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Chapter 2 Free Vibration of Single Degree of Freedom Systems -3 Bs = 5x% m st O v = [£ sá = pes ye = 442945 rad/sec = (E Si = 7:0497 Hy Taz 02] sec = 27 JE o Sma ca vF/27 ; - Tr o21 ICE (1) (Ci) new - 27/m = 27 dm - 2m — ) TIS sec (es 1.5 de E = 0» - (61) (Creu - am Jm cao! = 02970 sec. = 207 Lam [knew Vo5% ( 27 No5k (3) Wp = 62.832 rad/sec = JE V E = VE/c2. 832 when spring constant is reduced, «9, decreases. (Sm) new = 0:55 (9, = 34.5576 rad/sec = fere, - [t=200 nem [t- 200 *62:9I6 = 34:5576 [e-so0 = 0.55 2 tdos = (0:55) =0-3025 K= 1469534 Nfm 2 + dm = Jk/ca. 232 ; m= 4/ casar = 46: 9534 3947-8602 m = 02905 kg K= 00/1053 = 0000 N&n Up = Je = Je = (408)? % 10 +eg to = 63.2456 rad/sec To = ZE = 6.282 = 0:0993 sec too N Sn O ga2456 * 0089 55 Unit weight: 9000 N Natural frequency 5 rad/s - 10 rad/s Solution: 9000 m= 9.8 Let q, = 7.5 radísec m 90! kg= mg? = | 00 (7.52 = 51658.2 Nm = 4 k where k is the stifíness of the air spring. Thus k = St6562 , 12914.5 Nim x=Acos(Wt-4) , w=-0,A sn(MGt-d), x = -UPIA cos (Ut —4,) em E) uA=o1 me : T= q=2s% = 31416 rad/sec 0:4/69, = 0:03183 m d) x==(t=0)= A cs(-&) = 002 m costg) = o.02 = 0.G283 á = 51. o724º 6) % = A = k(t=0) = -0,A m(-g)= 04 sin(-sior2gs = 007779 m/sec (e) Xmas = cor A = (31416) (003123) = 0:314151 m/sed 56 Cart weight: 20 EN Steel Young's modulus E = 200 GPa N steel wire rope 1 mm diameter Solution: AE =(0.0012 (200 x 10º k=> = à (0.001) (200 x 10) = 17453.3 Nim 1 ti 9 A, E k= 7 = raso tro (5 - 145444 Nim 2 kq= +, = 31997.7 Nm Let x be measured from the unstretched length of the springs. The equation of motion is: mé=(,+k)(x+8)+WsnO where (k,+k)8!=Wsin 6 Le, mk+(k , k)x=0 Thus the natural frequency of vibration of the cart is given by 31997.7 x 9.8 o = (Bitk festa xos = 3.9597 rad/s n m 20 x 10 Weight of electronic chassis = 500 N. To be able to use the unit in a vibratory environment with a frequency range of O - 5 Hz, its natural frequency must be away from the frequency of the environment. Let the natural frequency be w = 10 Hz == 62.832 rad/sec. Since ka = a = 62.832 “ta m we have 58 kq=m ui = [e 500 | (62.832)? = 20.1857 (10!) N/m = 4 k 9.81 so that k = spring constant of each spring = 50,484.25 N/m. For a helical spring, 4 k= Gd' : 8nD Assuming the material of springs as steel with G = 80 (10º) Pa, n = 5 and d = 0,005 m, we find k = 50,464.95 = 8 (5) Dê This gives -3 D'= pad = 24,770.0 (1079) or D = 0,0291492 m = 2.91492 em (x) Ci) with springs & and: f—==— E | Let Yo. EN dp be deflections Lo q —) Pires. of beam at distanees w,b,? b frem fixed end. 2 16d, m= tm mta di : 2 dp 2 ve Ge (Bjo al) z A— *— 4 | q Er Ot ú Edo TS te e x= &s 2, = CEI (38- 2) > = = Eb o O x= bo H = SET (34-63 É x= t, H= Ep 3EI Vá tr ks (De + tam Go) a EI W = ' ( = ) 22 E where anã PO a(327) a Gl-a)+ a. (41) 4º (32-43º J/ E Eres Gl-a) + tea bt (3 0-6)? + 2 EI + (ii) without Springs +, and tea: +, 3EL On = JEr= - JEF 633) (a) = [4%/M (db) az /4/Cn+m) Initial conditions: º velocity of falling mass m = Va 23 pd (1 va W= 222) x=0 ot «tatic eguilibrium position . %o= alt=o)=. UERÊ = ão Conservation of momentum: (nem) bo =mv=m 298 dos 4 (rn) = To fagl +m complete solution: x(t)a Ag sin(t+ É) where Ases por (Ee - migr N mê 24 1 pasa 6 nz 24(M+m) and p= tus! (Ser (ões) Velocity of hammer 20 m/s (tas) Weight of hammer 50N Weight of anvil 500N Stiffness k = 20 EN/m Solution: (a) Velocity of hammer = v = 20 m/s, x = O at static equilibrium position. weight mg x =x(t=0=- o==(t=0) Ka 4k . . . my Conservation of momentum: (M + m)x,= mv, or X,= *(t=0)= M+m 4k Natural frequency: w, = q Ya M+m oq Complete solution: x(t) = Ag sim (w,t + 09) where E 1 E 1 252 2.2 a [oro | que, mil º n 16k (M+m)4k and egNM +m az -EE] O vv4k since v = 20 m/s, M = 500 kg, m= 500 kg, k=20 k'm,m= So kg, k = 20 EN'm 9.8 92.8 : 9.8 so» Tt [8 | 9.8 oome So=l4 (20000) (9.8) 9.8 ) 550 (4) (20000) | m 9.8 550 do = ra = tan-1(-0.01298) = —0.7436 deg o 9.8 (20) /4(20000) x=0 at static equilibrium position: x, = x(t=0) = 0. Conservation of momentum gives: Mxo=mvorx=x(t=0)= WE xt=0)=0 Complete solution: x(t) =A, sin (ot + 05) where ao das(a | Es om mv 50 (20) 9.8 o mar | V4kM 9.84 (20000) (500) n = 0.0505(m) and (E) do= tan[ 6 "| =tanl(0) =0 62 Ga) + =2fsme Neglect masses of finks. 48º pf a(ticateio imo) €) teg= + ($EE )= + arsnão = cos" B = (sãs ) Wa, = [tes = d&g tosec* From solution m Ww ( of£ problem 4. 6) 6) w,= Es since %eç= te. (23 y=Vê -(esinb-x) —Ccos6 =Vêco!0-x 42€xsinb —Lcos8 2 2€xsn6 =t cos 8 Ep 2lXMO peosê cos AVA E cos? 8 + € cost 8 sos 1 1 1 qiateçartçhs where 1 X 1 2€xsin8 saem 2 Feof5 4 4g08n) cem? xsin É a =x tan É cos 8 Thus ke can be expressed as eq = (ky + kz) tan? 8 Equation of motion: mX+kqx=0 E VACI V ACEITE , AT is 6-*) E: Natural frequency: 64 (3) (a) Neglect masses of rigid links. Let x = displacement of W. Springs are in series. 7 Equation of motion: Natual frequency: = *eg k “a = no 2m (b) Under a displacement of x of mass, each spring will be compressed by an an amount: Equation of motion: Natural frequency: um = A=fh= ax cos4s” Fa Fi + Fr= = dx cos 135º , uz E 4so F= force along x = F cos 45º + E 4135" tes vs * + cos45º + E, costas" F; sm = 2% (4, cost4sº + xy costt35') F tg = 2(M+ &). z t+ kz Eguation 0f motion: mX+(kitk)x=0 65 In the present example, (E5) and (E4) become k) COS GO! + 4; cos 240" + kg COS DA; + Ky COS 420! + Kp cof GOO” + K cos (360 4 203) = O k, Sin 60! + 2 Sin 240 + Hg sin das 4 fy Sin 420º + fg Sin coo” + kasin(Hd+ 244) = O bes 4 - ta + 2% cos zw; =0 . 243 cos 20403 = ka fer... (Es) ST ts VTka +2 ks sin 204; = O , Co Big sin 204; = sT (a = tD-(E9) Sguaring (85) and (Eç) and adding, gui= Gu) (1+3) ctazt Mato > te [mat Dividiag (E) by (Es), tan z&; = 2 67 o Jumper weight 720 N Rope length 60 m Rope stiffness 2 kN/m Solution: m= 20 kg, k= 2 kNím, 9.8 Veloeity of jumper as he falls through 60 m: mgh = 5 mvlorv=v2gh =2(9.8)(60) = 34.2928 m/s About static equilibrium position: x=x(t=0)=0, xo= x(t= 0) = 34.2928 m/s Response of jumper: x) =Açsin (mw t+d) where Ê 2/2 xo 34.2998 A= [mto] | = som => O -65m6m o 9.8 o, 2000 n and Ho WO, =tan! | z | =0 Po % Aerobat weight 540 N Natural frequency of vibration 10 rad/sec The natural