Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 18, Problem 18.18P

A standing wave is described by the wave function

y = 6 sin ( π 2 x ) cos ( 100 π t )

where x and y are in meters and t is in seconds. (a) Prepare graphs showing y as a function of x for five instants: t = 0, 5 ms, 10 ms, 15 ms, and 20 ms. (b) From the graph, identify the wavelength of the wave and explain how to do so. (c) From the graph, identify the frequency of the wave and explain how to do so. (d) From the equation, directly identify the wavelength of the wave and explain bow to do so. (e) From the equation, directly identify the frequency and explain how to do so.

(a)

Expert Solution
Check Mark
To determine

To draw: The graphs showing y as a function of x for five instants t=0 , t=5ms , t=10ms , t=15ms and t=20ms .

Answer to Problem 18.18P

The graph of y as a function of x at an instant t=0 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  1

Figure (1)

The graph of y as a function of x at an instant t=5ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  2

Figure (2)

The graph of y as a function of x at an instant t=10ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  3

Figure (3)

The graph of y as a function of x at an instant t=15ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  4

Figure (4)

The graph of y as a function of x at an instant t=20ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  5

Figure (5)

Explanation of Solution

Introduction:

The values of y varies in sinusoidal form. Initially it increases from zero to its maximum value and then decreases from maximum value to zero. This phenomenon gets repeated in periodic form.

Explanation:

Given info: The sinusoidal waves function is,

y=6sin(π2x)cos(100πt) . (1)

For t=0 :

Substitute 0 for t in the equation (1).

y=6sin(π2x)cos(100π×0)=6sin(π2x)cos(0)=6sin(π2x)×1=6sin(π2x)

The graph of y as a function of x at an instant t=0 is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  6

Figure (1)

For t=5ms :

Substitute 5ms for t in the equation (1).

y=6sin(π2x)cos(100π×(5ms))=6sin(π2x)cos(100π×(5×103s))=6sin(π2x)×cos(0.5π)=0

The graph of y as a function of x at an instant t=5ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  7

Figure (2)

For t=10ms :

Substitute 10ms for t in the equation (1).

y=6sin(π2x)cos(100π×10ms)=6sin(π2x)cos(100π×10×103s)=6sin(π2x)×cos(π)=6sin(π2x)

The graph of y as a function of x at an instant t=10ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  8

Figure (3)

For t=15ms :

Substitute 15ms for t in the equation (1).

y=6sin(π2x)cos(100π×15ms)=6sin(π2x)cos(100π×15×103s)=6sin(π2x)×cos(1.5π)=0

The graph of y as a function of x at an instant t=15ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  9

Figure (4)

For t=20ms :

Substitute 20ms for t in the equation (1).

y=6sin(π2x)cos(100π×20ms)=6sin(π2x)cos(100π×20×103s)=6sin(π2x)×cos(2π)=6sin(π2x)

The graph of y as a function of x at an instant t=20ms is shown below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 18, Problem 18.18P , additional homework tip  10

Figure (5)

(b)

Expert Solution
Check Mark
To determine
The wavelength of the wave using the graph of y as a function of x .

Answer to Problem 18.18P

The wavelength of the wave is 4m .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

The distance between the two-crest point or two trough point is called the wavelength of the wave.

From the figure (1), the distance between the two crest point is 4m . Therefore, the wavelength of the wave is,

λ=4m

Conclusion:

Therefore, the wavelength of the wave is 4m .

(c)

Expert Solution
Check Mark
To determine
The frequency of the wave using the graph of y as a function of x .

Answer to Problem 18.18P

The frequency of the wave is 50Hz .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

Compare the equation (1) with y=Asinkxcosωt .

ω=100π

The frequency of the wave is,

f=ω2π

Substitute 100π for ω in the above equation.

f=100π2π=50Hz

Conclusion:

Therefore, the frequency of the wave is 50Hz .

(d)

Expert Solution
Check Mark
To determine
The wavelength of the wave using the equation of the wave.

Answer to Problem 18.18P

The wavelength of the wave is 4m .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

Compare the equation (1) with y=Asinkxcosωt .

k=π2

The frequency of the wave is,

λ=2πk

Substitute π2 for k in the above equation.

