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Chapter 17: Problem 5

A cowboy stands on horizontal ground between two parallel vertical cliffs. He is not midway between the cliffs. He fires a shot and hears its echoes. The second echo arrives \(1.92 \mathrm{s}\) after the first and \(1.47 \mathrm{s}\) before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs. Take the speed of sound as \(340 \mathrm{m} / \mathrm{s}\) (a) What is the distance between the cliffs? (b) What If? If he can hear a fourth echo, how long after the third echo does it arrive?

Short Answer

Expert verified
Hence, the distance between the cliffs is 326.4 meters and the fourth echo would arrive 5.31 seconds after the third echo.

Step by step solution

01

Solving distance between cliffs

First, solve for the distance between the two cliffs. The second echo arrives 1.92 s after the first echo. In this interval, sound travels twice the distance between the cliffs. The speed of sound is given as \(340 \, m/s\). So, the distance between the cliffs is \(1.92s \times 340m/s / 2 = 326.4m\).
02

Solving time after third echo

The third echo actually arrives after the sound had time to travel four times the distance between the cliffs. Solving for this gives \(2 (1.92s + 1.47s) = 6.78s\). Therefore, a fourth echo from the first cliff would arrive \(6.78s - 1.47s = 5.31s\) after the third echo.

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Key Concepts

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

Speed of Sound
The speed of sound is a crucial factor in defining how fast sound waves travel through different mediums. Typically, in dry air at room temperature, this speed is roughly 340 meters per second (m/s). This velocity can change based on environmental conditions, such as humidity and temperature. In this exercise, the speed of sound is essential to calculate how quickly sound waves hit and reflect off the cliffs.
Sound waves can transport energy, such as the cowboy's gunshot noise, through the air. Knowing the speed allows us to compute how sound moves across given distances. This is vital in understanding how much time it takes for echoes to be heard.
  • Speed of sound: 340 m/s in this problem.
  • Is vital for calculating time and distance in echo-related problems.
Echo Timing
Echo timing refers to the measurement of time intervals between sounds and their subsequent echoes. An echo occurs when sound waves hit a surface, like a cliff, and bounce back, producing a delayed second sound.
In the problem, the time lag between each echo is crucial. For example, the cowboy heard the second echo 1.92 seconds after the first. After understanding this delay, we can calculate the distance to the reflecting surfaces using the known speed of sound.
  • Echoes are sound waves reflecting back to the listener.
  • Timing helps in determining distances using sound speed.
Proper echo timing allows us to measure how long it takes for a sound to travel a known distance and back, informing us about spatial arrangements.
Distance Calculation
Distance calculation is a fundamental part of analyzing sound wave behavior, particularly in echo scenarios. Knowing the time it takes for sound to travel to an object and return allows us to determine how far away that object is.
In this particular problem, the second echo occur 1.92 seconds after the first echo, which means sound traveled back and forth for that period. Since the sound speed is 340 m/s, it traveled twice the distance between the cliffs during this interval. By calculating this鈥攎ultiplying 1.92 seconds by 340 m/s, and dividing by 2鈥攚e find the distance as 326.4 meters between the cliffs.
Solving for the time after the third echo involves knowing sound traveled four times the cliff distance, requiring an understanding of how sound waves travel and reflect in space.
Sound Waves
Sound waves are vibrations traveling through a medium, like air, water, or solids. They are longitudinal waves, meaning their vibration direction is parallel to the wave's motion direction. Sound waves transmit energy, letting us hear sounds.
When a cowboy fires a gunshot, it sends out sound waves, which can then reflect off surfaces such as cliffs. These reflections produce echoes, enabling us to measure distances and timings based on when these echoes are heard again.
Understanding sound waves is essential in echo problems, as it helps visualize how sound travels, interacts with environments, and gets perceived. Recognizing sound wave behaviors reveals why and how echoes occur, allowing us to solve real-world acoustic problems.
  • Sends vibrations through mediums like air.
  • Produces echoes when reflecting off surfaces.

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Most popular questions from this chapter

Prove that sound waves propagate with a speed given by Equation \(17.1 .\) Proceed as follows. In Figure \(17.3,\) consider a thin cylindrical layer of air in the cylinder, with face area \(A\) and thickness \(\Delta x\). Draw a free-body diagram of this thin layer. Show that \(\Sigma F_{x}=m a_{x}\) implies that \(-[\partial(\Delta P) / \partial x] A \Delta x=\) \(\rho A \Delta x\left(\partial^{2} s / \partial t^{2}\right) .\) By substituting \(\Delta P=-B(\partial s / \partial x),\) obtain the wave equation for sound, \((B / \rho)\left(\partial^{2} s / \partial x^{2}\right)=\left(\partial^{2} s / \partial t^{2}\right) .\) To a mathematical physicist, this equation demonstrates the existence of sound waves and determines their speed. As a physics student, you must take another step or two. Substitute into the wave equation the trial solution \(s(x, l)=\) \(s_{\max } \cos (k x-\omega t) .\) Show that this function satisfies the wave equation provided that \(\omega / k=\sqrt{B / \rho} .\) This result reveals that sound waves exist provided that they move with the speed \(v=f \lambda=(2 \pi f)(\lambda / 2 \pi)=\omega / k=\sqrt{B / \rho}\)

A train is moving parallel to a highway with a constant speed of \(20.0 \mathrm{m} / \mathrm{s} .\) A car is traveling in the same direction as the train with a speed of \(40.0 \mathrm{m} / \mathrm{s} .\) The car horn sounds at a frequency of \(510 \mathrm{Hz},\) and the train whistle sounds at a frequency of \(320 \mathrm{Hz}\). (a) When the car is behind the train, what frequency does an occupant of the car observe for the train whistle? (b) After the car passes and is in front of the train, what frequency does a train passenger observe for the car horn?

Note: Use the following values as needed unless otherwise specified: the equilibrium density of air at \(20^{\circ} \mathrm{C}\) is \(\rho=1.20 \mathrm{kg} / \mathrm{m}^{3} .\) The speed of sound in air is \(v=\) \(343 \mathrm{m} / \mathrm{s} .\) Pressure variations \(\Delta P\) are measured relative to atmospheric pressure, \(1.013 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) Problem 70 in Chapter 2 can also be assigned with this section. A bat (Fig. P17.7) can detect very small objects, such as an insect whose length is approximately equal to one wavelength of the sound the bat makes. If a bat emits chirps at a frequency of \(60.0 \mathrm{kHz},\) and if the speed of sound in air is \(340 \mathrm{m} / \mathrm{s},\) what is the smallest insect the bat can detect? PICTURE CANT COPY

As a certain sound wave travels through the air, it produces pressure variations (above and below atmospheric pressure) given by \(\Delta P=1.27 \sin (\pi x-340 \pi t)\) in SI units. Find (a) the amplitude of the pressure variations, (b) the frequency, (c) the wavelength in air, and (d) the speed of the sound wave.

With particular experimental methods, it is possible to produce and observe in a long thin rod both a longitudinal wave and a transverse wave whose speed depends primarily on tension in the rod. The speed of the longitudinal wave is determined by the Young's modulus and the density of the material as \(\sqrt{Y / \rho} .\) The transverse wave can be modeled as a wave in a stretched string. A particular metal rod is \(150 \mathrm{cm}\) long and has a radius of \(0.200 \mathrm{cm}\) and a mass of \(50.9 \mathrm{g} .\) Young's modulus for the material is \(6.80 \times 10^{10} \mathrm{N} / \mathrm{m}^{2} .\) What must the tension in the rod be if the ratio of the speed of longitudinal waves to the speed of transverse waves is \(8.00 ?\)
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