Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 1 Introduction 1.1 Equation (a) of the problem statement is used to solve for h as Q& h= (a) A(T − T∞ ) The Principle of Dimensional Homogeneity is used to determine the dimensions of the heat transfer coefficient Using the F-L-T system dimensions of the quantities in Equation (a) are ⎡F⋅ L⎤ Q& = ⎢ (b) ⎣ T ⎥⎦ [A] = L2 (b) [T − T∞ ] = [Θ] (c) Thus from Equations (a)-(d) the dimensions of the heat transfer coefficient are [h] = ⎡⎢ F ⋅ L ⎤⎥ ⎣T ⋅ Θ ⋅ L ⎦ ⎡ F ⎤ (d) =⎢ ⎥ ⎣T ⋅ Θ ⋅ L⎦ N while Possible units for the heat transfer coefficient using the SI system are m⋅s⋅K [] [ ] possible units using the English system are lb ft ⋅ s ⋅ R 1.2 The Reynolds number is defined as Re = ρVD µ (a) The dimensions of the quantities on the left-hand side of Equation (a) are obtained using Table 1.2 as (b) [ρ ] = ⎡⎢ M3 ⎤⎥ ⎣L ⎦ [V] = ⎡⎢ L ⎤⎥ (c) ⎣T ⎦ [D] = [L] (d) [µ ] = ⎡⎢ M ⎤ ⎥ ⎣L ⋅ T⎦ Substituting Equations (b)-(e) in Equation (a) leads to Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ (e) Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter ⎡M L ⎤ ⎢ ⋅ ⋅ L⎥ [Re] = ⎢ L MT ⎥ ⎢ ⎥ ⎣⎢ L ⋅ T ⎦⎥ ⎡ M ⋅ L3 ⋅ T ⎤ =⎢ ⎥ ⎣M ⋅ L ⋅ T⎦ = [1] Equation (f) shows that the Reynolds number is dimensionless (f) 1.3 The capacitance of a capacitor is defined by i (a) dv dt The dimension of i is that of electric current, which is a basic dimension The dimensions of electric potential are obtained from Table 1.2 as [v] = ⎡⎢ F ⋅ L ⎤⎥ (b) ⎣ i⋅T ⎦ Thus the dimensions of the time rate of change of electric potential are ⎡ dv ⎤ ⎡ F ⋅ L ⎤ (c) ⎢⎣ dt ⎥⎦ = ⎢⎣ i ⋅ T ⎥⎦ Use of Equation (c) in Equation (a) leads to ⎡ ⎤ ⎢ i ⎥ [C ] = ⎢ F ⋅ L ⎥ ⎢ ⎥ ⎣⎢ i ⋅ T ⎦⎥ C= ⎡i2 ⋅ T2 ⎤ =⎢ ⎥ ⎣ F⋅L ⎦ (d) 1.4 (a) The natural frequency of a mass-spring system is k ωn = (a) m where m is mass with dimension [M] and k is stiffness with dimensions in the M-L-T ⎡M⎤ system of ⎢ ⎥ Thus the dimensions of natural frequency are ⎣T ⎦ Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 1 ⎤ ⎡ M ⎛ ⎞ ⎢⎜ ⎟ ⎥ ⎢⎜ T ⎟ ⎥ [ω n ] = ⎢ ⎜M⎟ ⎥ ⎢⎜ ⎟ ⎥ ⎠ ⎥ ⎢⎣⎝ ⎦ ⎡1⎤ (b) =⎢ ⎥ ⎣T ⎦ (b) The natural frequency of the system is 100 Hz, which for calculations must be converted to r/s, cycles ω n = 20 s r ⎞ ⎛ cycles ⎞⎛ ⎟ = ⎜ 20 ⎟⎜⎜ 2π s ⎠⎝ cycles ⎟⎠ ⎝ = 125.7 r s (c) Equation (a) is rearranged as k = mω n2 Substitution of known values into Equation (d) leads to (d) r⎞ ⎛ k = (0.1 kg )⎜125.7 ⎟ ⎝ = 1.58 x10 s⎠ N m (e) 1.5 (a) The mass of the carbon nanotube is calculated as m = ρAL = ρ πr L ( ) kg ⎞ ⎛ = ⎜1300 ⎟π (0.34 x10 m ⎠ ⎝ - 23 3.78x10 kg (b) Conversion between TPa and psi leads to N E = 1.1 TPa = 1.1x1012 m −9 )( m 80 x10 −9 m N ⎞⎛ lb ⎞⎛ m ⎞ ⎛ = ⎜1.1x1012 ⎟⎜ 0.225 ⎟⎜ ⎟ N ⎠⎝ 3.28 ft ⎠ m ⎠⎝ ⎝ lb = 1.60 x10 in (c) Calculation of the natural frequency leads to ) ⎛ ft ⎞ ⎜ ⎟ ⎝ 12 in ⎠ Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter ω = 22.37 EI ρAL4 ⎛ 12 N ⎞ π 0.34 x10 −9 m ⎜1.1x10 ⎟ m ⎠4 ⎝ = 22.37 kg ⎞ ⎛ −9 80 x10 −9 m ⎜1300 ⎟π 0.34 