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Solutions to B Problems CHAPTER 2 E(t) = 0 t0 dot) = s Referring to Equation (2-2), we obtain F(s) = do [e(t)) = dy It 6-2) = ARE B-2-2. (a) E (t) =0 t0 2] = 2 do [t2] BE Referring to Equation (2-2), we obtain P(s) = do [E(t)] = À) [tZeat)= —2 (s + a)3 E(t) = 0 t0 cos 2cWt cos 3Wt = 4(cos 5Swt + cosuwt) F(s) = 45 [E(t)] = dy [blcos St t + cos 4/t)] a (s2 + 13w2)s s s = 2 (== +25? + =) (s2 + 25 w2)(s2 + 4?) B-2-5. The function f(t) can be written as E(t)=(t-a) i(t-a) The Laplace transform of f(t) is eras z. F(s) = 0; [E(t)] =) [lt -a) lt -a)]= B-2-6. f(t)=ci(t-a)-ci(t-b) The Laplace transform of f(t) is ns et. o (eras — ebs) F(s) =c = ES The limiting value of lim F(s) a>0 F(s) as a approaches zero is 24(1 - as ertas - eras) = lim a+0 a3s2 = 2a(1 - as e-tas — eras) = lim a>0 É a3s2 2 24(-s e8S + -BSÉ qtas + 5 eras) = lim a>0 3a2s2 de etas + aê ertas + Eras) = 1im da A a>0 ad 2 qa às 8 [a etas ,S cas ,as(syçtas se-as| = lim a>0 2as | -as -kas -as qa (4 e -ase = 46) = lim a>0 qd a da E» 1 -29 6488 5 AS + as 5- CRS +45 es = lim a+0 z =-28 -s+45=s5 B-2-9. f(oo) = 1im f(t) = lim sF(s) tj=2: 00: s>0 =iim SME +2) (5x2 140 E s(s +) 1 BRENO: e(0+) = lim e(t)=lim —S2C+2 3 t>0+ soc S(s + 1)(s +3) B-2-11. Define A y=x Then Es y(0+) = x(0+) The initial value of y can be obtained by use of the initial value theorem as follows: y(0+) = lim sy(s) s>00 Since A Y(s) = daly(t)] = de lx(t)] = sX(s) - x(0+) ve obtain y(0+) = lim sY(s) = lim s[sX(s) - x(0+)] s>0 s>0 = lim [s2x(s) - sx(0+)] 3200 B-2-12. Note that AR E O] = sP(s) - £(0) 3 [5 o] = s2p(s) - se(0) - £(0) dt2 Define a2 = SÉ f(t) g(t) 2 Then See)=f|Eg(t)|= " E = - d ag 91t))= no(s) - a(o) = s[s2r(s) - s£(0) - £(0)] - (0) = e3r(s) - s2r(0) - sE(O) - E(0) B-2-13. se? T ) f(t) est a o () o (EA) ca (ce) " b ( acfta = a-S 0 " 3 s(s+1 - ( ) e ( Es) can thus be written as Fls) = 8. de s S+1 s+2Z and the inverse Laplace transfomm of E(s) is £(t)=6-9 et+3e2t B-2-16. EEB 16 43 (a) F,(s) Ea Eu E The inverse Laplace transform of F,(s) is E(t)=6+3 (8) hi F(s) = asda te nho as 2 tas ado 20 PM a ea E +12 where 5s +2 =5+2..3 ER 12 =-1 p = 58 +2 ==10+2. 2 s+2 = 4 1 =-2 » E (843) -5(s+1)- (5s+2 2 1 dsjs+1 qi? (s +31) =) - (10 +2) 43 Ele) can thus be written as Fa(s) ES) 8 3 + + s+1 (s +2)2 8 +2 and the inverse Laplace transform of Fols) is ft) = 3 tra et sao B-2-17. . 282 +45 +5 ad 5 Eles s(s+1) s+1'5(5+1) =2+ Deo red E pe s+1 s s+1 SRI s The inverse Laplace transform of F(s) is att) =2 6(t)-3et+5 B-2-18. pt) = Ra t+ 244 s2 s s2 The inverse Laplace transform of F(s) is F(t)= D(t)+2+4 - B-2-19. B-2-19 Et) É z O Eee E s2 +2s+10 (s+1)2+432 Es, s+1 o 3 (s+1)2+32 (s+1)2+32 Hence f(t) = et cos 3% - E et sin 3 B-2-20. p(s) =-S2 +28 +5 as, [3 s2(s +1) RR where a=-S2+28+5 =5 s+1 amu) B-2-22. Di 1 5 r(s) = - ( a ) RS) 2 aa ipê The inverse Laplace transform