Ch. 11--Vibraory Moion Chaper 11 VIBRATORY MOTION Noe: There are wo areas of ineres when discussing oscillaory moion: he mahemaical characerizaion of vibraing srucures ha generae waves and he ineracion of waves wih oher waves and wih heir surroundings. We will examine he former opic in his chaper, he laer in he nex chaper. A.) Vibraory Moion--Basic Conceps: 1.) For any srucure o vibrae periodically, here mus be a resoring force on he body. A resoring force is a force ha is consanly aemping o accelerae he objec back oward is equilibrium posiion. 2.) The easies way o examine vibraory moion is wih an example. We will use a spring sysem: a.) Consider he mass aached o he spring shown in Figure 11.1. When he spring is neiher compressed nor elongaed, he mass feels no force and is, hence, in a sae of equilibrium. spring fricionless surface Noe: For his and similar sysems, he coordinae axis used o define mass posiion has is origin (i.e., x = 0) defined a he body's equilibrium posiion. x = 0 mass a equilibrium (no spring force being applied) FIGURE 11.1 b.) I has been experimenally observed ha if an "ideal" spring (i.e., one of hose myhical ypes ha loses no energy during oscillaion) is displaced a disance x (see Figure 11.2), he force F required o displace he spring will be proporional o he displacemen x. Pu anoher way, if a mass is aached o he spring and he spring is displaced 333
a disance x, he spring will exer a force F on he mass when released. Tha force will be proporional o he spring's displacemen from is equilibrium posiion. Defining k as he proporionaliy consan (unis: n/m), his force is: F = -k ( x). spring compressed x = 0 mass displaced a disance " x" x = 0 direcion of force applied on mass by spring FIGURE 11.2 Called Hooke's Law, his relaionship and he moion i describes is called "simple harmonic moion." Noe 1: Because x is measured from equilibrium (i.e., from x = 0), a displacemen x-unis-long will equal x = x final - x iniial = x - 0. In oher words, x = x. As such, HOOKE'S LAW IS ALWAYS WRITTEN: F = - kx. Noe 2: Be sure you undersand which force Hooke's Law alludes o: i is he force he spring applies o he mass, no he force he mass applies o he spring. Noe 3: The negaive sign in fron of he kx erm insures ha he force is always direced back oward he equilibrium posiion. To see his, assume he spring in Figure 11.2 has a spring consan of 2 n/m and is displaced a disance x = -.6 meers. The force equaion yields: F = - [(2 n/m) (-.6 m)] = +1.2 ns. The direcion of he spring's force on he mass is posiive, jus as common sense would dicae. Wihou he negaive sign on he righ hand side of he force equaion, he mahemaics would no accuraely model he siuaion. 334
Ch. 11--Vibraory Moion 3.) Here are some DEFINITIONS needed for he discussion of vibraory sysems: a.) Periodic moion: Any moion ha repeas iself hrough ime. b.) Simple harmonic moion: Periodic moion whose force funcion is of he form -kx, where k is a consan and x is he displacemen of he srucure a some arbirary poin in ime. c.) Frequency ( ν): The number of cycles swep hrough per-uniime; he MKS unis are cycles per second (i.e., herz, abbreviaed Hz). The symbol used for frequency-- ν --is he Greek leer nu. Imporan Noe: Cycles is no echnically a uni. In many exs, herz is defined as inverse seconds (i.e., 1/seconds). We will use boh, depending upon he siuaion. d.) Period (T): The ime required o sweep hrough one complee cycle. The unis are seconds per cycle (or jus seconds). Noe ha he period and frequency of a body's moion are inversely relaed. Tha is: T = 1 / ν. e.) Displacemen (x): The disance a vibraing objec is from is equilibrium posiion a