Exam (elaborations) TEST BANK FOR Engineering Vibration 3rd Edition By Daniel J. Inman (Solution Manual)

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) 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 = 9.81m/ s 2 ) k = mg/ x μ = !ki n = 86.164 The sample standard deviation in computed stiffness is: ! = (ki " μ) 2 i=1 n # n "1 = 0.164 1.2 Derive the solution of m˙x˙ + kx = 0 and plot the result for at least two periods for the case with ωn = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s. Solution: Given: m!x!+ kx = 0 (1) Assume: x(t) = aert . Then: rt x! = are and rt x ar e 2 !! = . Substitute into equation (1) to get: mar2ert + kaert = 0 mr2 + k = 0 r = ± k m i Thus there are two solutions: x1 = c1e k m i ! " # $ % & t , and x2 = c2e ' k m i ! " # $ % & t where (n = k m = 2 rad/s The sum of x1 and x2 is also a solution so that the total solution is: it it x x x c e c e 2 2 2 1 2 1 ! = + = + Substitute initial conditions: x0 = 1 mm, v0 = 5 mm/s x(0) = c1 + c2 = x0 = 1!c2 = 1" c1 , and v(0) = x! (0) = 2ic1 " 2ic2 = v0 = 5 mm/s !"2c1 + 2c2 = 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns): "2c1 + 2 " 2c1 = 5 i!c1 = 1 2 " 5 4 i, and c2 = 1 2 + 5 4 i Therefore the solution is: x = 1 2 ! 5 4 i " # $ % & ' e2it + 1 2 + 5 4 i " # $% & ' e!2it Using the Euler formula to evaluate the exponential terms yields: x = 1 2 ! 5 4 i " # $ % & ' (cos2t + i sin2t ) + 1 2 + 5 4 i " # $ % & ' (cos2t ! i sin 2t ) ( x(t ) = cos2t + 5 2 sin2t = 3 2 sin(2t + 0.7297) Using Mathcad the plot