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Chapter 23: Problem 17

A car moving at \(20 \mathrm{~m} / \mathrm{s}\) with its horn blowing \((f=1200 \mathrm{~Hz})\) is chasing another car going at \(15 \mathrm{~m} / \mathrm{s}\) in the same direction. What is the apparent frequency of the horn as heard by the driver being chased? Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). This is a Doppler problem. Draw the observer-to-source arrow; that's the positive direction (see Fig. 23-1). Both the source and the observer are moving in the negative direction. Hence, we use \(-v_{o}\) and \(-v_{s}\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(1200 \mathrm{~Hz}) \frac{340-15}{340-20}=1.22 \mathrm{kHz} $$ Because the source is approaching the observer, the latter will measure an increase in frequency.

Short Answer

Expert verified
The apparent frequency is approximately 1.22 kHz.

Step by step solution

01

Identify the Problem Type

This is a Doppler Effect problem involving a source and an observer moving in the same direction. We'll use the Doppler Effect formula to find the apparent frequency.
02

Establish the Known Variables

The frequency of the horn, \(f_s\), is \(1200 \text{ Hz}\). The speed of the source car, \(v_s\), is \(20 \text{ m/s}\). The speed of the observer car, \(v_o\), is \(15 \text{ m/s}\). The speed of sound, \(v\), is \(340 \text{ m/s}\).
03

Apply the Doppler Effect Formula

The Doppler Effect formula for frequency observed by a moving observer from a moving source is:\[ f_o = f_s \frac{v + v_o}{v - v_s} \]Since both vehicles are moving in the same direction, we're using the formula with \(-\) in the denominator for \(v_s\) and \(-\) in the numerator for \(v_o\).
04

Substitute Known Values into the Formula

Substitute the known values into the equation:\[ f_o = 1200 \times \frac{340 - 15}{340 - 20} \]
05

Calculate the Apparent Frequency

First, calculate the values in the numerator and the denominator:\[ f_o = 1200 \times \frac{325}{320} \]Now, calculate the apparent frequency:\[ f_o = 1200 \times 1.015625 = 1218.75 \text{ Hz} \]
06

Round to Three Significant Figures

The apparent frequency can be rounded to three significant figures:\[ f_o \approx 1.22 \text{ kHz} \]

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Key Concepts

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

Apparent Frequency
The apparent frequency refers to the frequency of a wave as observed in a different frame of reference. In this case, the observer hears a different frequency than the one emitted by the source due to their relative motion. The Doppler Effect is responsible for this change in frequency.
For example, imagine a car honking its horn while moving towards another car. The stationary driver hears a higher pitch (frequency) than what the horn actually emits. This is because the sound waves are compressed due to the approaching motion, causing the observer to receive waves more frequently.
When both the source (the honking car) and the observer (the second car) are moving, the formula for the apparent frequency becomes:
  • \( f_o = f_s \frac{v + v_o}{v - v_s} \)
In this case:
  • \( f_s \) is the source frequency, 1200 Hz
  • \( v \) is the velocity of sound, 340 m/s
  • \( v_o \) is the velocity of the observer, 15 m/s
  • \( v_s \) is the velocity of the source, 20 m/s
By using the proper values, the apparent frequency can be calculated, showing how much it increases when the source approaches the observer.
Sound Waves
Sound waves are vibrations that travel through a medium, like air. They are longitudinal waves, meaning the displacement of the medium is parallel to the direction of the wave's movement. These waves are responsible for the transmission of sound from a source to an observer.
Each wave consists of alternating regions of compression and rarefaction. Compression is where air molecules are close together, and rarefaction is where they are spread apart. The frequency of these compressions and rarefactions determines the pitch of the sound we hear.
In everyday situations, sound waves allow us to hear everything from a gentle whisper to the blare of a car horn. In the Doppler Effect, the motion of the sound source or the observer changes the frequency of the waves as perceived by the observer. This is why we might hear a change in pitch or frequency, known as the apparent frequency, especially if the source is moving quickly or approaching us.
Velocity of Sound
The velocity of sound refers to the speed at which sound waves travel through a medium. This speed can vary depending on factors like the medium's density and temperature. In air at room temperature, sound travels at approximately 340 meters per second.
Different materials have different sound velocities. For example:
  • Sound travels faster in water than in air.
  • In solids, like metals, sound can travel even faster because molecules are packed more tightly, allowing waves to transmit more efficiently.
Velocity plays a crucial role in many calculations involving sound, such as determining the apparent frequency using the Doppler Effect formula. In our previous example with the cars, the sound velocity in air was a critical factor in finding out how the frequency changed as one car chased another.
Ultimately, understanding velocity allows us to comprehend how sound behaves in different environments, which is important for applications in acoustics, audio technology, and even navigation systems.