a given poin in ime. Displacemen is a ime varying quaniy whose unis are in meers or cenimeers or whaever he disance unis are for he sysem. f.) Ampliude (A): The maximum displacemen x max of an oscillaing body. Assuming he vibraory moion does no lose energy, he ampliude of he moion remains consan--i does no vary wih ime. Ampliudes are measured from equilibrium and have he same unis as displacemen. Noe: I is ineresing o observe ha because he force funcion for a spring is proporional o he spring's displacemen (F = -kx), he period and, hence, frequency of a given spring/mass sysem will be a consan. Why? An oscillaion wih a very small displacemen will have a very small disance o ravel during one period, bu i will also have a very small spring force o moivae i. An oscillaion wih a very large displacemen will have a very large disance o cover during one period, bu i will have a very large spring force o help i along. The ne resul: wheher you have big oscillaions or small oscillaions, i akes he same amoun of ime o oscillae hrough one cycle. 335
B.) The Mahemaics of Simple Harmonic Moion: 1.) We would like o derive an expression ha defines he displacemen of a vibraing objec from equilibrium as a funcion of ime--i.e., x(). To generae he appropriae equaion, we will examine he vibraory moion of a mass aached o a spring (see Figure 11.3), using Newon's Second Law o evaluae he moion. spring elongaed x = 0 x = 0 direcion of force applied o mass by spring mass displaced a disance "x" a.) A free body diagram is shown in Figure 11.4. Summing he forces in he horizonal, and leaving he sign of he acceleraion embedded wihin he ma, we ge: y N FIGURE 11.3 f.b.d. on mass F x : kx -kx = ma a + (k/m)x = 0. x b.) We know ha he acceleraion and velociy are relaed by a = dv/d, and ha he velociy and displacemen are mg FIGURE 11.4 relaed by v = dx/d. As such, i is rue ha a = d 2 x/d 2, where he noaion used is mean o convey he second derivaive of he posiion wih respec o ime. 336
Ch. 11--Vibraory Moion c.) Subsiuing a = d 2 x/d 2 ino he force expression yields: d 2 x d + k 2 m x = 0. d.) Wha does his equaion really say? I suggess ha here exiss a funcion x such ha when you add is second derivaive o a consan imes iself, you ALWAYS ge zero. The quesion is, "Wha funcion will do he job?" The answer is, "A sine wave." 2.) The mos general expression for a sine wave (see Figure 11.5a on he nex page) is: x() = A sin ( ω +φ), where A is he ampliude of he displacemen (i.e., is maximum possible value); ω is a consan called he angular frequency whose unis are radians/second and whose significance will become clear laer; and φ is anoher consan called he phase shif whose unis are in radians and whose significance will also be discussed laer. 3.) Using he Calculus on our sine funcion, we find ha if: x() = A sin ( ω +φ). a.) The velociy of he moion will be: v() = dx/d = ωa cos ( ω +φ). b.) The acceleraion of he moion will be: a() = dv/d = - ω 2 A sin ( ω +φ). c.) The graphs of all hree of hese funcions are found in Figures 11.5a, b, and c. 337
Noe: Noice ha he horizonal axis is no labeled in ime bu raher in ω. Sine waves are funcions of angles. Angles mus have argumens in angular measure (radians in his case). Tha means he expression x = A sin makes no sense as wrien--you can' have a sine argumen whose unis are in ime. To ge around he problem, we modify he ime variable by muliplying by ω radians per second. d.) The maximum value for boh a sine and a cosine funcion is 1. This means: x() A A v() A posiion vs. ime x() = A sin ( + 0) FIGURE 11.5a velociy vs. ime v max = ωa and a max = ω 2 A (where a max is a magniude). A v() = A cos ( + 0) 4.) Analyzing he graphs: a.) Look a he firs long, verical doed line spanning Figures 11.5a, b, and c: i.) The graphs sugges ha when he displacemen x is maximum-and-posiive (i.e., as far o he righ of equilibrium as i ges), he acceleraion is maximumand-negaive. a() 2 A FIGURE 11.5b acceleraion vs. ime Noe: A negaive ampliude value (-ω 2 A = -10 m/s 2, for insance) does no signify a minimum. The fac ha -10 is smaller han +10 on a number line is no relevan here. The 2 A 2 a() = - A sin ( + 0) FIGURE 11.5c 338
