Oscillations 15-1 SIMPLE HARMONIC MOTION. Learning Objectives After reading this module, you should be able to...

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1 C H A P T E R 5 Oscillaions 5- SIMPLE HARMONIC MOTION Learning Objecives Afer reading his module, you should be able o Disinguish simple harmonic moion from oher ypes of periodic moion. 5.0 For a simple harmonic oscillaor, apply he relaionship beween posiion and ime o calculae eiher if given a value for he oher Relae period T, frequency f, and angular frequency v Idenify (displacemen) ampliude m, phase consan (or phase angle) f, and phase v f Sech a graph of he oscillaor s posiion versus ime, idenifying ampliude m and period T From a graph of posiion versus ime, velociy versus ime, or acceleraion versus ime, deermine he ampliude of he plo and he value of he phase consan f On a graph of posiion versus ime describe he effecs of changing period T, frequency f, ampliude m, or phase consan f Idenify he phase consan f ha corresponds o he saring ime ( 0) being se when a paricle in SHM is a an ereme poin or passing hrough he cener poin Given an oscillaor s posiion () as a funcion of ime, find is velociy v() as a funcion of ime, idenify he velociy ampliude v m in he resul, and calculae he velociy a any given ime. Key Ideas The frequency f of periodic, or oscillaory, moion is he number of oscillaions per second. In he SI sysem, i is measured in herz: Hz s. The period T is he ime required for one complee oscillaion, or cycle. I is relaed o he frequency by T /f. In simple harmonic moion (SHM), he displacemen () of a paricle from is equilibrium posiion is described by he equaion m cos(v f) (displacemen), in which m is he ampliude of he displacemen, v f is he phase of he moion, and f is he phase consan. The angular frequency v is relaed o he period and frequency of he moion by v p/t pf. Differeniaing () leads o equaions for he paricle s SHM velociy and acceleraion as funcions of ime: 5.0 Sech a graph of an oscillaor s velociy v versus ime, idenifying he velociy ampliude v m. 5. Apply he relaionship beween velociy ampliude v m, angular frequency v, and (displacemen) ampliude m. 5. Given an oscillaor s velociy v() as a funcion of ime, calculae is acceleraion a() as a funcion of ime, idenify he acceleraion ampliude a m in he resul, and calculae he acceleraion a any given ime. 5.3 Sech a graph of an oscillaor s acceleraion a versus ime, idenifying he acceleraion ampliude a m. 5.4 Idenify ha for a simple harmonic oscillaor he acceleraion a a any insan is always given by he produc of a negaive consan and he displacemen jus hen. 5.5 For any given insan in an oscillaion, apply he relaionship beween acceleraion a, angular frequency v, and displacemen. 5.6 Given daa abou he posiion and velociy v a one insan, deermine he phase v f and phase consan f. 5.7 For a spring bloc oscillaor, apply he relaionships beween spring consan and mass m and eiher period T or angular frequency v. 5.8 Apply Hooe s law o relae he force F on a simple harmonic oscillaor a any insan o he displacemen of he oscillaor a ha insan. v v m sin(v f) (velociy) and a v m cos(v f) (acceleraion). In he velociy funcion, he posiive quaniy v m is he velociy ampliude v m. In he acceleraion funcion, he posiive quaniy v m is he acceleraion ampliude a m. A paricle wih mass m ha moves under he influence of a Hooe s law resoring force given by F is a linear simple harmonic oscillaor wih and v A m T p A m (angular frequency) (period). 43

2 44 CHAPTER 5 OSCILLATIONS Wha Is Physics? Our world is filled wih oscillaions in which objecs move bac and forh repeaedly. Many oscillaions are merely amusing or annoying, bu many ohers are dangerous or financially imporan. Here are a few eamples: When a ba his a baseball, he ba may oscillae enough o sing he baer s hands or even o brea apar.when wind blows pas a power line, he line may oscillae ( gallop in elecrical engineering erms) so severely ha i rips apar, shuing off he power supply o a communiy. When an airplane is in fligh, he urbulence of he air flowing pas he wings maes hem oscillae, evenually leading o meal faigue and even failure. When a rain ravels around a curve, is wheels oscillae horizonally ( hun in mechanical engineering erms) as hey are forced o urn in new direcions (you can hear he oscillaions). When an earhquae occurs near a ciy, buildings may be se oscillaing so severely ha hey are shaen apar. When an arrow is sho from a bow, he feahers a he end of he arrow manage o snae around he bow saff wihou hiing i because he arrow oscillaes. When a coin drops ino a meal collecion plae, he coin oscillaes wih such a familiar ring ha he coin s denominaion can be deermined from he sound. When a rodeo cowboy rides a bull, he cowboy oscillaes wildly as he bull jumps and urns (a leas he cowboy hopes o be oscillaing). The sudy and conrol of oscillaions are wo of he primary goals of boh physics and engineering. In his chaper we discuss a basic ype of oscillaion called simple harmonic moion. Heads Up. This maerial is quie challenging o mos sudens. One reason is ha here is a rucload of definiions and symbols o sor ou, bu he main reason is ha we need o relae an objec s oscillaions (somehing ha we can see or even eperience) o he equaions and graphs for he oscillaions. Relaing he real, visible moion o he absracion of an equaion or graph requires a lo of hard wor. + m m 0 Figure 5- A paricle repeaedly oscillaes lef and righ along an ais, beween ereme poins m and m. Simple Harmonic Moion Figure 5- shows a paricle ha is oscillaing abou he origin of an ais, repeaedly going lef and righ by idenical amouns. The frequency f of he oscillaion is he number of imes per second ha i complees a full oscillaion (a cycle) and has he uni of herz (abbreviaed Hz), where herz Hz oscillaion per second s. (5-) The ime for one full cycle is he period T of he oscillaion, which is T. (5-) f Any moion ha repeas a regular inervals is called periodic moion or harmonic moion. However, here we are ineresed in a paricular ype of periodic moion called simple harmonic moion (SHM). Such moion is a sinusoidal funcion of ime. Tha is, i can be wrien as a sine or a cosine of ime. Here we arbirarily choose he cosine funcion and wrie he displacemen (or posiion) of he paricle in Fig. 5- as () m cos(v f) (displacemen), (5-3) in which m, v, and f are quaniies ha we shall define. Freeze-Frames. Le s ae some freeze-frames of he moion and hen arrange hem one afer anoher down he page (Fig. 5-a). Our firs freeze-frame is a 0 when he paricle is a is righmos posiion on he ais. We label ha coordinae as m (he subscrip means maimum); i is he symbol in fron of he cosine

3 5- SIMPLE HARMONIC MOTION 45 A A paricle oscillaes lef and righ in simple harmonic moion. The speed is zero a he ereme poins. The speed is greaes a he midpoin. m 0 + m m 0 + m = 0 = 0 v = T/4 = T/4 v = T/ = T/ = 3T/4 = 3T/4 v = T = T v m 0 + m m 0 + m (a) (b) Roaing he figure reveals ha he moion forms a cosine funcion. m This is a graph of he moion, wih he period T indicaed. Displacemen m 0 m T Time () 0 (d) The speed is zero a ereme poins. (c) m 0 T/ T Displacemen (e) m 0 m Time () The speed is greaes a = 0. Figure 5- (a) A sequence of freeze-frames (aen a equal ime inervals) showing he posiion of a paricle as i oscillaes bac and forh abou he origin of an ais, beween he limis m and m.(b) The vecor arrows are scaled o indicae he speed of he paricle.the speed is maimum when he paricle is a he origin and zero when i is a m. If he ime is chosen o be zero when he paricle is a m, hen he paricle reurns o m a T, where T is he period of he moion.the moion is hen repeaed. (c) Roaing he figure reveals he moion forms a cosine funcion of ime, as shown in (d). (e) The speed (he slope) changes.

