Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 10, Problem 40P

(a)

To determine

Whether the maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds or low-frequency sounds for a given vibration amplitude.

(a)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds for a given vibration amplitude.

Explanation of Solution

Write the expression for the maximum velocity.

vm=ωA (I)

Here, vm is the maximum velocity, ω is the angular frequency and A is the vibration amplitude

Write the expression for ω.

ω=2πf (II)

Here, f is the frequency

Put equation (II) in equation (I).

vm=2πfA (III)

Equation (III) implies that vmf.

Write the expression for the maximum acceleration.

am=ω2A (IV)

Here, am is the maximum acceleration

Put equation (II) in equation (IV).

am=(2πf)2A=4π2f2A (V)

Equation (V) implies that amf2.

Conclusion:

Since maximum velocity is proportional to frequency and the maximum acceleration is proportional to square of the frequency, maximum velocity and acceleration will occur for high frequency sounds.

Therefore, the maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds for a given vibration amplitude.

(b)

To determine

The maximum velocity and acceleration of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 Hz.

(b)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 Hz is 1.3×106 m/s and the maximum acceleration is 1.6×104 m/s2.

Explanation of Solution

Equation (III) can be used to find the maximum velocity and equation (V) can be used to find the maximum acceleration.

Conclusion:

Substitute 20.0 Hz for f and 1.0×108 m for A in equation (III) to find vm.

vm=2π(20.0 Hz)(1.0×108 m)=1.3×106 m/s

Substitute 20.0 Hz for f and 1.0×108 m for A in equation (V) to find am.

am=4π2(20.0 Hz)2(1.0×108 m)=1.6×104 m/s2

Therefore, the maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 Hz is 1.3×106 m/s and the maximum acceleration is 1.6×104 m/s2.

(c)

To determine

The maximum velocity and acceleration of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 kHz.

(c)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 kHz is 0.0013 m/s and the maximum acceleration is 160 m/s2.

Explanation of Solution

Equation (III) can be used to find the maximum velocity and equation (V) can be used to find the maximum acceleration.

Conclusion:

Substitute 20.0 kHz for f and 1.0×108 m for A in equation (III) to find vm.

vm=2π(20.0 kHz)(1.0×108 m)=2π(20.0×103 Hz)(1.0×108 m)=0.0013 m/s

Substitute 20.0 kHz for f and 1.0×108 m for A in equation (V) to find am.

am=4π2(20.0 kHz)2(1.0×108 m)=4π2(20.0×103 Hz)2(1.0×108 m)=160 m/s2

Therefore, the maximum velocity of the eardrum for vibrations of amplitude 1.0×108 m at frequency of 20.0 kHz is 0.0013 m/s and the maximum acceleration is 160 m/s2.

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
Define operational amplifier
A bungee jumper plans to bungee jump from a bridge 64.0 m above the ground. He plans to use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 6.00 m above the water. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00 m length of the cord, his body weight stretches it by 1.55 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the bridge. (a) What length of cord should he use? Use subscripts 1 and 2 respectively to represent the 5.00 m test length and the actual jump length. Use Hooke's law F = KAL and the fact that the change in length AL for a given force is proportional the length L (AL = CL), to determine the force constant for the test case and for the jump case. Use conservation of mechanical energy to determine the length of the rope. m (b) What maximum acceleration will he…
9 V 300 Ω www 100 Ω 200 Ω www 400 Ω 500 Ω www 600 Ω ww 700 Ω Figure 1: Circuit symbols for a variety of useful circuit elements Problem 04.07 (17 points). Answer the following questions related to the figure below. A What is the equivalent resistance of the network of resistors in the circuit below? B If the battery has an EMF of 9V and is considered as an ideal batter (internal resistance is zero), how much current flows through it in this circuit? C If the 9V EMF battery has an internal resistance of 2 2, would this current be larger or smaller? By how much? D In the ideal battery case, calculate the current through and the voltage across each resistor in the circuit.

