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Chapter 12: Q6P (page 328)

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.80 s apart. How far away did the impact occur? See Table 12-1.

Short Answer

Expert verified

The distance between the person and impact occur is \(309.82\;{\rm{m}}\).

Step by step solution

01

Understanding the concept of distance travelled by a moving system

In this problem, the distance travelled by the moving system is determined by the product of the speed of the moving system and the time of the journey.

02

Given data

The time difference between the sound travel in air and concrete is \[\Delta t = 0.80\;{\rm{s}}\].

03

Evaluating the total distance travelled by the sound by using the relation among speed, distance, and time

The standard value for the speed of sound in air and concrete from Table-12.1 are \({v_{{\rm{air}}}} = 343\;{\rm{m/s}}\) and \({v_{{\rm{concrete}}}} = 3000\;{\rm{m/s}}\) respectively.

The total distance travelled by the sound in air and concrete is equal.

The time taken by the sound to travel in air is calculated below:

\(\begin{aligned}{c}{d_{{\rm{air}}}} = {d_{{\rm{concrete}}}}\\{v_{{\rm{air}}}}{t_{{\rm{air}}}} = {v_{{\rm{concrete}}}}{t_{{\rm{concrete}}}}\\{v_{{\rm{air}}}}{t_{{\rm{air}}}} = {v_{{\rm{concrete}}}}\left( {{t_{{\rm{air}}}} - \Delta t} \right)\\{t_{{\rm{air}}}} = \frac{{{v_{{\rm{concrete}}}}}}{{{v_{{\rm{concrete}}}} - {v_{{\rm{air}}}}}}\left( {\Delta t} \right)\end{aligned}\)

Here, \({d_{{\rm{air}}}}\) and \({d_{{\rm{concrete}}}}\) are the distance travelled by the sound in air and concrete, \({t_{{\rm{air}}}}\) and \({t_{{\rm{concrete}}}}\) are the time taken by the sound in air and concrete.

The total distance travelled by the sound is calculated below:

\(\begin{aligned}{c}d = {v_{{\rm{air}}}}{t_{{\rm{air}}}}\\ = {v_{{\rm{air}}}}\left[ {\frac{{{v_{{\rm{concrete}}}}}}{{{v_{{\rm{concrete}}}} - {v_{{\rm{air}}}}}}\left( {\Delta t} \right)} \right]\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c} = 343\;{\rm{m/s}}\left[ {\frac{{3000\;{\rm{m/s}}}}{{3000\;{\rm{m/s}} - 343\;{\rm{m/s}}}}\left( {0.80\;{\rm{s}}} \right)} \right]\\ = 309.82\;{\rm{m}}\end{aligned}\)

Hence, the distance between the person and impact occur is \(309.82\;{\rm{m}}\).

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Most popular questions from this chapter

Question: The 鈥渁lpenhorn鈥� (Fig. 12鈥�42) was once used to send signals from one Alpine village to another. Since lower frequency sounds are less susceptible to intensity loss, long horns were used to create deep sounds. When played as a musical instrument, the alpenhorn must be blown in such a way that only one of the overtones is resonating. The most popular alpenhorn is about \({\bf{3}}{\bf{.4}}\,{\bf{m}}\) long, and it is called the \({{\bf{F}}^{\bf{\# }}}\) horn. What is the fundamental frequency of this horn, and which overtone is close to\({{\bf{F}}^{\bf{\# }}}\)? (See Table12鈥�3.) Model as a tube open at both ends.

Question: (II) (a) Determine the length of an open organ pipe that emits middle C (262 Hz) when the temperature is 18掳C. (b) What are the wavelength and frequency of the fundamental standing wave in the tube? (c) What are \(\lambda \)and f in the traveling sound wave produced in the outside air?

Question: (II) A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length?

Explain how a tube might be used as a filter to reduce the amplitude of sounds in various frequency ranges. (An example is a car muffler.)

Question: The intensity at the threshold of hearing for the human ear at a frequency of about \({\bf{1000}}\,{\bf{Hz}}\)is \({{\bf{I}}_{\bf{0}}}{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 12}}}}\,{\bf{W/}}{{\bf{m}}^{\bf{2}}}\), for which\({\bf{\beta }}\), the sound level, is \({\bf{0}}\,{\bf{dB}}\). The threshold of pain at the same frequency is about\({\bf{120}}\,{\bf{dB}}\), or \({\bf{I = 1}}{\bf{.0}}\,{\bf{W/}}{{\bf{m}}^{\bf{2}}}\)corresponding to an increase of intensity by a factor of \({\bf{1}}{{\bf{0}}^{{\bf{12}}}}\)By what factor does the displacement amplitude,\({\bf{A}}\), vary?

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