EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 16, Problem 74PQ

The total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J.

  1. a. What is the kinetic energy of the system when the position of the oscillator is 0.750 cm?
  2. b. What is the potential energy of the system at this position?
  3. c. What is the position for which the potential energy of the system is equal to its kinetic energy?
  4. d. For a simple harmonic oscillator, what, if any, are the positions for which the kinetic energy of the system exceeds the maximum potential energy of the system? Explain your answer.

Chapter 16, Problem 74PQ, The total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J. a. What is the

FIGURE P16.73

(a)

Expert Solution
Check Mark
To determine

The kinetic energy of the system.

Answer to Problem 74PQ

The kinetic energy of the system is 0.469J .

Explanation of Solution

Write an expression for the total energy of the system.

  E=12kymax2                                                                                                               (I)

Here, E is the total energy of the system, k is the force constant and ymax is the maximum amplitude.

Rewrite the equation (I) to find k.

  k=2Eymax2                                                                                                                (II)

Write an expression for the potential energy of the system.

  U=12ky2                                                                                                             (III)

Here, U is the potential energy and y is the position.

Write an expression for the kinetic energy of the system.

  K=EU                                                                                                             (IV)

Here, K is the kinetic energy of the system.

Substitute equation (I) and (III) in equation (IV).

      K=12kymax212ky2                                                                                      (V)

Conclusion:

Substitute 0.500J for E and 3.00cm for ymax in equation (II) to find k.

    k=2(0.500J)((3.00cm)(1m102cm))2=1.00J(3.00×102m)2=1.11×103N/m

Substitute 1.11×103N/m for k, 3.00cm for ymax and 0.750cm for y in equation (V) to find K.

  K=12(1.11×103N/m)((3.00cm)(1m102cm))212(1.11×103N/m)((0.750cm)(1m102cm))2=12(1.11×103N/m)(3.00×102m)212(1.11×103N/m)(0.750×102m)2=0.469J

Thus, the kinetic energy of the system is 0.469J .

(b)

Expert Solution
Check Mark
To determine

The potential energy of the system.

Answer to Problem 74PQ

The potential energy of the system is 3.13×102J.

Explanation of Solution

Write an expression for the potential energy of the system.

  U=12ky2                                                                                                               (III)

Conclusion:

Substitute 1.11×103N/m for k and 0.750cm for y in equation (III) to find U.

  U=12(1.11×103N/m)((0.750cm)(1m102cm))2=12(1.11×103N/m)(0.750×102m)2=3.13×102J

Thus, the potential energy of the system is 3.13×102J.

(c)

Expert Solution
Check Mark
To determine

The position at which the potential energy of the system is equal to the kinetic energy.

Answer to Problem 74PQ

The position at which the potential energy of the system is equal to the kinetic energy is 2.12cm .

Explanation of Solution

The potential energy will be half of the total energy if the potential energy and kinetic energy are same.

Write the expression for the potential energy

  U=12E                                                                                                                (VI)

Substitute equation (I) and (III) in equation (VI).

    12ky2=12(12kymax2)                                                                                           (VII) 

Rewrite the equation (VII) to find y.

    y=ymax2                                                                                                              (VIII)

Conclusion:

Substitute 3.00cm for ymax to find y.

    y=3.00cm2=3.00cm1.41=2.12cm

Thus, the position at which the potential energy of the system is equal to the kinetic energy is 2.12cm .

(d)

Expert Solution
Check Mark
To determine

The possibility of presence of a position for a simple harmonic oscillator at which the kinetic energy of the system exceeds the total potential energy of the system.

Answer to Problem 74PQ

No position exists for a simple harmonic oscillator at which the kinetic energy of the system exceeds the total potential energy of the system.

Explanation of Solution

The total mechanical energy is conserved for the system. The maximum potential energy is equal to the total energy of the system. The total energy of the system is the sum of kinetic energy and potential energy.

Since the total energy conserved, the total energy will be a constant. The kinetic energy can also attain a maximum that equal to the total energy. Thus, the kinetic energy will never exceed the maximum potential energy.