Ch. 11--Vibraory Moion value -ω 2 A is he larges acceleraion possible; he negaive sign simply ells you he direcion in which i is a maximum. ii.) Back o he "firs verical line:" When x is a is exremes, he velociy of he body is zero. This makes sense. A he exremes he body sops before beginning back in he opposie direcion. b.) Look a he second long, verical doed line spanning Figures 11.5a, b, and c: i.) The graphs sugges ha when x is zero (i.e., he body is a equilibrium), he acceleraion is zero. This makes sense. A equilibrium he force applied o he body by he spring is zero, hence zero acceleraion would be expeced. ii.) When x is a equilibrium, he velociy is a posiive or negaive maximum, depending upon which direcion he body is moving. This also makes sense inuiively. Only when every bi of acceleraion has been exhaused in moivaing he mass back oward he equilibrium poin will he velociy be a is maximum. Tha occurs a equilibrium. 5.) If we go back o Newon's Second Law wih his informaion: a.) Subsiuing our sine-relaed x() and a() funcions back ino our force expression (i.e., a + (k/m) x = 0), we ge: [-w 2 A sin ( ω +φ)] + (k/m) [A sin ( ω +φ)] = 0. b.) Noing ha he A's and he sine funcions cancel, we end up wih: - ω 2 + (k/m) = 0 ω = (k/m) 1/2. 6.) Evidenly, he funcion x() = A sin ( ω + f) saisfies he equaion a + (k/m) x = 0 as long as ω = (k/m) 1/2. a.) BIG GENERAL POINT: If you can manipulae a Newon's Second Law equaion ino he form: acceleraion + (some consan) (displacemen) = 0, you know for cerain ha: 339
i.) The moion will be simple harmonic moion (versus some oher form of oscillaory moion); and ii.) The angular frequency ω of he moion will be equal o he square roo of he consan in fron of he displacemen variable in he manipulaed N.S.L. equaion. In he case of a spring, Newon's Second Law yielded: a + (k/m) x = 0, and he consan in fron of he posiion variable k/m was found o be such ha: ω = (k/m) 1/2. b.) You migh no hink much of his revelaion now, bu i is going o be very useful laer when we examine oher kinds of vibraing sysems. Before we can look a hese oher ypes of vibraing sysems, hough, we need o make some sense ou of he angular frequency ω and he phase shif φ erms. C.) Angular Frequency ( ω ): 1.) Look a he POSITION VERSUS TIME graph of a vibraing body (Figure 11.6). How can I ell you where he body is in is moion a a given poin in ime? How, for insance, can I ell you ha he body is a, say, Poin A in Figure 11.6? There are hree ways o do he deed. Each is useful in is own way; each is lised below: x Poin A ime FIGURE 11.6 a.) The firs way has already been discussed. I could simply say, "The body is a x = A." In ha case, I am giving you he "linear displacemen" of he body a he poin-in-ime of ineres. Though simple, i is no very useful if we wan o know somehing abou how fas he oscillaions are aking place. Tha is, if I know he ime i akes for he body o ge from x = 0 o x = A, dividing he ime ino 340
