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Chapter 16: Problem 43

A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than to the He fires a gun and, after a while, hears three echoes. The second echo arrives \(1.6 \mathrm{s}\) after the first, and the third echo arrives \(1.1 \mathrm{s}\) after the second. Assuming that the speed of sound is \(343 \mathrm{m} / \mathrm{s}\) and that there are no reflections of sound from the ground, find the distance between the cliffs.

Short Answer

Expert verified
The distance between the cliffs is 686 meters.

Step by step solution

01

Understand the Echoes

The hunter hears three echoes. The first echo is from the nearest cliff, the second one comes after 1.6 seconds, and the third one arrives 1.1 seconds after the second echo. This means the second echo is from the farther cliff, and the third echo is a reflection back from the original side.
02

Calculate Distance for Each Echo

The time taken for each echo gives the time sound travels to a cliff and back. - First echo: Sound travels to the first cliff and back. If the distance to the first cliff is denoted as \(d_1\), the total travel time is \( \frac{2d_1}{343} = t_1\). - Second echo: Sound travels to the second cliff and back. If the distance to the second cliff is \(d_2\), \( \frac{2d_2}{343} = t_1 + 1.6\).
03

Analyze Third Echo Timing

The third echo comes 1.1 seconds after the second. This third echo is a result of the sound going from the second cliff to the first cliff and back. The time taken for this journey (from the second cliff, to the first cliff, and back to the second cliff) can be calculated: \( \frac{d_2 + d_1 + d_2}{343} = t_1 + 1.6 + 1.1\)
04

Establish Equations

Now we have three key relationships:- **Echo 1:** \( t_1 = \frac{2d_1}{343} \)- **Echo 2:** \( t_1 + 1.6 = \frac{2d_2}{343} \)- **Echo 3:** Time from second to the first and back: \( t_1 + 2.7 = \frac{2d_2 + 2d_1}{343} \)
05

Solve the Equations

First, solve for \(d_1\) using the first equation and \(t_1 = \frac{2d_1}{343}\) implies \(d_1 = \frac{343t_1}{2}\). Substitute into the second equation and solve for \(d_2\): \(d_2 = \frac{343(t_1 + 1.6)}{2}\). Place these results into the third equation to confirm the solution matches.
06

Calculate the Distance Between Cliffs

Using the distance expressions found, the total distance \(D\) between the cliffs is \(d_1 + d_2\). Use the expressions:\[d_1 = \frac{343t_1}{2}\] \[d_2 = \frac{343(t_1 + 1.6)}{2}\]Calculate \(D = d_1 + d_2 = \frac{343t_1}{2} + \frac{343(t_1 + 1.6)}{2}\). Simplifying, you find \(D = 343(t_1 + 0.8)\). Solve for the time by using the additional equations.

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Key Concepts

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

Echo Calculation
In physics, an echo is a sound that reflects off a surface back to the listener, much like how light bounces off a mirror. When solving echo problems, you tackle the challenge of determining how far sound has traveled based on timing. Timing is crucial because it determines both when and how far the sound waves travel.

When the hunter fires his gun, the sound waves move towards the cliff, reflect, and then travel back to the hunter. This is the journey of the echo. The time for an echo gives us a clue about the total distance covered 鈥� going to the cliff and coming back. Understanding this concept helps in the breakdown of the echoes based on their arrival times:
  • The first echo returns from the nearest cliff.
  • The second and third echoes involve the farther cliff, either reflecting directly or bouncing between the cliffs.
This sequential timing of the echoes helps us calculate specific distances when combined with the speed of sound.
Sound Speed
The speed of sound is a fundamental concept in solving echo-related problems. At room temperature, sound travels through air at approximately 343 meters per second (m/s). This speed, however, can change with varying conditions like temperature or medium.

For calculations, the speed of sound allows us to connect time and distance in the problem. Sound travels a specific distance in a given amount of time, making the speed of sound essential for deriving distances. To calculate how far sound has traveled, the formula used is:
distance = speed \( \times \) time
This is directly applied to each echo observed by the hunter, allowing us to compute the travel path of the echoes using the time differences. By knowing the speed, you can turn echo timings into distances, crucial to understanding the layout of the cliffs.
Distance Between Cliffs
Determining the distance between cliffs involves interpreting different echo timings and deducing the various distances involved. The total distance calculation depends on understanding the paths that each sound reflection (or echo) might take.

The overall goal is to find the distance between the two cliffs. Use the echo delay times to set up equations that reflect the sequential journeys of the sound waves:
  • The hunter hears the first echo from the nearest cliff.
  • The second echo comes from the farther cliff, and the time difference helps us to calculate this separate distance.
  • The third echo is more complex as it includes the sound traveling across both distances, thus requires combining two separate cliff distances.
Solving these equations by combining knowledge of echo reflections and the speed of sound lets you find the total distance separating the cliffs in a logical manner. Ultimately, understanding each part of this journey allows precise calculation of geographic separations.

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Most popular questions from this chapter

At a distance of \(3.8 \mathrm{m}\) from a siren, the sound intensity is \(3.6 \times 10^{-2} \mathrm{W} / \mathrm{m}^{2} .\) Assuming that the siren radiates sound uniformly in all directions, find the total power radiated.

Two sources of sound are located on the \(x\) axis, and each emits power uniformly in all directions. There are no reflections. One source is positioned at the origin and the other at \(x=+123 \mathrm{m}\). The source at the origin emits four times as much power as the other source. Where on the \(x\) axis are the two sounds equal in intensity? Note that there are two answers.

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is \(1522 \mathrm{m} / \mathrm{s}\). When the dolphin is swimming directly away at \(8.0 \mathrm{m} / \mathrm{s},\) the marine biologist measures the number of clicks occurring per second to be at a frequency of \(2500 \mathrm{Hz}\). What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?

A car is parked \(20.0 \mathrm{m}\) directly south of a railroad crossing. A train is approaching the crossing from the west, headed directly east at a speed of \(55.0 \mathrm{m} / \mathrm{s}\). The train sounds a short blast of its \(289-\mathrm{Hz}\) horn when it reaches a point \(20.0 \mathrm{m}\) west of the crossing. What frequency does the car's driver hear when the horn blast reaches the car? The speed of sound in air is \(343 \mathrm{m} / \mathrm{s}\). (Hint: Assume that only the component of the train's velocity that is directed toward the car affects the frequency heard by the driver.

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does not change when the sound enters the water. The wavelength of the sound is \(2.74 \mathrm{m}\) in the air, and the temperature of both the air and the water is \(20^{\circ} \mathrm{C} .\) What is the wavelength in the water?
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