Consider two identical oscillators, each with spring constant \(k\) and mass \(m\), in simple harmonic motion. One oscillator is started with initial conditions \(x_{0}\) and \(v_{j}\) the other starts with slightly different conditions, \(x_{0}+\delta x\) and \(v_{0}+\delta v_{1}\) a) Find the difference in the oscillators' positions, \(x_{1}(t)-x_{2}(t)\) for all t. b) This difference is bounded; that is, there exists a constant \(C\) independent of time, for which \(\left|x_{1}(t)-x_{2}(t)\right| \leq C\) holds for all \(t\). Find an expression for \(C\). What is the best bound, that is, the smallest value of \(C\) that works? (Note: An important characteristic of chaotic systems is exponential sensitivity to initial conditions; the difference in position of two such systems with slightly different initial conditions grows exponentially with time. You have just shown that an oscillator in simple harmonic motion is not a chaotic system.)

Short Answer

Expert verified
Answer: The best bound for the absolute difference in positions of the two oscillators is 2A, where A is the amplitude of the oscillation.

Step by step solution

01

Find the general position functions of the oscillators

In simple harmonic motion, the position function is given by: $$ x(t) = A \cos(\omega t + \phi) $$ where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase shift. The angular frequency is given by: $$ \omega = \sqrt{\frac{k}{m}} $$ For the first oscillator with initial conditions \(x_0\) and \(v_0\), we can find \(A\) and \(\phi\) as follows: $$ x_0 = A \cos(\phi) $$ and $$ v_0 = -A\omega \sin(\phi) $$ For the second oscillator with initial conditions \(x_0 + \delta x\) and \(v_0 + \delta v_1\), we will have a similar set of equations but with a slight difference: $$ x_0 + \delta x = A \cos(\phi + \delta \phi) $$ and $$ v_0 + \delta v_1 = -A\omega \sin(\phi + \delta \phi) $$ where \(\delta \phi\) accounts for the difference in phase shift between the two oscillators.
02

Find the difference in positions of the oscillators

Now that we have the general position functions for both oscillators, we can find the difference between them: $$ x_1(t) - x_2(t) = A[\cos(\omega t + \phi) - \cos(\omega t + \phi + \delta \phi)] $$ Applying the sum-to-product formula: $$ x_1(t) - x_2(t) = 2A \sin\left(\frac{\delta \phi}{2}\right) \sin\left(\omega t + \phi + \frac{\delta \phi}{2}\right) $$
03

Find the expression for the constant C and the best bound

We want to find a constant \(C\) such that the absolute difference in positions is bounded by it for all time, t: $$ \left|x_1(t)-x_2(t)\right| \leq C $$ From the difference in positions found in the previous step, we have: $$ \left|2A \sin\left(\frac{\delta \phi}{2}\right) \sin\left(\omega t + \phi + \frac{\delta \phi}{2}\right)\right| \leq C $$ Since the maximum value of the sine function is 1, we can say that the best bound is when the two sine functions are at their maximum, which corresponds to the amplitude, \(2A\): $$ C = 2A $$ Thus, the best bound for the absolute difference in positions of the two oscillators is \(2A\), where \(A\) is the amplitude of the oscillation. This result shows that the simple harmonic oscillator is not a chaotic system since the difference in their positions does not grow exponentially with time but is bounded by a constant value.

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 Vaia!

Key Concepts

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

Understanding the Oscillator in Physics
An oscillator in physics is a system that experiences repeated cycles or oscillatory motion, typically about an equilibrium position. This term encompasses a broad range of physical systems, from pendulums and springs to electronic circuits.

In the context of mechanics, a classical example of an oscillator is a mass attached to a spring, which moves back and forth when displaced from its equilibrium position. This type of motion is known as simple harmonic motion (SHM), characterized by its sinusoidal nature and constant frequency, regardless of amplitude.

