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Chapter 23: Problem 59

Two cars are heading straight at each other with the same speed. The horn of one \((f=3.0 \mathrm{kHz})\) is blowing, and is heard to have a frequency of \(3.4 \mathrm{kHz}\) by the people in the other car. Find the spee at which each car is moving if the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
Each car is moving at approximately 21.25 m/s.

Step by step solution

01

Identify the Given Information

We have two cars moving toward each other. One car emits sound at a frequency \( f_s = 3.0 \, \text{kHz} \) which the other car hears as \( f' = 3.4 \, \text{kHz} \). The speed of sound \( v \) is given as \( 340 \, \text{m/s} \).
02

Apply the Doppler Effect Formula

For objects moving towards each other, the formula for the observed frequency \( f' \) is given by: \[ f' = \frac{v + v_o}{v - v_s} \cdot f_s \]where \( v_o \) is the speed of the observer and \( v_s \) is the speed of the source. Since the cars move at the same speed \( v_c \), we have \( v_o = v_s = v_c \).
03

Substitute Values into the Formula

Substitute the known values into the formula:\[ 3.4 \times 10^3 = \frac{340 + v_c}{340 - v_c} \times 3.0 \times 10^3 \]
04

Simplify the Equation

Divide both sides by \( 3.0 \times 10^3 \):\[ \frac{3.4}{3.0} = \frac{340 + v_c}{340 - v_c} \] which simplifies to:\[ \frac{17}{15} = \frac{340 + v_c}{340 - v_c} \]
05

Solve the Equation for Car Speed \( v_c \)

Cross multiply to solve for \( v_c \):\[ 17(340 - v_c) = 15(340 + v_c) \]Solve for \( v_c \):\[ 5780 - 17v_c = 5100 + 15v_c \]\[ 680 = 32v_c \]\[ v_c = \frac{680}{32} \approx 21.25 \, \text{m/s} \]
06

Conclusion

The speed at which each car is moving is approximately \( 21.25 \, \text{m/s} \).

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through a medium, usually air, but they can also travel through water, solids, and gases. They are longitudinal waves, which mean they cause the particles of the medium to move back and forth in the same direction as the wave travels.
As sound moves through air, it compresses and rarefies the air molecules, creating a wave that our ears perceive as sound.
  • Sound waves can vary in intensity, frequency, and amplitude.
  • The speed at which sound waves travel is influenced by the medium, with sound traveling faster in liquids and solids than in gases.
  • Sound waves can't travel through a vacuum as there are no particles to vibrate.
Understanding how sound waves work is crucial for analyzing concepts like the Doppler Effect, where sound is perceived differently due to motion.
Frequency
Frequency refers to how frequently the particles in a medium vibrate when a wave passes through. It is measured in Hertz (Hz), indicating the number of cycles of a wave passing through a point in one second.
Higher frequency sound waves are perceived as high-pitched sounds, while lower frequencies are perceived as low-pitched sounds.
  • A typical human ear can hear frequencies between 20 Hz and 20,000 Hz.
  • Frequency changes when the source of sound or the observer moves; this is a key aspect of the Doppler Effect.
  • In the given problem, an original frequency of 3.0 kHz changes to 3.4 kHz due to the motion of the cars.
Properly understanding frequency helps in solving problems involving calculating perceived sound, especially when relative motion is involved.
Speed of Sound
The speed of sound is how fast sound waves travel through a medium. In air at room temperature, it is about 340 meters per second (m/s).
The speed of sound is affected by temperature, humidity, and the medium it travels through. For example, sound moves faster through warmer air due to increased energy and faster collisions of molecules.
  • At sea level, sound travels at about 343 m/s in air at 20 degrees Celsius.
  • In this problem, it is given as 340 m/s to simplify calculations.
  • The speed of sound is an essential factor in determining how sound is perceived when either the source or the observer is moving.
Thus, knowing how to work with the speed of sound is essential for understanding the motion-related change in sound frequency.
Observer and Source Motion
When the source of a sound and an observer move relative to each other, it affects how the sound is heard, a phenomenon known as the Doppler Effect. When they move closer, the frequency appears to increase (higher pitch), and when moving apart, it decreases (lower pitch).
Both the source and observer moving towards each other maximizes the change in frequency.
  • In the problem, both cars are moving towards each other at the same speed, which enhances the perceived frequency change.
  • This relative motion is a key element in calculating the exact frequency heard by the observer due to changes in the sound wave as it compresses or stretches.
  • Understanding the movement dynamics allows one to use the correct formulas to find the speed of the cars interacting in this Doppler scenario.
Recognizing observer and source motion is critical in predicting and explaining sound frequency perception changes.

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Most popular questions from this chapter

Determine the speed of sound in carbon dioxide \((M=44 \mathrm{~kg} / \mathrm{kmo}\) \(y=1.30\) ) at a pressure of \(0.50\) atm and a temperature of \(400^{\circ} \mathrm{C}\).

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 \(u\), and have to find \(a_{0}\). From \(I=2 \pi^{2} f^{2} \rho v a_{0}^{2}\) $$ a_{0}=\frac{1}{\pi f} \sqrt{\frac{1}{2 p v}}=\frac{1}{\left(800 \mathrm{~s}^{-1} \pi\right)} \sqrt{\frac{0.54 \mathrm{~W} / \mathrm{m}^{2}}{(2)\left(1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)(340 \mathrm{~m} / \mathrm{s})}}=9.9 \times 10^{-6 \mathrm{~m}} \mathrm{~m}=9.9 \mu \mathrm{m} $$

A machine produces a sound level of \(80 \mathrm{~dB}\) at the location of a detector. What would be the new sound level at the detector if an identical machine at the same distance was now turned on adding to the tumult? [Hint: Intensity levels do not simply add, whereas intensities do.]

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 \(u_{s}=0 .\) Hence, use \(+u_{o}\) and So (a) \(f_{f}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=544 \mathrm{~Hz}\) With the car leaving in the negative direction use \(-\mathrm{u}_{o}\) and (b) \(f_{n}=f_{0} \frac{v \pm v_{o}}{v \mp v_{3}}=(500 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}-30.0 \mathrm{~m} / \mathrm{s}}{340 \mathrm{~m} / \mathrm{s}-0}=456 \mathrm{~Hz}\)

Find the speed of sound in a diatomic ideal gas that has a density of \(3.50 \mathrm{~kg} / \mathrm{m}^{3}\) and a pressure of \(215 \mathrm{kPa}\). Using Eq. (23.2) $$ v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{(1.40)\left(215 \times 10^{3} \mathrm{~Pa}\right)}{3.50 \mathrm{~kg} / \mathrm{m}^{3}}}=293 \mathrm{~m} / \mathrm{s} $$ We used the fact that \(\gamma \approx 1.40\) for a diatomic ideal gas, as discussed in Chapter \(20 .\)
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