Mechanical Vibration(2th Ed) Solution - Inman

Problems and Solutions Section 1.1 (1.1 through 1.17) 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) x(m) Solution: Free-body diagram:

kx

10 1.14

11 1.25

12 1.37

13 1.48

14 1.59

15 1.71

16 1.82

From the free-body diagram and static equilibrium:

kx = mg ( g = 9.81m / s 2 )

k m mg

20

k = mg / x

µ=

Σki = 86.164 n

Standard deviation in computed stiffness:

m

15

σ=

∑ (k

i =1

n

i

− µ )2 = 0.164

n −1

10 0 1 x 2

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

1.2

˙˙ Derive the solution of mx + 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: Assume: x (t ) = ae :

rt

m&& + kx = 0 x

(1)

Then: x = are rt and && = ar 2 e rt & x Substitute into equation (1): mar 2 e rt + kae rt = 0 mr 2 + k = 0 r=± Thus there are two solutions: k i m

 k  i t − m  

x1 = c1e where

 k  i t   m 

, and x2 = c2 e

k = ωn = 2 rad/s m The sum of x1 and x2 is also a solution so that the total solution is:

x = x1 + x 2 = c1e 2it + c 2 e −2it

Substitute initial conditions: x0 = 1 mm, v0 =

5 mm/s

x(0) = c1 + c2 = x0 = 1 ⇒ c2 = 1 − c1 ˙ v(0) = x(0) = 2ic1 − 2ic2 = v0 = 5 −2c1 + 2c2 = 5 i −2c1 + 2 − 2c1 = 5 i 1 − 2 1 c2 = + 2 c1 =

Therefore the solution is:

5 i 4 5 i 4

1 5  2 it  1 5  −2 it x= − i e +  + i e 4  4  2 2 Using the Euler formula to evaluate the exponential terms yields : 1 1 5  5  x= − i (cos 2t + i sin 2t ) +  + i (cos 2t − i sin 2t ) 4  4  2 2 5 3 sin2t = sin(2t + 0.7297 ) 2 2 Using Mathcad the plot is: x = cos2t...