# hw2sol - HW2 solution 1 3 Einitial = Efinal(4 gnu = mghf(5...

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Unformatted text preview: HW2 solution January 23, 2008 1 3) Einitial = Efinal (4) gnu)? = mghf (5) From the ﬁgure, hf = L(1 — cos(0maz) 1 2 im’ui = mgL(1 — 008(9mam) (6) v? cos(0mam) = —29L + 1 (7) 9mm = = 7.5° (8) 13) mat = —mg sin(6) (9) (12:17 . W — —g sm(0) dza: . Lat—2- : ~g sm(6) m» C) Given conditions are, When we compare sin(0m,n) with 0mm. sin(0maz) z (9me (11) sin(0.13) — 0.13 = 0.000366 (12) [sin(0.13) — 0.131 m : 2K9 (2) The difference is only about 0.2%. Thus, we can use vinitial = 0-5m/8 V (3) the small angle approximation. given cositions are, m = 2K9 k = 500N/m a = g/3 a) At equilibrium, F=—mg+ka:=0 mg=kx _B‘l 3' k = b) In this case, the net force is not zero. F=ma=—mg+kx 9 mg 2 —mg + km 4 5mg = ks: 4mg 3 — 3k = (14) (15) (16) (17) (18) (19) (20) (21) (22) (23) (24) (25) C) The solution of this harmonic oscillator is, a:(t) = Asin(27rft + (15) = Acos(27rft + a5) (26) The amplitude, A, is (24) - (19). _ 4mg ﬂ ’4‘ 3]: k (27) _ ﬂ _ 3k (28) = 0.013 m (29) The frequency is, Jag 53: (so) We can get the phase from the initial condition. 23(0) : Asin(¢) (31) = A 008W) (32) sin(¢) = 1 (33) ¢ = 921’ (34) 008W) = (35) ¢' =IEJ (36) You can use either sin or cos. 3 In this problem, the value does not have to be the same as the values in this solution. As long as, you got the right forms of equations and show how you ﬁnd all the variables, you will get credit. a) For this part, you can use either ﬁgure 1 or ﬁgure 2. In this solution, I used the second plots because it is easier to ﬁnd the phases from the plots on the ﬁgure 2. Figure 1: Figure 1 Figure 2: ﬁgure 2 3.0.1 Body Temperature(Tb) If we can approximate the time variation of the body temperature of the rat, Tb, by a harmonic oscillation, Tb will satisfy the simple harmonic equation, clsz dtz 27r = —w2Tb, ,where w = — T (37) Here T is the period. The general solution of this equation is, mt) = no sin(%_7—rt + (to) + c (38) - From the plot, T m 1 day Tw m (37.5 — 36.5)/2 2 05°C C m (37.5 + 36.5)/2 = 370°C At t = 0 day, Tb(0) z 36.75°C'. From the (38), Sin(¢o) = (36.75 -— = —-0.5 7r ¢:_6 Thus, the solution is, Tb(t) = 0.5°Csin(27rt — 675) + 37°C (39) 3.0.2 Heat Production(Hpmd) This is similar to the ﬁrst part. Hprod satisﬁes the equation, 277 T JVII' : —w2Hp,.od, ,where w = The general solution of this equation is, 21r 7;)1'011 Hprod(t) : Hprodo Sin( t + (bprod) + Cprod (41) From the plot, Tpmd z lday Hprodo z (2.3 — = 0.4W 0pm, 3 (2.3 + 1.5) /2 = 1.9W At t = 0 day, Hpmd z 1.9. From the (41), sin(¢o) z (1.9 — Cprod)/Hprodo = 0 as m 0 Thus, the solution is, Hprod(t) = 0.5W X sin(27rt) + 1.9W (42) 3.0.3 Heat loss (H1033) 27r H1053“) : HlossO t + duo”) + Class (43) From the plot, 7:088 z Hlosao z (0.8 — 0.4)/2 2 0.2W 010,, z (0.8 + 0.4)/2 = 0.6W At t = 0 day, H10“ z 0.5. From the (43), Sin<¢0) % (0'5 — Closs)/HlosaO = —0.5 7r (1’ ~ “5 Thus, the solution is, Hloss(t) = 0.2W x sin(21rt — g) + 0.6W (44) b) The phase difference between Hprod and H10” is sim— ply, <45) 4 Enefgy Plot Figure 3: ﬁgure 3 Here the equation of each plot is, K E = 2(cos(t))2 (46) PE 2 2(sin(t))2 (47) Etotal = 2(COS(t))2 + 2(Sin(?5))2 (48) a) As we see from the ﬁgure above, the Total Energy plot is a ﬂat line. b) See the ﬁgure 3. C) From the graph, the period of the KE and PE oscillation is about 11'. However, the harmonic oscillation period is 27r from the eq. (46).(period of cos(t) is 21r.) The period of (sin(a:1:))2 is half of the period of sin(aa:). 5 See the ﬁgure in the problem #1 a) K = ém‘g)? (49) 1 d0 = EMU/EV (50) b) plug (61) and (62) into (59) U = Mgh (52) = MgL(1 — cos(9)) (53) A08~<ﬁ5¢mnmL i ( /(.§_ 2 _ 1p Then, Total Energy is (51) + (53). 