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# SolSec 2.2 - Problems and Solutions Section 2.2(2.16...

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Problems and Solutions Section 2.2 (2.16 through 2.31) 2.16 Calculate the constants A and ! for arbitrary initial conditions, x 0 and v 0 , in the case of the forced response given by Equation (2.37). Compare this solution to the transient response obtained in the case of no forcing function (i.e. F 0 = 0). Solution: From equation (2.37) x ( t ) = Ae ! "# n t sin( # d t + \$ ) + X cos( # t ! % ) & ˙ x ( t ) = ! "# n Ae ! "# n t sin( # d t + \$ ) + A # d e ! "# n t cos( # d t + \$ ) ! X # sin( # t ! % ) Next apply the initial conditions to these general expressions for position and velocity to get: x (0) = A sin ! + X cos " ˙ x (0) = # \$% n A sin ! + A % d cos ! + X % sin " Solving this system of two equations in two unknowns yields: ! = tan " 1 ( x 0 " X cos # ) \$ d v 0 + ( x 0 " X cos # ) %\$ n " X \$ sin # & ( ) * + A = x 0 " X cos # sin ! Recall that X has the form X = F 0 / m ( ! n 2 " ! 2 ) 2 + (2 #! n ! ) 2 and \$ = tan " 1 2 #! n ! ! n 2 " ! 2 % & ( ) * Now if F 0 = 0, then X = 0 and A and φ from above reduce to: ! = tan " 1 x 0 # d v 0 + x 0 \$# n % & ( ) * A = x 0 sin ! = ( v 0 + \$# n x 0 ) 2 + ( x 0 # d ) 2 # d 2 These are identical to the values given in equation (1.38).

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2.17 Show that Equations (2.28) and (2.29) are equivalent by verifying Equations (2.29) and (2.30). Solution: From equation (2.28) and expanding the trig relation yields x p = X cos( ! t " # ) = X cos ! t cos # + sin ! t sin # [ ] = ( X cos # ) A s ! " # \$ # cos ! t + ( X sin # ) B s !" # \$ # sin ! t Now with A s and B s defined as indicated, the magnitude is computed: X = A s 2 + B s 2 and B s A s = X sin ! X cos ! " ! = tan # 1 B s A s \$ % & ( ) 2.18 Plot the solution of Equation (2.27) for the case that m = 1 kg, ! = 0.01, ! n = 2 rad/s. F 0 = 3 N, and ! = 10 rad/s, with initial conditions x 0 = 1 m and v 0 = 1 m/s. Solution: The particular solution is given in equations (2.36) and (2.37). Substitution of the values given yields: x p = 0.03125cos(10 t + 8.333 ! 10 " 3 ) . Then the total solution has the form: x ( t ) = Ae ! 0.02 t sin(2 t + " ) + 0.03125cos(10 t + 0.008333) = e ! 0.02 t A sin2 t + B cos2 t ( ) + 0.03125cos(10 t + 0.008333) Differentiating then yields ! x ( t ) = ! 0.02 e ! 0.02 t A sin2 t + B cos2 t ( ) + sin(2 t + " ) + 2 e ! 0.02 t A cos2 t ! B sin2 t ( ) ! 0.3125sin(10 t + 0.008333) Apply the initial conditions to get: x (0) = 1 = B + 0.03125cos(0.00833) ! B = 0.969 ! x (0) = 1 = " 0.02 B + 2 A " 0.3125sin(0.00833) ! A = 0.489 So the solution and plot become (using Mathcad):

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2.19 A 100 kg mass is suspended by a spring of stiffness 30 × 10 3 N/m with a viscous damping constant of 1000 Ns/m. The mass is initially at rest and in equilibrium. Calculate the steady-state displacement amplitude and phase if the mass is excited by a harmonic force of 80 N at 3 Hz. Solution: Given m = 100kg, k =30,000 N/m, c = 1000 Ns/m, F 0 = 80 N and ω = 6 π rad/s: f 0 = F 0 m = 80 100 = 0.8 m/s 2 , ! n = k m = 17.32 rad/s " = c 2 km = 0.289 X = 0.8 17.32 2 + 36 # 2 ( ) 2 + 2(0.289)(17.32)(6 # ) ( ) 2 = 0.0041 m Next compute the angle from ! = tan " 1 188.702 " 55.323 # \$ % & ( Since the denominator is negative the angle must be found in the 4 th quadrant. To find this use Window 2.3 and then in Matlab type atan2(188.702,-55.323) or use the principle value and add π to it. Either way the phase is θ =1.856 rad.
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