PS 01 Solutions

PS 01 Solutions - Physics 2214 Problem Set#1(Due by 3:00 pm...

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Unformatted text preview: Physics 2214 Problem Set #1 (Due by 3:00 pm, Thursday 9/3/2009) 1. At the right is a plot of F(x) versus x, the force experienced by a particle in the x direction as a function of its position x. (a) At which point or points may the particle undergo simple harmonic motion, when it is initially displaced by a small Ax? Explain briefly. ' (b) Which point or points correspond to an unstable equilibrium position? (c) At which point(s) does the potential energy have a (local) minimum? ‘ o r r ‘y +1— 34 La SH‘M wkenaltik laced-ibyk ‘0.) posf'h‘ow X , l FCx) ust x . mud” pa“ flags“ tare ai‘flh‘i’pmw wt’t‘k a. . M3415“; siege, . y . r . ‘ Paws Te? we saw, lukewarm; ‘ ‘7 l l be Elie—re «WM ‘equzlltbrlcjwf BP‘aflox. aégfic'kbn,whew ‘l'bya objau‘l’ ts dbplaaeéum Elie Puffivc Xfio‘irecfiom , The—Force mos‘l' (can: a- U c. Pc’mfin Tn hag negah‘ue z-Jlru’fion’b pull .‘l't {dc I, . ‘. . 4 ‘. :‘ ‘ M edi‘b'en. CpflVcTJ¢\ when flare .obéc til" H’s uch nu; 9 ~ ' J ls Affilaccd lame, he @VQXsclwe‘c-houfhae—‘F‘Drce, mu: . e. fpovszfiVc $69))“ :L'l'loaglblb ll? eqUt libn‘um 17m: I’h’bw ., . ‘ l; ' belt is! anks‘l‘n NA ums’l’alolc e dil§hnym ‘PM‘L'hQVl... b) Pa;fl§‘b?cé {:3 efo. a‘fD’S‘o flag—ti: eqqui‘ltbtguw 96mth L 501' 15} Theelojcd’k. ATs‘pVLLcJ. \Vt'hme o; tfiVfbK-Atménbn.) '11.; L{Aimee is positive «MA.» ~\\\ pull 'hme Olod’eb mun?! 53m!“ in lllelOM. infleobded’ ,Cs.‘ 77(9l4LC‘K1 flead. ue_ ix 5;? Mcfiou :mefiwe. is he “by; ayLAwI “la-564W pu ,‘\‘:'\ae_ 0333* gamer: QWc-ZfS'YDm usilibnom . : 7 p V ' {gal M‘xmmcmeccurs F;F(x)=Q. Anal. , Q ‘wgaebpe, 55:} F(x) as. x is “esoth‘ve..- occur; at, Spams: B MAG? :' ' ‘ ‘ ‘ * i _ ‘ a , . 2. Two objects, 1 and 2, oscillate with the same frequency but with different amplitudes, phase ‘ constants, and equilibrium points. (a) Determine the values of all constants in the equations mm = xeql + A1 cos (wt + 4901) and x20) = xeqz + A2 cos (mt + (1)02) from the graphs of x1(t) and x20) shown below. , \ , -——x1 (mm) 'A ’ ‘ ‘ i, I ,l \‘ 'J ‘ - -xz I ' . ‘ 1% mp" eel/ow +11. 11¢: qmbmm “vglgfigxféii: x; a on gift/315% ThaTjrchr‘cha-rl. ‘ ’ ,i‘s ‘ . Bfimwcx (Mammy q} minimum) of? ¢Cblv¢> s‘i'nosmal. , , ,xe'qtéiiiLOW‘WMm ;‘§ W . ._; _ l lampl'x‘l‘vagrpr', , BT14; vie—r‘h'veaal Aklavi 4e The equill brim undue, «FM 1‘» “Ne, Maximumgvdue, «Flue, stuuuio‘: I ; J a f lo5d'Wlll’Vl1 i ‘ ‘ J JT (P .Th 0- Ul r—F I inc 1» is ‘Foovid $1»ch pen‘a ) ) a m: siéusiiafwei X? 1'? , , t . ; i ‘ 1 Majx'r; , where T, 359m: Mbrfzouhl, armm bell-ween aAgaaén‘l—yedb: ' A l.:®"$ ’ ’ ‘ ‘ i ' l ' ‘ ' -—-t- ’ ‘ 'I ' » s 1 z i v - ' \MQS; Weaha‘ue‘, var-=— :-:., Trg": VGA/S (sh-9,5: SMEH {Quay Wclwquflkflz‘rt C's respee‘li'ho‘lke. 130 was: ‘— (Doll ‘3. ' T a i“; $9M . gmmfle‘ All“? I“ +OMACQ1‘59‘IMQ r3010~x|‘s)lw..w as 0.85s é;;:=a« :wmrwwli? by "as: mm; 5:29 2w avg-ewe m 1 \l ' I"l l l : i? ‘s : i0 , 9s @nhwucol) 1 / ‘_ v w i ‘ iii—7% I (b) Find the maximum velocity and maximum acceleration of object 1_ 1» (c) Show that the total mechanical energym—the sum of kinetic and potential energies—is constant for either object (d) Object 2 leads object l—that is, 262 reaches its maximum (or zero crossing or minimum) before x1 does by a time interval At. What is the numerical value of At? by’lMequafion Objev’l'l it‘s; _ ‘ I . w ' . ‘Xlfl’exaél,+A.:[email protected]+QJ ‘ J j ‘ L ' i . 