solution1 - A (fl) .aL—W— ((05 km] : ficé {MW} J; M4...

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Unformatted text preview: A (fl) .aL—W— ((05 km] : ficé {MW} J; M4 M £01 ~60” 4 LC :: y... a) "Q, 2/ 2' fit / 41/6516 1/:jiédeCJaj .._. («R i '— L , “C dxéle “(7/ Z/QoLZCr-‘42/Li— Q4 7669(1727 j l 2: 2;” $200164“; W®5bmj:~¥@fighb/ Mr} [V CWJVLAA‘L’fi “(7+ 4L 76; gall/WW Z'dulxm‘fw/L J2 ‘ 1. 2736M“) :"méf’wfi 70; ivfmc'fi‘m/ . 3.6/34777/3— “’é’HWL :~/(7(,3~}c21,/2 )0; z» “74%? “HI/1 PM i (L) of .' ' a; (3.4m TQOSK’K/ : «aim/m rr a» m) m _/ a, da #506; ~ H 2. L 2/ ‘ EDprk’x—L— e / ~— iC mm + [4v(2<»——~?v0/e”°(“ N0 7 «- LI'Y‘fq" COML’"”“I"0VL g1 7M9 JWO/wm’fm w/Wv fan/V.— (fyfnvalwoz LIC S." WAC—(1W KLWUAH (WLI'AWW g [ EffieefiAmf‘fI‘A/L Wer (MMagee/uncv/I‘m; tom!" WWW. 2 . C’Wflflak W 04797:“)567” {EX/:—7C 1: “(623/ (GM :Lcwé'v :[oJéV-za/[y I 8van (Y ) (My? 2 C0§é¢/il‘ £05 [2/ T L 2%“ “W W = :3ch 1:) (MW c van/ a. _o(7c 66/56 2M W IMF! 3M {4762: ~Ly‘4fi/‘7 —_——-— am“ 044, :79 WHOM :9 beach 672 0, .fl ‘ (2,! 56-4731” 1/7/3277 [fitlm/arzd "twp/I] ’00 '. 9° W 2. w 1., 0° 4 L W 56’4fi/fidméz : £24ldbge”?él : (’je $0197 3A2, /W ’- rao 00 9b are, 15000 M 4&4 Va, Wqu/flmez (Jun 1! In“, " "’° / ~ g ’ EV .. ' M , gofrge } (PM? 9; e fdrd Z Palathmbw w fig/«405:9 0 00 @4207 ‘- 2’ Ivar)?“ “f if 1 g /\ M 61/2;— FMW’MJL’ E 3k on f o 3 <7 ,Mfflgdg :A [5:0 c '/ l’blélll PS #1, Problem #3 Let's define some things first. amu :: 1.6605-]0_ 27kg A2: 10— mm . ‘ N I k1 .: 481-n—1 mH.= lamu m0 := 35.53mu mH'mCl — - “1;: ____ p1 =1.615x 10 27kg 13:: 6-10 201 mH+ a) Here are my functions, one with a fixed k, the other sol can vary k. 1 2 Vx:=-—k-x () 2] 1 2 v(x,k :2 —k-x ) 2 -5-10—” 0 5-10—11 As l increase the value for k, the potential gets steeper. That is, the curvature increases. This is easily understood by noting that the curvature (our definition,_ the second derivative) is always going to be twioe the value of k. b) We know from lecture the following relations: x(t) := A- sin(m 't) 13:: —-k-A E2: 6-10_ 2OJ A2: 2-E-k1 A=7.597x 10‘ 9-N 2 PSI C? a) :: -—- co = 5.457 x 1014Hz x(t) :2 A-sin(u)~t) H1 ‘ 1-10—8 510-9 x(t) 0 ~5-10_9 *1 “10—8 - ~ 4 ‘14 ‘14 0 1-10 ‘4 2-10 ‘ 310 4-10 t c) The period is one over the frequency, but we have to take (0 out of radians 1 .. w:l v=8.686x 1013112 T:=— T=1.151X10 143 2-7: v or, converting this to a Unit we think about freq uentiy, the femtosecond f5? 10— 15s 1 =11.513fs . “I 1(k) 2: 21: -—— k T(1.1-k1)=10.977f3 101.9111) = 12.136 fs t(11) 2: 21: £- iii (3;) 1(11-111) = 12.075 fs 1(0.9-p1) = 10.952fs When the spring constant is increased, it pulls harder on the attached mass (as pictured in the reduced mass sense). Therefore, it makes sense that it would turn the mass around faster, giving it a shorter period. When the mass is increased, it's harder to reverse the momentum, giving it a longer period. ' 2 2 I 3 WY) 3: Ekl')’ —y-y 1 :2 _.k . 30/) 2 1 y The graph is now skewed to the positive, but still has an overall parabolic nature (at least over ther range shown) that is a little stretched out. One might guess that the wider potential would give lower frequencies of oscillation as compared to the harmonic potential. 4? A lmwsyv3w/ IMZIO’C’M / C=2'7782<108M5J, hré'éllxlb’}?u.§ fi;él.8fl/WV . —— a "7 n 0 5le Y—%/ [3:11V33/63XID J (EM/07% M / (1:914! Energy : [Mira/$2.5 32497036 .17 D Find/WI; .; 24 MD”) ~7738?.MDL4 0/ 3I/«é32r/077 " 45> MAM”)? :: M’UH g7 A J flack ; low“ / )mqfly 3 fl 2 /'”‘””D ’1 I" V [my flwly fle/ZS'QBK/Dfié‘flmh/ 'VtCS—O')C )e : 4‘85é K/D’/LW WW»! I; “Greta I'S "ll IICCMIL TC ' “ a hob a» ovqu [/3 o 4149/ EZCW’fiz/lgfit Mill/7W5 7L $720 cf warclebflh/ W) W7” 0 7M, (1%; (fl/WW weak/gleam x: m» rm WW“ / ‘ ...
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solution1 - A (fl) .aL—W— ((05 km] : ficé {MW} J; M4...

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