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While sitting in your car by the side of a country road, you see your friend, who happens to have an identical car with an identical horn, approaching you. You blow your horn, which has a frequency of \(260 \mathrm{~Hz}\); your friend begins to blow his horn as well, and you hear a beat frequency of \(6.0 \mathrm{~Hz}\). How fast is your friend approaching you?

Short Answer

Expert verified
Your friend is approaching at approximately 7.74 m/s.

Step by step solution

01

Understand the Given Information

The given problem involves two cars with identical horns. Both cars emit a sound frequency of \(260 \text{ Hz}\). A beat frequency of \(6 \text{ Hz}\) is observed when the other car approaches. Let's recall that the beat frequency is the difference between the frequencies of two waves.
02

Recognize the Doppler Effect

The observed beat frequency occurs due to the Doppler effect, which causes the frequency of a wave perceived by an observer to change when the source or the observer is moving. The formula for the Doppler effect with sound is used to calculate the perceived frequency: \[ f' = f \left(\frac{v + v_o}{v - v_s}\right) \]where \(f'\) is the observed frequency, \(f\) is the source frequency, \(v\) is the speed of sound in air, \(v_o\) is the speed of the observer, and \(v_s\) is the speed of the source vehicle.
03

Set Up the Equation

Here, the observer's car is stationary, so \(v_o = 0\). The speed of sound \(v\) in air is approximately \(343 \text{ m/s}\). The friend approaching is the source, so we can adjust the Doppler equation:\[ f' = 260 \left(\frac{343}{343 - v_s}\right) \]The beat frequency is 6 Hz, which means the observed frequency \(f'\) is either 266 Hz or 254 Hz.
04

Solve for Source Speed, Case 1

First calculate assuming the observed frequency is higher (approaching car):\[ f' = 260 + 6 = 266 \]Plug the values into the Doppler effect equation:\[ 266 = 260 \left(\frac{343}{343 - v_s}\right) \]Solving for \(v_s\), \[ 266 = 260 \left(\frac{343}{343 - v_s}\right) \]Re-arranging, \[ 266(343 - v_s) = 260 \times 343 \]\[ 91138 - 266v_s = 89080 \]\[ 266v_s = 2058 \]\[ v_s \approx 7.74 \text{ m/s} \]
05

Confirm with Case 2

Check by assuming the frequency heard could be lower (in case of moving away):\[ f' = 254 \]Plug the values into the Doppler effect equation:\[ 254 = 260 \left(\frac{343}{343 + v_s}\right) \]Solving for \(v_s\) gives results that do not fit the context of approaching cars.
06

Conclusion

Based on calculations from both scenarios, the speed at which your friend is approaching is approximately \(7.74 \text{ m/s}\). This makes sense logically, as the observed frequency is actually higher indicating approach.

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Key Concepts

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

Understanding Sound Frequency
Sound frequency refers to the number of vibrations or cycles per second that a sound wave produces. It is measured in Hertz (Hz), which indicates the number of cycles per second. The sound frequency of a car horn, for example, might be 260 Hz, as given in this exercise. The frequency affects the pitch of the sound. Higher frequencies result in sounds with a higher pitch, while lower frequencies produce deeper, bass-like sounds.

In our scenario, two cars with identical horns have the same sound frequency of 260 Hz. However, changes in the pitch we perceive can occur when one of the cars moves, due to the Doppler Effect. This concept explains how motion can affect the frequencies we actually hear.
Exploring Beat Frequency
Beat frequency arises when two sound waves of similar frequencies interfere with each other. This interference results in a new sound wave that has a frequency equal to the difference between the original two frequencies. The beat frequency is what we perceive as variations in the sound, like a periodic rising and falling.

