1.00 \mathrm{kHz}\) waves moves toward a stationary listener at onehalf the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare with that in part (a)? Did you expect to get the same answer in both cases? Explain on physical grounds why the two answers differ.

Short Answer

Expert verified
(a) 2000 Hz; (b) 1500 Hz. The answers differ due to the asymmetrical nature of the Doppler effect.

Step by step solution

01

Identify Known Values

The speed of sound in air is approximately \( v = 343 \, \text{m/s} \). The frequency of the source is \( f_s = 1.00 \, \text{kHz} = 1000 \, \text{Hz} \). The speed of the source (or listener) is half the speed of sound, so \( v_s = v/2 = 171.5 \, \text{m/s} \).
02

Calculate Frequency for Part (a)

For part (a), the source moves towards the listener. The Doppler effect formula is used here: \( f' = \left( \frac{v}{v - v_s} \right) f_s \). Plugging in the values, we get:\[f' = \left( \frac{343}{343 - 171.5} \right) \times 1000 = \left( \frac{343}{171.5} \right) \times 1000 = 2000 \, \text{Hz}.\]
03

Calculate Frequency for Part (b)

For part (b), the listener moves towards the stationary source. The formula used is: \( f' = \left( \frac{v + v_l}{v} \right) f_s \), where \( v_l = v/2 = 171.5 \, \text{m/s} \). Thus,\[f' = \left( \frac{343 + 171.5}{343} \right) \times 1000 = \left( \frac{514.5}{343} \right) \times 1000 = 1500 \, \text{Hz}.\]
04

Compare and Explain Differences

The frequency heard by the listener in part (a) is \( 2000 \, \text{Hz} \), and in part (b) it is \( 1500 \, \text{Hz} \). These values differ because when the source moves, the wavelength changes, effectively altering the wave itself. In part (b), only the perceived number of waves per second changes, due to relative motion, not the actual wave structure. This illustrates the asymmetrical nature of the Doppler effect in sound.

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through a medium like air, water, or solids. They are longitudinal waves, meaning the particles in the medium move back and forth in the same direction as the wave travels. Unlike transverse waves, which include light waves, sound waves require a medium to move.
Sound travels at different speeds depending on the medium. In air, sound speed is about 343 meters per second (m/s) at room temperature. The wave's frequency determines the pitch of the sound, and this is where the concept of the Doppler effect becomes important. As the source or listener moves, the frequency of sound waves that are heard changes due to relative motion.
The differences in sound speed and wave behavior are crucial for understanding physical phenomena, such as the Doppler effect. Watching ocean waves can provide a visual analogy; the waves look different depending on whether you stand still or move towards them.
Frequency Calculation
The Doppler effect changes how we perceive the frequency of waves when there is relative motion between the source and the observer. When calculating frequency changes due to the Doppler effect, it's essential to understand some key equations.
For a moving source moving towards a listener:
  • The observed frequency, \( f' \), is given by: \[ f' = \left( \frac{v}{v - v_s} \right) f_s \]
  • Where \( v \) is the speed of sound, \( v_s \) is the speed of the source, and \( f_s \) is the source frequency.
For a moving listener moving towards a stationary source:
  • The formula changes to: \[ f' = \left( \frac{v + v_l}{v} \right) f_s \]
  • Here, \( v_l \) represents the listener's speed.
These calculations allow determining how the observer's perception of the frequency changes. The key is to note whether it is the source or the listener in motion.
Motion of Source and Listener
The motion of the sound source or the listener affects how sound waves are perceived. Understanding their movement is crucial to calculating the correct frequency the listener hears.
1. **When the source moves:** The sound waves either compress or stretch. If the source moves towards the listener, the waves compress, resulting in a higher frequency. This is why a siren sounds higher-pitched as it approaches you.
2. **When the listener moves:** The listener intercepts more waves per second than if they were stationary. However, in this case, the wave's length does not change, but the frequency does as seen by the listener.
This difference in behavior explains why the frequency calculations differ. Motion affects wave properties differently depending on whether it's the source or the listener that's moving.
Physics Problems
Physics problems involving concepts like the Doppler effect challenge our understanding of wave behavior. Solving these problems requires careful application of formulas and understanding how physical principles interact.
The steps to approach such problems generally include:
  • Identifying known values and variables, such as the speed of sound and source frequency.
  • Applying the relevant formulas based on whether the source or listener is moving.
  • Performing substitutions and mathematical operations to find the unknown frequency.
Each problem helps reinforce physics concepts, like wave properties and the impact of motion. Additionally, recognizing why different scenarios give different results—such as variations in perceived frequency—enhances your understanding and appreciation of physics.

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Most popular questions from this chapter

One of the \(63.5-\mathrm{cm}\) -long strings of an ordinary guitar is tuned to produce the note \(\mathrm{B}_{3}\) (frequency \(245 \mathrm{~Hz}\) ) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is \(344 \mathrm{~m} / \mathrm{s},\) find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the speed of light. (a) FM radio. Find the wavelength of an FM radio station signal broadcasting at a frequency of \(104.5 \mathrm{MHz}\). (b) X-rays. X-rays have a wavelength of about 0.10 nm. What is their frequency? (c) The Big Bang. Microwaves with a wavelength of \(1.1 \mathrm{~mm}\), left over from soon after the Big Bang. have been detected. What is their frequency? (d) Sunburn. Sunburn (and skin cancer) is caused by ultraviolet light waves having a frequency of around \(10^{16} \mathrm{~Hz}\). What is their wavelength? (e) SETI. It has been suggested that extraterrestrial civilizations (if they exist) might try to communicate by using electromagnetic waves having the same frequency as that given off by the spin flip of the electron in hydrogen, which is \(1.43 \mathrm{GHz}\). To what wavelength should we tune our telescopes in order to search for such signals? (f) Microwave ovens. Microwave ovens cook food with electromagnetic waves of frequency around \(2.45 \mathrm{GHz}\). What wavelength do these waves have?

With what tension must a rope with length \(2.50 \mathrm{~m}\) and mass \(0.120 \mathrm{~kg}\) be stretched for transverse waves of frequency \(40.0 \mathrm{~Hz}\) to have a wavelength of \(0.750 \mathrm{~m} ?\)

A piano tuner stretches a steel piano wire with a tension of \(800 \mathrm{~N}\). The wire is \(0.400 \mathrm{~m}\) long and has a mass of \(3.00 \mathrm{~g}\). (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies yp to \(10.000 \mathrm{~Hz}\) ?

Standing sound waves are produced in a pipe that is \(1.20 \mathrm{~m}\) long. For the fundamental frequency and the first two overtones, determine the locations along the pipe (measured from the left end) of the displacement nodes if (a) the pipe is open at both ends; (b) the pipe is closed at the left end and open at the right end.

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