/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 As the drawing shows, one microp... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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).

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The tension in a string is \(15 \mathrm{~N},\) and its linear density is \(0.85 \mathrm{~kg} / \mathrm{m} .\) A wave on the string travels toward the \(-x\) direction; it has an amplitude of \(3.6 \mathrm{~cm}\) and a frequency of \(12 \mathrm{~Hz}\) What are the (a) speed and (b) wavelength of the wave? (c) Write down a mathematical expression (like Equation 16.3 or 16.4 ) for the wave, substituting numbers for the variables \(A, f,\) and \(\lambda\)

An aircraft carrier has a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) relative to the water. A jet is catapulted from the deck and has a speed of \(67.0 \mathrm{~m} / \mathrm{s}\) relative to the water. The engines produce a \(1550-\mathrm{Hz}\) whine, and the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). What is the frequency of the sound heard by the crew on the ship?

The sonar unit on a boat is designed to measure the depth of fresh water \(\left(\rho=1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}, B_{\mathrm{ad}}=2.20 \times 10^{9} \mathrm{~Pa}\right) .\) When the boat moves into seawater \(\left(\rho=1025 \mathrm{~kg} / \mathrm{m}^{3}, B_{\mathrm{ad}}=2.37 \times 10^{9} \mathrm{~Pa}\right),\) the sonar unit is no longer calibrated properly. In seawater, the sonar unit indicates the water depth to be \(10.0 \mathrm{~m}\). What is the actual depth of the water?

Two blocks are connected by a wire that has a mass perunit length of \(8.50 \times 10^{-4} \mathrm{~kg} / \mathrm{m} .\) One block has a mass of \(19.0 \mathrm{~kg},\) and the other has a mass of 42.0 kg. These blocks are being pulled across a horizontal frictionless floor by a horizontal force \(\overrightarrow{\mathrm{P}}\) that is applied to the less massive block. A transverse wave travels on the wire between the blocks with a speed of \(352 \mathrm{~m} / \mathrm{s}\) (relative to the wire). The mass of the wire is negligible compared to the mass of the blocks. Find the magnitude of \(\overrightarrow{\mathbf{P}}\).

A recording engineer works in a soundproofed room that is \(44.0 \mathrm{~dB}\) quieter than the outside. If the sound intensity in the room is \(1.20 \times 10^{-10} \mathrm{~W} / \mathrm{m}^{2}\), what is the intensity outside?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Space Physics

Read Explanation

Torque and Rotational Motion

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Thermodynamics

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.