λ=2π(π2)=4m

Conclusion:

Therefore, the wavelength of the wave is 4m .

(e)

Expert Solution
Check Mark
To determine
The frequency of the wave using the equation of the wave.

Answer to Problem 18.18P

The frequency of the wave is 50Hz .

Explanation of Solution

Given info: The sinusoidal waves function is y=6sin(π2x)cos(100πt) .

Compare the equation (1) with y=Asinkxcosωt .

ω=100π

The frequency of the wave is,

f=ω2π

Substitute 100π for ω in the above equation.

f=100π2π=50Hz

Conclusion:

Therefore, the frequency of the wave is 50Hz .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A standing wave is described by the wave function                                    y = 6 sin (π/2 x) cos (100πt)where x and y are in meters and t is in seconds. (a) Prepare graphs showing y as a function of x for five instants: t = 0, 5 ms, 10 ms, 15 ms, and 20 ms. (b) From the graph, identify the wavelength of the wave and explain how to do so. (c) From the graph, identify the frequency of the wave and explain how to do so. (d) From the equation, directly identify the wavelength of the wave and explain how to do so. (e) From the equation, directly identify the frequency and explain how to do so.
A wave traveling along the x axis is described mathematically by the equation y = 0.21 sin(4.1zt + 0.65rx), where y is the displacement (in meters), t is in seconds, and x is in meters. a) What is the frequency of the wave? b) What is the wavelength of the wave?
The function y(x, t) = A cos(kx - wt) describes a traveling wave on a taut string with the x-axis parallel to the string. If the wavelength of the wave λ = 0.8 m and w = 17.1л/s, what is the speed of the traveling wave? Express your answer in m/s, to at least one digit after the decimal point.

