/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A harmonic oscillator has angula... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A .\) (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

Short Answer

Expert verified
(a) Magnitude: displacement \( A/\sqrt{2} \), velocity \( A\omega/\sqrt{2} \). (b) Occurs twice per cycle, \( T/2 \) between. (c) 25% potential, 75% kinetic.

Step by step solution

01

Understanding the System's Total Energy

The total energy of a harmonic oscillator is given by the formula for mechanical energy, which combines both kinetic energy and potential energy: \[ E = rac{1}{2}kA^2 = rac{1}{2}mv^2 + U \] where \( U = rac{1}{2}kx^2 \) is the potential energy. This expression sums up the energy stored as potential energy (when the spring is stretched or compressed) and kinetic energy (when the particle is in motion). The potential energy \( U \) is zero at equilibrium.
02

Equating Kinetic and Potential Energies

For the energies to be equal, we set the kinetic energy equal to the potential energy: \[ rac{1}{2}mv^2 = rac{1}{2}kx^2 \] which simplifies by canceling common terms to: \[ mv^2 = kx^2 \] Given that \( k = m\omega^2 \), substitute it into the equation to find: \[ m^2v^2 = m\omega^2x^2 \] leading to: \[ v^2 = \omega^2x^2 \] Thus, \( v = \omega x \) shows the velocity required when kinetic and potential energies are equal.
03

Magnitude of Displacement and Velocity at Energy Equivalence

Using the derived equation where displacement \( x = A/\sqrt{2} \) when both kinetic and potential energies are equal: \[ rac{1}{2}kx^2 = rac{1}{2}k(A/\sqrt{2})^2 = rac{1}{4}kA^2 \].Therefore, the velocities are equal, so \( v = v_{max}/\sqrt{2} \) or \( v = A\omega/\sqrt{2} \).
04

Frequency of Energy Equivalence

As the harmonic oscillator oscillates, it passes through the energy equilibrium condition twice each cycle: once as it moves away from equilibrium and once towards it. Therefore, this equilibrium condition occurs twice per cycle, and given one complete cycle of period \( T = \frac{2\pi}{\omega} \), each occurrence is separated by half the period, or \( \frac{T}{2} \).
05

Energy Distribution at Specific Displacement

At a displacement \( x = A/2 \), the potential energy \( U \) is: \[ U = rac{1}{2}k(A/2)^2 = rac{1}{8}kA^2 = rac{1}{4}E \]where \( E \) is the total energy.Thus, the remaining \( rac{3}{4}E \) is kinetic energy.The fractions are 25% potential energy and 75% kinetic energy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency
Angular frequency is a key concept when discussing harmonic oscillators. It is usually denoted by the Greek letter \( \omega \).
In a harmonic oscillator like a simple pendulum or a mass-spring system, angular frequency tells us how quickly an oscillation occurs.
This frequency is given in radians per second. How do you compute it? The formula is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the object attached to the spring.
High angular frequency means the oscillator moves more cycles per second, producing faster oscillations. A low angular frequency results in slower oscillations.
  • Angular frequency affects the oscillator's period: the time it takes to complete one full cycle.

