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A source traveling to the right at a speed of \(10.00 \mathrm{~m} / \mathrm{s}\) emits a sound wave at a frequency of \(100.0 \mathrm{~Hz}\). The sound wave bounces off of a reflector, which is traveling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). What is the frequency of the reflected sound wave detected by a listener back at the source?

Short Answer

Expert verified
Answer: The frequency of the reflected sound wave detected by the listener back at the source is approximately \(99.5 \thinspace \text{Hz}\).

Step by step solution

01

Find the Frequency of the Sound Wave at the Reflector

First, we need to find the frequency of the sound wave when it reaches the reflector. We will use the Doppler effect formula for moving source and moving observer: $$f' = f \frac{v \pm v_o}{v \pm v_s}$$ where \(f'\) is the observed frequency, \(f\) is the source frequency, \(v\) is the speed of sound in the medium, \(v_o\) is the speed of the observer (reflector) relative to the medium, \(v_s\) is the speed of the source relative to the medium, and the signs depend on the motion direction. In this case, since the source is moving toward the reflector and the reflector is moving toward the source, we can use the positive signs: $$f' = f \frac{v + v_o}{v - v_s}$$ In our example, we have \(f = 100.0 \thinspace \text{Hz}\), \(v = 343 \thinspace \text{m/s}\) (speed of sound in air at room temperature), \(v_o = 5.00 \thinspace \text{m/s}\), and \(v_s = 10.00 \thinspace \text{m/s}\). Now we can find \(f'\): $$f' = 100 \thinspace \text{Hz} \frac{343+5}{343-10} \Rightarrow f' \approx 109.1 \thinspace \text{Hz}$$ So, the frequency of the sound wave at the reflector is approximately \(109.1 \thinspace \text{Hz}\).
02

Find the Frequency of the Reflected Sound Wave at the Listener

Now that we have the frequency of the sound wave at the reflector, we can use the Doppler effect formula again to find the frequency of the reflected sound wave as it is detected by the listener back at the source. This time, since the listener is static, and the reflector is moving to the left, we will use a '+' for \(v_o\) and a '-' for \(v_s\) in the formula: $$f'' = f' \frac{v + v_s}{v - v_o}$$ With \(f' = 109.1 \thinspace \text{Hz}\), \(v = 343 \thinspace \text{m/s}\), \(v_o = -5.00 \thinspace \text{m/s}\), and \(v_s = -10.00 \thinspace \text{m/s}\), we can find the frequency \(f''\): $$f'' = 109.1 \thinspace \text{Hz} \frac{343 - 5}{343 + 10} \Rightarrow f'' \approx 99.5 \thinspace \text{Hz}$$ So, the frequency of the reflected sound wave detected by the listener back at the source is approximately \(99.5 \thinspace \text{Hz}\).

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Key Concepts

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

Sound Wave Frequency
Sound wave frequency refers to the number of oscillations per second of a sound wave. It is measured in hertz (Hz). In simpler terms, it is how many times a wave crest passes a fixed point each second. Think of it as the musical note—higher frequencies sound like a high note, while lower frequencies resemble a deep sound. Frequencies audible to the human ear range from 20 Hz to 20,000 Hz. For the Doppler Effect, knowing the frequency of sound waves is crucial as it directly affects how sound is perceived when there is relative motion between the source of sound and an observer. When a source emits sound waves while moving, the frequency can appear higher or lower depending on the movement direction—toward or away from the observer.

In our problem, a source with common frequency of 100.0 Hz generates sound waves. These waves undergo changes in frequency as perceived by reflectors and listeners, influenced by their motion and the surrounding medium.
Moving Source
When discussing a moving source in sound waves, it relates to the Doppler Effect. A source moving relative to an observer causes a shift in the frequency of waves reaching the observer. If the source approaches the observer, the frequency seems higher—imagine hearing an ambulance siren getting sharper as it nears you. Conversely, if the source moves away, the frequency appears lower—the siren pitch decreases as it passes by and moves away. This shift occurs due to compression and elongation of the sound waves in front and behind the moving source, respectively.

In the given exercise, the sound source moves to the right at 10.00 m/s. It affects how the sound waves interact with other objects, like a moving reflector, changing the perceived frequency at various points in its path.
Velocity of Medium
The velocity of medium for sound reflects how fast sound waves travel through it. The medium's characteristics, like temperature, pressure, and density, significantly impact sound speed. Typically, in air at room temperature, this velocity is around 343 m/s. The speed of sound influences how frequency shifts due to motion of source or observer are calculated mathematically.

In Doppler Effect situations, it's vital to consider the velocity of the medium because the speed differences between moving objects and sound in medium affect the calculations. It explains how sound frequency changes are perceived in the moving medium—whether it’s air, water, or any other material sound travels through. In our example, the velocity of sound in air is assumed constant at 343 m/s, a typical value used in similar problem scenarios.
Reflected Sound
Reflected sound is the phenomenon when sound waves bounce back upon encountering a surface or object, like a wall or, in our case, a moving reflector. This reflection can also be subject to the Doppler Effect, where the motion of the reflector changes the pitch and frequency of the reflected waves. Imagine standing across a lake and shouting—what you hear reflected back, echoes are similar reflections of sound waves.

In the example, the sound wave initially emitted from the moving source reaches a moving reflector. Upon reflection, the frequency of these waves changes again before being detected by a listener. Calculations consider both the source and reflector's motions, leading to the final detected frequency of approximately 99.5 Hz.

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Most popular questions from this chapter

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.

If you blow air across the mouth of an empty soda bottle, you hear a tone. Why is it that if you put some water in the bottle, the pitch of the tone increases?

A (somewhat risky) way of telling if a train that cannot be seen or heard is approaching is by placing your ear on the rail. Explain why this works.

A string of a violin produces 2 beats per second when sounded along with a standard fork of frequency \(400 . \mathrm{Hz}\). The beat frequency increases when the string is tightened. a) What was the frequency of the violin at first? b) What should be done to tune the violin?

You are driving along a highway at \(30.0 \mathrm{~m} / \mathrm{s}\) when you hear a siren. You look in the rear-view mirror and see a police car approaching you from behind with a constant speed. The frequency of the siren that you hear is \(1300 \mathrm{~Hz}\). Right after the police car passes you, the frequency of the siren that you hear is \(1280 \mathrm{~Hz}\). a) How fast was the police car moving? b) You are so nervous after the police car passes you that you pull off the road and stop. Then you hear another siren, this time from an ambulance approaching from behind. The frequency of its siren that you hear is \(1400 \mathrm{~Hz}\). Once it passes, the frequency is \(1200 \mathrm{~Hz}\). What is the actual frequency of the ambulance's siren?

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