In our example, the originally identical sound frequencies from the car horns create a beat frequency of 6 Hz. This means that when the two sound waves interfere with each other, they generate a new frequency that's noticeable to our ears. These auditory beats can be a tool for identifying variations in sound frequencies, just as observed in this exercise with cars moving and modifying perceived sound.
The Role of Speed of Sound
The speed of sound is an important constant in physics that denotes how fast sound travels through a medium, typically air at 20°C, where it's about 343 meters per second. This speed can change based on temperature, altitude, and the medium through which sound travels.

Understanding the speed of sound helps explain how quickly sound waves can reach an observer. In our exercise, using 343 m/s for the speed of sound allows for calculating how the motion of your friend’s car changes the frequency you perceive. It plays a critical role in applying the Doppler Effect formula to solve for your friend's speed as they approach with their car horn.
Wave Perception and the Doppler Effect
Wave perception refers to how humans experience variations in sound waves, particularly when the source of sound or the listener is in motion. The Doppler Effect is a key concept here, detailing how motion affects sound perception. When you or the sound source moves, the frequency observed differs from the actual frequency emitted.

The Doppler Effect formula: \[ f' = f \left(\frac{v + v_o}{v - v_s}\right) \]explains this relationship. Here, \(f'\) is the frequency observed by the listener, \(f\) is the frequency of the source, \(v\) is the speed of sound, \(v_o\) is the observer's speed, and \(v_s\) is the source's speed.

In our example with the cars, the Doppler Effect causes your friend’s horn frequency to be perceived higher as the car approaches. This alteration in wave perception leads to a beat frequency, helping to calculate the friend's approaching speed.

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Most popular questions from this chapter

In the larynx, sound is produced by the vibration of the vocal cords. The diagram in Figure 12.44 is a cross section of the vocal tract at one instant in time. Air flows upward (in the \(+z\) direction) through the vocal tract, causing a transverse wave to propagate vertically upward along the surface of the vocal cords. In a typical adult male, the thickness of the vocal cords in the direction of airflow is \(d=2.0 \mathrm{~mm} .\) High-speed photography shows that for a frequency of vibration of \(f=125 \mathrm{~Hz}\), the wave along the surface of the vocal cords travels upward at a speed of \(v=375 \mathrm{~cm} / \mathrm{s}\). Take \(t\) to be time, \(z\) to be displacement in the \(+z\) direction, and \(\lambda\) to be wavelength. What is the wavelength of the wave that travels on the surface of the vocal cords when they are vibrating at frequency \(f ?\) A. \(2.0 \mathrm{~mm}\) B. \(3.3 \mathrm{~mm}\) C. \(0.50 \mathrm{~cm}\) D. \(3.0 \mathrm{~cm}\)

A railroad train is traveling at \(30.0 \mathrm{~m} / \mathrm{s}\) in still air. The frequency of the note emitted by the train whistle is \(262 \mathrm{~Hz}\). What frequency is heard by a passenger on a train moving in the opposite direction to the first at \(18.0 \mathrm{~m} / \mathrm{s}\) and (a) approaching the first; and (b) receding from the first?

The range of audible frequencies is from about \(20 \mathrm{~Hz}\) to \(20.000 \mathrm{~Hz}\). What is the range of the wavelengths of audible sound in air? (b) Visible light. The range of visible light extends from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\). What is the range of visible frequencies of light? (c) Brain surgery. Surgeons can remove brain tumors by using a cavitron ultrasonic surgical aspirator, which produces sound waves of frequency \(23 \mathrm{kHz}\). What is the wavelength of these waves in air? (d) Sound in the body. What would be the wavelength of the sound in part (c) in bodily fluids in which the speed of sound is \(1480 \mathrm{~m} / \mathrm{s},\) but the frequency is unchanged?

A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pitch. The first-chair player's note is exactly \(440 \mathrm{~Hz}\), and 2.8 beats per second are heard. What are the two possible frequencies of the other player's note?

(a) By what factor must the sound intensity be increased to raise the sound intensity level by \(13.0 \mathrm{~dB} ?\) (b) Explain why you don't need to know the orieinal sound intensit

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