Ch. 11--Vibraory Moion he displacemen will give only he average velociy over he moion--a none-oo-useful commodiy in mos cases. b.) Anoher possibiliy is o say, "The body is one quarer of a cycle hrough is moion." In ha case, I am giving you he cyclic displacemen of he body. If I addiionally ell you how long i akes o achieve ha posiion wihin is cyclic moion, we can divide he ime ino ha cyclic displacemen and come up wih an expression for he body's frequency in cycles per second. For oscillaory moion, frequency measuremens are very useful. 1/4 cycle or /2 radians 1/2 cycle or radians 1 cycle or 2 radians FIGURE 11.7 c.) Anoher more exoic possibiliy is o say, "The body is /2 radians hrough is moion." In his case, I would be giving you he angular displacemen of he body. This very peculiar way of characerizing he posiion of a vibraing body is made simply by looking a he graph of a sine funcion (see Figure 11.7). Noice ha a body having compleed one full cycle has moved hrough an angular displacemen of 2 radians. Following logically, a body having moved hrough one-half cycle has displaced an angular measure of radians and a body having moved hrough onequarer cycle has displaced /2 radians. In oher words, if you undersand he language, an angular displacemen can ell you where a body is in is moion jus as well as a cyclic displacemen can. If we furher divide his radian-displacemen by he ime required o ge o ha posiion, we end up wih an expression for he body's angular frequency ω in radians per second. 2.) One cycle is he equivalen of an angular measure of 2 radians. Tha means ha oscillaory moion whose frequency is 1 cycle/second has an angular frequency of 2 radians/second. Expanding his, i becomes obvious ha he relaionship beween frequency ν and angular frequency ω is: ω = 2 ν. 341
3.) Reconsidering x() = A sin ( ω + φ ), he angular frequency ω governs how fas he funcion, hence body posiion, changes. Large ω means i akes very lile ime for ω o incremen by 2 (i.e., move hrough one cycle), which means he period of he funcion and he body's moion is small. This corresponds o a high frequency oscillaion. A small ω does jus he opposie. D.) Phase Shif ( φ): 1.) A ypical sine wave funcion is characerized by he graph shown in Figure 11.8 and is mahemaically wrien as: x() = A sin ( ω). a.) This expression predics ha a = 0, x = 0 (i.e., pu in = 0 and you ge x = 0!). Wha's more: x b.) Jus afer = 0, he value of A sin ( ω ) is posiive and ges larger as ime proceeds, jus as he graph shows. BIG NOTE: The displacemen variable x() is measured from EQUILIBRIUM. Tha means ha if you look a a graph of x() and see ha he variable is geing larger over a paricular ime inerval (eiher large in a posiive sense or larger in a negaive sense), i means ha he funcion is modeling moion ha is moving AWAY FROM equilibrium. FIGURE 11.8 c.) The problem arises when we do no wan he body o be a equilibrium (x = 0) a = 0. For insance, wha do we do if we wan i o be a x = A when we sar he clock (i.e., a = 0)? Dealing wih such problems is exacly wha he phase shif φ is designed o do. I allows one o make compensaions in he mah so ha a sine funcion can be used o characerize oscillaory moion ha doesn' assume x = 0 a = 0. 2.) Easy Example: Assume we define he posiion of an oscillaing body as x = +A a = 0. How can we use a sine funcion o characerize ha moion? a.) Noice ha if we shif he verical axis of he sine wave shown in Figure 11.9a (nex page) by /2 radians (see Figure 11.9b for he shifed 342
