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Chapter 16: Problem 38

As the drawing shows, one microphone is located at the origin, and a second microphone is located on the \(+y\) axis. The microphones are separated by a distance of \(D=1.50 \mathrm{~m}\). A source of sound is located on the \(+x\) axis, its distances from microphones 1 and 2 being \(L_{1}\) and \(L_{2}\), respectively. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). The sound reaches microphone 1 first, and then, 1.46 ms later, it reaches microphone 2 . Find the distances \(L_{1}\) and \(L_{2}\)

Short Answer

Expert verified
The distances are approximately \( L_1 = 1.991 \, \mathrm{m} \) and \( L_2 = 2.493 \, \mathrm{m} \).

Step by step solution

01

Understand the relation between time delay and distances

The difference in time it takes for the sound to reach the two microphones can be expressed using the formula:\[ \Delta t = \frac{L_{2} - L_{1}}{v} \]where \( \Delta t \) is the time delay, \( L_{1} \) and \( L_{2} \) are the distances to microphones 1 and 2 respectively, and \( v \) is the speed of sound.
02

Convert time delay to seconds

Convert the time delay from milliseconds to seconds:\[ 1.46 \, \mathrm{ms} = 1.46 \times 10^{-3} \, \mathrm{s} \]
03

Setup equation for the time delay

Substitute the known values into the equation:\[ 1.46 \times 10^{-3} = \frac{L_{2} - L_{1}}{343} \]
04

Solve for \(L_{2} - L_{1}\)

Rearrange the equation and solve for \( L_{2} - L_{1} \):\[ L_{2} - L_{1} = 1.46 \times 10^{-3} \times 343 \]\[ L_{2} - L_{1} = 0.50178 \, \mathrm{m} \]
05

Apply the Pythagorean theorem

The microphones form a right triangle with the source of the sound, where:\[ L_{2}^2 = L_{1}^{2} + 1.50^2 \]
06

Solve the system of equations

We have two equations now:1. \( L_{2} - L_{1} = 0.50178 \)2. \( L_{2}^2 = L_{1}^{2} + 1.50^2 \)Using these, substitute \( L_{2} = L_{1} + 0.50178 \) into the second equation:\[ (L_{1} + 0.50178)^2 = L_{1}^{2} + 1.50^2 \]Expand:\[ L_{1}^2 + 2 \cdot 0.50178 \cdot L_{1} + 0.50178^2 = L_{1}^{2} + 2.25 \]Simplify:\[ 0.50178 \cdot 2 \times L_{1} + 0.251785 = 2.25 \]Solve for \(L_{1}\):\[ 1.00356 \cdot L_{1} = 2.25 - 0.251785 \]\[ 1.00356 \cdot L_{1} = 1.998215 \]\[ L_{1} \approx 1.991 \mathrm{~m} \]
07

Calculate \( L_{2} \) using \( L_{1} \)

Use the relation \( L_{2} = L_{1} + 0.50178 \):\[ L_{2} = 1.991 + 0.50178 \]\[ L_{2} \approx 2.493 \mathrm{~m} \]

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Key Concepts

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

Sound Propagation
Sound propagation refers to the movement of sound waves through a medium, usually air. Sound waves are mechanical waves that require a medium to travel. They are created by a vibrating source, causing the molecules within the medium to oscillate and transmit energy from one location to another.
When a sound wave is generated, it travels outward from the source in all directions. The speed and direction of these waves can change due to environmental conditions such as temperature, humidity, and wind. However, under standard conditions, sound travels at a predictable speed. Understanding sound propagation is crucial for solving problems involving sound delay and distance, like the one in our example.
  • Sound travels as longitudinal waves, meaning that the displacement of the medium is parallel to the direction of wave propagation.
  • The frequency and wavelength of a sound wave determine its pitch and perceived sound quality.
  • Different materials affect the speed and intensity of sound propagation.
Speed of Sound
The speed of sound is the rate at which sound waves pass through a medium. In air, under normal conditions, this speed is approximately 343 meters per second (m/s). This velocity is influenced by various factors including the medium, temperature, and atmospheric pressure.
Knowing the speed of sound allows us to calculate distances and travel times of sound waves between two points. In the problem at hand, the sound travels from a source to two microphones separated by a certain distance, and knowing the speed helps determine the time it takes for the sound to reach each microphone.
Factors affecting the speed of sound include:
  • Temperature: As the temperature increases, the speed of sound in air also increases.
  • Medium: Sound travels faster in liquids and solids compared to gases because particles in these states are closer together.
  • Humidity: Higher humidity can increase the speed of sound since air becomes less dense when humidity is high.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry that relates the lengths of the sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem is given by the formula:\[ c^2 = a^2 + b^2 \]
In our exercise, the microphones and the sound source form a right triangle. The microphones are positioned such that the distance between them and the source must be calculated using this theorem.
  • Allows us to solve for unknown distances when given two known distances in a right-angled triangle.
  • Useful in problems involving spatial relationships and distances.
  • Assists in converting the problem of sound distance into a solvable mathematical equation.
Time Delay Calculation
Time delay calculation is a key part of identifying how sound travels over distances with varying times. It is the time taken for sound to travel from one point to another, and can be crucial in finding spatial positions.
In exercises such as this, where microphones receive sound at different times, the time delay (\( \Delta t \)) between the two receptions can be used to determine the differing distances (\( L_2 \) and \( L_1 \)) the sound traveled.
This formula\[ \Delta t = \frac{L_2 - L_1}{v} \]helps calculate the additional distance sound travels to reach the second microphone. Converting milliseconds to seconds allows us to accurately use the speed of sound (\( v \)) in these calculations.
Calculating the time delay involves:
  • Identifying the time it takes for sound to reach different points from a source.
  • Utilizing known speeds of sound to deduce distances.
  • Ensuring accurate unit conversions between time measurements (milliseconds to seconds).

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Most popular questions from this chapter

Multiple-Concept Example 4 presents one method for modeling this type of problem. Civil engineers use a transit theodolite when surveying. A modern version of this device determines distance by measuring the time required for an ultrasonic pulse to reach a target, reflect from it, and return. Effectively, such a theodolite is calibrated properly when it is programmed with the speed of sound appropriate for the ambient air temperature. (a) Suppose the round-trip time for the pulse is \(0.580 \mathrm{~s}\) on a day when the air temperature is \(293 \mathrm{~K},\) the temperature for which the instrument is calibrated. How far is the target from the theodolite? (b) Assume that air behaves as an ideal gas. If the air temperature were \(298 \mathrm{~K}\), rather than the calibration temperature of \(293 \mathrm{~K}\), what percentage error would there be in the distance measured by the theodolite?

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 other. 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.

A longitudinal wave with a frequency of \(3.0 \mathrm{~Hz}\) takes \(1.7 \mathrm{~s}\) to travel the length of a \(2.5-\mathrm{m}\) Slinky (see Figure \(16-3\) ). Determine the wavelength of the wave.

In a discussion person \(\mathrm{A}\) is talking \(1.5 \mathrm{~dB}\) louder than person \(\mathrm{B},\) and person \(\mathrm{C}\) is talking \(2.7 \mathrm{~dB}\) louder than person \(\mathrm{A}\). What is the ratio of the sound intensity of person \(\mathrm{C}\) to the sound intensity of person \(\mathrm{B} ?\)

At provides some pertinent background for this problem. A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant \(9.00 \mathrm{~m} / \mathrm{s}\), and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away?
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