Chapter 11 Simple Harmonic Motion 11.1 Introduction: Periodic Motion There are two basic ways to measure time: by duration or periodic motion. Early clocks measured duration by calibrating the burning of incense or wax, or the flow of water or sand from a container. Our calendar consists of years determined by the motion of the sun; months determined by the motion of the moon; days by the rotation of the earth; hours by the motion of cyclic motion of gear trains; and seconds by the oscillations of springs or pendulums. In modern times a second is defined by a specific number of vibrations of radiation, corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom (see Section 1.3). Sundials calibrate the motion of the sun through the sky including seasonal corrections. A clock escapement is a device that can transform continuous movement into discrete movements of a gear train. The early escapements used oscillatory motion to stop and start the turning of a weight-driven rotating drum. Soon, complicated escapements were regulated by pendulums, the theory of which was first developed by the physicist Christian Huygens in the mid 17th century. The accuracy of clocks was increased and the size reduced by the discovery of the oscillatory properties of springs by Robert Hooke. By the middle of the 18th century, the technology of timekeeping advanced to the point that William Harrison developed timekeeping devices that were accurate to one second in a century. One of the most important examples of periodic motion is Simple Harmonic Motion, in which some physical quantity varies sinusoidally. Suppose a function of time has the form of sine wave function, () sin(2 / ) sin(2 )ytA tTA ftππ== (11.1.1) where is the amplitude (maximum value). The function 0A > ( )yt varies between and , since a sine function varies between AA− 1+ and 1−. A graph of function vs. time might look like that shown in Figure 13.1. Figure 13.1 Sinusoidal function of time 10/20/2006 13-1The sine function is periodic in time. This means that the value of the function at time t will be exactly the same at a later time ttT′=+ , where T is the period. That the sine function satisfies the periodic condition can be seen from 222( ) sin ( ) sin 2 sin ( )ytT A tT A t A t ytTTTππππ⎡⎤⎡⎤⎡⎤+= + = + = =⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦. (11.1.2) The frequency, f, is defined to be 1/fT≡. (11.1.3) The SI unit of frequency is inverse seconds, 1s−⎡⎤⎣⎦, or . The angular frequency of oscillation is defined to be hertz [Hz] 2/ 2Tfωππ≡= (11.1.4) and is measured in radians per second. One oscillation per second, 1 , corresponds to an angular frequency of Hz12radsπ−⋅. (Unfortunately, the same symbol ω is used for angular velocity in circular motion. For uniform circular motion the angular velocity is equal to the angular frequency but for non-uniform motion the angular velocity will not be constant but the angular frequency for simple harmonic motion is a constant by definition.) 11.2 Simple Harmonic Motion: Object-Spring System Our first example of a system that demonstrates simple harmonic motion is an object-spring system on a frictionless surface, shown in Figure 13.2 Figure 11.2 object spring system The object is attached to one end of a spring. The other end of the spring is attached to a wall at the right in the figure. Assume that the object undergoes one-dimensional motion. The spring has a spring constant . The spring is initially stretched a distance from the equilibrium position and is given an initial speed in the positive k00x >0vx-direction. 10/20/2006 13-2Choose the origin at the equilibrium position and choose the positive x-direction to the left in the figure. In the figure, corresponds to an extended spring, to a compressed spring. Define 0x > 0x <( )xt to be the position of the object with respect to the equilibrium position. The force acting on the spring is a linear restoring force, (Figure 11.3). xFk=− x Figure 11.3: Force law for object-spring system Newton’s Second law in the x-direction becomes 22dxkx mdt−= . (11.2.1) This equation of motion, Equation (11.2.1), is called the simple harmonic oscillator (SHO) equation. Since the spring force depends on the distance x, the acceleration is not constant. Equation (11.2.1)is a second order linear differential equation, in which the second derivative of the dependent variable is proportional to the dependent variable, 22dxxdt∝− . (11.2.2) In this case, the constant of proportionality is , /km 22dx kxdt m=− . (11.2.3) Equation (11.2.3) can be solved from energy considerations (see the Appendix to this chapter) or other advanced techniques but instead we shall first guess the solution and then verify that the guess satisfies the SHO differential equation.1 1 “Guess” is perhaps too weak a term to use. We are using an “ansatz,” a proposed solution based on our knowledge of the system and noticing that the sinusoidal function in Figure 11.1 has the property represented in Equation (11.2.2) 10/20/2006 13-3We are looking for a position function such that the second time derivative of the position (acceleration) of the object is proportional to the negative of the position of the object. Since the sine and cosine functions both satisfy this property, we make a preliminary guess that our position function is given by () cos((2 / )) cos( )xtA TtA tπω==, (11.2.4) where as in Equation (11.1.4) 2/Tωπ≡is the angular frequency (as of yet, undetermined). In Equation (11.2.4), the constant is not necessarily the amplitude of the motion; is the amplitude for the case , AA00x >00v=. We shall now find the condition the angular frequency ω must satisfy in order to insure that the function in (11.2.4) solves the simple harmonic oscillator equation (11.2.1). The first and second derivatives of the position function are given by 2222sin( )cos( ) .dxAtdtdxAtdtωωxωωω=−=− =− (11.2.5) Substitute the second derivative, the second expression in (11.2.5), and the position function, Equation (11.2.4), into the SHO Equation (11.2.1), giving 2cos( ) cos( )kAt Amtωω−=−ω. (11.2.6) Equation (11.2.6) is valid for all times provided that kmω= . (11.2.7) The period of oscillation is then 22mTkππω== . (11.2.8) One possible solution is then () cos() sin .kxt A
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