Ch. 11--Vibraory Moion version), we end up wih a graph ha gives us x = +A a = 0 (Figure 11.9c). In oher words, adding /2 o he sine's angle will do for us exacly wha we wan. x b.) Mahemaically, we are suggesing ha he correc funcion is: x() = A sin ( ω + /2). original origin x new axis posiion FIGURE 11.9a i.) To check, we know wha he funcion's value should be a = 0: i should be x = +A. axis shifed o righ by /2 radians (wih his axis, x=+a a =0) ii.) Plugging = 0 ino he funcion we are esing yields: x(=0) = A sin ( ω (0) + /2) = A sin ( /2) = A. Our funcion works! x +A final graph of: x() = A sin ( + /2) FIGURE 11.9b c.) The moral: The phase shif ells us how much we have o ranslae (shif) he verical axis o define he correc displacemen x a = 0. FIGURE 11.9c Noe: A "+" phase shif shifs he axis o he righ; a "-" phase shif shifs he axis o he lef. 3.) In he case above, i was obvious ha he shif needed o be /2 radians. Unforunaely, no all problems are his easy. How does one deermine he phase shif for more complex siuaions? 343
a.) Assume you know where an oscillaing body is supposed o be a = 0. The key o deermining he general sine wave funcion ha will fi he siuaion lies in evaluaing he displacemen equaion x() a a known poin in ime (preferably a = 0), hen solving ha expression for he appropriae φ. The process will be formally presened using he relaively easy case cied above. We will hen ry he approach on more difficul problems. 4.) Advanced Example #1: Assume ha a = 0, x = +A. a.) Puing ha informaion ino our displacemen expression yields x() = A sin ( ω +φ) A = A sin ( ω(0) + φ). b.) Dividing by A and muliplying w by zero gives us: 1 = sin ( φ) This is exacly wha we expeced. φ = sin -1 (1) = 1.57 (i.e., /2). c.) Knowing φ for one poin in ime means we know i for all poins in ime ( φ is a consan for he moion). Puing i back ino our general algebraic expression for he displacemen gives us: x() = A sin ( ω + 1.57). Noe: In mos problems, you will have already deermined boh A and w. Tha is, boh will have numeric values. As an example, if A = 2 meers and ω = 7.5 radians/second, he finished expression will look like: x() = 2 sin (7.5 + 1.57). 5.) Example #2: Deermine he general algebraic expression for he displacemen of a spring-mass sysem whose posiion a = 0 is (3/4)A going away from equilibrium (see Figure 11.10a and 11.10b on he nex page). 344
Ch. 11--Vibraory Moion a.) In general: x() = A sin ( ω +φ 1 ). x = 0 direcion of moion a =0 b.) Subsiuing = 0 and x = (3/4)A ino our general equaion yields: (3/4)A a =0 (3/4)A = A sin ( ω(0) +φ 1 ). c.) Dividing by A and muliplying ω by zero gives us: 3/4 = sin ( φ 1 ), which implies ha φ 1 is he angle whose sine is 3/4, or φ 1 = sin -1 (3/4) =.848 radians. original axis A (3/4)A shifed maximum displacemen (A) o be achieved a some laer ime FIGURE 11.10a d.) Puing our value for φ back ino our general algebraic expression for he displacemen gives us: push axis 0 radians so ha a =0, x=+(3/4)a FIGURE 11.10b x() = A sin ( ω +.848). e.) By shifing he axis of he sine wave by.848 radians (see Figure 11.10b), we ge a graph ha has he body's posiion equal o.75a a = 0 and ha addiionally has he displacemen proceeding away from equilibrium jus afer = 0. 6.) Example #3--a lile differen wis: Deermine he general algebraic expression for he displacemen of a spring/mass sysem whose posiion a = 0 is (3/4)A going oward equilibrium. 345
Noe: This is almos exacly he same as he problem in #5. The only difference is in he direcion of he moion jus afer = 0. Proceeding hrough he seps: a.) In general: x() = A sin ( ω +φ 2 ). b.) Subsiuing = 0 and x = (3/4)A ino our general equaion yields: (3/4)A = A sin ( ω(0) +φ 2 ). c.) Dividing by A and muliplying ω by zero gives us: 3/4 = sin ( φ 2 ), which implies ha φ 2 is he angle whose sine is 3/4, or φ 2 = sin -1 (3/4) =.848 radians. original axis A (3/4)A shifed axis x() x = 0 (3/4)A a =0 maximum displacemen(a) o be achieved a some laer ime push axis 0 radians so ha 2 a =0, x=+(3/4)a direcion of moion a =0 FIGURE 11.11a FIGURE 11.11b d.) THE SNAG: This suggess ha he angle φ 2 equals he angle φ 1, which clearly can' be he case (see Figure 11.12). Wha we need is an angle ha predics moion ha proceeds back oward equilibrium jus afer = 0... no an angle ha predics moion ha proceeds away from equilibrium afer = 0. x() (3/4)A 0 1 a his ime, body moving away from equilibrium 0 2 a his ime, body moving oward equilibrium 346 FIGURE 11.12