x10 m ⎠ ⎝ r = 1.73x1010 s Converting to Hz gives r ⎞⎛ cycle ⎞ ⎛ ω = ⎜1.73 x1010 ⎟⎜ ⎟ s ⎠⎝ 2π r ⎠ ⎝ = 2.75 x10 Hz ( ( ) )( ) 1.6 The power of the motor is calculated as 900 kW ⋅ hr 24 hr = 37.5 kW The power is converted to English units using the conversions of Table 1.1 P = 37.5 x10 W N⋅m = 37.5x10 s ⎛ 0.225 lb ⎞ ⎛ 3.28 ft ⎞ N⎜ ⎟ ⋅ m⎜ ⎟ N ⎠ ⎝ m ⎠ ⎝ = 37.5 x10 s ft ⋅ lb = 2.77 x10 s Conversion to horsepower leads to ⎛ ⎞ ⎜ ft ⋅ lb ⎜ hp ⎟⎟ P = 2.77 x10 s ⎜ 550 ft ⋅ lb ⎟ ⎜ ⎟ s ⎝ ⎠ = 50.3 hp P= 1.7 The conversion of density from English units to SI units is Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ (a) (b) (c) Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ ρ = 1.94 = 1.94 Chapter slugs ft ⎞⎛ 3.28 ft ⎞ slugs ⎛ kg ⎜ ⎟⎜ ⎟ ⎜ ft ⎝ 0.00685 slugs ⎟⎠⎝ m ⎠ = 9.99 x10 3 kg m3 (a) 1.8 The constant acceleration of the train is a = −6 m s2 (a) The velocity is obtained using Equation (a) as v (t ) = −6t + C (b) The constant of integration is evaluated by requiring km v(t = 0) = 180 hr km ⎛ 1000 m ⎞⎛ hr ⎞ = 180 ⎜ ⎟⎜ ⎟ hr ⎝ km ⎠⎝ 3600 s ⎠ m = 50 (c) s Using Equation (c) in Equation (b) leads to m v (t ) = −6t + 50 (d) s The train stops when its velocity is zero, = −6t + 50 t = 8.33 s (e) The distance traveled is obtained by integrating Equation (d) and assuming x(0)=0, leading to (f) x(t ) = −3t + 50t The distance traveled before the train stops is x(8.33) = −3(8.33) + 50(8.33) (g) = 208.3 m 1.9 The differential equation for the angular velocity of a shaft is dω + ct ω = T (a) J dt Each term in Equation (a) has the same dimensions, those of torque or [F ⋅ L] The ⎡1⎤ dimensions of angular velocity are ⎢ ⎥ Thus the dimensions of ct are ⎣T ⎦ Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter ⎤ ⎡ ⎢F⋅ L⎥ [ct ] = ⎢ ⎥ ⎥ ⎢ ⎣⎢ T ⎦⎥ = [F ⋅ L ⋅ T ] 1.10 The equation for the torque applied to the armature is T = K a ia i f Equation (a) is rearranged as T Ka = ia i f The dimensions of torque are [F ⋅ L] thus the dimensions of the constant are [K a ] = ⎡⎢ F ⋅2L ⎤⎥ ⎣ i ⎦ The equation for the back emf is v = K vi f ω Equation (d) is rearranged as v Kv = ifω (b) (a) (b) (c) (d) (e) ⎡F⋅ L⎤ ⎡1⎤ and the dimensions of angular velocity are ⎢ ⎥ The dimensions of voltage are ⎢ ⎥ ⎣ i⋅T ⎦ ⎣T ⎦ The dimensions of the constant K v are ⎡ F⋅L ⎤ ⎢ ⎥ [K v ] = ⎢ i ⋅1T ⎥ ⎢ i ⎥ ⎣⎢ T ⎦⎥ ⎡F⋅ L⎤ (f) =⎢ ⎥ ⎣ i ⎦ It is clear from Equations (c) and (f) that the dimensions of [K a ] and [K v ] are the same These dimensions are the same as those of inductance (Table 1.2) 1.11 (a) The dimensions of Q& are determined from Equation (a) Q& = σAε (T − Tb4 ) [ ][ ] ⎡ F ⋅ L ⎤ ⎡F ⋅ L⎤ ⎢⎣ L2 ⋅ T ⋅ Θ ⎥⎦ L Θ = ⎢⎣ T ⎥⎦ (b) The differential equations governing the temperature in the body is dT ρc + σε (T − Tb ) = dt The perturbation in temperature in the radiating body is defined by Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ (a) (b) (c) Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter Tb = Tbs + Tb1 This leads to a perturbation in