Fall 04 ME 39 Mechanical Engineering Analsis Eam # Soluions Direcions: Open noes (including course web posings). No books, compuers, or phones. An calculaor is fair game. Problem Deermine he posiion of he mass as a funcion of ime for a spring-mass ssem wih he following properies: mass, M = kg viscous damping, C = 3 kg/s spring modulus, k = N/m forcing funcion, F() = 4e - N d iniial condiions: (=0) =0 m and 0. m/s 0 Soluion: We begin b wriing our general for a mass-spring ssem ODE d d M C k F() Seing F() and rearranging d C d k 4 - e M M M where a = C/M = 3/= 3 and b = k/m = / = and 4/M = 4/ = 4 This is a nd order, liner, consan coefficien, nonhomogeneous ODE. We know ha he soluion is composed of a homogenous par and a paricular par or () = h + p For he homogenous soluion we will use he e m mehod and for he paricular soluion we use he mehod of undeermined coefficiens. Beginning wih he homogenous soluion. Then we can follow our sep b sep procedure. Sep Saisfied Sep Sep 3 m + 3m + = 0 m (9) 4() m (9) 4()
ME 39 Mechanical Engineering Analsis Fall 04 So ha a 4b > 0 Sep 4 () C e h C e Sep 5 No applicable Sep 6 No applicable Sep 7 Skip unil we add he paricular soluion Now we appl he mehod of undeermined coefficiens Sep Since our forcing funcion is 4e -, we would like o assume he following funcion for our paricular soluion p () = Ae - The problem is ha his is he same form as one of our homogeneous soluions, so we mus modif his o p () = Ae - Sep Subsiuing ino our differenial equaion 4( )Ae 3A( )e Ae 4e Sep 3 Collecing like erms of e - : -4A +3A = 4 e - : 4A -6A + A = 0 Sep 4 Solving for he undeermined coefficiens A = -4 Sep 5 Assemble ou oal soluion () Ce Ce 4e
ME 39 Mechanical Engineering Analsis Fall 04 Sep 6 Appl our condiions ( 0) 0 C C C = -C d C e Ce 4( ) d 0. C C 4 0 Subsiuing for C 4. C C C = 4. C = -4. Our final soluion becomes () 4.e 4.e 4e e Problem Two wells are drilled for an oil field. One well pumps a a rae given b 0. m 5e ou, while he oher well pumps a a rae of m ou, m where m is he mass of oil in he field. If a = he mass in he oil field is M, deermine he ime variaion of mass in he oil field, m(). Hin: Remember our form o inegrae b pars is u ()v ()d u()v() u()v() d Soluion: We recognize his as a conservaion of mass problem, so our general equaion is: dm m in - m ou inflows ouflows We have no inflows and wo ouflows, so his reduces o dm 0. - m - m ou, ou, 5e m Rearranging dm 0. m 5e We noe ha i is no separable, here is no variable ransformaion, bu since i is linear so we can use he Inegraing Facor Mehod. Hence we need o find F and G, so we ma ransform he equaion o he form d F()m G() 3
ME 39 Mechanical Engineering Analsis Fall 04 where F ep ep ln() Then 0. 0. G F 5e 5e Then our equaion becomes d 0. m 5e which is in separae and inegrae form. Separaing 0. m 5e Inegraing m 5 e 0. 0. e 5 0. C 0. e C 0. 0. 0. 5e 5e C Solving for m 0. 5 0. C m 5e e Appling he condiion 0. 0. M 5e 5e C Solving for C 0. C M 50e Then our final soluion becomes 0. 5 0. M 50e m 5e e 0. Problem 3 Obain he soluion o d u du du 4 8u 0 wih u(=0) = 0 and d d d 0 Soluion: We begin b rearranging he equaion d u du 7 4u 0 d d This is a linear, nd order, consan coefficien, homogeneous ODE, so we can use he e m mehod. Following our sep b sep procedure: Sep Saisfied wih a = 7 and b = 4 4
ME 39 Mechanical Engineering Analsis Fall 04 Sep Our m equaion becomes m 7m 4 0 Sep 3 Solving for he roos of m. 7 m m 7 wih a 4b < 0 Sep 4 No applicable (7) (7) 4(4) 3.5 4(4) 3.5 7 7 Sep 5 For a 4b < 0, our general soluion becomes 3.5 u() e Acos.33 Bsin.33 Sep 6 No applicable Sep 7 Appl he iniial condiions o obain he arbirar consans. 3.5(0) u( 0) 0 e Acos.33(0) Bsin.33(0) A = 0 du 3.5 e.33bcos.33 3.5e 3.5 Bsin.33 d du 3.5(0) 3.5(0 e.33bcos.33(0) 3.5e ) Bsin.33(0) d 0 B = /.33 = 0.756 Our final soluion is hen 3.5 u() 0.756e sin.33 Problem 4 Given A (a, 8) and B (6,3) Deermine a so ha hese vecors are perpendicular o each oher. Soluion: The definiion of he do produc in dimensions gives A B a b a b Since A and B are o be perpendicular we mus have A B a b a b 0 5
ME 39 Mechanical Engineering Analsis Fall 04 Solving for b a b a b Subsiuing ( 8)(3) a 4 6 6