Chapter 18 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

Ch. 18 - When two tuning forks are sounded at the same...Ch. 18 - A tuning fork is known to vibrate with frequency...Ch. 18 - An archer shoots an arrow horizontally from the...Ch. 18 - As oppositely moving pulses of the same shape (one...Ch. 18 - Prob. 18.10OQCh. 18 - Suppose all six equal-length strings of an...Ch. 18 - Assume two identical sinusoidal waves are moving...Ch. 18 - Prob. 18.1CQCh. 18 - When two waves interfere constructively or...Ch. 18 - Prob. 18.3CQCh. 18 - What limits the amplitude of motion of a real...Ch. 18 - Prob. 18.5CQCh. 18 - An airplane mechanic notices that the sound from a...Ch. 18 - Despite a reasonably steady hand, a person often...Ch. 18 - Prob. 18.8CQCh. 18 - Does the phenomenon of wave interference apply...Ch. 18 - Two waves are traveling in the same direction...Ch. 18 - Two wave pulses A and B are moving in opposite...Ch. 18 - Two waves on one string are described by the wave...Ch. 18 - Two pulses of different amplitudes approach each...Ch. 18 - A tuning fork generates sound waves with a...Ch. 18 - The acoustical system shown in Figure OQ18.1 is...Ch. 18 - Two pulses traveling on the same string are...Ch. 18 - Two identical loudspeakers are placed on a wall...Ch. 18 - Two traveling sinusoidal waves are described by...Ch. 18 - Why is the following situation impossible? Two...Ch. 18 - Two sinusoidal waves on a string are defined by...Ch. 18 - Two identical sinusoidal waves with wavelengths of...Ch. 18 - Two identical loudspeakers 10.0 m apart are driven...Ch. 18 - Prob. 18.14PCh. 18 - Two sinusoidal waves traveling in opposite...Ch. 18 - Verify by direct substitution that the wave...Ch. 18 - Two transverse sinusoidal waves combining in a...Ch. 18 - A standing wave is described by the wave function...Ch. 18 - Two identical loudspeakers are driven in phase by...Ch. 18 - Prob. 18.20PCh. 18 - A string with a mass m = 8.00 g and a length L =...Ch. 18 - The 64.0-cm-long string of a guitar has a...Ch. 18 - The A string on a cello vibrates in its first...Ch. 18 - A taut string has a length of 2.60 m and is fixed...Ch. 18 - A certain vibrating string on a piano has a length...Ch. 18 - A string that is 30.0 cm long and has a mass per...Ch. 18 - In the arrangement shown in Figure P18.27, an...Ch. 18 - In the arrangement shown in Figure P17.14, an...Ch. 18 - Review. A sphere of mass M = 1.00 kg is supported...Ch. 18 - Review. A sphere of mass M is supported by a...Ch. 18 - Prob. 18.31PCh. 18 - Review. A solid copper object hangs at the bottom...Ch. 18 - Prob. 18.33PCh. 18 - The Bay of Fundy, Nova Scotia, has the highest...Ch. 18 - An earthquake can produce a seiche in a lake in...Ch. 18 - High-frequency sound can be used to produce...Ch. 18 - Prob. 18.37PCh. 18 - Prob. 18.38PCh. 18 - Calculate the length of a pipe that has a...Ch. 18 - The overall length of a piccolo is 32.0 cm. The...Ch. 18 - The fundamental frequency of an open organ pipe...Ch. 18 - Prob. 18.42PCh. 18 - An air column in a glass tube is open at one end...Ch. 18 - Prob. 18.44PCh. 18 - Prob. 18.45PCh. 18 - A shower stall has dimensions 86.0 cm 86.0 cm ...Ch. 18 - Prob. 18.47PCh. 18 - Prob. 18.48PCh. 18 - As shown in Figure P17.27, water is pumped into a...Ch. 18 - As shown in Figure P17.27, water is pumped into a...Ch. 18 - Two adjacent natural frequencies of an organ pipe...Ch. 18 - Why is the following situation impossible? A...Ch. 18 - A student uses an audio oscillator of adjustable...Ch. 18 - An aluminum rod is clamped one-fourth of the way...Ch. 18 - Prob. 18.55PCh. 18 - Prob. 18.56PCh. 18 - In certain ranges of a piano keyboard, more than...Ch. 18 - Prob. 18.58PCh. 18 - Review. A student holds a tuning fork oscillating...Ch. 18 - An A-major chord consists of the notes called A,...Ch. 18 - Suppose a flutist plays a 523-Hz C note with first...Ch. 18 - A pipe open at both ends has a fundamental...Ch. 18 - Prob. 18.63APCh. 18 - Two strings are vibrating at the same frequency of...Ch. 18 - Prob. 18.65APCh. 18 - A 2.00-m-long wire having a mass of 0.100 kg is...Ch. 18 - The fret closest to the bridge on a guitar is 21.4...Ch. 18 - Prob. 18.68APCh. 18 - A quartz watch contains a crystal oscillator in...Ch. 18 - Review. For the arrangement shown in Figure...Ch. 18 - Prob. 18.71APCh. 18 - Two speakers are driven by the same oscillator of...Ch. 18 - Review. Consider the apparatus shown in Figure...Ch. 18 - Review. The top end of a yo-yo string is held...Ch. 18 - On a marimba (Fig. P18.75), the wooden bar that...Ch. 18 - A nylon siring has mass 5.50 g and length L = 86.0...Ch. 18 - Two train whistles have identical frequencies of...Ch. 18 - Review. A loudspeaker at the front of a room and...Ch. 18 - Prob. 18.79APCh. 18 - Prob. 18.80APCh. 18 - Prob. 18.81APCh. 18 - A standing wave is set up in a string of variable...Ch. 18 - Two waves are described by the wave functions...Ch. 18 - Prob. 18.84APCh. 18 - Review. A 12.0-kg object hangs in equilibrium from...Ch. 18 - Review. An object of mass m hangs in equilibrium...Ch. 18 - Review. Consider the apparatus shown in Figure...Ch. 18 - Prob. 18.88CP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
University Physics Volume 1
Physics
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:OpenStax - Rice University
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
An Introduction to Physical Science
Physics
ISBN:9781305079137
Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres
Publisher:Cengage Learning
What Are Sound Wave Properties? | Physics in Motion; Author: GPB Education;https://www.youtube.com/watch?v=GW6_U553sK8;License: Standard YouTube License, CC-BY