  • Because \( \omega \) is linked to the stiffness \( k \), a stiffer spring increases the angular frequency.
In our scenario, when kinetic and potential energies are equal, the role of angular frequency becomes evident as it governs how quickly these energies interchange.
Understanding angular frequency helps explain not only the speed but the rhythm of oscillations in harmonic motion.
Therefore, it is crucial to grasp this term when tackling problems related to harmonic oscillators.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion.
In a harmonic oscillator, like a mass attached to a spring, kinetic energy is vital for understanding how the system moves. At any given time, the kinetic energy \( KE \) can be expressed as: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity.
During oscillation, the velocity, and thus the kinetic energy, continually changes. Here are some key points:
  • The kinetic energy is maximum when the object passes through the equilibrium position, because that's when its speed is greatest.
  • It drops to zero at the extreme positions where the object momentarily stops before reversing direction.
When the potential and kinetic energies are equal, it signifies a unique point in the oscillation cycle where neither predominates.
This condition provides insights into the forces acting and the displacement at which this balance occurs.
Understanding the gatekeeper role of kinetic energy, one can better analyze how energy is conserved and exchanged during harmonic motion.
Potential Energy
Potential energy is the energy stored within a system due to its position or configuration.
In the context of a harmonic oscillator, this is usually the elastic potential energy stored in the spring as it is compressed or stretched.
We calculate the potential energy \( U \) as: \( U = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from equilibrium. Points to ponder include:
  • Potential energy is highest at the points of maximum displacement in simple harmonic motion.
  • When an object is in motion through equilibrium, the potential energy momentarily reaches zero as the spring is neither compressed nor stretched.
In harmonic oscillators, potential energy and kinetic energy constantly convert into each other in a beautiful dance.
At certain displacements, such as half the amplitude, these energies are crucial for determining how energy is partitioned between motion and position.
Understanding where the potential energy peaks and ebbs helps unravel the dynamics of harmonic oscillators.
Mechanical Energy
Mechanical energy in any harmonic oscillator is the sum of its kinetic and potential energies. It's a measure of the total energy of the system, representing the ability to do work.
For a harmonic oscillator, total mechanical energy \( E \) is given by: \( E = \frac{1}{2}kA^2 \). Notably:
  • This energy remains constant over time, assuming no external forces (like friction) are at play.
  • The interplay of kinetic and potential energy ensures conservation of total mechanical energy.
In every cycle, these energies shift between potential and kinetic forms, reflecting a transforming yet consistent sum.
When the displacement is at half the amplitude, the distribution between these energies reveals fractions of energy stored and released, advancing our understanding of energy dynamics in harmonic systems.
Recognizing this constancy in total mechanical energy reaffirms the principle of energy conservation and reinforces the predictability of harmonic oscillation patterns.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 1.80 -kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.53. The center of gravity of the rod was located by balancing and is 0.200 \(\mathrm{m}\) from the pivot. When the rod is set into small-amplitude oscillation, it makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.

A glider is oscillating in \(\mathrm{SHM}\) on an air track with an amplitude \(A_{1 .}\) You slow it so that its amplitude is halved. What happens to its (a) period, frequency, and angular frequency; (b) total mechanical energy; (c) maximum speed; (d) speed at \(x=\pm A_{1} / 4 ;\) (e) potential and kinetic energies at \(x=\pm A_{1} / 4 ?\)

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is \(0.090 \mathrm{m},\) it takes the block 2.70 s to travel from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} .\) If the amplitude is doubled, to \(0.180 \mathrm{m},\) how long does it take the block to travel (a) from \(x=0.180 \mathrm{m}\) to \(x=-0.180 \mathrm{m}\) and (b) from \(x=0.090 \mathrm{m}\) to \(x=-0.090 \mathrm{m} ?\)

CP A child with poor table manners is sliding his \(250-\mathrm{g}\) dinner plate back and forth in \(\mathrm{SHM}\) with an amplitude of 0.100 \(\mathrm{m}\) on a horizontal surface. At a point 0.060 \(\mathrm{m}\) away from equilibrium, the speed of the plate is 0.400 \(\mathrm{m} / \mathrm{s}\) . (a) What is the period? (b) What is the displacement when the speed is 0.160 \(\mathrm{m} / \mathrm{s} ?\) (c) In the center of the dinner plate is a 10.0 -g carrot slice. If the carrot slice is just on the verge of slipping at the endpoint of the path, what is the coefficient of static friction between the carrot slice and the plate?

CALC A slender, uni- form, metal rod with mass \(M\) is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring with force constant \(k\) is attached to the lower end of the rod, with the other end of the spring attached to a rigid support. If the rod is displaced by a small angle \(\theta\) from the vertical (Fig. \(P 14.97\) ) and released, show that it moves in angular SHM and calculate the period. (Hint: Assume that the angle \(\theta\) is small enough for the approximations \(\sin \theta \approx \theta\) and \(\cos \theta \approx 1\) to be valid. The motion is simple harmonic if \(d^{2} \theta / d t^{2}=-\omega^{2} \theta,\) and the period is then \(T=2 \pi / \omega .\) )

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.