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# SolSec1_1 - Problems and Solutions Section 1.1(1.1 through...

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Problems and Solutions Section 1.1 (1.1 through 1.19) 1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static) displacement is recorded below. Plot the data and calculate the spring's stiffness. Note that the data contain some error. Also calculate the standard deviation. m (kg) 10 11 12 13 14 15 16 x (m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82 Solution: Free-body diagram: m k kx mg Plot of mass in kg versus displacement in m Computation of slope from mg / x m (kg) x (m) k (N/m) 10 1.14 86.05 11 1.25 86.33 12 1.37 85.93 13 1.48 86.17 14 1.59 86.38 15 1.71 86.05 16 1.82 86.24 0 1 2 10 15 20 m x From the free-body diagram and static equilibrium: kx = mg ( g = 9.81 m / s 2 ) k = mg / x μ = ! k i n = 86.164 The sample standard deviation in computed stiffness is: ! = ( k i " ) 2 i = 1 n # n " 1 = 0.164

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1.2 Derive the solution of m ˙ ˙ x + kx = 0 and plot the result for at least two periods for the case with ω n = 2 rad/s, x 0 = 1 mm, and v 0 = 5 mm/s. Solution: Given: 0 = + kx x m ! ! (1) Assume: x ( t ) = ae rt . Then: rt are x = ! and rt e ar x 2 = ! ! . Substitute into equation (1) to get: mar 2 e rt + kae rt = 0 mr 2 + k = 0 r = ± k m i Thus there are two solutions: x 1 = c 1 e k m i ! " # \$ % & t , and x 2 = c 2 e k m i ! " # \$ % & t where ( n = k m = 2 rad/s The sum of x 1 and x 2 is also a solution so that the total solution is: it it e c e c x x x 2 2 2 1 2 1 ! + = + = Substitute initial conditions: x 0 = 1 mm, v 0 = 5 mm/s x 0 ( ) = c 1 + c 2 = x 0 = 1 ! c 2 = 1 " c 1 , and v 0 ( ) = ! x 0 ( ) = 2 ic 1 " 2 ic 2 = v 0 = 5 mm/s !" 2 c 1 + 2 c 2 = 5 i . Combining the two underlined expressions (2 eqs in 2 unkowns): " 2 c 1 + 2 " 2 c 1 = 5 i ! c 1 = 1 2 " 5 4 i , and c 2 = 1 2 + 5 4 i Therefore the solution is: x = 1 2 ! 5 4 i " # \$ % & e 2 it + 1 2 + 5 4 i " # \$ % & e ! 2 it Using the Euler formula to evaluate the exponential terms yields: x = 1 2 ! 5 4 i " # \$ % & cos2 t + i sin2 t ( ) + 1 2 + 5 4 i " # \$ % & t ! i t ( ) ( x ( t ) = t + 5 2 t = 3 2 sin 2 t + 0.7297 ( )
Using Mathcad the plot is: x t cos . 2 t . 5 2 sin . 2 t 0 5 10 2 2 x t t

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1.3 Solve m ˙ ˙ x + kx = 0 for k = 4 N/m, m = 1 kg, x 0 = 1 mm, and v 0 = 0. Plot the solution. Solution: This is identical to problem 2, except v 0 = 0. ! n = k m = 2 rad/s " # \$ % & . Calculating the initial conditions: x 0 ( ) = c 1 + c 2 = x 0 = 1 ! c 2 = 1 " c 1 v 0 ( ) = ˙ x 0 ( ) = 2 ic 1 " 2 ic 2 = v 0 = 0 ! c 2 = c 1 c 2 = c 1 = 0.5 x t ( ) = 1 2 e 2 it + 1 2 e " 2 it = 1 2 cos2 t + i sin2 t ( ) + 1 2 t " i t ( ) x ( t )= cos (2 t ) The following plot is from Mathcad: Alternately students may use equation (1.10) directly to get x ( t ) = 2 2 (1) 2 + 0 2 2 sin(2 t + tan ! 1 [ 2 " 1 0 ]) = 1sin(2 t + # 2 ) = t x t cos . 2 t 0 5 10 1 1 x t t
1.4 The amplitude of vibration of an undamped system is measured to be 1 mm. The phase shift from t = 0 is measured to be 2 rad and the frequency is found to be 5 rad/s. Calculate the initial conditions that caused this vibration to occur. Assume the response is of the form x ( t ) = A sin( !

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