Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 103P

(a)

To determine

To calculate:The wave speed.

(a)

Expert Solution
Check Mark

Answer to Problem 103P

The wave speed (v) is 10.0m/s .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

The wave speed can be calculated as:

  v=FTμ

Where,

  v= Velocity of the wave.

  FT= tension in the wave.

  μ= Mass per unit length of the rope.

Calculation:

Let, the Δm represents the mass of the segment of the length Δx=1.00mm .

From the given data, wave speed can be found out as:

  v=FTμ

Where,

  v= Velocity of the wave.

  FT= tension in the wave.

  μ= Mass per unit length of the rope.

Now, substituting the all values of the linear density and tension in the equation:

  v= F T μv= 10.0N 0.100kg/mv=10.0m/s

Conclusion:

Thus, the wave speed (v) is 10.0m/s .

(b)

To determine

To calculate: The wavelength.

(b)

Expert Solution
Check Mark

Answer to Problem 103P

The wavelength (λ) is 2.00m .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

The wave speed can be calculated as:

  v= f λ

Where,

Sound’s speed: v .

Frequency of wave: .

The wavelength: λ .

Calculation:

To evaluate the wavelength, expression used is:

  v= f λ

Where,

Sound’s speed: v .

Frequency of wave: .

The wavelength: λ .

After substituting the values of the

Now, substituting the all values of the linear density and tension in the equation:

  λ=vλ=10.0m/s5.00s -1λ=2.00m

Conclusion:

Thus, the wavelength (λ) is 2.00m .

(c)

To determine

To calculate:The maximum transverse linear momentum.

(c)

Expert Solution
Check Mark

Answer to Problem 103P

The maximum transverse linear momentum (pmax) is 1.26×10-4kg.m/s .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

To calculate the maximum transverse linear momentum the expression used is:

  pmax=Δmvmax

Calculation:

Initially, relate the max transverse linear momentum of the 1.00mm segment to the max transverse speed of the wave.

  pmax=Δmvmaxpmax=μΔxAωpmax=2πfμΔxA

Substitute the numerical values to evaluate pmax:

  pmax=2πfμΔxApmax=2π(5.00s -1)(0.100kg/m)×(1 .00×10 -3m)(0.0400m)pmax=1.257×10-4kg.m/spmax=1.26×10-4kg.m/s

Conclusion:

Thus, the maximum transverse linear momentum (pmax) is 1.26×10-4kg.m/s .

(d)

To determine

To calculate: The maximum net force.

(d)

Expert Solution
Check Mark

Answer to Problem 103P

The maximum net force (Fmax) is 3.95mN .

Explanation of Solution

Given:

Mass per unit length of the rope = 0.100kg/m .

Tension = 10.0N .

Frequency = 5.00 cycles per second.

Amplitude = 40.0mm .

Formula used:

To calculate the maximum net forcethe expression used is:

  Fmax=Δmamax

Calculation:

The max net force acting on the segment is the product of the mass of th segment and its max acceleration:

  Fmax=ΔmamaxFmax=μΔxAω2Fmax=ωpmaxFmax=2πfpmax

Substitute the numerical values to evaluate Fmax :

  Fmax=2πfpmaxFmax=2π(5.00s -1)(1 .257×10 -4kg.m/s)Fmax=3.95mN

Conclusion:

Thus, the maximum net force (Fmax) is 3.95mN .

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
Question 3 of 17 L X L L T 0.5/ In the figure above, three uniform thin rods, each of length L, form an inverted U. The vertical rods each have a mass m; the horizontal rod has a mass 3m. NOTE: Express your answer in terms of the variables given. (a) What is the x coordinate of the system's center of mass? xcom L 2 (b) What is the y coordinate of the system's center of mass? Ycom 45 L X Q Search MD bp N
Sketch the harmonic on graphing paper.
Exercise 1: (a) Using the explicit formulae derived in the lectures for the (2j+1) × (2j + 1) repre- sentation matrices Dm'm, (J/h), derive the 3 × 3 matrices corresponding to the case j = 1. (b) Verify that they satisfy the so(3) Lie algebra commutation relation: [D(Î₁/ħ), D(Î₂/h)]m'm₁ = iƊm'm² (Ĵ3/h). (c) Prove the identity 3 Dm'm,(β) = Σ (D(Ρ)D(Ρ))m'¡m; · i=1

Chapter 15 Solutions

Physics for Scientists and Engineers, Vol. 1

Ch. 15 - Prob. 11PCh. 15 - Prob. 12PCh. 15 - Prob. 13PCh. 15 - Prob. 14PCh. 15 - Prob. 15PCh. 15 - Prob. 16PCh. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - Prob. 19PCh. 15 - Prob. 20PCh. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 23PCh. 15 - Prob. 24PCh. 15 - Prob. 25PCh. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - Prob. 28PCh. 15 - Prob. 29PCh. 15 - Prob. 30PCh. 15 - Prob. 31PCh. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - Prob. 34PCh. 15 - Prob. 35PCh. 15 - Prob. 36PCh. 15 - Prob. 37PCh. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Prob. 40PCh. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - Prob. 45PCh. 15 - Prob. 46PCh. 15 - Prob. 47PCh. 15 - Prob. 48PCh. 15 - Prob. 49PCh. 15 - Prob. 50PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Prob. 54PCh. 15 - Prob. 55PCh. 15 - Prob. 56PCh. 15 - Prob. 57PCh. 15 - Prob. 58PCh. 15 - Prob. 59PCh. 15 - Prob. 60PCh. 15 - Prob. 61PCh. 15 - Prob. 62PCh. 15 - Prob. 63PCh. 15 - Prob. 64PCh. 15 - Prob. 65PCh. 15 - Prob. 66PCh. 15 - Prob. 67PCh. 15 - Prob. 68PCh. 15 - Prob. 69PCh. 15 - Prob. 70PCh. 15 - Prob. 71PCh. 15 - Prob. 72PCh. 15 - Prob. 73PCh. 15 - Prob. 74PCh. 15 - Prob. 75PCh. 15 - Prob. 76PCh. 15 - Prob. 77PCh. 15 - Prob. 78PCh. 15 - Prob. 79PCh. 15 - Prob. 80PCh. 15 - Prob. 81PCh. 15 - Prob. 82PCh. 15 - Prob. 83PCh. 15 - Prob. 84PCh. 15 - Prob. 85PCh. 15 - Prob. 86PCh. 15 - Prob. 87PCh. 15 - Prob. 88PCh. 15 - Prob. 89PCh. 15 - Prob. 90PCh. 15 - Prob. 91PCh. 15 - Prob. 92PCh. 15 - Prob. 93PCh. 15 - Prob. 94PCh. 15 - Prob. 95PCh. 15 - Prob. 96PCh. 15 - Prob. 97PCh. 15 - Prob. 98PCh. 15 - Prob. 99PCh. 15 - Prob. 100PCh. 15 - Prob. 101PCh. 15 - Prob. 102PCh. 15 - Prob. 103PCh. 15 - Prob. 104P
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
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: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
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
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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