Complete solutions to Exercise 14(b) 1. Very similar to EXAMPLE 4. We have same characteristic equation:

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1 Soluions 4(b) Complee soluions o Exercise 4(b). Very similar o EXAMPE 4. We have same characerisic equaion: 5 i Ae = + Be By using he given iniial condiions we obain he simulaneous equaions A+ B= 0 5A B= 6 Solving hese gives A = and B =. Hence he soluion is 5 i e = e. Use (4.8) wih he appropriae value of k x= Acos + Bsin Because k = (a) [ ] (b) x= Acos( 4) + Bsin ( 4) [ Because k = 4] x= Acos + Bsin Because k = x= Acos 5 + Bsin 5 Because k = 5. Characerisic equaion is r + km= 0, (using he characerisic equaion variable r because m represens mass in his problem). r k + = 0 m By (4.8), x = Acos( ω)+ Bsin( ω) where ω = km (c) (d) 4. Characerisic equaion is m m = = 0 By (4.8) v= Acos ( 0 ) Bsin ( 0 ) + ( ) Subsiuing = 0, v = ino ( ) = Acos()+ 0 Bsin() 0 gives A = [Remember cos( 0)= and sin()= 0 0 ] dv Differeniaing ( ) and subsiuing he second condiion =0, = 0 ; d dv = ( 0 ) Asin ( 0 ) + ( 0 ) Bcos ( 0 ) d 0 = 0+ 0 B gives B= Subsiuing A = and B = ino ( ) gives ( v= cos 0 ) + sin ( 0 ) π Using (4.76) wih a = and b = gives v = cos ( 0 ) 4 (4.76) acos θ + bsin θ = a + b cos ( θ α) where α= an ( b a ) (4.8) If r + k =0 hen y = Acos( kx) + Bsin ( kx)

2 Soluions 4(b) 5. By EXAMPE 6 where k = y = Acos kx + Bsin kx (*) PEI. Subsiuing he boundary condiion x =, y = e gives e= Acos+ Bsin ( ) Subsiuing x =, y = e ino (*) gives k k e= Acos+ Bsin = AcosBsin ( ) by (4.5) by (4.50) Adding ( ) and ( ) yields e= Acos e A= = esec k by (4.) cos e Subsiuing A = cos k ino ( ) gives e e= cos Bsin k + cos e= e+ Bsin, hence B= 0 Subsiuing A = esec k and B = 0 ino (*) yields y = eseccos( kx) 6. Characerisic equaion is r + ωζr + ω = 0 (using r, since m represens mass). This is a quadraic equaion wih variable r. Puing a =, b = ωζ and c = ω ino (.6) gives b± b 4ac (.6) x = a sec cos x = ( x) (4.) (4.50) sin ( x) = sin ( x) (4.5) cos( x) = cos( x)

3 Soluions 4(b) ωζ ± 4ω ζ 4ω ± r = = ωζ 4ω ζ ωζ ± ω ζ = = ± ωζ ω ζ [ ] Cancelling s ( ) [ Taking Ou ] ( ) and r r = ω ζ ± ζ ω r = ω ζ + ζ = ω ζ ζ = ω ζ + ζ Since we have real and differen roos we use (4.4) x = Ae r + Be r where r and r are as above. 7. The characerisic equaion is same as soluion 6: m + ωζ m+ ω = 0 By soluion o quesion 6 ( ) m = ω ζ ± ζ For ζ =, m =ω (equal roos) so by (4.5) x = A+ B e ω Subsiuing he firs iniial condiion, when = 0, x = 5; 0 5 = e ( A+ B.0) gives A=5 [Remember e 0 =] Differeniaing x = ( A + B)e ω by using he produc rule, (6.), yields x e ω ω = ω A+ B + e B. Subsiuing he oher iniial condiion when = 0, x Ý = 0; = ωe A+ B.0 + e. B [ ] B = ωa= 5ω Because A= 5 Subsiuing A = 5 and B = 5ω ino x = ( A + B)e ω gives ω x = e 5+ 5ω ( ω ) x = 5e ω + {Taking Ou 5] The following is he MAPE oupu wih θ = ω > x:=5*exp(-hea)*(+hea); x := 5 e ( θ ) ( + θ ) > plo(x,hea=0..0);

4 Soluions 4(b) 4 8. Since we have he same characerisic equaion as soluion 6 so ( ) m = ω ζ ± ζ ( ( )) j = ω ζ ± ζ = ω ζ ± ζ ζ < ζ Because and so < m= ωζ ± jωβ where β = ζ Since we have complex roos so by (4.6) ζω x= e Acos( βω) + Bsin ( βω) (*) (4.6) If m = α ± jβ hen y = e αx A cos βx [ + B sin( βx) ]

5 Soluions 4(b) 5 Subsiuing he iniial condiion, when = 0, x = 0; 0 = e 0 Acos 0 + Bsin 0, gives A=0 because e 0 =, cos 0 = and sin 0 0 = Subsiuing A = 0 ino (*) gives ζω x = Be βω sin Subsiuing he oher iniial condiion, when = 0, x = ωβ means we need o differeniae x = Be ζω sin( βω). x Be ζω ζω = ζω sin βω + Bβωe cos βω Subsiuing = 0 and x = ωβ ino his ωβ = 0 + Bωβ which gives B = Subsiuing A = 0, B = ino (*) gives x = e ζω sin βω where β = ζ 9. Dividing he characerisic equaion by C gives m + RC m + C = 0 Subsiuing a =, b = RC and c = ino (.6) gives C 4 ± RC R C C 4R C m = = ± RC R C 4R C RC R C = ± 4R C = ± RC RC 4R C m = ± (*) RC Case (a) = 4CR ; Subsiuing = 4CR ino (*) gives m = [Equal Roos] RC By (4.5) RC v = ( A+ B)e Case (b) > 4CR ; Using (*) gives wo roos m and m [Disinc Roos] m RC RC By (4.4) v = Ae m + Be m Case (c) < 4CR ; 4 4 R C, = + m = R C (.6) b± b 4ac m = a (4.4) If m and m hen y = Ae m x + Be m x (4.5) Equal roos m hen y = ( A+ Bx)e

6 Soluions 4(b) 6 4R C 4R C m = ± = ± RC RC RC RC 4RC m= ± j RC RC [ Complex Roos] 4R C e α = and β =, hen subsiuing hese ino (4.6) RC RC gives ( β ) sin ( β ) α v= e Acos + B 0. Same differenial equaion as quesion 9. The characerisic equaion is given by Cm + m 0 R + = The roos of his equaion are given by soluion 9 4R C m = ± (*) RC Subsiuing R = 0 0 and C = 0 9 gives 9 4 ( 0 0 ) 0 m = 9 ± ( 0 0 ) ( 0 ) = 50 0 ± 0.4 ( 50 0 ) ( 50 0 ) m = ± 0.4 Since 0.4, ( 50 0 ) ( 50 0 ) 0.4 < m= ± j. Equaing he imaginary par of his o he imaginary par of he roos given in he quesion ( 50 0 )± j( 0 0 ) gives 0.4 j( 50 0 ) = j( 0 0 ) 0.4 = 0.6 Dividing by 50 0 Squaring boh sides 0.4 = = 0.6 Solving his equaion gives = 0.94 (correc o hree d.p.) Hence = 0.94 H.

7 Soluions 4(b) 7. Dividing hrough by C we have dv dv v + + = 0 d RC d C m + m+ = 0 RC C Equaing wih m + ζωm + ω = 0 ω = C gives ω = C Equaing he m erms gives ζω = RC and subsiuing ω = C ζ = C RC C ζ = = RC R C we have