Concept explainers
The total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J.
- a. What is the kinetic energy of the system when the position of the oscillator is 0.750 cm?
- b. What is the potential energy of the system at this position?
- c. What is the position for which the potential energy of the system is equal to its kinetic energy?
- 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.
FIGURE P16.73
(a)
The kinetic energy of the system.
Answer to Problem 74PQ
The kinetic energy of the system is
Explanation of Solution
Write an expression for the total energy of the system.
Here,
Rewrite the equation (I) to find
Write an expression for the potential energy of the system.
Here,
Write an expression for the kinetic energy of the system.
Here,
Substitute equation (I) and (III) in equation (IV).
Conclusion:
Substitute
Substitute
Thus, the kinetic energy of the system is
(b)
The potential energy of the system.
Answer to Problem 74PQ
The potential energy of the system is
Explanation of Solution
Write an expression for the potential energy of the system.
Conclusion:
Substitute
Thus, the potential energy of the system is
(c)
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
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
Substitute equation (I) and (III) in equation (VI).
Rewrite the equation (VII) to find
Conclusion:
Substitute
Thus, the position at which the potential energy of the system is equal to the kinetic energy is
(d)
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?
Chapter 16 Solutions
Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2
- A spring of mass ms and spring constant k is attached to an object of mass M and set into simple harmonic motion on a frictionless, horizontal table. All portions of the spring are assumed to oscillate in phase, and the velocity of each segment dx of the spring with mass dm can be assumed to be proportional to the distance x of that segment from point A in Figure P16.25. a. What is the kinetic energy of the system at the instant the object is moving with speed v? b. What is the frequency of oscillation of the system? FIGURE P16.25arrow_forwardAn object-spring system moving with simple harmonic motion has an amplitude A. (a) What is the total energy of the system in terms of k and A only? (b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this situation, using only the variables for the mass m, velocity v, spring constant k, and position x. (c) Using the results of parts (a) and (b) and the conservation of energy equation, find the positions x of the object when its kinetic energy equals twice the potential energy stored in the spring. (The answer should in terms of A only.)arrow_forwardThe equations listed in Table 2.2 give position as a function of time, velocity as a function of time, and velocity as a function of position for an object moving in a straight line with constant acceleration. The quantity vxi appears in every equation. (a) Do any of these equations apply to an object moving in a straight line with simple harmonic motion? (b) Using a similar format, make a table of equations describing simple harmonic motion. Include equations giving acceleration as a function of time and acceleration as a function of position. State the equations in such a form that they apply equally to a blockspring system, to a pendulum, and to other vibrating systems. (c) What quantity appears in every equation?arrow_forward
- (a) If frequency is not constant for some oscillation, can the oscillation be simple harmonic motion? (b) Can you mink of any examples of harmonic motion where the frequency may depend on the amplitude?arrow_forward(a) If frequency is not constant for some oscillation, can the oscillation be SHM? (b) Can you think of any examples of harmonic motion where the frequency may depend on the amplitude?arrow_forward(a) A hanging spring stretches by 35.0 cm when an object of mass 450 g is hung on it at rest. In this situation, we define its position as x = 0. The object is pulled down an additional 18.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later? (b) Find the distance traveled by the vibrating object in part (a), (c) What If? Another hanging spring stretches by 35.5 cm when an object of mass 440 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 18.0 cm and released from rest to oscillate without friction. Find its position 84.4 s later, (d) Find the distance traveled by the object in part (c). (e) Why are the answers to parts (a) and (c) so different when the initial data in parts (a) and (c) are so similar and the answers to parts (b) and (d) are relatively close? Does this circumstance reveal a fundamental difficulty in calculating the future?arrow_forward
- A block of mass m rests on a frictionless, horizontal surface and is attached to two springs with spring constants k1 and k2 (Fig. P16.22). It is displaced to the right and released. Find an expression for the angular frequency of oscillation of the resulting simple harmonic motion. FIGURE P16.22 Problems 22 and 81.arrow_forwardA restaurant manager has decorated his retro diner by hanging (scratched) vinyl LP records from thin wires. The records have a mass of 180 g, a diameter of 12 in., and negligible thickness. The records oscillate as torsion pendulums. a. Records hung from a small hole near their rims have a period of roughly 3.5 s (Fig. P16.41A). What is the torsion spring constant of the wire? b. If a record is hung from its center hole using a wire of the same torsion spring constant (Fig. P16.41B), what is its period of oscillation? FIGURE P16.41arrow_forwardA spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency , the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (c) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s.arrow_forward
- A simple pendulum is 5.00 in long. (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s2? (b) What is its period if the elevator is accelerating downward at 5.00 m/s2? (c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s2?arrow_forwardA wooden block (m = 0.600 kg) is connected to a spring and undergoes simple harmonic motion with an amplitude of oscillation of 0.075 m. The frequency of the motion is 12.50 Hz. a. What is the spring constant? b. What is the maximum speed of the block? c. What is the speed of the block when it is 0.015 m away from the equilibrium position?arrow_forward
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning