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(a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half 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 to that in part (a)? Explain on physical grounds why the two answers differ.

Short Answer

Expert verified
(a) 2000 Hz, (b) 1500 Hz. Frequencies differ due to wave compression by a moving source.

Step by step solution

01

Identify the Parameters and Formula for Part (a)

In part (a), the sound source is moving at half the speed of sound, which we'll denote as \( v_s = \frac{V}{2} \), where \( V \) is the speed of sound in the air. The source frequency is \( f_s = 1000 \) Hz. The listener is stationary. We use the Doppler effect formula for a moving source: \( f' = \left( \frac{V}{V - v_s} \right) f_s \).
02

Plug Values into the Formula for Part (a)

Substitute \( v_s = \frac{V}{2} \) and \( f_s = 1000 \) Hz into the formula: \( f' = \left( \frac{V}{V - \frac{V}{2}} \right) \times 1000 = \left( \frac{V}{\frac{V}{2}} \right) \times 1000 = 2 \times 1000 = 2000 \) Hz.
03

Identify the Parameters and Formula for Part (b)

In part (b), the source is stationary, and the listener is moving toward the source at half the speed of sound, \( v_l = \frac{V}{2} \). The source frequency is the same \( f_s = 1000 \) Hz. We use the Doppler effect formula for a moving listener: \( f' = \left( \frac{V + v_l}{V} \right) f_s \).
04

Plug Values into the Formula for Part (b)

Substitute \( v_l = \frac{V}{2} \) and \( f_s = 1000 \) Hz into the formula: \( f' = \left( \frac{V + \frac{V}{2}}{V} \right) \times 1000 = \left( \frac{3V/2}{V} \right) \times 1000 = 1.5 \times 1000 = 1500 \) Hz.
05

Compare and Explain the Differences

The listener hears a higher frequency in part (a) (2000 Hz) than in part (b) (1500 Hz). The difference arises because when the source moves, it compresses the sound waves in front of it, significantly increasing the frequency more than when the listener moves toward the source.

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Key Concepts

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

Wave Frequency
Wave frequency is a fundamental concept when discussing sound waves and the Doppler Effect. It refers to the number of complete wave cycles that occur per second, measured in Hertz (Hz). When you think of wave frequency, imagine how often the waves of sound hit your ear every second.
Similar to how the beat of music changes, frequency determines the pitch of the sound you hear. A higher frequency means a higher pitch. In problems involving the Doppler Effect, like our exercise, the wave frequency is affected by the relative motion between the sound source and the listener. The source frequency in the example is 1000 Hz, where the motion will adjust the wave frequency that the listener perceives. Different scenarios, such as sources or listeners moving, change this frequency, which is key to understanding how the Doppler Effect operates.
Sound Waves
Sound waves are vibrations that travel through the air (or other media) and are perceived as sound when they reach our ears. Think of them as ripples on a pond after you toss in a stone; they move outward from the source of the sound.
These waves are longitudinal, meaning the vibration of the particles in the medium (like air) happens in the same direction as the wave travels. As such, when the source of the sound or the listener is moving, the nature of these waves can change – this is vital when examining the Doppler Effect. By measuring these changes, we can understand shifts in perceived frequency due to movement, illustrating the dynamics between the source and listener.
Stationary Source
A stationary source is one that is not moving relative to the medium through which the sound travels. In our exercise, this was the case for part (b). The sound waves produced by this stationary source will have a consistent frequency and wavelength as they travel through the medium.
This means when a listener approaches the stationary source, the waves will still compress because the listener is moving toward them, increasing the frequency perceived by the listener, even though the source itself doesn't change position. This highlights the principle that relative motion affects how sound is heard. The frequency perceived by the listener in this scenario was 1500 Hz, due to their movement toward the source.
Moving Source
When a source moves, it significantly alters the frequency perceived by a listener; this is a direct observance of the Doppler Effect. In part (a) of our exercise, the sound source moves toward the stationary listener at half the speed of sound. As it moves, it compresses the sound waves in its path, causing the waves to reach the listener more frequently.
This compression increases the wave frequency, making the sound pitch higher. The listener perceives the frequency as 2000 Hz, which is a doubling of the original frequency due to the continued motion of the source toward the observer. This effect is more noticeable than when the listener moves, showing how the source's speed and direction directly impact the perceived sound frequency.
Moving Listener
A moving listener is an individual or object moving toward or away from a sound source. In part (b) of our exercise, the listener moves toward a stationary source at half the speed of sound. As a result, the sound waves reach the listener more quickly compared to if they were stationary.
This action increases the frequency of waves received by the listener's ears, and therefore raises the pitch of the sound. However, the increase is not as significant as when the source is the one moving. The perceived frequency in this context was 1500 Hz, which is smaller than the frequency perceived when the source was moving. This demonstrates how the Doppler Effect operates differently depending on whether the source or the listener is the one in motion.

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Most popular questions from this chapter

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?

A 75.0-cm-long wire of mass 5.625 g is tied at both ends and adjusted to a tension of 35.0 N. When it is vibrating in its second overtone, find (a) the frequency and wavelength at which it is vibrating and (b) the frequency and wavelength of the sound waves it is producing.

At point \(A, 3.0 \mathrm{~m}\) from a small source of sound that is emitting uniformly in all directions, the sound intensity level is \(53 \mathrm{~dB}\). (a) What is the intensity of the sound at \(A ?\) (b) How far from the source must you go so that the intensity is one-fourth of what it was at \(A\) ? (c) How far must you go so that the sound intensity level is one-fourth of what it was at \(A ?\) (d) Does intensity obey the inverse-square law? What about sound intensity level?

Example 16.1 (Section 16.1) showed that for sound waves in air with frequency 1000 Hz, a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produces a pressure amplitude of 3.0 \(\times\) 10\(^{-2}\) Pa. (a) What is the wavelength of these waves? (b) For 1000-Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? (c) For what wavelength and frequency will waves with a displacement amplitude of 1.2 \(\times\) 10\(^{-8}\) m produce a pressure amplitude of 1.5 \(\times\) 10\(^{-3}\) Pa?

(a) \(\textbf{Whale communication.}\) Blue whales apparently communicate with each other using sound of frequency 17 Hz, which can be heard nearly 1000 km away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 m/s? (b) \(\textbf{Dolphin clicks.}\) One type of sound that dolphins emit is a sharp click of wavelength 1.5 cm in the ocean. What is the frequency of such clicks? (c) \(\textbf{Dog whistles.}\) One brand of dog whistles claims a frequency of 25 kHz for its product. What is the wavelength of this sound? (d) \(\textbf{Bats.}\) While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39 kHz and 78 kHz. What is the range of wavelengths of this sound? (e) \(\textbf{Sonograms.}\) Ultrasound is used to view the interior of the body, much as x rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.0 mm across if the speed of sound in the tissue is 1550 m/s?

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