4 46 CHAPTER 5 OSCILLATIONS Displacemen a ime Phase Ampliude Time Angular Phase frequency consan or phase angle Figure 5-3 A handy guide o he quaniies in Eq. 5-3 for simple harmonic moion. p rad () = m cos( ω + φ ) 3 p rad p rad 0 funcion in Eq In he ne freeze-frame, he paricle is a bi o he lef of m. I coninues o move in he negaive direcion of unil i reaches he lefmos posiion, a coordinae m. Thereafer, as ime aes us down he page hrough more freeze-frames, he paricle moves bac o m and hereafer repeaedly oscillaes beween m and m. In Eq. 5-3, he cosine funcion iself oscillaes beween and l.the value of m deermines how far he paricle moves in is oscillaions and is called he ampliude of he oscillaions (as labeled in he handy guide of Fig. 5-3). Figure 5-b indicaes he velociy of he paricle wih respec o ime, in he series of freeze-frames. We ll ge o a funcion for he velociy soon, bu for now jus noice ha he paricle comes o a momenary sop a he ereme poins and has is greaes speed (longes velociy vecor) as i passes hrough he cener poin. Menally roae Fig. 5-a counerclocwise by 90, so ha he freeze-frames hen progress righward wih ime. We se ime 0 when he paricle is a m. The paricle is bac a m a ime T (he period of he oscillaion), when i sars he ne cycle of oscillaion. If we filled in los of he inermediae freezeframes and drew a line hrough he paricle posiions, we would have he cosine curve shown in Fig. 5-d. Wha we already noed abou he speed is displayed in Fig. 5-e. Wha we have in he whole of Fig. 5- is a ransformaion of wha we can see (he realiy of an oscillaing paricle) ino he absracion of a graph. (In WileyPLUS he ransformaion of Fig. 5- is available as an animaion wih voiceover.) Equaion 5-3 is a concise way o capure he moion in he absracion of an equaion. More Quaniies. The handy guide of Fig. 5-3 defines more quaniies abou he moion. The argumen of he cosine funcion is called he phase of he moion. As i varies wih ime, he value of he cosine funcion varies. The consan f is called he phase angle or phase consan. I is in he argumen only because we wan o use Eq. 5-3 o describe he moion regardless of where he paricle is in is oscillaion when we happen o se he cloc ime o 0. In Fig. 5-, we se 0 when he paricle is a m. For ha choice, Eq. 5-3 wors jus fine if we also se f 0. However, if we se 0 when he paricle happens o be a some oher locaion, we need a differen value of f. A few values are indicaed in Fig For eample, suppose he paricle is a is lefmos posiion when we happen o sar he cloc a 0.Then Eq. 5-3 describes he moion if f p rad.to chec, subsiue 0 and f p rad ino Eq See, i gives m jus hen. Now chec he oher eamples in Fig The quaniy v in Eq. 5-3 is he angular frequency of he moion.to relae i o he frequency f and he period T, le s firs noe ha he posiion () of he paricle mus (by definiion) reurn o is iniial value a he end of a period. Tha is, if () is he posiion a some chosen ime, hen he paricle mus reurn o ha same posiion a ime T. Le s use Eq. 5-3 o epress his condiion, bu le s also jus se f 0 o ge i ou of he way. Reurning o he same posiion can hen be wrien as m cos v m cos v( T). (5-4) The cosine funcion firs repeas iself when is argumen (he phase, remember) has increased by p rad. So, Eq. 5-4 ells us ha v( T) v p or vt p. Thus, from Eq. 5- he angular frequency is m 0 + m Figure 5-4 Values of f corresponding o he posiion of he paricle a ime 0. v p The SI uni of angular frequency is he radian per second. pf. (5-5)

5 5- SIMPLE HARMONIC MOTION 47 The ampliudes are differen, bu he frequency and period are he same. The ampliudes are he same, bu he frequencies and periods are differen. Displacemen ' m m 0 m ' m Displacemen T m 0 m T' T' (a) (b) Figure 5-5 In all hree cases, he blue curve is obained from Eq. 5-3 wih f 0. (a) The red curve differs from he blue curve only in ha he red-curve ampliude m is greaer (he red-curve eremes of displacemen are higher and lower). (b) The red curve differs from he blue curve only in ha he red-curve period is T T/ (he red curve is compressed horizonally). (c) The red curve differs from he blue curve only in ha for he red curve f p/4 rad raher han zero (he negaive value of f shifs he red curve o he righ). (c) Displacemen m m 0 φ = 0 φ = 4 _ π This negaive value shifs he cosine curve righward. This zero gives a regular cosine curve. We ve had a lo of quaniies here, quaniies ha we could eperimenally change o see he effecs on he paricle s SHM. Figure 5-5 gives some eamples. The curves in Fig. 5-5a show he effec of changing he ampliude. Boh curves have he same period. (See how he peas line up?) And boh are for f 0. (See how he maima of he curves boh occur a 0?) In Fig. 5-5b, he wo curves have he same ampliude m bu one has wice he period as he oher (and hus half he frequency as he oher). Figure 5-5c is probably more difficul o undersand. The curves have he same ampliude and same period bu one is shifed relaive o he oher because of he differen f values. See how he one wih f 0 is jus a regular cosine curve? The one wih he negaive f is shifed righward from i.tha is a general resul: negaive f values shif he regular cosine curve righward and posiive f values shif i lefward. (Try his on a graphing calculaor.) Checpoin A paricle undergoing simple harmonic oscillaion of period T (lie ha in Fig. 5-) is a m a ime 0. Is i a m,a m, a 0, beween m and 0, or beween 0 and m when (a).00t, (b) 3.50T, and (c) 5.5T? The Velociy of SHM We briefly discussed velociy as shown in Fig. 5-b, finding ha i varies in magniude and direcion as he paricle moves beween he ereme poins (where he speed is momenarily zero) and hrough he cenral poin (where he speed is maimum). To find he velociy v() as a funcion of ime, le s ae a ime derivaive of he posiion funcion () in Eq. 5-3: v() d() d d d [ m cos(v f)] or v() v m sin(v f) (velociy). (5-6) The velociy depends on ime because he sine funcion varies wih ime, beween he values of and. The quaniies in fron of he sine funcion

