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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
Displacement is \( \pm \frac{A}{\sqrt{2}} \) and velocity is \( \pm \omega \frac{A}{\sqrt{2}} \) when KE = U. This occurs twice per cycle, every \( \frac{\pi}{\omega} \). At \( x = \frac{A}{2} \), the potential energy is \( \frac{1}{4}E \) and kinetic energy is \( \frac{3}{4}E \).

Step by step solution

01

Understand the System Properties

The system we are analyzing is a harmonic oscillator. The total mechanical energy, consisting of kinetic energy (KE) and potential energy (U), is constant throughout its motion.- Total Energy: \[ E = KE + U \] - For a harmonic oscillator, the total energy is given by: \[ E = \frac{1}{2}kA^2 \] where \( k \) is the spring constant and \( A \) is the amplitude of oscillation.
02

Equation for Displacement and Velocity

The expressions for displacement and velocity as functions of time \( t \) for a harmonic oscillator are:- Displacement: \[ x(t) = A \cos(\omega t) \]- Velocity: \[ v(t) = -A \omega \sin(\omega t) \]Both equations are essential for finding the displacement and velocity at specific energy conditions.
03

Equate Kinetic and Potential Energy

Kinetic Energy \( KE \) and Potential Energy \( U \) are given by:- Kinetic Energy: \[ KE = \frac{1}{2}mv^2 \] - Potential Energy: \[ U = \frac{1}{2}kx^2 \]We need \( KE = U \). Let's set the equations equal and solve for \( x \) when this condition is fulfilled: \[ \frac{1}{2}kx^2 = \frac{1}{2}mv^2 \] Simplifying gives: \[ kx^2 = mv^2 \] Recognizing that \( m = \frac{k}{\omega^2} \), thus: \[ x^2 = \frac{k}{k/\omega^2} \cdot v^2 / \omega^2 \] \[ x^2 = v^2 / \omega^2 \] \[ x = v / \omega \]
04

Solve for Displacement When KE = U

From the previous step, we substitute periodic properties: \[ x^2 + (A \omega \sin(\omega t) / \omega)^2 = A^2 \cos^2(\omega t) \] \[ x^2 = A^2 \cos^2(\omega t) = \frac{A^2}{2} \] Where \( \cos^2(\frac{\pi}{4}) = \sin^2(\frac{\pi}{4}) = \frac{1}{2} \)Thus, displacement magnitude: \[ x = \pm \frac{A}{\sqrt{2}} \] Velocity magnitude is thus: \[ v = \pm \omega \cdot \frac{A}{\sqrt{2}} \]
05

Determine Frequency of Energy Equality

KE = U happens twice in each cycle, once as the oscillator moves toward maximum displacement and once as it returns.The oscillator must cover half a cycle (\( T / 2 \)), where \( T \) is the period: \[ T = \frac{2\pi}{\omega} \]Time between occurrences is: \[ \frac{T}{2} = \frac{\pi}{\omega} \]
06

Energy Fractions at Specific Displacement

At displacement \( x = \frac{A}{2} \):- Potential Energy: \[ U = \frac{1}{2}k \left( \frac{A}{2} \right)^2 = \frac{1}{8}kA^2 \] Check total energy and fractions: \[ E = \frac{1}{2}kA^2 \]- Fraction of Total Energy as Potential: \[ \text{Fraction } U = \frac{\frac{1}{8}kA^2}{\frac{1}{2}kA^2} = \frac{1}{4} \] - Fraction of Total Energy as Kinetic: \[ \text{Fraction } KE = 1 - \frac{1}{4} = \frac{3}{4} \]
07

Summarize the Outcomes

At points where kinetic and potential energies are equal, the displacement is \( \pm \frac{A}{\sqrt{2}} \) and velocity is \( \pm \omega \frac{A}{\sqrt{2}} \). This occurs twice per cycle, occurring every \( \frac{\pi}{\omega} \) interval. When displacement is \( \frac{A}{2} \), potential energy is \( \frac{1}{4} \) of total energy, with kinetic the remaining \( \frac{3}{4} \).

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Key Concepts

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

Displacement and Velocity Analysis
The harmonic oscillator, a key model in physics, describes periodic motion such as a spring or pendulum. Its displacement and velocity change predictably over time. Displacement, represented by \( x(t) = A \cos(\omega t) \), measures how far the system has moved from its equilibrium position at any time \( t \). Velocity is given by \( v(t) = -A \omega \sin(\omega t) \), which represents the speed and direction of the motion.