Ch. 11--Vibraory Moion Pu a lile differenly, i is clear from he skech ha here are wo phase shifs ha can pu x = (3/4)A a = 0. The firs (i.e., φ 1 ) is he one we used in #5. I corresponds o he siuaion when, jus afer = 0, he moion proceeds away from equilibrium (look a he graph--he value for x ges more posiive as ime progresses). The second phase shif ( φ 2 ) is he one we wan here. I makes x = (3/4)A a = 0, and i also has he displacemen going back oward equilibrium as ime progresses. x() e.) To deermine φ 2, we need o use he symmery of he sine funcion (see Figure 11.13). Noice from he figure ha he phase shif ( φ 2 ) is equal o -.848 radians, or 2.29 radians. Using his, he final expression becomes: (3/4)A 0 0 1 1 (from symmery) 0 = - 0 2 1 x() = A sin ( ω + 2.29). FIGURE 11.13 f.) Boom line: Before deciding if he angle your calculaor produces is correc, make a skech of a sine wave and decide wheher you need φ 1 or φ 2. 7.) Example #4: Deermine a general algebraic expression for he displacemen of an oscillaing body whose posiion a = 0 is (-3/4)A going away from equilibrium (see Figures 11.14a). Assume also ha A =.6 meers and ω = 12 rad/sec. direcion of a.) Using he same approach as before: x = (-3/4)A a =0 x = 0 FIGURE 11.14a 347
b.) A negaive phase shif moves he axis o he lef. Again, here are wo posiions where an axis can be placed so ha a = 0, x = (-3/4)A (see Figure 11.14b). The firs, corresponding o an angular shif of he axis of φ 3, has he body moving oward equilibrium jus afer = 0; he second, corresponding o an angular shif of he axis of φ 4, has he body moving ( ) ( ) x() = A sin ω +φ 3 (-3/4)(.6) = (.6) sin ω(0) +φ 3-3/4 = sin ( φ 3 ) φ 3 = sin -1 (-3/4) = -.848 radians. shifed axis original axis 0 3 (-3/4)A 0 4 where 0 4 = - + 0 3 away from equilibrium jus afer = 0. In our example, he appropriae angular shif is φ 4. The displacemen expression is, herefore: x() = (.6) sin ( 12 + 2.29). FIGURE 11.14b 8.) The echnique for deermining phase shifs is simple. Pu he = 0 value for displacemen ino x() = A sin ( ω + φ ), solve algebraically for φ, and your calculaor will crank ou a number for you. a.) If he calculaor's number is posiive, shif he axis o he righ. If he number is negaive, shif he axis o he lef. b.) The only hing ricky abou he operaion: in almos all cases here will be wo possible axes (i.e., shif angles) ha will correspond o he required = 0 displacemen. Deermine which is appropriae by noing wheher he moion is proceeding away from equilibrium or oward equilibrium. Tha informaion will dicae wheher you can use 348
Ch. 11--Vibraory Moion your calculaor-provided phase shif value or wheher you will have o add or subrac. c.) Whaever he case, you should end up wih an expression ha looks somehing like x() = 2 sin (7.5 + 1.57). E.) Energy in a Vibraing Sysem: 1.) Consider he moion of a mass aached o a vibraing spring: a.) A he exremes, he body's velociy is zero (i's a a urn-around poin), is posiion is a maximum (i.e., x = A), and all he energy in he sysem is poenial energy. Tha is, a he exremes: E oal = U (x max ). b.) The poenial energy funcion for a spring sysem is (1/2)kx 2. This means: E oal = U (x max ) 2 = (1/2)kx max = (1/2)kA 