6 48 CHAPTER 5 OSCILLATIONS Displacemen Velociy Acceleraion + m 0 m v + ω m 0 ω m a + ω m 0 ω m (a) (b) (c) Figure 5-6 (a) The displacemen () of a paricle oscillaing in SHM wih phase angle f equal o zero. The period T mars one complee oscillaion. (b) The velociy v() of he paricle. (c) The acceleraion a() of he paricle. T Ereme values here mean... zero values here and... ereme values here. deermine he een of he variaion in he velociy, beween v m and v m. We say ha v m is he velociy ampliude v m of he velociy variaion. When he paricle is moving righward hrough 0, is velociy is posiive and he magniude is a his greaes value. When i is moving lefward hrough 0, is velociy is negaive and he magniude is again a his greaes value. This variaion wih ime (a negaive sine funcion) is displayed in he graph of Fig. 5-6b for a phase consan of f 0, which corresponds o he cosine funcion for he displacemen versus ime shown in Fig. 5-6a. Recall ha we use a cosine funcion for () regardless of he paricle s posiion a 0.We simply choose an appropriae value of f so ha Eq. 5-3 gives us he correc posiion a 0. Tha decision abou he cosine funcion leads us o a negaive sine funcion for he velociy in Eq. 5-6, and he value of f now gives he correc velociy a 0. The Acceleraion of SHM Le s go one more sep by differeniaing he velociy funcion of Eq. 5-6 wih respec o ime o ge he acceleraion funcion of he paricle in simple harmonic moion: a() dv() d d d [ v m sin(v f)] or a() v m cos(v f) (acceleraion). (5-7) We are bac o a cosine funcion bu wih a minus sign ou fron. We now he drill by now. The acceleraion varies because he cosine funcion varies wih ime, beween and. The variaion in he magniude of he acceleraion is se by he acceleraion ampliude a m, which is he produc v m ha muliplies he cosine funcion. Figure 5-6c displays Eq. 5-7 for a phase consan f 0, consisen wih Figs. 5-6a and 5-6b. Noe ha he acceleraion magniude is zero when he cosine is zero, which is when he paricle is a 0. And he acceleraion magniude is maimum when he cosine magniude is maimum, which is when he paricle is a an ereme poin, where i has been slowed o a sop so ha is moion can be reversed. Indeed, comparing Eqs. 5-3 and 5-7 we see an eremely nea relaionship: a() v (). (5-8) This is he hallmar of SHM: () The paricle s acceleraion is always opposie is displacemen (hence he minus sign) and () he wo quaniies are always relaed by a consan (v ). If you ever see such a relaionship in an oscillaing siuaion (such as wih, say, he curren in an elecrical circui, or he rise and fall of waer in a idal bay), you can immediaely say ha he moion is SHM and immediaely idenify he angular frequency v of he moion. In a nushell: In SHM, he acceleraion a is proporional o he displacemen bu opposie in sign, and he wo quaniies are relaed by he square of he angular frequency v. Checpoin Which of he following relaionships beween a paricle s acceleraion a and is posiion indicaes simple harmonic oscillaion: (a) a 3, (b) a 5, (c) a 4, (d) a /? For he SHM, wha is he angular frequency (assume he uni of rad/s)?

7 5- SIMPLE HARMONIC MOTION 49 The Force Law for Simple Harmonic Moion Now ha we have an epression for he acceleraion in erms of he displacemen in Eq. 5-8, we can apply Newon s second law o describe he force responsible for SHM: F ma m( v ) (mv ). (5-9) The minus sign means ha he direcion of he force on he paricle is opposie he direcion of he displacemen of he paricle.tha is, in SHM he force is a resoring force in he sense ha i fighs agains he displacemen, aemping o resore he paricle o he cener poin a 0. We ve seen he general form of Eq. 5-9 bac in Chaper 8 when we discussed a bloc on a spring as in Fig.5-7.There we wroe Hooe s law, F, (5-0) m = 0 + m Figure 5-7 A linear simple harmonic oscillaor. The surface is fricionless. Lie he paricle of Fig. 5-, he bloc moves in simple harmonic moion once i has been eiher pulled or pushed away from he 0 posiion and released. Is displacemen is hen given by Eq m for he force acing on he bloc. Comparing Eqs. 5-9 and 5-0, we can now relae he spring consan (a measure of he siffness of he spring) o he mass of he bloc and he resuling angular frequency of he SHM: mv. (5-) Equaion 5-0 is anoher way o wrie he hallmar equaion for SHM. Simple harmonic moion is he moion of a paricle when he force acing on i is proporional o he paricle s displacemen bu in he opposie direcion. The bloc spring sysem of Fig. 5-7 is called a linear simple harmonic oscillaor (linear oscillaor, for shor), where linear indicaes ha F is proporional o o he firs power (and no o some oher power). If you ever see a siuaion in which he force in an oscillaion is always proporional o he displacemen bu in he opposie direcion, you can immediaely say ha he oscillaion is SHM. You can also immediaely idenify he associaed spring consan. If you now he oscillaing mass, you can hen deermine he angular frequency of he moion by rewriing Eq. 5- as v A m (angular frequency). (5-) (This is usually more imporan han he value of.) Furher, you can deermine he period of he moion by combining Eqs. 5-5 and 5- o wrie T p A m (period). (5-3) Le s mae a bi of physical sense of Eqs. 5- and 5-3. Can you see ha a siff spring (large ) ends o produce a large v (rapid oscillaions) and hus a small period T? Can you also see ha a large mass m ends o resul in a small v (sluggish oscillaions) and hus a large period T? Every oscillaing sysem, be i a diving board or a violin sring, has some elemen of springiness and some elemen of ineria or mass. In Fig. 5-7, hese elemens are separaed: The springiness is enirely in he spring, which we assume o be massless, and he ineria is enirely in he bloc, which we assume o be rigid. In a violin sring, however, he wo elemens are boh wihin he sring. Checpoin 3 Which of he following relaionships beween he force F on a paricle and he paricle s posiion gives SHM: (a) F 5, (b) F 400, (c) F 0, (d) F 3?

8 40 CHAPTER 5 OSCILLATIONS Sample Problem 5.0 Bloc spring SHM, ampliude, acceleraion, phase consan A bloc whose mass m is 680 g is fasened o a spring whose spring consan is 65 N/m. The bloc is pulled a disance cm from is equilibrium posiion a 0 on a fricionless surface and released from res a 0. (a) Wha are he angular frequency, he frequency, and he period of he resuling moion? KEY IDEA The bloc spring sysem forms a linear simple harmonic oscillaor, wih he bloc undergoing SHM. Calculaions: The angular frequency is given by Eq. 5-: 9.8 rad/s. (Answer) The frequency follows from Eq. 5-5, which yields f v p The period follows from Eq. 5-, which yields T f 0.64 s 640 ms..56 Hz (b) Wha is he ampliude of he oscillaion? KEY IDEA (Answer) (Answer) Wih no fricion involved, he mechanical energy of he spring bloc sysem is conserved. Reasoning: The bloc is released from res cm from is equilibrium posiion, wih zero ineic energy and he elasic poenial energy of he sysem a a maimum. Thus, he bloc will have zero ineic energy whenever i is again cm from is equilibrium posiion, which means i will never be farher han cm from ha posiion. Is maimum displacemen is cm: m cm. (Answer) (c) Wha is he maimum speed v m of he oscillaing bloc, and where is he bloc when i has his speed? KEY IDEA v A m 65 N/m A 0.68 g 9.78 rad/s p rad 9.78 rad/s.56 Hz.6 Hz. The maimum speed v m is he velociy ampliude v m in Eq Calculaion: Thus, we have v m v m (9.78 rad/s)(0. m). m/s. (Answer) This maimum speed occurs when he oscillaing bloc is rushing hrough he origin; compare Figs. 5-6a and 5-6b, where you can see ha he speed is a maimum whenever 0. (d) Wha is he magniude a m of he maimum acceleraion of he bloc? KEY IDEA The magniude a m of he maimum acceleraion is he acceleraion ampliude v m in Eq Calculaion: So, we have a m v m (9.78 rad/s) (0. m) m/s. (Answer) This maimum acceleraion occurs when he bloc is a he ends of is pah, where he bloc has been slowed o a sop so ha is moion can be reversed. A hose ereme poins, he force acing on he bloc has is maimum magniude; compare Figs. 5-6a and 5-6c, where you can see ha he magniudes of he displacemen and acceleraion are maimum a he same imes, when he speed is zero, as you can see in Fig. 5-6b. (e) Wha is he phase consan f for he moion? Calculaions: Equaion 5-3 gives he displacemen of he bloc as a funcion of ime. We now ha a ime 0, he bloc is locaed a m. Subsiuing hese iniial condiions, as hey are called, ino Eq. 5-3 and canceling m give us cos f. (5-4) Taing he inverse cosine hen yields f 0 rad. (Answer) (Any angle ha is an ineger muliple of p rad also saisfies Eq. 5-4; we chose he smalles angle.) (f) Wha is he displacemen funcion () for he spring bloc sysem? Calculaion: The funcion () is given in general form by Eq Subsiuing nown quaniies ino ha equaion gives us () m cos(v f) (0. m) cos[(9.8 rad/s) 0] 0. cos(9.8), (Answer) where is in meers and is in seconds. Addiional eamples, video, and pracice available a WileyPLUS

