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Chapter 23: Problem 22

Three seconds after a gun is fired, the person who fired the gun hears an echo. How far away was the surface that reflected the sound of the shot? Use \(340 \mathrm{~m} / \mathrm{s}\) for the speed of sound.

Short Answer

Expert verified
The reflecting surface is 510 meters away.

Step by step solution

01

Understand the Problem

When the gun is fired, the sound travels to the reflective surface and back to the shooter in 3 seconds. We need to find the distance to the surface that reflected the sound.
02

Define Known Values

The speed of sound is given as \(340 \mathrm{~m/s}\), and the total time for the sound to travel to the surface and back is 3 seconds.
03

Calculate Total Distance

Using the formula for distance \( \text{Distance} = \text{Speed} \times \text{Time} \), the total distance traveled by the sound is: \[ 340 \mathrm{~m/s} \times 3 \mathrm{~s} = 1020 \mathrm{~m} \]. This distance includes the trip to the surface and back.
04

Determine the One-Way Distance

Since 1020 meters is the total distance for the round trip, the distance to the reflective surface is half of this value. Thus, \[ \text{One-way Distance} = \frac{1020 \mathrm{~m}}{2} = 510 \mathrm{~m} \].

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Key Concepts

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

Understanding Echoes
When you hear an echo, what you're actually hearing is the reflection of sound off a surface. An echo occurs when sound waves hit a barrier, like walls, mountains, or buildings, and bounce back to the source. This phenomenon is not only fascinating but also used in practical applications such as sonar and measuring distances.
  • An echo is only heard if the reflected sound reaches your ear after enough time has passed for you to distinguish it from the original sound.
  • For instance, you need a distance of at least 17 meters from the source to hear a clear echo, as the sound speed in air is about 340 meters per second (m/s).
To solve echo-related problems, you need to understand that the sound travels to the surface and back. This journey is called a round trip and it helps in distance calculations.
Basics of Distance Calculation
Distance calculation involves determining how far sound has traveled in a given period. The fundamental formula for calculating distance is:
  • Distance = Speed x Time
For the problem at hand, sound travels from the source to the reflective surface and back. Here鈥檚 what you need to remember:
  • The speed of sound is a constant (340 m/s in air).
  • The time considered is the total time for the round trip.
By applying the formula, you multiply the speed of sound by the time to calculate the whole distance for both going to the surface and returning. Then, divide this by two to find the one-way distance to the wall.
Problem Solving in Physics: Step by Step
Approaching physics problems systematically is key to arriving at the correct solution. Let's look at the necessary steps:
  • First, comprehend the problem closely. Identify the task and what's being asked.
  • Break down what you know from the problem: Is it the speed, the time, or other useful values?
  • Apply relevant formulas, like the distance formula, to find what you need.
  • Check your work by reviewing calculations and ensuring they make sense in the context.
Following these steps ensures that you see the problem accurately and do not miss key details. Physics problems often involve layers of information that need clear identification and understanding to solve correctly.

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Most popular questions from this chapter

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

In an experiment to determine the speed of sound, two observers, \(\mathrm{A}\) and \(\mathrm{B}\), were stationed \(5.00 \mathrm{~km}\) apart. Each was equipped with a gun and a stopwatch. Observer-A heard the report of B's gun \(15.5 \mathrm{~s}\) after seeing its flash. Later, A fired his gun and \(\mathrm{B}\) heard the report \(14.5 \mathrm{~s}\) after seeing the flash. Determine the speed of sound and the component of the speed of the wind along the line joining \(\mathrm{A}\) to \(\mathrm{B}\).

A disk has 40 holes around its circumference and is rotating at \(1200 \mathrm{rpm}\). Determine the frequency and wavelength of the tone produced by the disk when a jet of air is blown against it. The temperature is \(15^{\circ} \mathrm{C}\).

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 speed at which each car is moving if the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\).

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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