2M 2M M E = lML2(—(£—0-)2 + M L(1 — cos(6)) = const — I‘{L :l: — + k] (63) 2 dt 9 ' ) 2M V 2M M (54 2 c) = Ace-(ﬁaiv<ﬁw)2—ﬁ>t[z% :l: HHﬁ-V — ﬁ 2 2 KE =PEa (55) L_ _E_ ‘/_F_2_£ 1 d0 max ma: +4M k 2M3FP (2M M+k] §ML2(~— 2 = MgL(1 — cos(6mam)) (56) dt man: (533%: = sqrt(21g)(1 — cos(9maz)) (57) z 0 (65) We can ﬁnd the maxinum by taking derivative of 32,-?)me with respect to 6mm. mac: _ (£3)Sin(0max) dam ' sqrt(2,g)(1 — comm» (58) «Law “LL” 2 0 when gm” = 7r. This shows that Let the solution of (59) be, d(%)ma¢ is maximum when 0mm = 1r. 6 x(t) = Age“??? sin(w1t + <15) (66) k F2 dzx day: an = —- — -— (67) __ _ = 4M 3 Mdtz +th+km 0 (59) m a) We will verify that (67) is the solution of (59). Let the solution of (59) be, me) = Ace'<ﬁw*¢<rh>”-n‘ >t (so) 5‘: = Ace—56: ( — L smelt + ¢) We will verify that (60) is the solution of (59). dt 21‘; + wl cos(w1t + (15) (68) ix. 2 _A0(__F_ i “(L 2 ~ £)e-(§‘hiV(§%)2—ME )t dt 2M 2M M (61) (12\$ _ = 435 _ .12. dzz AO( F :l: (/( 1‘ 2 k )2e—<-5e,i(/(mf Tag): «it? A” ( MWICOSW1H a» .__.. _—_— _ _._..._ _ _. 2 1-. dt2 2M 2M M (62) + “my _ w?) sin(w1t + ¢)) (69) Plug (69) and (70) into (59). b) Aoe_ ﬁzz <M( _ Lwl cos<w1t + 45) Let the solution of (75) be, M a:(t) = 1405* 81110“ t + 05) (81) + ((5177):?) Sin<w1t+¢>l Then, 1 F _ _ . t t k ' t + ( 2M 81n(w1 + ¢) + M cos(w1 + ‘1») + 81110“ EqﬁDAoe—Et? ( _ i Sin(wlt + ‘75) + “)1 0050017? + (M) plug (68) into the equaion dza: t W1 —-—::Ae”F ——cos(wt+¢) -re k F2 k I‘2 2 0 ( T 1 2A“ m [(‘PV E " 4M2 +F “n? _ 4M2)°°s(”\$t+_f) 1 k F2 2 — w?) sin(w1t + 45)) (83) 2 . + (_ 3134— + k 7 M ( E _ 4M2) )S‘“(“’1t + w] Plug (85) and (86) into (75). (71) t (4)1 Age—2‘7 cos(w1t + <15) = 0 (72) + {9217? — wf} sin(w1t + 45)} l 7 + I‘{~§T— sin(w1t + <23) + an cos(w1t + ¢)} + k s' t + 84 Ma(t) + 1w) + mt) = 0 (73) mm ‘m ( ) a) _ c Mwl Let the solution of (75) be, = A06 ’7 H- T + F011} 005(W1t + <15) hV—l x(t) = Age—'0‘ (74) a M F . Then, + { (m2 — w'fM — 5; + k} s1n(w1t + ¢)] (85) due _ a? = —CAoe 0‘ (75) 0 b ( ) dz‘” 2 -—Ct = 86 W = 0 A06 (76) In order for the equation to be zero plug (77) and (78) into (75) for any time, t, a must be 0 and b must be 0. Age—C‘[MC'2 — PC + k] = 0 {.MT“’1 + pm} : 0 (87) Since Aoe’Ct aé 0, constants in the brakets must be zero. (77> —+ <88) 4M0? — PC + k = 0 (78) plug (91) into b 1‘ :l: x/I‘i — 4kM r2 2 1“2 —-——2‘1W—‘-— (79) m—wlM—W—l—k—O (89) = fat (2;)2—5 <80) w= %-\$ (90) 8 given conditions are, m = 0.375Kg (91) k = 10N/m (92) 1" = 0.1 N sec/m (93) a) If I‘ > v4kM, the oscillation is overdamped. On the other hand, if I‘ < V4kM, the oscillation is under— damped. v4kM = V4 x 10 x 0.375 N sec/m (94) = 3.87 N sec/m (95) Thus, v4kM > I‘ (96) This is underdamped oscillation. 13) Then, the solution is given by (67). a:(t) = Aw‘ille? sin(w1t + <15) (97) k F2 (4)1 = T—n- — m—z The amplitude of this solution is, A(t) = Age-2% (99) When A(t) = %A(0), t value is, \$200) = 21(1) (100) \$140 = Ace-2% (101) :- = e— n (102) 111%) = _§FA14 (103) t = EPA-{1:12 (104) = (105) when E(t) = 1/2 E(O), Time, t’ is, \$1110) = E(t’) 1 I 21M13, = gage-51?:- 1 _ I‘t’ I‘t’ _ 2 7 = ——-- 2 e 111(2) M t’ = gln2 = 9 Given conditions are, k =6.4N/m f0 =1.7N m =0.15Kg Am) =O.44m (106) (107) (108) (109) (110) (111) (112) (113) (114) From the equation of motion of driven oscillator, F =ma = —kx + f0 sin(Qt) m— + kit = f0 sin(Qt) andx(t) 2 A(Q) sin(Qt) plug a:(t) into the equation (115) (110) ( — mA(ﬂ)f22 + 1114(9) — f0) sin(Qt) = 0 .9 — mA(Q)92 + kA(Q) — f0 = 0 _ fo/m 14(9) —|k/m - 92) _ fo/m —lW2 — ml When (1)2 > 92, _ f0 92 “ _mA(Q) “’2 = 16.9 ra,di:e1.112/sec2 Q (117) (118) (119) (120) (121) (122) as f <— n1A(f» ' w — 68.4 radiang/sec2 (I — g = HZ \*-v/ (124) (125) ...
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