9 5- ENERGY IN SIMPLE HARMONIC MOTION 4 Sample Problem 5.0 Finding SHM phase consan from displacemen and velociy A 0, he displacemen (0) of he bloc in a linear oscillaor lie ha of Fig. 5-7 is 8.50 cm. (Read (0) as a ime zero. ) The bloc s velociy v(0) hen is 0.90 m/s, and is acceleraion a(0) is 47.0 m/s. (a) Wha is he angular frequency v of his sysem? KEY IDEA Wih he bloc in SHM, Eqs. 5-3, 5-6, and 5-7 give is displacemen, velociy, and acceleraion, respecively, and each conains v. Calculaions: Le s subsiue 0 ino each o see wheher we can solve any one of hem for v.we find (0) m cos f, (5-5) v(0) v m sin f, (5-6) and a(0) v m cos f. (5-7) In Eq. 5-5, v has disappeared. In Eqs. 5-6 and 5-7, we now values for he lef sides, bu we do no now m and f. However, if we divide Eq. 5-7 by Eq. 5-5, we nealy eliminae boh m and f and can hen solve for v as v a(0) A (0) 47.0 m/s A m 3.5 rad/s. (Answer) (b) Wha are he phase consan f and ampliude m? Calculaions: We now v and wan f and m. If we divide Eq. 5-6 by Eq. 5-5, we eliminae one of hose unnowns and reduce he oher o a single rig funcion: Solving for an f, we find an f v(0) v(0) 0.90 m/s (3.5 rad/s)( m) This equaion has wo soluions: f 5 and f 80 ( 5 ) 55. Normally only he firs soluion here is displayed by a calculaor, bu i may no be he physically possible soluion. To choose he proper soluion, we es hem boh by using hem o compue values for he ampliude m. From Eq. 5-5, we find ha if f 5, hen m v(0) (0) v m sin f m cos f v an f. (0) cos f m cos( 5 ) m. We find similarly ha if f 55, hen m m. Because he ampliude of SHM mus be a posiive consan, he correc phase consan and ampliude here are f 55 and m m 9.4 cm. (Answer) Addiional eamples, video, and pracice available a WileyPLUS 5- ENERGY IN SIMPLE HARMONIC MOTION Learning Objecives Afer reading his module, you should be able o For a spring bloc oscillaor, calculae he ineic energy and elasic poenial energy a any given ime. 5.0 Apply he conservaion of energy o relae he oal energy of a spring bloc oscillaor a one insan o he oal energy a anoher insan. 5. Sech a graph of he ineic energy, poenial energy, and oal energy of a spring bloc oscillaor, firs as a funcion of ime and hen as a funcion of he oscillaor s posiion. 5. For a spring bloc oscillaor, deermine he bloc s posiion when he oal energy is enirely ineic energy and when i is enirely poenial energy. Key Ideas A paricle in simple harmonic moion has, a any ime, ineic energy K mv and poenial energy U. If no fricion is presen, he mechanical energy E K U remains consan even hough K and U change. Energy in Simple Harmonic Moion Le s now eamine he linear oscillaor of Chaper 8, where we saw ha he energy ransfers bac and forh beween ineic energy and poenial energy, while he sum of he wo he mechanical energy E of he oscillaor remains consan. The

10 4 CHAPTER 5 OSCILLATIONS Energy E U() + K() U() poenial energy of a linear oscillaor lie ha of Fig. 5-7 is associaed enirely wih he spring. Is value depends on how much he spring is sreched or compressed ha is, on ().We can use Eqs. 8- and 5-3 o find U() m cos (v f). (5-8) E 0 T/ T (a) U() + K() U() K() As ime changes, he energy shifs beween he wo ypes, bu he oal is consan. Cauion: A funcion wrien in he form cos A (as here) means (cos A) and is no he same as one wrien cos A, which means cos(a ). The ineic energy of he sysem of Fig. 5-7 is associaed enirely wih he bloc. Is value depends on how fas he bloc is moving ha is, on v(). We can use Eq. 5-6 o find K() mv mv m sin (v f). (5-9) If we use Eq. 5- o subsiue /m for v, we can wrie Eq. 5-9 as K() mv m sin (v f). (5-0) Energy K() m 0 + m (b) As posiion changes, he energy shifs beween he wo ypes, bu he oal is consan. Figure 5-8 (a) Poenial energy U(), ineic energy K(), and mechanical energy E as funcions of ime for a linear harmonic oscillaor. Noe ha all energies are posiive and ha he poenial energy and he ineic energy pea wice during every period. (b) Poenial energy U(), ineic energy K(), and mechanical energy E as funcions of posiion for a linear harmonic oscillaor wih ampliude m. For 0 he energy is all ineic, and for m i is all poenial. The mechanical energy follows from Eqs. 5-8 and 5-0 and is E U K Checpoin 4 m cos (v f) m sin (v f) m [cos (v f) sin (v f)]. For any angle a, cos a sin a. Thus, he quaniy in he square braces above is uniy and we have E U K m. (5-) The mechanical energy of a linear oscillaor is indeed consan and independen of ime. The poenial energy and ineic energy of a linear oscillaor are shown as funcions of ime in Fig. 5-8a and as funcions of displacemen in Fig. 5-8b.In any oscillaing sysem, an elemen of springiness is needed o sore he poenial energy and an elemen of ineria is needed o sore he ineic energy. In Fig. 5-7, he bloc has a ineic energy of 3 J and he spring has an elasic poenial energy of J when he bloc is a.0 cm. (a) Wha is he ineic energy when he bloc is a 0? Wha is he elasic poenial energy when he bloc is a (b).0 cm and (c) m? Sample Problem 5.03 Many all buildings have mass dampers, which are ani-sway devices o preven hem from oscillaing in a wind. The device migh be a bloc oscillaing a he end of a spring and on a lubricaed rac. If he building sways, say, easward, he bloc also moves easward bu delayed enough so ha when i finally moves, he building is hen moving bac wesward. Thus, he moion of he oscillaor is ou of sep wih he moion of he building. Suppose he bloc has mass m g and is designed o oscillae a frequency f 0.0 Hz and wih ampliude m 0.0 cm. SHM poenial energy, ineic energy, mass dampers (a) Wha is he oal mechanical energy E of he spring bloc sysem? KEY IDEA The mechanical energy E (he sum of he ineic energy K of he bloc and he poenial energy U mv of he spring) is consan hroughou he moion of he oscillaor. Thus, we can evaluae E a any poin during he moion. Calculaions: Because we are given ampliude m of he oscillaions, le s evaluae E when he bloc is a posiion m,