2. c.) Assuming here is no energy loss during he moion, he ampliude of he moion remains consan and he oal energy of he sysem is conserved. The energy flows back and forh beween being poenial and kineic, bu he sum of he wo is always equal o (1/2)kA 2. F.) A summary example: 1.) You have a spring hanging from he ceiling. You know ha if you elongae he spring by 3 meers, i will ake 330 ns of force o hold i a ha elongaed posiion. The spring is hung and a 5 kg mass is aached. The sysem is allowed o reach equilibrium; hen is displaced an addiional 1.5 meers and released. For his sysem, wha is he: a.) Spring consan? b.) Angular frequency? 349
c.) Ampliude? d.) Frequency? e.) Period? f.) Toal energy? g.) Maximum velociy of he mass? h.) Posiion of he mass a maximum velociy? i.) Maximum acceleraion of he mass? j.) Posiion of he mass a maximum acceleraion? k.) General algebraic expression for he posiion of he mass as a funcion of ime, assuming ha a = 0 he body's posiion is locaed a y = -A/4 going away from equilibrium? 2.) Soluions: a.) F/x = 110 n/m; b.) (k/m) 1/2 = 4.7 rad/sec; c.) 1.5 m (from observaion); d.) ω / 2 =.75 hz; e.) 1 / ν = 1.33 sec/cycle; f.) (1/2)kA 2 = 123.75 joules; g.) ωa = 7.05 m/s; h.) a equilibrium posiion; i.) ω 2 A = 33.135 m/s 2 ; j.) a he exremes; k.) eiher x() = 1.5 sin (4.7 + 3.39) or x() = 1.5 sin (4.7-2.89). G.) Anoher Kind of Vibraory Moion--The Pendulum: 1.) Consider a swinging pendulum bob of mass m a he end of a sring of lengh L posiioned a an arbirary angle θ as shown in Figure 11.15. Wha is he sysem's frequency, period of oscillaion, angular frequency, ec.? Poin P a.) We will begin he same way we did wih he spring. If he Newon's Second Law equaion for his siuaion maches he form: 0 L acc. + (consan) disp. = 0, mg FIGURE 11.15 350
Ch. 11--Vibraory Moion we know he moion will be simple harmonic and we know ha he consan will numerically equal he angular frequency squared. b.) The only difference beween his siuaion and he spring siuaion is ha in his case, he pendulum bob is moving in a roaional sense around he sring's poin of aachmen P. The version of N.S.L. ha is applicable here, herefore, is he roaional version. c.) Figure 11.16a shows he f.b.d. geomery free body diagram for he se- Poin P Poin P oward r = L sin 0 up. Figure L 11.16b shows line of T mg ha he orque 0 T abou Poin P 0 due o he ension T is zero (he ension force mg passes hrough mg Poin P), and he orque due o graviy is FIGURE 11.16a FIGURE 11.16b mg (L sin θ ) (in his case, r is L sin θ ). Remembering ha he momen of ineria for a poin mass is I pmass = ml 2, he roaional counerpar o Newon's Second Law yields: Γ p : which implies: -mg (L sin θ) = I α = (ml 2 ) a α + (g/l) sin θ = 0. d.) This is no he form for which we were hoping. Forunaely, if θ is small and measured in radians, sin θ = θ (pu your calculaor in radian mode and see wha sin (.02) is--you should find ha i is.01999999--.02 o a good approximaion). 351
e.) Making he small angle approximaion, we ge: for θ <<: α + (g/l) θ = 0. f.) Running a parallel from our spring experience, we know ha he oscillaion's angular frequency mus be: ω = (g/l) 1/2. g.) Wih he angular frequency ω, we can deermine general algebraic expressions for he moion's frequency ( ω /2 ) and period (1/ ν ). 2.) Reieraion: If you are ever asked o deermine eiher he period or frequency of an exoic oscillaory sysem, use N.S.L. and see if you can pu he resuling equaion of moion ino he form: acc. + (consan)(displ.) = 0. If you can do so, he moion will be simple harmonic in naure and he angular frequency will equal he square roo of he consan. From here you can easily deermine he moion's frequency and/or period. QUESTIONS 11.1) A spring/mass se-up oscillaing in he verical is found o vibrae wih an ampliude of.5 meers and a period of.3 seconds per cycle. If he mass is 1.2 kg, deermine: a.) The frequency of oscillaion; b.) The angular frequency; c.) The spring consan; d.) The maximum velociy (in general, where does his happen); e.) The maximum acceleraion (in general, where does his happen); f.) How much energy is wrapped up in he sysem? 352
