Chapter 4 Oscillatory Motion

1 Chapter 4 Oscillatory The Important Simple Harmonic MotionIn this Chapter we consider systems which have a Motion whichrepeats itself in time, that is,it isperiodic. In particular we look at systems which have some coordinate(say,x) whichhas a sinusoidal dependence on time. A graph this kind of Motion is shown inFig. Suppose a particle has a periodic, sinusoidal Motion on thexaxis, and its motiontakes it betweenx= +Aandx= A. Then the general expression forx(t) isx(t) =Acos( t+ )( )Ais called theamplitudeof the Motion . For reasons which will become clearer later, iscalled theangular frequency. We say that a mass which has a Motion of the type givenin Eq.

2 Undergoessimple harmonic we see that when the timetincreases by an amount2 , the argument of thecosine increases by 2 and the value ofxwill be the same. So the motionrepeats itselfafter a time interval2 , which we denote asT, theperiodof the Motion . The number oftxFigure :Plot simple harmonic Motion . (tandxaxes are unspecified!)6970 Chapter 4. Oscillatory Motion oscillations per time is given byf=1T, called thefrequencyof the Motion :T=2 f=1T= 2 ( )Rearranging we have a formula for in terms offorT: = 2 f=2 T( )Though (angular frequency) andf(frequency) are closely related (with just a factor of2 between them, we need to be careful to distinguish them; to help in this, we normallyexpress in units ofradsandfin units ofcycles, or Hz (Hertz).)

3 However, the real dimensionsof both are1sin the SI (t) we get the velocity of the particle:v(t) =dxdt= Asin( t+ )( )and its acceleration:a(t) =dvdt= 2 Acos( t+ )( )We note that the maximum values ofvandaare:vmax= Aamax= 2A( )The maximum speed occurs in themiddleof the oscillation. (The slope greatestin size whenx= 0.) The magnitude of the acceleration is greatest at theendsof theoscillation (whenx= A).Comparing Eq. and Eq. we see thatd2xdt2= 2x( )which is the same asa(t) = 2x(t). Using and and some trig we can also arrive ata relation between the speed|v(t)|of the mass and its coordinatex(t):|v(t)|= A|sin( t+ )|= A 1 cos2( t+ )= A 1 (x(t)A)2.

4 ( )We could also arrive at this relation using energy conservation (as discussed below). Note,if we are givenxwe canonlygive the absolute value ofvsince there aretwopossibilities forvelocity at eachx(namely a pair). THE IMPORTANT STUFF71mkxFigure :Massmis attached to horizontal spring of force constantk; it slides on a frictionless surface! Mass Attached to a SpringSuppose a massmis attached to the end of a spring of force constantk(whose other end isfixed) and slides on a frictionless surface. This system is illustrated in Fig. Then if wemeasure the coordinatexof the mass from the place where it would be if the spring were atits equilibrium length, Newton s 2ndlaw givesFx= kx=max=md2xdt2,and then we haved2xdt2= kmx.

5 ( )Comparing Eqs. and we can identify 2withkmso that = km( )From the angular frequency we can find the periodTand frequencyfof the Motion :T=2 = 2 mkf=1T=12 km( )It should be noted that (and henceTandf) does not depend on the amplitudeAof the Motion of the mass. In reality, of course if the Motion of the mass is too large thenthen spring will not obey Hooke s Law so well, but as long as the oscillations are small the period is the same for all the lab, it s much easier to work with a mass bobbing up and down on averticalspring. One can (and should!) ask if we can still use the same formulae forTandf, or ifgravity (g) enters in somehow. In fact, the same formulae (Eq.)