the temperature of the receiving body defined as T = Ts + T1 Substitution of equations (d) and (e) in Equation (c) leads to d ρc (Ts + T1 ) + σε (Ts + T1 )4 − (Tbs + Tb1 )4 = dt Simplifying Equation (f) gives ⎡ ⎛ T ⎞4 Tb1 ⎞ ⎤ dT1 4⎛ ⎟⎟ ⎥ = ρc + σε ⎢Ts ⎜⎜1 + ⎟⎟ − Tbs ⎜⎜1 + dt ⎢⎣ ⎝ Ts ⎠ ⎝ Tbs ⎠ ⎥⎦ Expanding the nonlinear terms, keeping only through the linear terms and noting Ts = Tbs [ ] ρc ⎡ ⎛ T ⎞ ⎛ T ⎞⎤ dT1 + σε ⎢Ts4 ⎜⎜ ⎟⎟ − Tbs4 ⎜⎜ b1 ⎟⎟⎥ = dt ⎝ Tbs ⎠⎦ ⎣ ⎝ Ts ⎠ ρc dT1 + 4σεTs3T1 = 4σεTbs3 Tb1 dt (d) (e) (f) (g) that (h) 1.12 The differential equation is linearized by using the small angle assumption which implies sin θ ≈ θ and cosθ ≈ Using these approximations in the differential equation leads to the linearized approximation as && & mL θ + cL θ + kL2θ = (a) 1.13 The differential equation is linearized by using the small angle assumption which implies sin θ ≈ θ and cosθ ≈ Using these approximations in the differential equation leads to the linearized approximation as L && ⎛ ⎞ mL θ + ⎜ mg + &y& ⎟θ = L&x& (a) ⎝ ⎠ 1.14 The nonlinear differential equations governing the concentration of the reactant and temperature are dC A V + (q + αVe − E /( RT ) )C A = qC Ai (a) dt dT ρqc p Ti − ρqc p T − Q& + λVαe − E /( RT ) C A = ρVc p (b) dt The reactor is operating at a steady-state when a perturbation in flow rate occurs according to q = q s + q p (t ) (c) The flow rate perturbation induces perturbations in concentration and temperature according to Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter C A = C As + C Ap (t ) (d) T = Ts + T p (t ) (e) The steady-state conditions are defined by setting time derivatives to zero in Equation (a) leading to (q ) + αVe − E /( RTs ) C A s = qs C Ai ρqc T − ρq c T − Q& + λVαe − E /( RTs )C s p i s p s (f ) =0 As (g ) Substitution of Equations (d) and (e) into Equations (a) and (b) leads to dC Ap − E / [R (Ts +T p ) ] (C As + C Ap ) = (q s + q p )C Ai + q s + q p + αVe V dt ( ) ρ (q s + q p )c p Ti − ρ (q s + q p )c p (Ts + T p ) − Q& + λVαe [ − E / R (Ts +T p ) ] (C As (h) ) + C Ap = ρVc p dT p dt (i ) It is noted from Equation (f) of Example (1.6) that a linearization of the exponential terms in Equations (h) and (i) is − e E R (T s + T p ) =e − E RT s + E − E RT s e Tp RTs2 Use of Equation (j) in Equations (h) and (i) and rearrangement leads to E − ⎛ − RTE ⎞⎤ dC Ap ⎡ E RT s s ⎟ + ⎢ q s + q p + αV ⎜ e + e T V p ⎥ (C As + C Ap ) = (q s + q p )C Ai ⎜ ⎟⎥ dt RT ⎢⎣ s ⎝ ⎠⎦ E ⎡ − RTE E − RTs ⎤ s & + ρ (qs + q p )c p Ti − ρ (qs + q p )c p (Ts + T p ) − Q + λVα ⎢e e T p ⎥ (C As + C Ap ) RTs2 ⎢⎣ ⎥⎦ dT p = ρVc p dt Equations (g) and (h) are used to simplify Equations (k) and (l) to E − ⎛ E − RTE ⎞ dC Ap RTs s ⎟ V C Ap + αV ⎜ e T + q s C Ap + q p C As + q p C Ap + αVe p (C As + C Ap ) ⎜ RTs2 ⎟ dt ⎝ ⎠ = q p C Ai ρq p c p Ti − ρc p (q p Ts + q s T p + q p T p ) + λVαe = ρVc p − E RTs C Ap ⎡ E − RTE ⎤ e s T p ⎥ C As + C Ap + λVα ⎢ RT ⎦⎥ ⎣⎢ s ( dT p dt Neglecting products