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11.3.1.1: Illustrations

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    34143
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    Illustration 1: Representations of Simple Harmonic Motion

    In 1610 Galileo discovered four moons of Jupiter. Each moon seemed to move back and forth in what we would call simple harmonic motion. What was Galileo really seeing? He was seeing the essentially uniform circular motion of each moon, but he was looking at the motion edge-on. We can use what Galileo was experiencing to hand-wave some properties of simple harmonic motion by using an analogy with uniform circular motion. Consider the above animation (position is given in meters and time is given in seconds). Restart.

    First, let us look at position as a function of time. The point on the circle marked by the red ball is always at the same radius, \(R\). If we look at the y position as a function of time we see that \(y = R \cos (\omega t)\), and the \(x\) position is \(x = R \sin (\omega t)\). How do we know this? We can decompose the radius vector into components.

    What about the velocity? Well, we know it is tangent to the path of the ball and, since the motion is uniform, the magnitude of the velocity is constant and equal to \(\omega R\). We can break up the velocity vector into components. We get that \(v_{y} = \omega R \sin (\omega t)\) and \(v_{x} = -\omega R \cos (\omega t)\) and that both are functions of time. Watch the animation to convince yourself that this decomposition is correct as a function of time. If we know a bit of calculus, we can take the derivative of the position with respect to time. We again get that \(v_{y} = -\omega R \sin (\omega t)\) and \(v_{x} = \omega R \cos (\omega t)\).

    We also know that the acceleration is a constant, \(v^{2}/R\), and points toward the center of the circle. We can again decompose this acceleration as \(a_{y}=-\omega^{2} R\cos (\omega t)\) and \(a_{x}=-\omega^{2}R\sin (\omega t)\). Again, if we know a bit of calculus, we can take the derivative of the velocity with respect to time. We again find that \(a_{y}=-\omega^{2}R\cos (\omega t)\) and \(a_{x}=-\omega^{2}R\sin (\omega t)\). Note that since this is simple harmonic motion there must be a relationship between the position and the force. Since force must be a linear restoring force and, since force is also mass times acceleration, we must have that \(ma = - k x\) or that \(a(t) = - (k/m)\) and \(a(t) = -\omega^{2} x(t)\), which is the case if we compare our functions for \(y(t)\) and \(x(t)\) to \(a_{y}(t)\) and \(a_{x}(t)\).

    For simple harmonic motion we change two things, \(R\to A\) where \(A\) is called the amplitude, and we only consider one direction, in this example the \(y\) direction. This yields: \(y = A \cos (\omega t)\), \(v = -\omega A \sin (\omega t)\), and \(a=-\omega^{2} A\cos (\omega t)\). Simple harmonic motion requires a linear restoring force, an equilibrium position, and a displacement from equilibrium.

    Illustration 2: The Simple Pendulum and Spring Motion

    When we think about simple harmonic motion we think about a mass on a spring. This is the prototype motion and is the easiest to deal with since \(k\), the spring constant, is the proportionality factor between \(F\) and \(-x\). However, there is another standard example of simple harmonic motion that is all around us, that of pendulum motion. Restart. A pendulum is nothing more than a heavy object (the pendulum bob) hanging from a very light string (if the string's mass is large enough, we have a compound pendulum and the string must be considered). Consider Animation 1. Here the length of the string is \(15\text{ m}\) and the mass of the pendulum bob is \(1\text{ kg}\) (position is given in meters, angle is given in radians, and time is given in seconds). When we analyze the forces acting on the pendulum bob (drag the pendulum bob from its equilibrium position and press "play"), we find that the force of gravity and the force of tension act. The simplest way to analyze these forces is to consider their effect in the radial direction and the direction tangent to the circular path of the pendulum. The part of the gravitational force opposite to the tension must cancel the force of the tension in the string when the pendulum is at rest. However, when the pendulum bob is moving, the tension must be greater to provide the centripetal force required. This leaves the component of the force of gravity perpendicular to the tension and tangent to the path of the pendulum. Show Animation 1 with pendulum bob path. When we do the calculation, we find that the net force on the pendulum bob goes like

    \[F_{\text{tan}}=-mg\sin (\theta)\nonumber\]

    which at first glance does not look at all like simple harmonic motion. But what happens when the angle \(\theta\) is small? Well, \(\sin (\theta)\approx\theta\) for small enough \(\theta\); therefore, \(F_{\text{tan small angles}} = - mg\theta\).

