/*! 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 16 Two sound waves, from two differ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 16

Two sound waves, from two different sources with the same frequency, \(540 \mathrm{~Hz}\), travel in the same direction at \(330 \mathrm{~m} / \mathrm{s}\). The sources are in phase. What is the phase difference of the waves at a point that is \(4.40 \mathrm{~m}\) from one source and \(4.00 \mathrm{~m}\) from the other?

Short Answer

Expert verified
The phase difference is approximately 4.11 radians.

Step by step solution

01

Determine the wavelength of the sound waves

The wavelength \( \lambda \) can be found using the wave speed and frequency formulas. The formula to find the wavelength is:\[ \lambda = \frac{v}{f} \]where \( v = 330 \; \text{m/s} \) is the speed of sound and \( f = 540 \; \text{Hz} \) is the frequency of the sound waves. Plugging in these values gives:\[ \lambda = \frac{330}{540} = 0.611 \; \text{m} \]
02

Calculate the difference in path lengths

The difference in path lengths from the two sources is given by:\[ \Delta L = L_2 - L_1 \]where \( L_2 = 4.40 \; \text{m} \) and \( L_1 = 4.00 \; \text{m} \). So:\[ \Delta L = 4.40 - 4.00 = 0.40 \; \text{m} \]
03

Determine the phase difference

The phase difference \( \Delta \phi \) can be calculated using the equation:\[ \Delta \phi = \frac{2\pi \Delta L}{\lambda} \]Substitute \( \Delta L = 0.40 \; \text{m} \) and \( \lambda = 0.611 \; \text{m} \) into the equation:\[ \Delta \phi = \frac{2\pi \times 0.40}{0.611} \approx 4.11 \; \, \text{radians} \]

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.

Wavelength Calculation
Understanding wavelength is essential when dealing with sound waves. It tells us how long one complete wave cycle is. To calculate wavelength, we use the relationship between wave speed and frequency. This is an important formula:
  • Wavelength formula: \( \lambda = \frac{v}{f} \) where:
    • \(v\) is the speed of the wave in meters per second (m/s),
    • \(f\) is the frequency in hertz (Hz), which is the number of cycles per second.

To find the wavelength \( \lambda \) of a sound with a speed of 330 m/s and frequency of 540 Hz, simply plug in the values into the formula:
  • \( \lambda = \frac{330}{540} = 0.611 \, \text{m} \)
This means each sound wave stretches 0.611 meters long from the start to the end of one cycle.
Path Length Difference
When sound waves travel from two different sources, the path length difference comes into play. It represents how much farther one sound wave travels compared to the other. Calculating this helps us understand the interaction between these waves, like interference.

To find the path length difference \( \Delta L \), subtract the shorter path from the longer one:
  • Formula: \( \Delta L = L_2 - L_1 \)
  • Given: \( L_2 = 4.40 \, \text{m} \) and \( L_1 = 4.00 \, \text{m} \)

Substitute the values:
  • \( \Delta L = 4.40 - 4.00 = 0.40 \, \text{m} \)
The difference in distance that each wave travels is 0.40 meters. This small difference can significantly affect the phase at which waves meet.
Wave Speed and Frequency Relationship
Wave speed and frequency are intimately connected in wave dynamics. The speed of a wave is how fast it propagates through a medium. Frequency, however, shows how many wave cycles happen each second. These two factors directly influence a wave's wavelength when we consider sound waves.

The main relationship is:
  • \( v = f \times \lambda \)
  • This rearranges into the wavelength formula: \( \lambda = \frac{v}{f} \)

Using our example:
  • Speed of sound, \( v = 330 \, \text{m/s} \)
  • Frequency of the waves, \( f = 540 \, \text{Hz} \)

From this, we learned that a higher frequency results in a shorter wavelength, while a lower frequency gives a longer wavelength when speed is constant. This relationship helps us predict how changing frequency or speed impacts sound properties, which are crucial in many practical applications, such as acoustics and audio engineering.

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

A well with vertical sides and water at the bottom resonates at \(7.00 \mathrm{~Hz}\) and at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of \(1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.33 \times 10^{5} \mathrm{~Pa}\). How far down in the well is the water surface?

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a \(0.12\) s interval between; one sound comes through the rod and the other comes through the air along. side the rod. If the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\), what is the length of the rod?

Organ pipe \(A\), with both ends open, has a fundamental frequency of \(300 \mathrm{~Hz}\). The third harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe A. How long are (a) pipe \(A\) and (b) pipe \(B\) ?

A detector initially moves at constant velocity directly toward a stationary sound source and then (after passing it) directly from it. The emitted frequency is \(f .\) During the approach the detected frequency is \(f_{\text {app }}^{\prime}\) and during the recession it is \(f_{\text {rec }}^{\prime}\) If the frequencies are related by \(\left(f_{\text {app }}^{\prime}-f_{\text {rec }}^{\prime}\right) / f=0.500\), what is the ratio \(v_{D} / v\) of the speed of the detector to the speed of sound?

A whistle of frequency \(540 \mathrm{~Hz}\) moves in a circle of radius \(60.0 \mathrm{~cm}\) at an angular speed of \(15.0 \mathrm{rad} / \mathrm{s}\). What are the (a) lowest and (b) highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Fundamentals of Physics

Read Explanation

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation

Measurements

Read Explanation

Quantum Physics

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.