Solution Manual for Engineering Vibration, 5th Edition – (Inman, 2021) | All 8 Chapters Covered

All 8 Chapters Covered
ss ss ss




SOLUTION MANUAL
ss

,Problems ssand ssSolutions ssSection ss1.1 ss(1.1 ssthrough ss1.19)

1.1 The ssspring ssof ssFigure ss1.2 ssis sssuccessively ssloaded sswith ssmass ssand ssthe sscorresponding
ss(static) ssdisplacement ssis ssrecorded ssbelow. s s Plot ssthe ssdata ssand sscalculate ssthe

ssspring's ss stiffness. s s Note ssthat ssthe ssdata ss contain sssome ss error. s s Also sscalculate ssthe

ssstandard ssdeviation.




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 ssdiagram: From ssthe ssfree-body ssdiagram ssand ssstatic
ssequilibrium:

kx
kx ss= ss mg s s ss (g ss= ss 9.81ms/sss2)
k k s s = ss mgs/ ssx

ki
m  ss= ss = ss86.164
n
mg

20
The sssample ssstandard ssdeviation ssin
sscomputed ssstiffness ssis:

n

m 15  (k i s s − 2


 )
s s i=1
= ss 0.164
ss=
n ss−s1


10
0 1 2
x
Plot ssof ssmass ssin sskg ssversus ssdisplacement ssin ssm
Computation ssof ssslope ssfrom ssmg/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




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,1.2 Derive ssthe sssolution ssof ss m˙x˙ ss+ sskx ss= ss0 s s and ssplot ssthe ssresult ssfor ssat ssleast sstwo ssperiods ssfor ssthe sscase
with ssn ss= ss2 ssrad/s, ssx0 ss= ss1 ssmm, 5 mm/s.
ssand ssv0 ss=




Solution:

Given:
m˙x˙+ sskx ss= (1)
ss0


Assume: s s x(t) ss= ssaert ss. x˙ s s = and ˙x˙ ss= ssars2ert ss. s s Substitute ssinto ssequation ss(1) ssto
rt
s s Then: ssget: ss are


mar2ert s s + sskaert s s = ss0
mr2 ss + ssk ss = ss0
k
r ss = ss  i
m
Thus ssthere ssare sstwo sssolutions:
 k  k

   − 
 ss
x1 ss = ce
ss 1 i
 , s s and s s x = ce
2 ss ss iss t2

k
where ss n = ss2 s s rad/s
ss =
m

The sssum ssof ssx1 ssand ssx2 ssis ssalso ssa sssolution ssso ssthat ssthe sstotal sssolution ssis:

x ss= s s x s s + ssx = ssc s s e2it s s
+ ssc ss e−2it
1 2 1 2



Substitute ssinitial ssconditions: ssx0 ss= ss1 ssmm, ssv0 5ss= mm/s

x(s 0)s= ssc1 ss+ ssc2 s s = ssx0 s s = ss1ss ssc2 ss = ss1s− ssc1, s s and s s v ( 0 )s= ssx˙(0)s= 5 mm/s
ss2ic1 s s − ss2ic2 s s = ssv0 s s =



 ss−2c1 ss+ ss2c2 s s = 5 ssi. s s Combining ssthe sstwo ssunderlined ssexpressions ss(2 sseqs ssin ss2 ssunkowns):
1 1 ss
−2c1 + ss2 ss− = 5 ssi ss = s s 5 i, s s and = s s +5 i
s s c2 2 4
− 4
ss
ss2c1 ssc1


2

Therefore ssthe sssolution ssis:

 ss1  ss1 
5 
s s
5 −2it
x ss= ss  ss i 2it + ss ss i s s e
ss+
 2 s s e 4 
− 4   ss2
ss

Using ssthe ssEuler ssformula ssto ssevaluate ssthe ssexponential ssterms ssyields:
 ss1   ss1 
5 5
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, x ss= ss  ss − i

s s (cos2ts + ssissins2ts)+
ss
s
ss i s s (cos2t
s − ssissins2ts)
ss

 2 ss ss ss+ 4 
4   ss2
3 ss
 ssx(t)s ss= sscoss2t ss+ 5 sins2t ss = sin(2t ss+ ss0.7297)
2 2




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