    Drag the pendulum bob to a large angle and see how the two tangential forces (any angle vs. small angle) deviate at large angles. The motion of the pendulum is shown according to the actual force, \(F_{\text{tan}}= - mg \sin(\theta )\), and not the small angle approximation, \(F_{\text{net}} = - mg\theta\), although both are shown on the graph. Therefore the period of the pendulum is the actual period. When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.

    Since we are using radians, \(x =\theta L\), and the tangential force for small angles can be written as \(F_{\text{tan small angles}} = - (mg /L) x\), where the proportionality factor between \(F\) and \(-x\) is now \(mg /L\). For small enough angles (when \(\sin(\theta )\approx\theta\)) we have simple harmonic motion.

    Now consider both the motion of a pendulum and the motion of a mass attached to a spring by looking at Animation 2. In this animation the pendulum is the same as Animation 1 (the net force on the bob is shown as a green arrow), the spring has a spring constant of \(1.30666\text{ N/m}\), and the mass of the red ball attached to the spring is \(2\text{ kg}\) (the net force on the red ball is represented by the blue arrow). It may seem strange that we have chosen such an oddly precise value for the spring constant. Drag the pendulum to about \(0.15\) radians and drag the mass on the spring to some initial amplitude (it does not matter what this value is, but for simplicity chose \(2.3\text{ m}\)) and play the animation. What do you notice about the graph? Do you see why the spring constant was carefully chosen? These values were chosen to tune the motion of the two systems to be the same:

    \[\omega_{\text{mass-spring}}=(k/m)^{0.5}=\omega_{\text{pendulum}}=(k_{\text{effective}}/m)^{0.5}=(g/L)^{0.5}\nonumber\]

    Now reset this animation and drag the pendulum bob to \(0.75\) radians and the mass on the spring to \(10.3\text{ m}\) and play the animation. What happens now? By looking at Animation 1 can you say why this is? Notice as time goes on, that the two motions now deviate from each other. Large-amplitude pendulum motion is no longer simple harmonic motion.

    Illustration authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.
    Script authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.

    Illustration 3: Energy and Simple Harmonic Motion

    In this Illustration we shall look at energy and simple harmonic motion of both a pendulum and a mass on a spring. We shall consider small-amplitude motion for the pendulum since this will yield simple harmonic motion (see Illustration 16.2 for details). In addition, like Illustration 16.2, we have chosen the mass of the pendulum bob to be \(1\text{ kg}\) and the length of the pendulum to be \(15\text{ m}\), while choosing the mass of the ball on the spring to be \(2\text{ kg}\) and the spring constant to be \(1.30666\text{ N/m}\) (position is given in meters, angle is given in radians, and time is given in seconds). Restart. These values tune the motion of the two systems to be the same:

    \[\omega_{\text{mass-spring}}=(k/m)^{0.5}=\omega_{\text{pendulum}}=(k_{\text{effective}}/m)^{0.5}=(g/L)^{0.5}\nonumber\]

    In the following animations we will show graphs of the kinetic and potential energy of the mass-spring system but will not show the kinetic and potential energy of the pendulum. However, the kinetic and potential energy of the pendulum will look the same with exactly half the kinetic and potential energy (and therefore half the total energy) of the mass-spring system. Why half? For the mass-spring system, kinetic energy is (\(1/2 mv^{2}\)) and the potential energy is (\(1/2 kx^{2}\)), and for the pendulum the kinetic energy of the bob is (\(1/2 mv^{2}\)) and the potential energy is (\(1/2 k_{\text{effective}}x^{2}\)). In this Illustration, since the mass on the spring is twice the mass of the pendulum bob, the mass-spring system will always have twice as much kinetic energy as the pendulum bob. Since the spring constant for the mass-spring system is twice the effective spring constant for the pendulum (\(k_{\text{effective}} = m_{\text{pendulum}}g/L = 0.6533\text{ N/m}\)), the mass-spring system will always have twice as much potential energy as the pendulum bob.