An essential feature of an oscillator is its ability to store and transfer energy between different forms. For a mass-spring system, energy alternates between kinetic energy of the moving mass and potential energy stored in the spring's compression or extension. The interplay of these energy forms ensures that the motion continues until external forces, like friction, dampen the oscillations, leading to eventual cessation.Through understanding oscillators, students can grasp foundational concepts across various disciplines including mechanics, acoustics, and electrical engineering.
Demystifying Angular Frequency
The term angular frequency refers to the rate at which an object travels through its cycle in oscillatory motion, such as simple harmonic motion, and is denoted by the Greek letter omega (\f\(\fomega\f\)).In the equation \f\(x(t) = A \times \fcos(\fomega t + \fphi)\f\), the angular frequency \f\(\fomega\f\) is connected to the physical properties of the oscillator by the formula \f\(\fomega = \fsqrt{\frac{k}{m}}\f\), where \f\(k\f\) is the spring constant and \f\(m\f\) is the mass of the object.For students, it's useful to recognize that angular frequency is expressed in radians per second and is directly linked to the more familiar concept of frequency (\f\(f\f\)), which counts the number of cycles per second (Hertz). The two are related by the equation \f\(\fomega = 2\fpi f\f\), which bridges the gap between linear and angular measurements in oscillatory systems.By grasping angular frequency, students can predict how quickly an oscillator completes its cycle and delve deeper into understanding wave phenomena, resonance, and even quantum mechanics.
The Role of Phase Shift in Oscillatory Motion
The phase shift, denoted as \f\(\fphi\f\), is an essential concept in understanding oscillatory motion, particularly in determining the starting point or initial angle of the sine wave representing the motion.In the context of our problem with two oscillators, the phase shift affects the initial conditions of an object's oscillation, causing a 'shift' in the position of the waveform along the time axis. For example, when two oscillators with slightly different phase shifts (\f\(\fphi\f\) and \f\(\fphi + \f\fdelta\fphi\f\)) are compared, their motion is out of sync. This misalignment is particularly noticeable when drawing a comparison between their positions over time, as depicted by their sinusoidal waveforms.A clear understanding of phase shift is crucial for comprehending phenomena such as interference patterns in waves, the synchronization of multiple oscillators (like in clocks or musical instruments), and the modulation of signals in communications. For students tackling problems involving oscillatory motion, recognizing the impact of phase shift can be the key to solving complex problems involving wave superposition or resonance.

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 small mass, \(m=50.0 \mathrm{~g}\), is attached to the end of a massless rod that is hanging from the ceiling and is free to swing. The rod has length \(L=1.00 \mathrm{~m} .\) The rod is displaced \(10.0^{\circ}\) from the vertical and released at time \(t=0\). Neglect air resistance. What is the period of the rod's oscillation? Now suppose the entire system is immersed in a fluid with a small damping constant, \(b=0.0100 \mathrm{~kg} / \mathrm{s},\) and the rod is again released from an initial displacement angle of \(10.0^{\circ}\). What is the time for the amplitude of the oscillation to reduce to \(5.00^{\circ}\) ? Assume that the damping is small Also note that since the amplitude of the oscillation is small and all the mass of the pendulum is at the end of the rod, the motion of the mass can be treated as strictly linear, and you can use the substitution \(R \theta(t)=x(t),\) where \(R=1.0 \mathrm{~m}\) is the length of the pendulum rod.

A \(3.00-\mathrm{kg}\) mass attached to a spring with \(k=140 . \mathrm{N} / \mathrm{m}\) is oscillating in a vat of oil, which damps the oscillations. a) If the damping constant of the oil is \(b=10.0 \mathrm{~kg} / \mathrm{s}\), how long will it take the amplitude of the oscillations to decrease to \(1.00 \%\) of its original value? b) What should the damping constant be to reduce the amplitude of the oscillations by \(99.0 \%\) in 1.00 s?

With the right choice of parameters, a damped and driven physical pendulum can show chaotic motion, which is sensitively dependent on the initial conditions. Which statement about such a pendulum is true? a) Its long-term behavior can be predicted. b) Its long-term behavior is not predictable. c) Its long-term behavior is like that of a simple pendulum of equivalent length. d) Its long-term behavior is like that of a conical pendulum. e) None of the above is true.

The period of a pendulum is \(0.24 \mathrm{~s}\) on Earth. The period of the same pendulum is found to be 0.48 s on planet \(X,\) whose mass is equal to that of Earth. (a) Calculate the gravitational acceleration at the surface of planet \(X\). (b) Find the radius of planet \(\mathrm{X}\) in terms of that of Earth.

A block of wood of mass \(55.0 \mathrm{~g}\) floats in a swimming pool, oscillating up and down in simple harmonic motion with a frequency of \(3.00 \mathrm{~Hz}\). a) What is the value of the effective spring constant of the water? b) A partially filled water bottle of almost the same size and shape as the block of wood but with mass \(250 . g\) is placed on the water's surface. At what frequency will the bottle bob up and down?

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.

Sign-up for free