of perturbations Equations (m) and (n) are rearranged as Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ ( j) (k ) (l) (m) ) (n) Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ V dC Ap dt + q s C Ap + q p C Ap + αVe − E RTs Chapter ⎛ E − RTE ⎞ C Ap + αV ⎜ e s T p ⎟C As ⎜ RTs2 ⎟ ⎝ ⎠ = q p C Ai − q p C As ρVc p dT p dt ( o) − ρq p c p Ti + ρc p (q p Ts + qs T p ) + λVαe − E RTs C Ap ⎡ E − RTE ⎤ + λVα ⎢ e s T p ⎥C As = ⎥⎦ ⎢⎣ RTs 1.15 The specific heat is related to temperature by c p = A1 + A2T 1.5 + A3T 2.6 (a) The transient temperature is the steady-state temperature plus a perturbation, T = Ts + T p Substituting Equation (b) into Equation (a) leads to 1.5 2.6 c P = A1 + A2 (Ts + T p ) + (Ts + T p ) 1.5 ( p) (b) 2.6 Tp ⎞ ⎛ Tp ⎞ 2.6 ⎛ (c) = A1 + A2Ts ⎜⎜1 + ⎟⎟ + A3Ts ⎜⎜1 + ⎟⎟ ⎝ Ts ⎠ ⎝ Ts ⎠ Using the binominal expansion to linearize Equation (c) leads to Tp ⎞ Tp ⎞ 1.5 ⎛ 2.6 ⎛ c p = A1 + A2Ts ⎜⎜1 + 1.5 ⎟⎟ + A3Ts ⎜⎜1 + 2.6 ⎟⎟ (d) Ts ⎠ Ts ⎠ ⎝ ⎝ The differential equation for the time-dependent temperature is dT 1 cp + T = T∞ (e) dt R R Substituting Equations (b) and (d) into Equation (e) along with T∞ = T∞s + T∞p leads to 1.5 T ⎞⎤ d ⎞ ⎛ 1 ⎟⎟ + A3Ts 2.6 ⎜⎜1 + 2.6 p ⎟⎟⎥ (Ts + T p ) + (Ts + T p ) = (T∞s + T∞p ) (f) Ts ⎠⎥⎦ dt R R ⎠ ⎝ dTs = and Ts = T∞s reduces Equation (f) to Noting that the steady-state is defined by dt ⎡ Tp ⎞ T p ⎞⎤ dT p 1 1.5 ⎛ 2.6 ⎛ + T p = T∞p (g) ⎢ A1 + A2Ts ⎜⎜1 + 1.5 ⎟⎟ + A3Ts ⎜⎜1 + 2.6 ⎟⎟⎥ T T dt R R ⎥ ⎢⎣ s ⎠⎦ s ⎠ ⎝ ⎝ Tp dT p > Problem1_23 Steady-state response of series LRC circuit Input resistance in ohms >> 100 R= 100 Input capacitance in farads >>0.2e-6 14 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ C= 2.0000e-007 Input inductance in henrys >> 0.5 L= 0.5000 Input source frequency in r/s >> 2000 om = 2000 Input source amplitude in V >> 120 V0 = 120 Natural freqeuncy in r/s = omn = 3.1623e+003 Dimensionless damping ratio = zeta = 0.0316 Phase angle in rad= phi = -1.5042 Steady-state amplitude in A = I= 15 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Chapter Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 0.0798 >> The resulting steady-state plot is 1.24 The MATALB file Prolbem1_24.m is listed below % Problem1_24.m %(a) Input two five by five matrices disp('Please input matrix A by row') for i=1:5 for j=1:5 str={['Enter A(',num2str(i),num2str(j),')']}; disp(str) A(i,j)=input('>> '); end end disp('Please input matrix B by row') for i=1:5 for j=1:5 str={['Enter B(',num2str(i),num2str(j),')']}; disp(str) B(i,j)=input('>> '); end end A 16 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ B % (b) =A+B C=A+B % (c) D=A*B D=A*B % (d) det(A) detA=det(A) % eigenvalues and eigenvectors of A [x,Y]=eigs(A); disp('Eigenvalues of A') Y disp('Matrix of eigenvalues of A') x A sample output from execution of the file is shown below >> clear >> Problem1_24 Please input matrix A by row 'Enter A(11)' >> 'Enter A(12)' >> 'Enter A(13)' >> 12 'Enter A(14)' >> -1 