    When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.

    Consider Animation 1, which shows the graph of kinetic and potential energy vs. position. What can you say about the total energy of the system? It is a constant and about \(1.89\text{ J}\). The energy starts out all potential and at the equilibrium position the energy is all kinetic. At maximum compression the energy is all potential again. Given that the total energy is kinetic plus potential, we have that

    \[E=0.5mv^{2}+0.5kx^{2}=0.5kx_{\text{max}}^{2}=0.5mv_{\text{max}}^{2}\nonumber\]

    Now consider Animation 2, which shows the graph of kinetic and potential energy vs. time. Notice how the two graphs are different in their functional form.

    The graphs in Animation 1 have the form of \(0.5 k x^{2}\) (the potential energy) and the form of \(A - 0.5 k x^{2}\) (the kinetic energy), where \(A\) is a constant, the total energy. In this animation the total kinetic energy is \(1.89\text{ J}\). The form of the kinetic energy can be understood from the energy function shown above. The potential energy is \(0.5 k x^{2}\), which is proportional to \(x^{2}\). The kinetic energy can be written in terms of the total energy and the potential energy as \(E - 0.5 k x^{2}\).

    The graphs in Animation 2 have the form of \(\cos^{2}\) (the potential energy) and the form of \(\sin^{2}\) (the kinetic energy) since both trigonometric functions are a function of time. Why? We know from simple harmonic motion that if the object is initially displaced from equilibrium with no initial velocity that

    \[x=x_{0}\cos(\omega t)\quad\text{and}\quad v=-\omega x_{0}\sin(\omega t)\nonumber\]

    Given the form of the kinetic energy and the potential energy, we have that

    \[KE(t)=0.5kx_{0}^{2}\sin^{2}(\omega t)\quad\text{and}\quad PE(t)=0.5kx_{0}^{2}\cos^{2}(\omega t)\nonumber\]

    where we used \(\omega^{2} = k/m\) to simplify the kinetic energy. Therefore the total energy will always add up to \(0.5 k x_{0}^{2}= 1.89\text{ J}\).

    Illustration authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.
    Script authored by Morten Brydensholt, Wolfgang Christian, and Mario Belloni.

    Illustration 4: Forced and Damped Motion

    A \(1\text{-kg}\) mass on a spring is shown (position is given in meters and time is given in seconds) initially at its equilibrium position. Various parameters related to the spring and the initial conditions of the motion are also given. Restart. Once a variable or the velocity box is changed, you must reinitialize your choice by clicking the "set values, then drag the ball" button. Once you have clicked this button, drag the ball to the initial position you desire (by default it starts at its equilibrium position) and then press "play." When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.

    We have thus far considered ideal motion of a mass on a spring: a perfect spring obeying Hooke's law and no additional varying force or damping. This Illustration will discuss what happens to a mass on a spring subject to a varying additional force and/or a damping force. Specifically, the damping force is \(-b v\) and the driving force is \(F_{0}\cos(\omega t)\).

    First, what is the natural frequency of oscillation of the mass? Look at the animation with no additional forces or damping. Drag the ball to \(3\text{ m}\) and let go. Pause the animation and measure the period (about \(4.45\) seconds from peak to peak). The frequency is one over this, or \(0.225\text{ Hz}\). The angular frequency is \(2\pi f\) or \(1.41\text{ rad/sec}\). Since the angular frequency squared (here \(2\)) is equal to the ratio of \(k/m\), we know that \(k = 2\text{ N/m}\).

    What happens to the motion of the mass when a driving force is turned on? Try it and find out. Vary the angular frequency of the driving force. What happens when the angular frequency of the oscillation is close to or far from that of the driving force? How sensitive is the motion to this parameter? When the natural and driving frequencies are the same, it is called resonance.