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Most popular questions from this chapter

A whisper has an intensity level of about \(15 \mathrm{~dB}\). What is the corresponding intensity of the sound?

An automobile moving at \(30.0 \mathrm{~m} / \mathrm{s}\) is approaching a factory whistle that has a frequency of \(500 \mathrm{~Hz}\). (a) If the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), what is the apparent frequency of the whistle as heard by the driver? \((b)\) Repeat for the case of the car leaving the factory at the same speed. This is a Doppler shift problem. Draw an arrow from observer to source; this is the positive direction. Here in part \((a)\) the observer is moving in the positive direction, and \(v_{s}=0\). Hence, use \(+v_{o}\) and \(s 0\) (a) \(f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=544 \mathrm{~Hz}\) With the car leaving in the negative direction use \(-v_{0}\) and (b) \(f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}-30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=456 \mathrm{~Hz}\)

Back in the days before computers, a single typist typing furiously could generate an average sound level nearby of \(60.0 \mathrm{~dB}\). What would be the decibel level in the vicinity if three equally noisy typists were working close to one another? If each typist emits the same amount of sound energy, then the final sound intensity \(I_{f}\) should be three times the initial intensity \(I_{i}\). We have and $$ \begin{array}{l} \beta_{f}=\log _{10}\left(\frac{I_{f}}{I_{0}}\right)=\log _{10} I_{f}-\log _{10} I_{0} \\ \beta_{i}=\log _{10} I_{i}-\log _{10} I_{0} \end{array} $$ Subtracting these yields the change in sound level in going from \(I_{i}\) to \(I_{f}=3 I_{P}\) $$ \beta_{f}-\beta_{i}=\log _{10} I_{f}-\log _{10} I_{i} $$ from which $$ \beta_{f}=\beta_{i}+\log _{10}\left(\frac{I_{f}}{I_{i}}\right)=60.0 \mathrm{~dB}+\log _{10} 3=60.5 \mathrm{~dB} $$ The sound level, being a logarithmic measure, rises very slowly with the number of sources.

An uncomfortably loud sound might have an intensity of \(0.54 \mathrm{~W} / \mathrm{m}^{2}\). Find the maximum displacement of the molecules of air in a sound wave if its frequency is \(800 \mathrm{~Hz}\). Take the density of air to be \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) and the speed of sound to be \(340 \mathrm{~m} / \mathrm{s}\). We are given \(I, f, \rho\), and \(v\), and have to find \(a_{0}\). From \(I=2 \pi^{2} f^{2} \rho v a_{0}^{2}\) $$ a_{0}=\frac{1}{\pi f} \sqrt{\frac{I}{2 \rho v}}=\frac{1}{\left(800 \mathrm{~s}^{-1} \pi\right)} \sqrt{\frac{0.54 \mathrm{~W} / \mathrm{m}^{2}}{(2)\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)(340 \mathrm{~m} / \mathrm{s})}}=9.9 \times 10^{-6} \mathrm{~m}=9.9 \mu \mathrm{m} $$

An organ pipe whose length is held constant resonates at a frequency of \(224.0 \mathrm{~Hz}\) when the air temperature is \(15^{\circ} \mathrm{C}\). What will be its resonant frequency when the air temperature is \(24^{\circ} \mathrm{C}\) ? The resonant wavelength must have the same value at each temperature because it depends only on the length of the pipe. (Its nodes and antinodes must fit properly within the pipe.) But \(\lambda=v / f\), and so \(v / f\) must be the same at the two temperatures. Consequently, $$ \frac{v_{1}}{224 \mathrm{~Hz}}=\frac{v_{2}}{f_{2}} \quad \text { or } \quad f_{2}=(224 \mathrm{~Hz})\left(\frac{v_{2}}{v_{1}}\right) $$ At temperatures near room temperature, \(v=\left(331+0.61 T_{c}\right) \mathrm{m} / \mathrm{s}\), where \(T_{c}\) is the Celsius temperature. Then $$ f_{2}=(224.0 \mathrm{~Hz})\left[\frac{331+(0.61)(24)}{331+(0.61)(15)}\right]=0.228 \mathrm{kHz} $$
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