of F(s) is e(t)=—L o (t -L sin cot) w? [o B-2-23. a b Fís) = = (1 - e-as) — o e-as a>o The inverse Laplace transform of F(s) is E(t)=ct-c(t-a)l(t-a)-bi(t-a) B-2-24. A MATIAB program to obtain partial-fraction expansions of the given function F(s) is given below. num=[0 0 0 0 1]; den=[1 3 2 0 0]; [r,p,k] = residue(num, den) r= =0.2500 1.0000 =0.7500 0.5000 10 From this computer output we obtain E -0.75 0. Flgite if E , 0.5 s4 + 353 + 282 s+2 s+1 s s2 The inverse Laplace transform of F(s) is f(t) = -0.25 e-2t + et - 0.75 + 0.5t B-2-25. A possible MATLAB program to obtain partial-fraction expansions of the given function F(s) is given below. num=[0 0 3 4 1; den=[1 2 5 8 10]; [r,p,k] = residue(num, den) r= 0.3661 - 0.4881i 0.3661 + 0.4881i -0.3661 - 0.0006i -0.3661 + 0.0006i p= 0.2758 + 1.9081i 0.2758 - 1.9081i -1.2758 + 1.0309i -1.2758 - 1.0309i From this computer output we obtain 22 +48 +1 s4+ 253 + 5s2 + 8s + 10 F(s) = 11 B-2-27. R+w x=t x(0)=0, *0)=0 The Laplace transform of this differential equation is sêx(s) +10 Zx(s) = — s Solving this equation for X(s), we obtain edad elis? + w 2) 2º 2+wW2)W,? The inverse Laplace transform of X(s) is x(t) = ER 1 PR (t - Da sin) t) This is the solution of the given differential equation. B-2-28. X+2+x=1 x(0)=0, x(0) =2 The Laplace transform of this differential equation is 2[s2x(s) - sx(0) - *(0)] + 2[sx(s) - x(0)] + X(s) = + Substitution of the initial conditions into this equation gives 2[s2x(s) - 2] + 2[sx(s)] + x(s) = 1 or (252 + 25 + 1)x(s) =4 + + Solving this last equation for X(s), we get 48 +1 s(252 + 2s +1) X(s) 4 FR 1 2822 +28+1 s(282 + 25 + 1) " 2 x 0.5 (s + 0.5)2 +0.25 sl(s + 0.5)2 + 0.25] " 13 = 4x 0.5 g À (8 + 0.5) +:0.5; (s+0.5)22+0.52 S (s+0.5)2+0.52 The inverse Laplace transform of X(s) gives x(t) = 4805 sino.5t +1 - O: cos 0.5t - e 0-5 sin 0.5t 1+3€0:5 sin 0.5t - e 95% cos 0.5t B-2-29. 2X +Mh+3x=0, x(0) = 3, x(0)-= 0 Taking the Laplace transform of this differential equation, we obtain 2[s2x(s) - sx(0) - x(0)] + 7[sxis) - x(0)] + 3x(s) = 0 By substituting the given initial conditions into this last equation, 2[s2x(s) - 38) + 7[sx(s) - 3] + 3x(s) = 0 or (282 + 75 + 3)x(s) = 6s + 21 Solving for X(s) yields ES +42 = 6s + 21 252 +75 +3 (2s+1)(s+3) x(s) = 72 Olou so oro 28+1 8+3 s+0.5 s+3 Finally, taking the inverse Laplace transform of X(s), we obtain -0.5t — = x(t) = 3.6 e 0.6 e B-2-30. X+x=sin 3, x(0) = 0, *(0) = 0 The Laplace transform of this differential equation is s2x(s) + x(s) => 3 s2 Fe 32 Solving this equation for X(s), we get 3 É 1 1 3 X(s) = “ = (s2 + 1)(s2 + 9) 8 s2+1 8 s2+9 14 CHAPTER 3 págel. JT =4mR2=5x100x0.52=12.5kg-m B-3-2. Assume that the body of known moment of inertia J, is turned through a small angle 6 about the vertical axis and then released. The equation of motion for the oscillation is Nou = ke where kK is the torsional spring constant