    There are three types of damped motion you should also investigate:

    • Under Damped: the damping is so small there are many oscillations before motion is stopped.
    • Over Damped: the damping is rather large; the motion takes a long time to get back to equilibrium.
    • Critically Damped: a special case in which the time to get back to equilibrium is minimized.

    Illustration 5: The Fourier Series, Qualitative Features

    We have thus far only looked at simple periodic motion that can be described by a single sine or cosine. This may seem like a horrible mistake. Most periodic functions are tremendously complicated. Have we been doing something wrong by focusing on only sines and cosines? Well, actually not. ANY periodic function can be represented as a sum of sines or cosines!   We find that sometimes we may need an infinite number of them, but nonetheless we are able to describe any periodic phenomena, no matter how complicated, in this way. Restart.

    Consider a sawtooth (position is given in meters) function that is periodic with \(L = 1\) (it is shown over two periods since it is easier to see the function this way). In the animation the amplitude is a function of \(x\), but it could have been a function of time. Select "play the Fourier series of the sawtooth." The gray function is the actual sawtooth, while the red function is the total approximate sawtooth from a Fourier series (if you did not change n, the animation shows the \(n = 1\) term only). Change \(n\), the number of sine functions that will be added together to approximate the sawtooth, and see how the red function changes. The green sine function is the current term that is added to get the total red function. On the right is a representation of the relative amount of each sine function as it is added to the total. You may add as many as \(35\) terms. Also note that at the point where the sawtooth kinks, there is always overshoot (this is called the Gibbs phenomenon).

    Now look at the square wave. It turns out that the \(n = 2,\: 4,\: 6,\ldots\) terms do not contribute to the sum. Verify this by watching the animation for \(n = 35\).

    When you get a good-looking graph, right-click on it to clone the graph and resize it for a better view.

    Illustration 6: Fourier Series, Quantitative Features

    Fourier's theorem states that any periodic function (whether periodic in position or time) can be represented as a sum of sine waves. We find that sometimes we may need an infinite number of sines, but nonetheless we may describe any periodic phenomena this way. In this animation we investigate odd periodic functions in position with Fourier's theorem. Restart.

    ANY odd periodic function of \(x\) (a period of \(L\) between \(0\) and \(L\) as opposed to between \(-L/2\) to \(L/2\)) can be described in terms of a Fourier series as:

    \[f(x)=\sum A_{n}\sin (n\ast 2\ast n\ast x/L)\nonumber\]

    where in this animation \(L = 1\). \(A_{n}\) is the result of an integral that represents the overlap between the original function and a particular Fourier component (one term in the Fourier series represented by the integer \(n\)). In order to get this to exactly work out, there must be a \(2/L\) (in our case just a factor of \(2\) since \(L = 1\) here) included in the integral. Verify that this is necessary by predicting \(A_{3}\) for the function \(\sin(3\ast 2\ast \text{pi}\ast x)\) and then use the animation as a check.

    Remember to use the proper syntax, such as \(-10+0.5\ast t\), \(-10+0.5\ast t\ast t\), and  \(-10+0.5\ast t\wedge 2\). Revisit Exploration 1.3.3 to refresh your memory.

    Try various odd functions to see the result of the integral, \(A_{n}\). Consider the following functions (you may copy and paste them in directly):

    the Sawtooth Wave in Illustration 5 the Square Wave in Illustration 5
    \((1-2\ast x)\ast\text{step}(1-x)+(1-2\ast (x-1))\ast\text{step}(x-1)\) \(\text{step}(0.5-x)-\text{step}(x-0.5)\ast\text{step}(1-x)+\text{step}(x-1)\)

    Table \(\PageIndex{1}\)

    Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.


    This page titled 11.3.1.1: Illustrations is shared under a CC BY-NC-ND license and was authored, remixed, and/or curated by Wolfgang Christian, Mario Belloni, Anne Cox, Melissa H. Dancy, and Aaron Titus, & Thomas M. Colbert.