Ch. 11--Vibraory Moion 11.2) A.25 kg mass sliding over a fricionless horizonal surface is aached o a spring whose spring consan is 500 n/m. If he spring's maximum velociy is 3 m/s, deermine he moion's: a.) Angular frequency; b.) Frequency; c.) Period; d.) Ampliude; e.) Toal energy; f.) Maximum force applied o he mass. 11.3) A body's moion is characerized by he expression: x() =.7 sin (14 -.35). Deermine he moion's: a.) Ampliude? b.) Angular frequency? c.) Frequency? d.) Posiion a = 3 seconds? e.) Posiion a = 3.4 seconds? f.) Velociy a = 0? g.) Acceleraion a = 0? 11.4) A pendulum consiss of a small, 2 kg weigh aached o a ligh sring of lengh 1.75 meers. The pendulum is se up on a disan plane and se in moion. Doing so, i is observed ha is period is 2 seconds per cycle. Wha is he acceleraion of graviy on he plane? 11.5) The Newon's Second Law equaion shown below came from he analysis of an exoic pendulum sysem oscillaing wih a small angular displacemen. I is: α + (12g/7L) θ = 0 a.) Given he informaion provided above, how can you ell ha he sysem oscillaes wih simple harmonic moion? b.) Wha is he sysem's heoreical frequency of oscillaion if he pendulum lengh is assumed o be 1.3 meers? 353
11.6) A 3 kg block is aached o a verical spring. The spring and mass are allowed o genly elongae unil hey reach equilibrium a disance.7 meers below heir iniial posiion. Once a equilibrium, he sysem is displaced an addiional.4 meers. A sopwach is hen used o rack he posiion of he mass as a funcion of ime. The clock is sared when he mass is a y = -.15 meers (relaive o equilibrium) moving away from equilibrium. Knowing all his, wha is: a.) The spring consan? b.) The oscillaion's angular frequency? c.) The oscillaion's ampliude? d.) The oscillaion's frequency? e.) The period? f.) The energy of he sysem? g.) The maximum velociy of he mass? h.) The posiion when a he maximum velociy? i.) The maximum acceleraion of he mass? j.) The posiion when a he maximum acceleraion? k.) A general algebraic expression for he posiion of he mass as a funcion of ime? 11.7) A unnel is dug hrough he earh from he Norh Pole o he Souh Pole. When done, Jack (he idio) goes for he hrill of his life and jumps ino he hole. The graviaional force on him is always direced oward he earh's cener, so Jack ends up oscillaing back and forh beween he wo poles. In he chaper on Graviaion, we derived an expression for he magniude of he graviaional force acing on a mass a disance r unis from he earh's cener, where r < r e wih r e being he earh's radius. Tailored o our siuaion, ha expression is: F J = Gm m e J 3 r r e where m e and r e are he mass and radius of he earh, respecively, m J is Jack's mass, and r is Jack's posiion along he y-axis (we are assuming he unnel is in he verical). Jack's faher misses him. As such, Papa has hired a surveillance saellie whose orbi is such ha every ime Jack's head emerges momenarily from he hole, he saellie and is cameras are direcly above o snap phoos. a.) For his o work, wha mus he saellie's period be? b.) Given he saellie's period, wha mus is orbial radius and velociy be? (This par is more a graviaion problem han a vibraory moion problem, bu i's good review.) 354