When analyzing situations where potential energy equals kinetic energy, both displacement and velocity are directly related. This occurs particularly when their contributions to the total energy, \( E = \frac{1}{2}kA^2 \), balance out. At such points, half the cycle, i.e., \( T/2 \), captures the condition perfectly.

Understanding these expressions aids in determining specific motion characteristics like the instances when motion energy distribution balances, which inform about the mechanical harmony inherent in oscillatory systems.
Energy Conservation in Oscillations
A pivotal concept in physics, energy conservation holds that the total energy of the harmonic oscillator remains constant. This system exchanges kinetic energy (KE) and potential energy (U) without losing total energy. Expressed as \( E = KE + U \), the constancy of total mechanical energy spans across its motion.

The system transitions between storing energy in the spring (elastic potential) and in movement (kinetic), balancing incessantly over each oscillation period. At specific points, this energy relationship can be adjusted mathematically to find particular balances like when \( KE = U \).

Recognizing when kinetic energy and potential energy are equal helps predict system behavior, emphasizing the beauty of undamped motion that cyclically returns energy between forms without external influence or total energy change.
Kinetic and Potential Energy Equivalence
In a harmonic oscillator, the points where the kinetic and potential energy are equal create unique motion insights. At these points, both energy forms share equally the total mechanical energy. This balance happens when the system is far enough from equilibrium to have significant potential but still maintains a velocity ensuring kinetic motion.

To achieve this, using the formula \( \frac{1}{2}mv^2 = \frac{1}{2}kx^2 \) equates the energies. When solved, these conditions yield the displacement \( x = \pm \frac{A}{\sqrt{2}} \) and the velocity \( v = \pm \omega \frac{A}{\sqrt{2}} \).

By exploring these conditions, we understand peculiar points in the oscillation cycle, notable twice every oscillation period, each half-cycle separated by \( \frac{\pi}{\omega} \), showcasing how energy symmetrically oscillates within the system.
Angular Frequency and Periodicity
Angular frequency \( \omega \) is critical in harmonic oscillators, defining how quickly the system completes its cycles. It relates directly to the periodicity of oscillations, dictating the time interval, or the period \( T = \frac{2\pi}{\omega} \), each complete cycle takes.

Within these cycles, specific times contrast significantly in terms of energy distribution. Energy situations like those described in potential-kinetic equivalence recur regularly, splitting the system cycle into distinctive phases.

With periods highly dependent on angular frequency, understanding \( \omega \) unveils rhythm and timing of oscillatory systems. It sets the stage for discriminating repetitive timescales like the occurrences of equal kinetic and potential energies, reinforcing phase transition anticipations and system regularity.

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Most popular questions from this chapter

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

A mass \(m\) is attached to one end of a massess spring with a force constant \(k\) and an unstretched length \(l_{0}\) . The other end of the spring is free to turn about a nail driven into a frictionless, horizontal surface (Fig. 13.44\() .\) The mass is made to revolve in a circle with an angular frequency of revolution \(\omega^{\prime}\) , (a) Calculate the length \(l\) of the spring as a function of \(\omega^{\prime} .\) (b) What happens to the result in part (a) when \(\omega^{\prime}\) approaches the natural frequency \(\omega=\sqrt{k} / m\) of the mass-spring system? (ff your result bothers you, remember that massless springs and frictionless surfaces don't exist as such, but are only approximate descriptions of real springs and surfaces. Also, Hooke's law is only an approximation of the way real springs behave; the greater the elongation of the spring, the greater the deviation from Hooke's law.)

A thrill-seeking cat with mass 4.00 \(\mathrm{kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates verticallyin SHM. The amplitude is \(0.050 \mathrm{m},\) and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; ( \((b)\) at its lowest point; (c) at its equilibrium position.

The two pendulums shown in Fig 13.34 each consist of a uniform solid ball of mass \(M\) supported by a massless string, but the ball for pendulum \(A\) is very tiny while the ball for pendulum \(B\) is much larger. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?

On a horizontal, frictionless table, an open-topped \(5.20-\mathrm{kg}\) box is attached to an ideal horizontal spring having force constant 375 \(\mathrm{N} / \mathrm{m}\) . Inside the box is a 3.44 \(\mathrm{kg}\) stone. The system is oscillaring with an amplitude of 7.50 \(\mathrm{cm} .\) When the box has reached its maximum speed, the stone is suddenly plucked vertically out of the box without touching the box. Find (a) the period and (b) the amplitude of the resulting motion of the box. (c) Without doing any calculations, is the new period greater or smaller than the original period? How do you know?

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