'Enter A(15)' >> 21 'Enter A(21)' >> 14 'Enter A(22)' >> -3 'Enter A(23)' >> 'Enter A(24)' >> 17 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Chapter Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 'Enter A(25)' >> -22 'Enter A(31)' >> 11 'Enter A(32)' >> 12 'Enter A(33)' >> 10 'Enter A(34)' >> -4 'Enter A(35)' >> 12 'Enter A(41)' >> 10 'Enter A(42)' >> 11 'Enter A(43)' >> 18 'Enter A(44)' >> 12 'Enter A(45)' >> 21 'Enter A(51)' >> 10 'Enter A(52)' >> 11 'Enter A(53)' >> 31 'Enter A(54)' >> 21 'Enter A(55)' 18 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ >> 11 Please input matrix B by row 'Enter B(11)' >> 21 'Enter B(12)' >> -21 'Enter B(13)' >> 21 'Enter B(14)' >> 10 'Enter B(15)' >> 'Enter B(21)' >> 'Enter B(22)' >> 'Enter B(23)' >> 'Enter B(24)' >> 'Enter B(25)' >> -5 'Enter B(31)' >> 16 'Enter B(32)' >> 12 'Enter B(33)' >> 11 'Enter B(34)' >> 18 'Enter B(35)' 19 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Chapter Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter >> 11 'Enter B(41)' >> 21 'Enter B(42)' >> 32 'Enter B(43)' >> 14 'Enter B(44)' >> 19 'Enter B(45)' >> 12 'Enter B(51)' >> 12 'Enter B(52)' >> 'Enter B(53)' >> -5 'Enter B(54)' >> 13 'Enter B(55)' >> 21 A= 14 11 10 10 -3 12 11 11 12 10 18 31 -1 21 -22 -4 12 12 21 21 11 20 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter B= 21 16 21 12 -21 12 32 21 11 14 -5 10 -5 18 11 19 12 13 21 -21 -1 24 43 20 33 21 32 26 30 -27 14 23 31 33 34 32 C= 22 22 27 31 22 D= 444 38 547 1090 1367 280 -474 -107 601 955 34 420 249 493 812 480 -122 418 969 1244 570 -299 353 818 859 detA = -1171825 Eigenvalues of A Y= 43.3949 0 0 -18.9247 0 0 0 0 -1.8896 -11.6121i 0 -1.8896 +11.6121i 0 10.3091 Matrix of eigenvalues of A x= -0.3664 -0.1632 -0.3379 - 0.4681i -0.3379 + 0.4681i 0.2658 21 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 0.1869 -0.2133 -0.6060 -0.6466 0.7510 -0.4463 -0.2256 0.3991 0.7187 0.7187 -0.4106 0.0528 + 0.2510i 0.0528 - 0.2510i -0.4042 -0.0845 - 0.0753i -0.0845 + 0.0753i 0.6726 -0.2465 + 0.1043i -0.2465 - 0.1043i 0.3808 1.25 A MATLAB file to calculate and plot Λ ( r , ζ ) is given below % Plots the function LAMBDA(r,zeta) as a function of r for several values of % zeta % Specify four values of zeta zeta1=0.1; zeta2=0.4; zeta3=0.8; zeta4=1.5; % Define values of r for calculations for i=1:400 r(i)=(i-1)*.01; % Calculate function LAMBDA1(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta1*r(i))^2)^0.5; LAMBDA2(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta2*r(i))^2)^0.5; LAMBDA3(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta3*r(i))^2)^0.5; LAMBDA4(i)=r(i)^2/((1-r(i)^2)^2+(2*zeta4*r(i))^2)^0.5; end plot(r,LAMBDA1,'-',r,LAMBDA2,'.',r,LAMBDA3,'-.',r,LAMBDA4,' ') xlabel('r') ylabel('\Lambda') str1=['\zeta=',num2str(zeta1)]; str2=['\zeta=',num2str(zeta2)]; str3=['\zeta=',num2str(zeta3)]; str4=['\zeta=',num2str(zeta4)]; legend(str1,str2,str3,str4) title('\Lambda vs r') The