1 w . ‘ Walmfimeoosf' VJ???) IéflLfiWeJériVo-five GIC'XJfi: , V‘Ci’B := "A3 alnCLdf'l'CDo.) 1 ' 1 “Coma-)4quM oelo 4'57 odors whenrfl’he'sinie Moran, F: ,1 vmwli‘bfll‘) ‘ Sulajfi'fi’fims valueS X’Y’om pax‘l’Ca) awe-St me =Q-S'mxawwlé3i 9.4-.ng V 1 \kflvd’éme 00;: agelerfi-‘Hom ‘1‘: The ‘Gécc‘t/Lésfivue Aeriwa'five ‘9‘L‘ . it ,‘ r ' o.(;t)i= 3+5 w- A; mam—ram 3v i The—.mOrx'imqm “gamma” :7; Was Qmui 'rAsieri ‘ a i a l i Sob'b'h’fims value-s ’Fbr A! ‘9, e0 par'l’cat) gives ‘ ‘ amm>Q.5fmmtaa-rd/g>2*=45amm/sz .’ , 1 , w ‘ i’l'b’l’B-l \mwticN Vi ‘ ‘ 1-90 ‘eéf’l a‘g’l’he. l‘scw‘ C) 33M: k? “33¢: anal Fifeihgigémaaiéls. . ‘ . , Elba-Kgth-UJQ, , ‘ Z l i : {Mkaflzl’ lika(fl"X'tlj 3 Here/bis wasp‘rmj Lomxl’avil", (Elam ‘h: “We, angular t'mqucnc)’ w b7 i ., Loan: . V We usa’hxc express id as &r x’ (f) anal, VJi’) me pail—00> (33 , fix.le 9v @wfiWJed> Fm pafibo) x',(:l:)=xeq\[email protected] 1 ' d“j QimfirfiArMJSlKQ'z‘wm) I V - , E , (I) =3 markups 9 i m [gm] 5%} Lx.t¢)~.xq\ ea]2~ w ‘ =12 Ml} AM 8""M‘t‘1’oblz’rik LA. meme] Smackéw‘m .t ' .‘ Em waisremoa«owed , I i : iAjL-um . , a W“ “‘1”th “M Mien «- 1qu at; (Jl’aloc: ‘ 1 nor am «mt-,1 :5 Wait. ~ ‘ 0‘ d) At t; We, ’HMe ih‘l'efoéJ waeén cm, yeah 6PcarueJ H . anA‘l’ke. chasm Peak. 0‘?COC’.U€ 5L (*0 rial/Fifi. 1 Re¢§1wsfflbm MaharaPh wietsfind V , i .' Alt; lip-“IE. 2048b: ~0.l’3.s : Q73»; 2— At 1(- Mailed ’\‘o the; til/mic Algae/tantra ACDEIflQfQZW , wt. + we ‘ * . ‘wtm our Mala" Albinfibmfle \ J (e) Find the relationship between At and the phase difference A4) = ¢02 _ 4101, Write this relationship in a general way (an algebraic expression involving At , A4), and other ConstantS); then check that it holds for the numerical values in this specific case. Q0 3. An object is attached to the end of a massless, nonlinear spring. The elastic potential energy as a function of the position x of the object is plotted below in black (solid line). For I V ' oscillations with small amplitude, the parabolic curve in red (dashed line) shows the E (approximate) potential energy curve that gives harmonic oscillations (but the curve is ‘ extended out to larger amplitudes, where it is not such a good approximation, so you can see I how it differs from the exact potential energy). (a) Estimate the numerical value of the spring constant k for small oscillations. (b) If the object has mass 0.015 kg, what is the angular frequency of small-amplitude oscillations? (c) Is the frequency of oscillation for larger amplitudes greater than, equal to, or less than the frequency for smaller amplitude, harmonic oscillations? Explain briefly. Potential Energy Exact versus Approximate 6.0 4.0 Potential energy (joules) 2‘0 0.0 xtm) l 3. (a) The parabolic approximation of the potential energy curve (colored red) has the form Upma) : Upgrapceq) + 1/2 k (x - xeqfl, (I) where k is the spring constant, xeq: 2.0 m is the equilibrium position of the object, and Upmmq) = -2.0 J is the value of the potential energy at equilibrium. Solving Equation 1 for the spring constant k gives )- Measuring from the graph, we see that at x m 3.0 m, Upa,a(x): 1.0 J. Substituting the above values into Equation (2), we obtain a value for the spring constant k : 1.0J—-2.0J k k :2(——L—é(3_0m_2.0m)) — 6.0N/m . Since the potential energy curve of the system matches the parabolic (red) curve for small oscillations, this is also the value of k for small oscillations of the actual system. L5”) - o I . Z (0 NT ‘ Problem :15 Cbfi'fn «"54 i . b) The, MSUlar' 'F‘mqoency no. 3‘; given. [07 i “ii. 2:6‘6% . ' w'): ~\i0.0\5’l¢g “a0 é E2 The restoring force for oscillations, F (x), is given by the negative gradient of the potential energy (PE) curve U(x): dU 17x) : ' dx For any given value of