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

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