of the string. This equation can be written as or where The period Tg of this oscillation is T -2X - 27 (1) o 9h) Já po e, Jo Next, we attach a rotating body of unknown moment of inertia J and measure the period T of oscillation. The equation for the period T is pus (2) pe J By eliminating the unknown torsional spring constant k from Equations (1) and (2), we obtain T= 2% 3% .27y7 T E: 2 ag (E 3 so E) (3) The unknowm moment of inertia J can therefore be determined by measuring the period of oscillation T and substituting it into Equation (3). 16 B-3-3. Define the vertical displacement of the ball as x(t) with x(0) = 0. The positive direction is downward. The equation of motion for the system is mk = mg vith initial conditions x(0) = 0 m and X(0) = 20 m/s. So we have x=g x =gt + x(0) TT x=bgt2+x(O +x(0)=b gt? + 20t x Assume that at t = t the ball reaches the ground. Then PM as 2 100 = & x 9.81 t,2 +20 t, from which we obtain = 2.915 t 15 s The ball reaches the ground in 2.915 s. B-3-4. Define the torque applied to the flywheel as T. The equation of motion for the system is J8=T, 0(0)=0, 6(0)=0 from which we obtain à vu 6=L ais By substituting numerical values into this equation, we have 20 x6.28=-L x5 50 Thus T = 1256 N-m B-355. J8=-T (T = braking torque) Integrating this equation, ô= -Te+ (o), — 6(0) = 100 rad/s Substituting the given numerical values, 20 = ==D » 15 + 100 17 The equivalent spring constant E is then obtained as 1 E a Next, consider the figure shom below. Note that A ABD and ACBE are similar. So we have [N or OC(OB + 4 OA) = OA(OB - 4 0C) Solving for &, we obtain A ps QRO Í. a =k oa + DB 1 Sie pal oB kk + B-3-8. (z)' The force £ due to the dampers is E =Di(y-*) +ba(y -%) = (by + Do)(y - x) In terms of the equivalent viscous friction coefficient Degr force £ is given by eos £ = beg(Y - X) beg = by + bz (b) The force £f due to the dampers is E=by(Z-x)=ba(y-2) (1) where z is the displacement of a point between damper by and damper bz. (Note that the same force is transmitted through the shaft.) From Equation (1), we have (by + b2)2 = boy + bjx or Ne NE (bay + bjx) (2) 19 In terms of the equivalent viscous friction coefficient begr force £ is given by £ = bag(Y - x) By substituting Equation (2) into Equation (1), we have . DRE . 1 . . £=ba(y -z) = baly - ETs (bzy + byx)] bjbz a . e (E) bj + bz Thus, o DE Fr bibz : £ = begly - x) = ba(y - 2) CE do E 2 Hence, bybz 2 Peq bra o GEE bj db B-3-9. Since the same force transmits the shaft, we have £=by(2-%)=bo(y- 2) + ba(y - 2) (1) where displacement z is defined in the figure below. ' HE x nas Po y In terms of the equivalent viscous friction coefficient, the force £ is given by £ = beglY - *) (2) From Equation (1) we have a bjZ + boz + baz = bjx + boy + bar or , ai “ , == 3 ts Es [bjx + (bz + b3)y] (3) By substituting Equation (3) into Equation (1), we have 20 