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
Plz don't use chatgpt pls will upvote
No chatgpt pls will upvote
look at answer  show all work step by step

Chapter 16 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

Ch. 16 - Prob. 5PQCh. 16 - Prob. 6PQCh. 16 - The equation of motion of a simple harmonic...Ch. 16 - The expression x = 8.50 cos (2.40 t + /2)...Ch. 16 - A simple harmonic oscillator has amplitude A and...Ch. 16 - Prob. 10PQCh. 16 - A 1.50-kg mass is attached to a spring with spring...Ch. 16 - Prob. 12PQCh. 16 - Prob. 13PQCh. 16 - When the Earth passes a planet such as Mars, the...Ch. 16 - A point on the edge of a childs pinwheel is in...Ch. 16 - Prob. 16PQCh. 16 - Prob. 17PQCh. 16 - A jack-in-the-box undergoes simple harmonic motion...Ch. 16 - C, N A uniform plank of length L and mass M is...Ch. 16 - Prob. 20PQCh. 16 - A block of mass m = 5.94 kg is attached to a...Ch. 16 - A block of mass m rests on a frictionless,...Ch. 16 - It is important for astronauts in space to monitor...Ch. 16 - Prob. 24PQCh. 16 - A spring of mass ms and spring constant k is...Ch. 16 - In an undergraduate physics lab, a simple pendulum...Ch. 16 - A simple pendulum of length L hangs from the...Ch. 16 - We do not need the analogy in Equation 16.30 to...Ch. 16 - Prob. 29PQCh. 16 - Prob. 30PQCh. 16 - Prob. 31PQCh. 16 - Prob. 32PQCh. 16 - Prob. 33PQCh. 16 - Show that angular frequency of a physical pendulum...Ch. 16 - A uniform annular ring of mass m and inner and...Ch. 16 - A child works on a project in art class and uses...Ch. 16 - Prob. 37PQCh. 16 - Prob. 38PQCh. 16 - In the short story The Pit and the Pendulum by...Ch. 16 - Prob. 40PQCh. 16 - A restaurant manager has decorated his retro diner...Ch. 16 - Prob. 42PQCh. 16 - A wooden block (m = 0.600 kg) is connected to a...Ch. 16 - Prob. 44PQCh. 16 - Prob. 45PQCh. 16 - Prob. 46PQCh. 16 - Prob. 47PQCh. 16 - Prob. 48PQCh. 16 - A car of mass 2.00 103 kg is lowered by 1.50 cm...Ch. 16 - Prob. 50PQCh. 16 - Prob. 51PQCh. 16 - Prob. 52PQCh. 16 - Prob. 53PQCh. 16 - Prob. 54PQCh. 16 - Prob. 55PQCh. 16 - Prob. 56PQCh. 16 - Prob. 57PQCh. 16 - An ideal simple harmonic oscillator comprises a...Ch. 16 - Table P16.59 gives the position of a block...Ch. 16 - Use the position data for the block given in Table...Ch. 16 - Consider the position data for the block given in...Ch. 16 - Prob. 62PQCh. 16 - Prob. 63PQCh. 16 - Use the data in Table P16.59 for a block of mass m...Ch. 16 - Consider the data for a block of mass m = 0.250 kg...Ch. 16 - A mass on a spring undergoing simple harmonic...Ch. 16 - A particle initially located at the origin...Ch. 16 - Consider the system shown in Figure P16.68 as...Ch. 16 - Prob. 69PQCh. 16 - Prob. 70PQCh. 16 - Prob. 71PQCh. 16 - Prob. 72PQCh. 16 - Determine the period of oscillation of a simple...Ch. 16 - The total energy of a simple harmonic oscillator...Ch. 16 - A spherical bob of mass m and radius R is...Ch. 16 - Prob. 76PQCh. 16 - A lightweight spring with spring constant k = 225...Ch. 16 - Determine the angular frequency of oscillation of...Ch. 16 - Prob. 79PQCh. 16 - A Two springs, with spring constants k1 and k2,...Ch. 16 - Prob. 81PQCh. 16 - Prob. 82PQ
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
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
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
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
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
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
SIMPLE HARMONIC MOTION (Physics Animation); Author: EarthPen;https://www.youtube.com/watch?v=XjkUcJkGd3Y;License: Standard YouTube License, CC-BY