Solution manual for engineering vibration 3rd edition by daniel j.Inman

SOLUTION MANUAL FOR
v v

,Problems and Solutions Section 1.1 (1.1 through 1.19)
v v v v v v v




1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static)dis
v v v v v v v v v v v v v v



placement is recorded below. Plot the data and calculate the spring's stiffness. Note that the
v v v v v v v v v v v v v v v



data contain some error. Also calculate the standard deviation.
v v v v v v v v




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: v
From the free-
v v



body diagram and staticequilibr
v v v v



kx ium:

kx  mg (g  9.81m/s2 )
v v v v v v v v

k
k  mg/ x
v v v v




m ki
 v v  86.164 v

n
mg

20
The sample standard deviation inc
v v v v v



omputed stiffness is: v v


n

m 15 (k  ) v
i v
2


 vv i1
 0.164
n 1
v v


v v




10
0 1 2
x
Plot of mass in kg versus displacement in m
v v v v v v v v



Computation of slope from mg/x v v v v



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 m˙x˙  kx  0 and plot the result for at least two periods for the case
v v v v v v v v v v v v v v v v v v v v




with n = 2 rad/s, x0 = 1 mm, and v0 =
v v v 5 mm/s. v v v v v v v v




Solution:

Given:
m!x! kx  0 (1) v v v v




Assume: x(t)  ae . Then:ge x!  are and !x! ar e . Substitute into equation (1) to
rt v
2 rt v v
rt
v v v v v v v v v v v v v v



t:
mar2ert  kaert  0 v v v v




mr2  k  0 v v v v




k 
r i
m 
v v




Thus there are two solutions:
v v v v

vvvkv   kv 
 m i  tv  m v
i t
 v v
x1  c ev v
1 
v , and x vv vv
2 c e v v
2 

k
 2 rad/s where n  v v v v


m
The sum of x1 and x2 is also a solution so that the total solution is:
v v v v v v v v v v v v v v v




x  x  x  c e2it  c e2it
v v v v
v v
v
v v
1 2 1 2



Substitute initial conditions: x0 = 1 mm, v0 = v v v v v v v v 5 mm/s

x0 c1  c2  x0  1 c2  1 c1, and v0  x!0 2ic1  2ic2  v0 
v v v v v v v v v v v v v v v v v

v v v v v v v v v v
5 mm/s
 2c1  2c2 
v v v v 5 i. Combining the two underlined expressions (2 eqs in 2 unkowns):
v v v v v v v v v v v




1 5 1 5
i, and c  
v v


2c  2  2c  5i  c   i v v
v
v v v v v v v v v v v v

1 1 1 2
2 4 2 4

Therefore the solution is: v v v




1 5  2it
x   1  5  2it
v v v
v v
v



2 i e  





4  4 
i e
2
v
v
vv v v vv v
v


Using the Euler formula to evaluate the exponential terms yields:
v v v v v v v v v



1 5  1 5  v v v v




x    i   cos2t  isin2t   i   cos2t  isin2t
2 4  2 4 
v v vvv v v v v v v v v vvv v v v v v v


v v vv v v v vv v




5 3
 x(t)  cos2t  sin2t  sin2t  0.7297
v v

v v v v v v v v v v



v 2 2

, Using Mathcad the plot is:
v v v v




5. v



x t
v cos 2.t v sin 2. t
v
v



2


2




xt
v


0 5 10


2

t