11 5-3 AN ANGULAR SIMPLE HARMONIC OSCILLATOR 43 where i has velociy v 0. However, o evaluae U a ha poin, we firs need o find he spring consan. From Eq. 5- (v /m) and Eq.5-5 (v pf ),we find mv m(pf ) ( g)(p) (0.0 Hz) N/m. We can now evaluae E as E K U mv 0 ( N/m)(0.0 m) J. 0 7 J. (Answer) (b) Wha is he bloc s speed as i passes hrough he equilibrium poin? Calculaions: We wan he speed a 0, where he poenial energy is U 0 and he mechanical energy is enirely ineic energy. So, we can wrie E K U mv J (.7 05 g)v 0, or v.6 m/s. (Answer) Because E is enirely ineic energy, his is he maimum speed v m. Addiional eamples, video, and pracice available a WileyPLUS 5-3 AN ANGULAR SIMPLE HARMONIC OSCILLATOR Learning Objecives Afer reading his module, you should be able o Describe he moion of an angular simple harmonic oscillaor. 5.4 For an angular simple harmonic oscillaor, apply he relaionship beween he orque and he angular displacemen u (from equilibrium). Key Idea 5.5 For an angular simple harmonic oscillaor, apply he relaionship beween he period T (or frequency f ), he roaional ineria I, and he orsion consan. 5.6 For an angular simple harmonic oscillaor a any insan, apply he relaionship beween he angular acceleraion a, he angular frequency v, and he angular displacemen u. A orsion pendulum consiss of an objec suspended on a wire. When he wire is wised and hen released, he objec oscillaes in angular simple harmonic moion wih a period given by I T p A, where I is he roaional ineria of he objec abou he ais of roaion and is he orsion consan of he wire. An Angular Simple Harmonic Oscillaor Figure 5-9 shows an angular version of a simple harmonic oscillaor; he elemen of springiness or elasiciy is associaed wih he wising of a suspension wire raher han he eension and compression of a spring as we previously had. The device is called a orsion pendulum, wih orsion referring o he wising. If we roae he dis in Fig. 5-9 by some angular displacemen u from is res posiion (where he reference line is a u 0) and release i, i will oscillae abou ha posiion in angular simple harmonic moion. Roaing he dis hrough an angle u in eiher direcion inroduces a resoring orque given by u. (5-) Here (Gree appa) is a consan, called he orsion consan, ha depends on he lengh, diameer, and maerial of he suspension wire. Comparison of Eq. 5- wih Eq. 5-0 leads us o suspec ha Eq. 5- is he angular form of Hooe s law, and ha we can ransform Eq. 5-3, which gives he period of linear SHM, ino an equaion for he period of angular SHM: We replace he spring consan in Eq. 5-3 wih is equivalen, he consan Fied end Suspension wire θ m Reference line + θ m Figure 5-9 A orsion pendulum is an angular version of a linear simple harmonic oscillaor. The dis oscillaes in a horizonal plane; he reference line oscillaes wih angular ampliude u m. The wis in he suspension wire sores poenial energy as a spring does and provides he resoring orque. 0

12 44 CHAPTER 5 OSCILLATIONS of Eq. 5-, and we replace he mass m in Eq. 5-3 wih is equivalen, he roaional ineria I of he oscillaing dis. These replacemens lead o T p A (orsion pendulum). (5-3) Sample Problem 5.04 Angular simple harmonic oscillaor, roaional ineria, period Figure 5-0a shows a hin rod whose lengh L is.4 cm and whose mass m is 35 g, suspended a is midpoin from a long wire. Is period T a of angular SHM is measured o be.53 s. An irregularly shaped objec, which we call objec X, is hen hung from he same wire, as in Fig. 5-0b, and is period T b is found o be 4.76 s. Wha is he roaional ineria of objec X abou is suspension ais? KEY IDEA The roaional ineria of eiher he rod or objec X is relaed o he measured period by Eq The consan, which is a propery of he wire, is he same for boh figures; only he periods and he roaional inerias differ. Le us square each of hese equaions, divide he second by he firs, and solve he resuling equaion for I b.the resul is I b I a T a p A I a Tb Ta ( g m ) g m. and T b p A I b. (4.76 s) (.53 s) (Answer) Calculaions: In Table 0-e, he roaional ineria of a hin rod abou a perpendicular ais hrough is midpoin is given as ml.thus, we have, for he rod in Fig. 5-0a, I a ml ( )(0.35 g)(0.4 m) g m. Now le us wrie Eq. 5-3 wice, once for he rod and once for objec X: Figure 5-0 Two orsion pendulums, consising of (a) a wire and a rod and (b) he same wire and an irregularly shaped objec. Suspension wire L (a) Rod (b) Objec X Addiional eamples, video, and pracice available a WileyPLUS 5-4 PENDULUMS, CIRCULAR MOTION Learning Objecives Afer reading his module, you should be able o Describe he moion of an oscillaing simple pendulum. 5.8 Draw a free-body diagram of a pendulum bob wih he pendulum a angle u o he verical. 5.9 For small-angle oscillaions of a simple pendulum, relae he period T (or frequency f ) o he pendulum s lengh L Disinguish beween a simple pendulum and a physical pendulum. 5.3 For small-angle oscillaions of a physical pendulum, relae he period T (or frequency f ) o he disance h beween he pivo and he cener of mass. 5.3 For an angular oscillaing sysem, deermine he angular frequency v from eiher an equaion relaing orque and angular displacemen u or an equaion relaing angular acceleraion a and angular displacemen u Disinguish beween a pendulum s angular frequency v (having o do wih he rae a which cycles are compleed) and is du/d (he rae a which is angle wih he verical changes) Given daa abou he angular posiion u and rae of change du/d a one insan, deermine he phase consan f and ampliude u m Describe how he free-fall acceleraion can be measured wih a simple pendulum For a given physical pendulum, deermine he locaion of he cener of oscillaion and idenify he meaning of ha phrase in erms of a simple pendulum Describe how simple harmonic moion is relaed o uniform circular moion.

13 5-4 PENDULUMS, CIRCULAR MOTION 45 Key Ideas A simple pendulum consiss of a rod of negligible mass ha pivos abou is upper end, wih a paricle (he bob) aached a is lower end. If he rod swings hrough only small angles, is moion is approimaely simple harmonic moion wih a period given by T p A I mgl (simple pendulum), where I is he paricle s roaional ineria abou he pivo, m is he paricle s mass, and L is he rod s lengh. A physical pendulum has a more complicaed disribuion of mass. For small angles of swinging, is moion is simple harmonic moion wih a period given by I T p (physical pendulum), A mgh where I is he pendulum s roaional ineria abou he pivo, m is he pendulum s mass, and h is he disance beween he pivo and he pendulum s cener of mass. Simple harmonic moion corresponds o he projecion of uniform circular moion ono a diameer of he circle. Pendulums We urn now o a class of simple harmonic oscillaors in which he springiness is associaed wih he graviaional force raher han wih he elasic properies of a wised wire or a compressed or sreched spring. The Simple Pendulum If an apple swings on a long hread, does i have simple harmonic moion? If so, wha is he period T? To answer, we consider a simple pendulum, which consiss of a paricle of mass m (called he bob of he pendulum) suspended from one end of an unsrechable, massless sring of lengh L ha is fied a he oher end, as in Fig. 5-a.The bob is free o swing bac and forh in he plane of he page, o he lef and righ of a verical line hrough he pendulum s pivo poin. The Resoring Torque. The forces acing on he bob are he force T : from he sring and he graviaional force F : g, as shown in Fig. 5-b, where he sring maes an angle u wih he verical. We resolve F : g ino a radial componen F g cos u and a componen F g sin u ha is angen o he pah aen by he bob.this angenial componen produces a resoring orque abou he pendulum s pivo poin because he componen always acs opposie he displacemen of he bob so as o bring he bob bac oward is cenral locaion. Tha locaion is called he equilibrium posiion ( u 0) because he pendulum would be a res here were i no swinging. From Eq. 0-4 ( r F), we can wrie his resoring orque as L(F g sin u), (5-4) where he minus sign indicaes ha he orque acs o reduce u and L is he momen arm of he force componen F g sin u abou he pivo poin. Subsiuing Eq. 5-4 ino Eq.0-44 ( Ia) and hen subsiuing mg as he magniude of F g,we obain L(mg sin u) Ia, (5-5) where I is he pendulum s roaional ineria abou he pivo poin and a is is angular acceleraion abou ha poin. We can simplify Eq. 5-5 if we assume he angle u is small, for hen we can approimae sin u wih u (epressed in radian measure). (As an eample, if u rad, hen sin u 0.087, a difference of only abou 0.%.) Wih ha approimaion and some rearranging, we hen have a mgl u. (5-6) I This equaion is he angular equivalen of Eq. 5-8, he hallmar of SHM. I ells us ha he angular acceleraion a of he pendulum is proporional o he angular displacemen u bu opposie in sign. Thus, as he pendulum bob moves o he righ, as in Fig. 5-a, is acceleraion o he lef increases unil he bob sops and Pivo poin θ L T s = Lθ F g sin θ This componen brings he bob bac o cener. L (a) (b) F g Figure 5- (a) A simple pendulum. (b) The forces acing on he bob are he graviaional force F : g and he force T : from he sring. The angenial componen F g sin u of he graviaional force is a resoring force ha ends o bring he pendulum bac o is cenral posiion. θ m m F g cos θ This componen merely pulls on he sring.