6 Doapply in this be more clear about the vertical mass spring system, we show such a system inFig. In (a), the spring is oriented vertically and has someunstretchedlength. (We areignoring the mass of the spring.) When a massmis attached to the end, the system will be72 Chapter 4. Oscillatory MOTIONmm(a)(b)(c)xFigure :(a) Unstretched vertical spring of force constantk(assumed massless). (b) Mass attached tospring is at equilibrium when the spring has been extended bya distancemg/k. (c) Mass will undergo smalloscillations about thenewequilibrium equilibrium when the spring has been extended by some lengthy; balancing forces on themass, this extension is given by:ky=mg= y= the mass is disturbed from its equilibrium position, itwill undergo harmonic oscil-lations which can be described by some coordinatex, wherexis measured from thenewequilibrium position of the end of the spring.

7 Then the Motion is just like that of thehorizontal , we note that for more precise work with arealspring mass system onedoesneedto take into account the mass of the spring. If the spring has atotal massms, one can showthat Eq. should be modified to: = km+ms3( )That is, we replace the value of the massmbymplusone thirdthe spring s Energy and the Simple Harmonic OscillatorFor the mass spring system, the kinetic energy is given byK=12mv2=12m 2A2sin2( t+ )( )and the potential energy isU=12kx2=12kA2cos2( t+ ).( )Using 2=kmin we then find that the total energy isE=K+U=12kA2[sin2( t+ ) + cos2( t+ )] THE IMPORTANT STUFF73and the trig identity sin2 + cos2 = 1 givesE=12kA2( )showing that the energy of the simple harmonic oscillator (as typified by a mass on a spring)is constant and is equal to the potential energy of the springwhen it is maximally extended(at which time the mass is motionless).

8 It is useful to use the principle of energy conservation to derive some general relations for1 dimensional harmonic Motion . (We will not use the particular parameters for the mass spring system, just the quantities contained in Eq. , which describes the Motion of a massmalong thexaxis. From Eq. we have the kinetic energy as a function oftimeK=12mv2=12m 2A2sin2( t+ )Now the maximum value of the kinetic energy is12m 2A2, which occurs whenx= 0. Sincewe are free to fix the zero point of the potential energy, wecan agree thatU(x) = 0 atx= 0. Then the total energy of the system must be equal to the maximum ( 0 valueof the kinetic energy:E=12m 2A2 Then using these expressions, the potential energy of the system isU=E K=12m 2A2 12m 2A2sin2( t+ ) =12m 2A2(1 sin2( t+ ))=12m 2A2cos2( t+ )=12m 2x2Of course, for the mass spring systemUis given by12kx2, which gives the relationm 2=k,or = km, which we ve already found.))

9 If we use the relationvmax= Athen the potentialenergy can be written asU(x) =12m 2x2=12mv2maxA2x2( ) Relation to Uniform Circular MotionThere is acorrespondencebetween simple harmonic Motion and uniform circular Motion ,which is illustrated in Fig. (a) and (b). In (a) a mass point moves in a horizontal circularpath with uniform circular Motion at a radiusR(for example, it might be glued to the edgeof a spinning disk of radiusR). Its angular velocity is , so its location is given by thetime varying angle , where (t) = t+ .74 Chapter 4. Oscillatory MOTIONq(t)Rxy+R0-Rxx(a)(b)Figure :(a) Mass point moves in a horizontal circle of radiusR. Theangular velocityof its Motion is.

10 A guy with a big nose (seen from above) is observing the Motion of the mass at the level of the sees only thexcoordinate of the point s Motion . (b) Motion of the mass as seen by the guy with the bignose. Theprojectionof the Motion is the same as simple harmonic Motion withangular frequency (a)CMPivotd(b)qI, MFigure :(a) Simple pendulum. (b) Physical (b) we show the Motion of the mass as it would be seen by someone lookingtoward the +ydirection at the level of the disk. Such an observer sees onlythe changingxcoordinate of the mass s Motion . Sincex=Rcos , the observed coordinate isx(t) =Rcos( (t)) =Rcos( t+ ),the same as Eq. The Motion of the corresponding (projected) harmonic oscillator hasanangular frequencyof and an amplitude The PendulumWe start with thesimple pendulum, which has just a small massmhanging from a stringof lengthLwhose mass we can ignore.