frequency of a vibrating rope is given by (see Problem 2.26): Solution: ae o = [Ter n mab where T = tension in rope, m = mass, and a and b are lengths of the rope on both sides of the mass. For the given data: 10= Bjos] = JT (0.013611) 0] (2) (4) which yields 100 T= 017 = 7346.9388 N 0.013611 68 (a) Let P = total spring force, F = centrifugal force acting on each ball. Equilibrium of moments about the pivot of bell crank lever (O) gives: 20 P | 12 F 8) “2 5) (1) When P = 10º |-L.| = 100 N, and 100 | — , 2rnN) o fio) fazn] =mrul = 7 = =0. F=mr nr ar [100 En 0.004471 Nº where N = speed of the governor in rpm. Equation (1) gives: 0.004471 Nº (0.2) = E (0.12) or N = 81.9140 rpm (b) Consider a small displacement of the ball arm about the vertical position. Equilibrium about point O gives: (m b?) é 4 (ka sin 9)a cos 8 =0 (2) For small vallues of 9, sin 9 = 8 and cos $ = 1, and hence Eq. (2) gives mb Brkat0=0 from which the natural frequency can be determined as 1 1 — ES 2 2 ka” |2 4 |012] 9.81 = = ———. —— = 37.585 An E E | (10) E 75 851 rad /sec 7% 100 DM rasa -=Iy2 a (em) so= dr o os=ho os K+ when each wire stretches by x,» let tre resulting vertical displacement of the Platform be x - os + x = hxS4 < pet + Es [| te + -4 | At + [E Elf tes) P-4] For small x, x is es Ema É 2£tx and disg = 1+2 and bene Nãs 1+ bx é [e (dê +48) *J- RE” prtential energy guia gives f%gr= + (L% x K team (Ej= as Mx + tez x=o am. tr = TIM UT = O, (Ceg/m A (251) Equation of motion: mt=DF, Le, (LAM)X=-2(Axpg) Le, sr ãx=o where À = cross-sectional area of the tube and xs p = density of mercury. Thus the natural frequency is given by: a k, =55 = =4 (m)º (25) = 986.9651 a)» ns ERR Na perto 1 k 2 k , = 1— |" = 408285; —— 1 = (nº (16.6668) = 657. 27 5 + o) 3 so E 4 UM ( ) = 657.9822 Using k, = AR we obtain 1 do MOO) .g8s9e51 m mm 2m le, A =9.5359 (10º) m (1) Also ' k; AE m+5000 4 (m+5000) 657.982 Le, — = = 6.3573 (10º dês a + 5000 (o) 2) Using Eqs. (1) and (2), we obtain A = 9.5359 (10º) m = 6.3573 (109) m + 31.7865 (1078) Le, 3.1786 (107º) m = 31.7865 (107º) Le, m = 10000.1573 kg (3) Equations (1) and (3) yield A = 9,5359 (109) m = 9,5359 (10º) (1000.1573) = 0.9536 (104) m? 73 Longitudinal Vibration: Let lu = part of weight w carried by lengih q of shast M=W-W = weight corred by length b z= Elongation of Length a= Ma | E= Youngs modulus AE ' y: shortening of length b= (W-wW (L-a) ! A= area, of cross-section RE Pos mdf Since x=%, w= Wíl-a) E) x= elongation or static deflection af Length a = Wa (tra) hEL Considering the shast fenot ith spring -mass system, SF densth au end mass W/3 os a on= [É = (Sire “fa We(l-a) Transverse Vibration: spring constant of a fixed — fixed becom with of$-center foad =k= ser 4º se1 | a b? aê (L-ay oe JE =S 30282 9% . aa? E Ea" cm nero moment of inertia Torsional vibration: 14 Llgwheel is given an angular deflection o, resisting torgues offered by fengths o and b are sro and sie. Total resisting torque= Mç= ss(+ Deo mM & = qe ar(ã+ 5) where JT= rd = polar moment of inertia 4, Gm [= [E (444 da % ma 3) where J = mass polar moment of inertia of the flyuheel. 74 