resulting output from execution of the m file is the following plot 22 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 1.26 The MATALB m file Problem1_26 which determines and plots the step response of an underdamped mechanical system is shown below % Problem1_26.m % Step response of an underdamped mechanical system % Input natural frequency and damping ratio clear disp('Step response of underdamped mechanical system') disp('Please input natural frequency in r/s') om=input('>> ') disp('Please input the dimensionless damping ratio') zeta=input('>> ') % Damped natural frequency omd=om*(1-zeta^2)^0.5; C1=zeta*om/omd; C2=1/om^2; C3=zeta*om; tf=10*pi/omd; dt=tf/500; for i=1:501 t(i)=(i-1)*dt; x(i)=C2*(1-exp(-C3*t(i))*(C1*sin(omd*t(i))-cos(omd*t(i)))); end plot(t,x) xlabel('t (s)') ylabel('x (m)') 23 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter str1=['Step response of underdamped mechancial system with \omega_n=',num2str(om),'and \zeta=',num2str(zeta)] title(str1) str2=['x(t)=',num2str(C2),'[1-e^-^',num2str(C3),'^t(',num2str(C1),'sin(',num2str(omd),'t)cos(',num2str(omd),'t))]'] text(tf/4,C2/2,str2) Output from execution of Problem1_26 follows >> Problem1_26 Step response of underdamped mechanical system Please input natural frequency in r/s >> 100 om = 100 Please input the dimensionless damping ratio >> 0.1 zeta = 0.1000 24 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ 1.27 The perturbation in liquid level is Chapter h(t ) = qR(1 − e − t /( RA) ) (a) (a) Since the argument of a transcendental function must be dimensionless the dimensions of the product of resistance and area must be time Thus the dimensions of ⎡T⎤ resistance must be ⎢ ⎥ ⎣L ⎦ (b) Note that the steady-state value of the liquid-level perturbation is qR The MATLAB file Problem1_27.m which calculates and plots h(t) from t=0 until h is within percent of its steady-state value is given below disp('Please enter resistance in s/m^2 ') R=input('>> ') % Final value of h hf=0.99*q*R; dt=0.01*R*A; h1=0; h(1)=0; t(1)=0; i=1; while h1> Please enter tank area in m^2 >> 100 A= 100 Please enter flow rate in m^3/s 25 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter >> 0.2 q= 0.2000 Please enter resistance in s/m^2 >> 15 R= 15 str1 = A=100 m^3/s str2 = R=15 s/m^2 str3 = q=0.2 m^3/s >> 26 Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ ... https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter 1.26 The MATALB m file Problem1_26 which determines and plots the step response. .. https://TestbankDirect.eu/ Solution Manual for System Dynamics and Response 1st Edition by Kelly Full file at https://TestbankDirect.eu/ Chapter str1=['Step response of underdamped mechancial system with omega_n=',num2str(om), 'and. .. density by p = Cρ γ Copyright © 2008 Nelson Thomson Learning All rights reserved Full file at https://TestbankDirect.eu/ (h) (a) (b) Solution Manual for System Dynamics and Response 1st Edition by Kelly