displacement x, the non-linear PE curve, U(x), has a steeper slope than the parabolic curve, Upampc). Therefore the restoring force for the non-linear oscillations is larger than for the parabolic potential, and the object on the non-linear spring will have a larger acceleration towards the equilibrium point. The non~linear oscillations must therefore have a shorter time period, and a higher frequency, fl than the linear oscillations. How does the frequency, f, of the non-linear system depend on amplitude? We know that for the linear system, the frequency, fix”, is independent of amplitude, and that for small oscillations, 4. Gas and oil pipelines must be designedito withstand earthquakes. The web site http://tapseisanl.gov/documents/docs/feis/volume7/3.4-2.pdf shows a seismic hazard map of the route of the Trans-Alaska Pipeline. The colors indicate the maximum horizontal ground acceleration in % of g (z 10 111/52) expected with at least 2% probability in 50 years (i.e., there is a l in 50 chance of an earthquake producmg an acceleration greater than the indicated value in 50 years.) (a) What is the largest expected ground acceleration along the pipeline's route? (b) If an earthquake produces an acceleration with the amplitude found in part (a) and has a frequency of 0.25 Hz, what is the amplitude of the back-and-forth motlon of the ground on which the pipeline rests? ' l T .thl Acadia—Malian is 15107., ,or _ od—G—Qf-S an map. V i a ’2. i V . g‘ Z i '2 b)§ft:1A,w cosQaSC) 1‘" 5 fiulhxiAlw i ‘ SO‘UU‘Q 89" A} . Atziigglw can) .. [-1 i _. . “Liar (0.15211.‘ Math Review First let's separate in) into its real and imaginary parts: £0) = Ae‘¢e‘”” =Ae"(“’”¢) :A cos(a)t + (15) + iA sin(a)t + (,5) f) Ream} m Re {A cosrcot + ¢) + iA smrcor + ¢) } = - to) (Art)! : Mencer + [Imem’ = WA cosrwt + ¢)]2 + [A smrwt + «52f = . , a 7‘) (a) Re{%} =Re{%[/1 cos(a)t + ¢) + iAsin(a)t+ {w} =Re { (-Awsin(wt+ ¢) + iAmcos(a)t + ¢)} = . d (b) :1?(Re{l(t)})= 6111029 { A cosmt + A + M mm“ W} ) = 2%“ 00W t W =. 7 SO we find that”"’\———k—A AA . a, ; r ‘ dx d 1 Re {:17} :g/Reea») This is not a coincidence - it doesn't matter whether you take the real part before or after you take the time derivative. 8) is a. aolgfifiovi +0 Apfilf-‘cfieufid aqua-H6“ kgg. {-mgmé =0 ’fialf‘a. ccr’iain value; (filmy m . To Ruck ’hvta deviate We so-ksfi‘i’d'i‘ wt): Aeie eiwt' j y i k x09: kA‘ei‘e e‘w’tawa img—Ejfi a —- m Audie-zeal”: AMtufi we get” Aeieeiwx Ck~quj flu. Wm be saw-sens“ an waves Germ two“ _ _v k A “ ~ “*5 ' e)“ ,_ - 1 16(1) : Ae‘m + ‘7’) follows a counterclockwise W , -— ' circular trajectory in the complex plane. The 7 t » center of the circle is at the origin, and the radius of the circle is A. The angular frequency of the circular motion is a). At a time t, the vector from the origin to the point gcft) makes an angle mt + (I) with the real axis. \\‘_—— ‘0) Let M) = a + ib :Aeig Then A ='\/a:+ b}, I and tan 6 = b/a. ' A ? So wehave ‘7 + lb 2 a + b exp i[tan'](b/a)] Re {2} H) LetfiiRBe’fl and§=Rceiz ' Then lBl/IQJ = C -‘ k Nowlet's evaluate [3/le R e'fl B / : _L_ I- g RM 21(RyRc)eW;= 1%. 1 )9.) COS t9 = [(cos 6+ ism 6) + (cos 6 - ism 6)] = (68+ e"0)/2 _ 1 sm 6 2 [(cos 6+ ism 6) - (cos 6 - mm 6)] = (eig- e‘i0)/2i. ‘ (‘5) ...
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This note was uploaded on 01/01/2010 for the course PHYS 2214 taught by Professor Giambattista,a during the Fall '07 term at Cornell.

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PS 01 Solutions - Physics 2214 Problem Set#1(Due by 3:00 pm...

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