Chapter 10 Solutions

Physics

Ch. 10.6 - Practice Problem 10.7 Energy at Maximum...Ch. 10.7 - Prob. 10.7CPCh. 10.7 - Prob. 10.8PPCh. 10.8 - Practice Problem 10.9 Pendulum on the Moon A...Ch. 10.8 - Prob. 10.8CPCh. 10.8 - Prob. 10.10PPCh. 10 - Prob. 1CQCh. 10 - Prob. 2CQCh. 10 - Prob. 3CQCh. 10 - Prob. 4CQCh. 10 - Prob. 5CQCh. 10 - Prob. 6CQCh. 10 - Prob. 7CQCh. 10 - Prob. 8CQCh. 10 - Prob. 9CQCh. 10 - Prob. 10CQCh. 10 - Prob. 11CQCh. 10 - Prob. 12CQCh. 10 - Prob. 13CQCh. 10 - Prob. 14CQCh. 10 - Prob. 15CQCh. 10 - Prob. 16CQCh. 10 - Prob. 17CQCh. 10 - Prob. 18CQCh. 10 - Prob. 1MCQCh. 10 - Prob. 2MCQCh. 10 - Prob. 3MCQCh. 10 - Prob. 4MCQCh. 10 - Prob. 5MCQCh. 10 - Prob. 6MCQCh. 10 - Prob. 7MCQCh. 10 - Prob. 8MCQCh. 10 - Prob. 9MCQCh. 10 - Prob. 10MCQCh. 10 - Prob. 11MCQCh. 10 - Prob. 12MCQCh. 10 - Prob. 13MCQCh. 10 - Prob. 14MCQCh. 10 - Prob. 15MCQCh. 10 - Prob. 16MCQCh. 10 - Prob. 17MCQCh. 10 - Prob. 18MCQCh. 10 - Prob. 19MCQCh. 10 - Prob. 20MCQCh. 10 - 1. A steel beam is placed vertically in the...Ch. 10 - Prob. 2PCh. 10 - 3. A man with a mass of 70 kg stands on one foot....Ch. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Prob. 10PCh. 10 - Prob. 11PCh. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - Prob. 16PCh. 10 - Prob. 15PCh. 10 - 17. The leg bone (femur) breaks under a...Ch. 10 - Prob. 18PCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43PCh. 10 - Prob. 44PCh. 10 - Prob. 45PCh. 10 - Prob. 46PCh. 10 - Prob. 47PCh. 10 - Prob. 48PCh. 10 - Prob. 49PCh. 10 - 50. The diaphragm of a speaker has a mass of 50.0...Ch. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Prob. 53PCh. 10 - Prob. 54PCh. 10 - Prob. 55PCh. 10 - Prob. 57PCh. 10 - Prob. 59PCh. 10 - 58. An object of mass 306 g is attached to the...Ch. 10 - Prob. 58PCh. 10 - Prob. 60PCh. 10 - Prob. 61PCh. 10 - An object moves in SHM. Its position as a function...Ch. 10 - Prob. 63PCh. 10 - Prob. 64PCh. 10 - Prob. 65PCh. 10 - Prob. 66PCh. 10 - Prob. 67PCh. 10 - Prob. 68PCh. 10 - Prob. 69PCh. 10 - Prob. 70PCh. 10 - Prob. 71PCh. 10 - 72. A grandfather clock is constructed so that it...Ch. 10 - Prob. 73PCh. 10 - Prob. 74PCh. 10 - Prob. 75PCh. 10 - Prob. 76PCh. 10 - Prob. 77PCh. 10 - Prob. 78PCh. 10 - Prob. 79PCh. 10 - Prob. 80PCh. 10 - Prob. 81PCh. 10 - Prob. 82PCh. 10 - Prob. 83PCh. 10 - Prob. 84PCh. 10 - Prob. 85PCh. 10 - Prob. 86PCh. 10 - Prob. 87PCh. 10 - Prob. 89PCh. 10 - Prob. 88PCh. 10 - Prob. 90PCh. 10 - Prob. 91PCh. 10 - Prob. 92PCh. 10 - Prob. 93PCh. 10 - Prob. 94PCh. 10 - Prob. 95PCh. 10 - Prob. 96PCh. 10 - Prob. 97PCh. 10 - Prob. 98PCh. 10 - Prob. 99PCh. 10 - 100. When the tension is 402 N, what is the...Ch. 10 - Prob. 101PCh. 10 - Prob. 105PCh. 10 - Prob. 103PCh. 10 - Prob. 102PCh. 10 - Prob. 104PCh. 10 - Prob. 106PCh. 10 - Prob. 107PCh. 10 - Prob. 108PCh. 10 - 109. The motion of a simple pendulum is...Ch. 10 - Prob. 110PCh. 10 - Prob. 111PCh. 10 - Prob. 112PCh. 10 - Prob. 113P
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
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
SIMPLE HARMONIC MOTION (Physics Animation); Author: EarthPen;https://www.youtube.com/watch?v=XjkUcJkGd3Y;License: Standard YouTube License, CC-BY