14 46 CHAPTER 5 OSCILLATIONS begins moving o he lef. Then, when i is o he lef of he equilibrium posiion, is acceleraion o he righ ends o reurn i o he righ, and so on, as i swings bac and forh in SHM. More precisely, he moion of a simple pendulum swinging hrough only small angles is approimaely SHM. We can sae his resricion o small angles anoher way: The angular ampliude u m of he moion (he maimum angle of swing) mus be small. Angular Frequency. Here is a nea ric. Because Eq. 5-6 has he same form as Eq. 5-8 for SHM, we can immediaely idenify he pendulum s angular frequency as being he square roo of he consans in fron of he displacemen: mgl v. A I In he homewor problems you migh see oscillaing sysems ha do no seem o resemble pendulums. However, if you can relae he acceleraion (linear or angular) o he displacemen (linear or angular), you can hen immediaely idenify he angular frequency as we have jus done here. Period. Ne, if we subsiue his epression for v ino Eq. 5-5 ( v p/t), we see ha he period of he pendulum may be wrien as O I T p (5-7) A mgl. All he mass of a simple pendulum is concenraed in he mass m of he pariclelie bob, which is a radius L from he pivo poin. Thus, we can use Eq (I mr ) o wrie I ml for he roaional ineria of he pendulum. Subsiuing his ino Eq. 5-7 and simplifying hen yield F g sin θ This componen brings he pendulum bac o cener. θ h C θ F g F g cos θ L T p A g We assume small-angle swinging in his chaper. (simple pendulum, small ampliude). (5-8) Figure 5- A physical pendulum. The resoring orque is hf g sin u. When u 0, cener of mass C hangs direcly below pivo poin O. The Physical Pendulum A real pendulum, usually called a physical pendulum, can have a complicaed disribuion of mass. Does i also undergo SHM? If so, wha is is period? Figure 5- shows an arbirary physical pendulum displaced o one side by angle u. The graviaional force F : g acs a is cener of mass C, a a disance h from he pivo poin O. Comparison of Figs. 5- and 5-b reveals only one imporan difference beween an arbirary physical pendulum and a simple pendulum. For a physical pendulum he resoring componen F g sin u of he graviaional force has a momen arm of disance h abou he pivo poin, raher han of sring lengh L. In all oher respecs, an analysis of he physical pendulum would duplicae our analysis of he simple pendulum up hrough Eq Again (for small u m ), we would find ha he moion is approimaely SHM. If we replace L wih h in Eq. 5-7, we can wrie he period as I T p A mgh (physical pendulum, small ampliude). (5-9) As wih he simple pendulum, I is he roaional ineria of he pendulum abou O. However, now I is no simply ml (i depends on he shape of he physical pendulum), bu i is sill proporional o m. A physical pendulum will no swing if i pivos a is cener of mass. Formally, his corresponds o puing h 0 in Eq Tha equaion hen predics T :, which implies ha such a pendulum will never complee one swing.

15 5-4 PENDULUMS, CIRCULAR MOTION 47 Corresponding o any physical pendulum ha oscillaes abou a given pivo poin O wih period T is a simple pendulum of lengh L 0 wih he same period T. We can find L 0 wih Eq. 5-8.The poin along he physical pendulum a disance L 0 from poin O is called he cener of oscillaion of he physical pendulum for he given suspension poin. Measuring g We can use a physical pendulum o measure he free-fall acceleraion g a a paricular locaion on Earh s surface. (Counless housands of such measuremens have been made during geophysical prospecing.) To analyze a simple case, ae he pendulum o be a uniform rod of lengh L, suspended from one end. For such a pendulum, h in Eq. 5-9, he disance beween he pivo poin and he cener of mass, is L. Table 0-e ells us ha he roaional ineria of his pendulum abou a perpendicular ais hrough is cener of mass is ml. From he parallel-ais heorem of Eq (I I com Mh ), we hen find ha he roaional ineria abou a perpendicular ais hrough one end of he rod is I I com mh ml m( L) ml. (5-30) 3 If we pu h L and I ml in Eq. 5-9 and solve for g, we find g 8p L 3T. (5-3) Thus, by measuring L and he period T, we can find he value of g a he pendulum s locaion. (If precise measuremens are o be made, a number of refinemens are needed, such as swinging he pendulum in an evacuaed chamber.) Checpoin 5 Three physical pendulums, of masses m 0,m 0, and 3m 0, have he same shape and size and are suspended a he same poin. Ran he masses according o he periods of he pendulums, greaes firs. 3 Sample Problem 5.05 Physical pendulum, period and lengh In Fig. 5-3a, a meer sic swings abou a pivo poin a one end, a disance h from he sic s cener of mass. (a) Wha is he period of oscillaion T? KEY IDEA The sic is no a simple pendulum because is mass is no concenraed in a bob a he end opposie he pivo poin so he sic is a physical pendulum. h O C P L 0 Calculaions: The period for a physical pendulum is given by Eq. 5-9, for which we need he roaional ineria I of he sic abou he pivo poin. We can rea he sic as a uniform rod of lengh L and mass m. Then Eq ells us ha I ml, and he disance h in 3 Eq. 5-9 is L. Subsiuing hese quaniies ino Eq. 5-9, (a) Figure 5-3 (a) A meer sic suspended from one end as a physical pendulum. (b) A simple pendulum whose lengh L 0 is chosen so ha he periods of he wo pendulums are equal. Poin P on he pendulum of (a) mars he cener of oscillaion. (b)

