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Chapter 12: Problem 60

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 6.99 m/s.

Step by step solution

01

Understand the Problem and Given Information

We know that the frequency of your horn is \(f_s = 260 \, \text{Hz}\). You hear the beat frequency, which is the difference in frequencies from two sources, as \(f_{beat} = 6.0 \, \text{Hz}\). The Doppler effect affects your friend's horn frequency as it approaches you.
02

Establish Relationship Due to Doppler Effect

Since your friend is approaching, the sound frequency of the approaching source increases. We'll denote your friend's observed frequency as \(f_o\). For beats against your horn, \(f_o\) can be either \(266 \, \text{Hz}\) or \(254 \, \text{Hz}\) (260 Hz 卤 beat frequency of 6 Hz). Given that your friend's car approaches, \(f_o\) should be greater than \(f_s\). Thus, \(f_o = 266 \, \text{Hz}\).
03

Apply the Doppler Effect Formula

The frequency of sound from a source approaching an observer is given by:\[ f_o = \frac{v + v_o}{v - v_s} \cdot f_s \]where \(v\) is the speed of sound in air (approximately 343 m/s), \(v_o\) is the observer's velocity (your velocity here is 0 m/s), \(v_s\) is the source's velocity (your friend's car), and \(f_s\) is the original frequency of 260 Hz. We need to solve for \(v_s\) when \(f_o = 266 \, \text{Hz}\).
04

Solve for Friend's Velocity

Since you are stationary (observer velocity \(v_o = 0\)), the formula simplifies to:\[ 266 = \frac{343}{343 - v_s} \cdot 260\]Rearrange and solve the equation:\[ 266(343 - v_s) = 343 \times 260 \ 91038 - 266v_s = 89180 \ 1858 = 266v_s \ v_s = \frac{1858}{266} \approx 6.99 \, \text{m/s} \]Your friend is approaching at approximately 6.99 m/s.

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Key Concepts

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

Beat Frequency
The concept of beat frequency occurs when two sound waves of slightly different frequencies interfere with each other. This interference produces a fluctuating sound whose frequency is the difference between the two original frequencies. This is known as the beat frequency.

In practical terms, if you have two horns sounding, each with frequencies close to each other, but not identical, you would hear a variation in sound intensity as the waves continuously go in and out of phase. This is the beat you've noticed: it has a frequency equal to the absolute value of the difference between the two frequencies.
  • If you're hearing a beat frequency of 6 Hz, this means the two frequencies could either add up to this difference above or below the stationary frequency. For example, if a horn has a frequency of 260 Hz, then possible heard frequencies could be 254 Hz or 266 Hz.
Understanding beat frequency helps in determining how fast objects move in sound-related experiments, like cars with horn sounds.
Sound Frequency
Sound frequency relates to the number of vibrations or cycles that occur per unit of time. It is often measured in Hertz (Hz), which corresponds to one cycle per second.

In the context of the problem, sound frequency directly ties into the idea of an object's motion. Our exercises demonstrate two horns with frequencies, where the motion of one horn affects its perceived frequency.
  • Your horn, for example, has a stationary sound frequency of 260 Hz.
  • Your friend's perceived horn frequency while moving is different due to the Doppler effect.
Recognizing changes in sound frequency with motion, especially important in physics, lays the foundation for understanding how sound behaves in the environment.
Observer Velocity
Observer velocity refers to the speed at which an observer is moving towards or away from a sound source. In Doppler effect phenomena, this velocity can significantly alter how the sound frequency is perceived.
  • A stationary observer will hear the frequency as it is emitted from the source.
  • An observer moving towards the source perceives a higher frequency due to compressed sound waves.
  • In the exercise, you acted as a stationary observer, meaning your observer velocity was 0 m/s, whereas your friend's movement caused a change in frequency perception.
This dependency on observer velocity highlights how movement affects sound frequency perceptions in everyday scenarios.
Speed of Sound
The speed of sound is the rate at which a sound wave travels through a medium. In air, at room temperature, it is typically around 343 meters per second.

This speed is a crucial factor in the Doppler effect formula and calculations.
  • It determines how quickly sound waves reach an observer from a moving source, such as a vehicle horn.
  • In the step-by-step exercise, the speed of sound acts as a reference that shifts frequencies based on relative motion between the observer and the source.
Understanding the speed of sound within various contexts allows for accurate calculations of changes in perceived sound frequencies when objects are in motion.

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Most popular questions from this chapter

\(\bullet\) Voiceprints. In this chapter, we have been concentrating on sinusoidal waves. But most waves in the real world are far more complicated. However, many complicated waves can be created by adding together sine waves of varying amplitude and frequency. When a singer, for example, sings a note, the pitch we hear is the fundamental frequency at which his or her larynx is vibrating. But the larynx also vibrates in other frequencies (the overtones) at the same time. So the sound we hear is a superposition of the fundamental frequency plus all the overtones. This set of all the frequencies (with their respective amplitudes) is called the person's voice print. (a) To see how this works, carefully graph a sine wave of frequency 440 Hz (concert \(A ),\) with time on the horizontal axis and displacement on the vertical axis. Let the amplitude be 1 unit. On the same set of axes, graph the first overtone of 880 \(\mathrm{Hz}\) , but with an amplitude of \(\frac{1}{2}\) unit. (b) Now add the two waves to find their superposition. Notice that the shape is no longer a Isine wave.

You're standing between two speakers that are driven by the same amplifier and are emitting sound waves with frequency 229 Hz. The two speakers are facing each other, 15 meters apart. (a) You begin walking away from one speaker toward the other one, and as you walk, you hear what sounds like beats, with a frequency of 2.50 Hz. How fast are you walking? (b) If the frequency of the sound emitted by the speakers increases to 573 Hz and you continue to walk at the same speed, what frequency of beats will you hear? [Hint: You can model this situation as a tube open at both ends; alternatively, you can treat it as a Doppler effect problem.]

A \(\mathrm{A} 75.0 \mathrm{cm}\) wire of mass 5.625 \(\mathrm{g}\) is tied at both ends and adjusted to a tension of 35.0 \(\mathrm{N} .\) When it is vibrating in its sec- ond 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.

\(\bullet\) (a) If the amplitude in a sound wave is doubled, by what factor does the intensity of the wave increase? (b) By what fac- tor must the amplitude of a sound wave be increased in order to increase the intensity by a factor of 9\(?\)

The portion of string between the bridge and upper end of the fingerboard (the part of the string that is free to vibrate) of a certain musical instrument is 60.0 cm long and has a mass of 2.00 g. The string sounds an \(\mathrm{A}_{4}\) note \((440 \mathrm{Hz})\) when played. (a) Where must the player put a finger (at what distance \(x\) from the bridge) to play a \(\mathrm{D}_{5}\) note \((587 \mathrm{Hz}) ?\) (See Figure \(12.40 . )\) For both notes, the string vibrates in its fundamental mode. (b) Without retuning, is it possible to play a G \(_{4}\) note \((392\) Hz \()\) on this string? Why or why not?
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