16 48 CHAPTER 5 OSCILLATIONS we find T p A L p A 3g I mgh p A 3 ml mg( L) (5-3) (5-33) ()(.00 m) p.64 s. (Answer) A (3)(9.8 m/s ) Noe he resul is independen of he pendulum s mass m. (b) Wha is he disance L 0 beween he pivo poin O of he sic and he cener of oscillaion of he sic? Calculaions: We wan he lengh L 0 of he simple pendu- lum (drawn in Fig. 5-3b) ha has he same period as he physical pendulum (he sic) of Fig. 5-3a. Seing Eqs. 5-8 and 5-33 equal yields T p L 0 L p (5-34) A g A 3g. You can see by inspecion ha L 0 3L (5-35) ( 3)(00 cm) 66.7 cm. (Answer) In Fig. 5-3a, poin P mars his disance from suspension poin O. Thus, poin P is he sic s cener of oscillaion for he given suspension poin. Poin P would be differen for a differen suspension choice. Addiional eamples, video, and pracice available a WileyPLUS Simple Harmonic Moion and Uniform Circular Moion In 60, Galileo, using his newly consruced elescope, discovered he four principal moons of Jupier. Over wees of observaion, each moon seemed o him o be moving bac and forh relaive o he plane in wha oday we would call simple harmonic moion; he dis of he plane was he midpoin of he moion. The record of Galileo s observaions, wrien in his own hand, is acually sill available. A. P. French of MIT used Galileo s daa o wor ou he posiion of he moon Calliso relaive o Jupier (acually, he angular disance from Jupier as seen from Earh) and found ha he daa approimaes he curve shown in Fig The curve srongly suggess Eq. 5-3, he displacemen funcion for simple harmonic moion.a period of abou 6.8 days can be measured from he plo, bu i is a period of wha eacly? Afer all, a moon canno possibly be oscillaing bac and forh lie a bloc on he end of a spring, and so why would Eq. 5-3 have anyhing o do wih i? Acually, Calliso moves wih essenially consan speed in an essenially circular orbi around Jupier. Is rue moion far from being simple harmonic is uniform circular moion along ha orbi.wha Galileo saw and wha you can see wih a good pair of binoculars and a lile paience is he projecion of his uniform circular moion on a line in he plane of he moion. We are led by Galileo s remarable observaions o he conclusion ha simple harmonic Angle (arc minues) (days) 40 Figure 5-4 The angle beween Jupier and is moon Calliso as seen from Earh. Galileo s 60 measuremens approimae his curve, which suggess simple harmonic moion. A Jupier s mean disance from Earh, 0 minues of arc corresponds o abou 0 6 m. (Based on A. P. French, Newonian Mechanics, W. W. Noron & Company, New Yor, 97, p. 88.)

17 5-4 PENDULUMS, CIRCULAR MOTION 49 y P is a paricle moving in a circle. y v ω m ω +φ y O m ω + φ () P' P ω + φ O v() P' P ω + φ O ω m P' a a() P (a) P is a projecion moving in SHM. (b) This relaes he velociies of P and P. (c) This relaes he acceleraions of P and P. Figure 5-5 (a) A reference paricle P moving wih uniform circular moion in a reference circle of radius m. Is projecion P on he ais eecues simple harmonic moion. (b) The projecion of he velociy v : of he reference paricle is he velociy of SHM. (c) The projecion of he radial acceleraion of he reference paricle is he acceleraion of SHM. a : moion is uniform circular moion viewed edge-on. In more formal language: Simple harmonic moion is he projecion of uniform circular moion on a diameer of he circle in which he circular moion occurs. Figure 5-5a gives an eample. I shows a reference paricle P moving in uniform circular moion wih (consan) angular speed v in a reference circle. The radius m of he circle is he magniude of he paricle s posiion vecor. A any ime, he angular posiion of he paricle is v f, where f is is angular posiion a 0. Posiion. The projecion of paricle P ono he ais is a poin P, which we ae o be a second paricle. The projecion of he posiion vecor of paricle P ono he ais gives he locaion () of P. (Can you see he componen in he riangle in Fig. 5-5a?) Thus, we find () m cos(v f), (5-36) which is precisely Eq Our conclusion is correc. If reference paricle P moves in uniform circular moion, is projecion paricle P moves in simple harmonic moion along a diameer of he circle. Velociy. Figure 5-5b shows he velociy v : of he reference paricle. From Eq. 0-8 (v vr), he magniude of he velociy vecor is v m ; is projecion on he ais is v() v m sin(v f), (5-37) which is eacly Eq The minus sign appears because he velociy componen of P in Fig. 5-5b is direced o he lef, in he negaive direcion of. (The minus sign is consisen wih he derivaive of Eq wih respec o ime.) Acceleraion. Figure 5-5c shows he radial acceleraion a : of he reference paricle. From Eq. 0-3 (a r v r), he magniude of he radial acceleraion vecor is v m ; is projecion on he ais is a() v m cos(v f), (5-38) which is eacly Eq Thus, wheher we loo a he displacemen, he velociy, or he acceleraion, he projecion of uniform circular moion is indeed simple harmonic moion.

18 430 CHAPTER 5 OSCILLATIONS 5-5 DAMPED SIMPLE HARMONIC MOTION Learning Objecives Afer reading his module, you should be able o Describe he moion of a damped simple harmonic oscillaor and sech a graph of he oscillaor s posiion as a funcion of ime For any paricular ime, calculae he posiion of a damped simple harmonic oscillaor Deermine he ampliude of a damped simple harmonic oscillaor a any given ime. Key Ideas The mechanical energy E in a real oscillaing sysem decreases during he oscillaions because eernal forces, such as a drag force, inhibi he oscillaions and ransfer mechanical energy o hermal energy. The real oscillaor and is moion are hen said o be damped. If he damping force is given by F : d b v :, where v : is he velociy of he oscillaor and b is a damping consan, hen he displacemen of he oscillaor is given by () m e b/m cos(v f), 5.4 Calculae he angular frequency of a damped simple harmonic oscillaor in erms of he spring consan, he damping consan, and he mass, and approimae he angular frequency when he damping consan is small. 5.4 Apply he equaion giving he (approimae) oal energy of a damped simple harmonic oscillaor as a funcion of ime. where v, he angular frequency of he damped oscillaor, is given by A m b 4m. If he damping consan is small (b m), hen v v, where v is he angular frequency of he undamped oscillaor. For small b, he mechanical energy E of he oscillaor is given by E() m e b/m. Rigid suppor Springiness, Mass m Vane Damping, b Figure 5-6 An idealized damped simple harmonic oscillaor. A vane immersed in a liquid eers a damping force on he bloc as he bloc oscillaes parallel o he ais. Damped Simple Harmonic Moion A pendulum will swing only briefly underwaer, because he waer eers on he pendulum a drag force ha quicly eliminaes he moion. A pendulum swinging in air does beer, bu sill he moion dies ou evenually, because he air eers a drag force on he pendulum (and fricion acs a is suppor poin), ransferring energy from he pendulum s moion. When he moion of an oscillaor is reduced by an eernal force, he oscillaor and is moion are said o be damped. An idealized eample of a damped oscillaor is shown in Fig. 5-6, where a bloc wih mass m oscillaes verically on a spring wih spring consan. From he bloc, a rod eends o a vane (boh assumed massless) ha is submerged in a liquid.as he vane moves up and down, he liquid eers an inhibiing drag force on i and hus on he enire oscillaing sysem. Wih ime, he mechanical energy of he bloc spring sysem decreases, as energy is ransferred o hermal energy of he liquid and vane. Le us assume he liquid eers a damping force F : d ha is proporional o he velociy v : of he vane and bloc (an assumpion ha is accurae if he vane moves slowly). Then, for force and velociy componens along he ais in Fig. 5-6, we have F d bv, (5-39) where b is a damping consan ha depends on he characerisics of boh he vane and he liquid and has he SI uni of ilogram per second. The minus sign indicaes ha F : d opposes he moion. Damped Oscillaions. The force on he bloc from he spring is F s. Le us assume ha he graviaional force on he bloc is negligible relaive o F d and F s. Then we can wrie Newon s second law for componens along he ais (F ne, ma ) as bv ma. (5-40)

19 + m m e b/m () 5-5 DAMPED SIMPLE HARMONIC MOTION (s) m m e b/m Figure 5-7 The displacemen funcion () for he damped oscillaor of Fig The ampliude, which is m e b/m, decreases eponenially wih ime. Subsiuing d/d for v and d /d for a and rearranging give us he differenial equaion m d d The soluion of his equaion is b d d 0. (5-4) () m e b/m cos(v f), (5-4) where m is he ampliude and v is he angular frequency of he damped oscillaor. This angular frequency is given by v A m b 4m. (5-43) If b 0 (here is no damping), hen Eq reduces o Eq. 5- ( v /m) for he angular frequency of an undamped oscillaor, and Eq. 5-4 reduces o Eq. 5-3 for he displacemen of an undamped oscillaor. If he damping consan is small bu no zero (so ha b m), hen v v. Damped Energy. We can regard Eq. 5-4 as a cosine funcion whose ampliude, which is m e b/m, gradually decreases wih ime, as Fig. 5-7 suggess. For an undamped oscillaor, he mechanical energy is consan and is given by Eq. 5- (E m). If he oscillaor is damped, he mechanical energy is no consan bu decreases wih ime. If he damping is small, we can find E() by replacing m in Eq. 5- wih m e b/m,he ampliude of he damped oscillaions.by doing so,we find ha E() me b/m, (5-44) which ells us ha, lie he ampliude, he mechanical energy decreases eponenially wih ime. Checpoin 6 Here are hree ses of values for he spring consan, damping consan, and mass for he damped oscillaor of Fig Ran he ses according o he ime required for he mechanical energy o decrease o one-fourh of is iniial value, greaes firs. Se 0 b 0 m 0 Se 0 6b 0 4m 0 Se b 0 m 0

20 43 CHAPTER 5 OSCILLATIONS Sample Problem 5.06 Damped harmonic oscillaor, ime o decay, energy For he damped oscillaor of Fig. 5-6, m 50 g, 85 N/m, and b 70 g/s. (a) Wha is he period of he moion? KEY IDEA Because b m 4.6 g/s, he period is approimaely ha of he undamped oscillaor. Calculaion: From Eq. 5-3, we hen have m T p (Answer) A p 0.5 g 0.34 s. A 85 N/m (b) How long does i ae for he ampliude of he damped oscillaions o drop o half is iniial value? KEY IDEA The ampliude a ime is displayed in Eq. 5-4 as m e b/m. Calculaions: The ampliude has he value m a 0.Thus, we mus find he value of for which m e b/m m. Canceling m and aing he naural logarihm of he equaion ha remains, we have ln on he righ side and ln(e b/m ) b/m on he lef side.thus, 5.0 s. (Answer) Because T 0.34 s, his is abou 5 periods of oscillaion. (c) How long does i ae for he mechanical energy o drop o one-half is iniial value? KEY IDEA From Eq. 5-44, he mechanical energy a ime is m e b/m. Calculaions: The mechanical energy has he value m a 0.Thus, we mus find he value of for which m e b/m ( m). If we divide boh sides of his equaion by m and solve for as we did above, we find m ln b m ln b (0.5 g)(ln ) g/s ()(0.5 g)(ln ) g/s.5 s. (Answer) This is eacly half he ime we calculaed in (b), or abou 7.5 periods of oscillaion. Figure 5-7 was drawn o illusrae his sample problem. Addiional eamples, video, and pracice available a WileyPLUS 5-6 FORCED OSCILLATIONS AND RESONANCE Learning Objecives Afer reading his module, you should be able o Disinguish beween naural angular frequency v and driving angular frequency v d For a forced oscillaor, sech a graph of he oscillaion ampliude versus he raio v d /v of driving angular frequency o naural angular frequency, idenify he approimae locaion of resonance, and indicae he effec of increasing he damping consan For a given naural angular frequency v, idenify he approimae driving angular frequency v d ha gives resonance. Key Ideas If an eernal driving force wih angular frequency v d acs on an oscillaing sysem wih naural angular frequency v, he sysem oscillaes wih angular frequency v d. The velociy ampliude v m of he sysem is greaes when v d v, a condiion called resonance. The ampliude m of he sysem is (approimaely) greaes under he same condiion. Forced Oscillaions and Resonance A person swinging in a swing wihou anyone pushing i is an eample of free oscillaion. However, if someone pushes he swing periodically, he swing has

21 5-6 FORCED OSCILLATIONS AND RESONANCE 433 forced, or driven, oscillaions. Two angular frequencies are associaed wih a sysem undergoing driven oscillaions: () he naural angular frequency v of he sysem, which is he angular frequency a which i would oscillae if i were suddenly disurbed and hen lef o oscillae freely, and () he angular frequency v d of he eernal driving force causing he driven oscillaions. We can use Fig. 5-6 o represen an idealized forced simple harmonic oscillaor if we allow he srucure mared rigid suppor o move up and down a a variable angular frequency v d. Such a forced oscillaor oscillaes a he angular frequency v d of he driving force, and is displacemen () is given by () m cos(v d f), (5-45) where m is he ampliude of he oscillaions. How large he displacemen ampliude m is depends on a complicaed funcion of v d and v. The velociy ampliude v m of he oscillaions is easier o describe: i is greaes when Ampliude b = 50 g/s (leas damping) b = 70 g/s b = 40 g/s ω d /ω Figure 5-8 The displacemen ampliude m of a forced oscillaor varies as he angular frequency v d of he driving force is varied. The curves here correspond o hree values of he damping consan b. v d v (resonance), (5-46) a condiion called resonance. Equaion 5-46 is also approimaely he condiion a which he displacemen ampliude m of he oscillaions is greaes.thus, if you push a swing a is naural angular frequency, he displacemen and velociy ampliudes will increase o large values, a fac ha children learn quicly by rial and error. If you push a oher angular frequencies, eiher higher or lower, he displacemen and velociy ampliudes will be smaller. Figure 5-8 shows how he displacemen ampliude of an oscillaor depends on he angular frequency v d of he driving force, for hree values of he damping coefficien b. Noe ha for all hree he ampliude is approimaely greaes when v d /v (he resonance condiion of Eq. 5-46). The curves of Fig. 5-8 show ha less damping gives a aller and narrower resonance pea. Eamples. All mechanical srucures have one or more naural angular frequencies, and if a srucure is subjeced o a srong eernal driving force ha maches one of hese angular frequencies, he resuling oscillaions of he srucure may rupure i. Thus, for eample, aircraf designers mus mae sure ha none of he naural angular frequencies a which a wing can oscillae maches he angular frequency of he engines in fligh. A wing ha flaps violenly a cerain engine speeds would obviously be dangerous. Resonance appears o be one reason buildings in Meico Ciy collapsed in Sepember 985 when a major earhquae (8. on he Richer scale) occurred on he wesern coas of Meico.The seismic waves from he earhquae should have been oo wea o cause eensive damage when hey reached Meico Ciy abou 400 m away. However, Meico Ciy is largely buil on an ancien lae bed, where he soil is sill sof wih waer. Alhough he ampliude of he seismic waves was small in he firmer ground en roue o Meico Ciy, heir ampliude subsanially increased in he loose soil of he ciy. Acceleraion ampliudes of he waves were as much as 0.0g, and he angular frequency was (surprisingly) concenraed around 3 rad/s. No only was he ground severely oscillaed, bu many inermediae-heigh buildings had resonan angular frequencies of abou 3 rad/s. Mos of hose buildings collapsed during he shaing (Fig. 5-9), while shorer buildings (wih higher resonan angular frequencies) and aller buildings (wih lower resonan angular frequencies) remained sanding. During a 989 earhquae in he San Francisco Oaland area, a similar resonan oscillaion collapsed par of a freeway, dropping an upper dec ono a lower dec. Tha secion of he freeway had been consruced on a loosely srucured mudfill. John T. Barr/Gey Images, Inc. Figure 5-9 In 985, buildings of inermediae heigh collapsed in Meico Ciy as a resul of an earhquae far from he ciy. Taller and shorer buildings remained sanding.