solution manual engineering vibration 5th Edition by Inman

All 8 Chapters Covered
m m m




SOLUTION MANUAL
m

,ProblemsmandmSolutionsmSectionm1.1m(1.1mthroughm1.19)

1.1 ThemspringmofmFigurem1.2mismsuccessivelymloadedmwithmmassmandmthemcorrespondingm(st
atic)mdisplacementmismrecordedmbelow.m Plotmthemdatamandmcalculatemthemspring'smstiffne
ss.m Notemthatmthemdatamcontainmsomemerror.m Alsomcalculatemthemstandardmdeviation.

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-bodymdiagram: Frommthemfree-
bodymdiagrammandmstaticmequilibrium:
kx
kxm mmgm m (gmm9.81mm/msm2)
k km mmgm/mx

ki
m mm m86.164
n
mg

20
Themsamplemstandardmdeviationminmco
mputedmstiffnessmis:
n

m 15  i
2


m m i1
m0.164



10
0 1 2
x
Plotmofmmassminmkgmversusmdisplacementminmm
Computationmofmslopemfrommmg/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 Derivemthemsolutionmofmm˙x˙mmkxmm0m andmplotmthemresultmformatmleastmtwomperiodsmformthemcase
withmnm=m2mrad/s,mx0m=m1mmm,mandm 5 mm/s.
v0m=

Solution:

Given:
mxmmkxmm0 (1)
Assume:m x(t)mmae m.m The
rt
xm mare and xmmarm e m.m Substitutemintomequationm(1)mto
rt 2 rt

n:mget:
mar2ertm mkaertm m0
mr2mmkmm0
k
rm  m  i
m
Thusmtheremaremtwomsolutions:
 m  m
m
m  
x1m mc1e ,m andm x2m mc2e m



k
wherem n m m2m rad/s
 m

Themsummofmx1mandmx2mismalsomamsolutionmsomthatmthemtotalmsolutionmis:

xmm xm mx mcm e2itm mcm e2it
1 2 1 2



Substituteminitialmconditions:mx0m=m1mmm,mv0m=5 mm/s

xm0mmc1mmc2m mx0m m1mmc2m m1mmc1,m andm v  0  mmxm0mm2ic1m m2 5 mm/s
ic2m mv0m  




m2c1mm2c2m 5mi.m Combiningmthemtwomunderlinedmexpressionsm(2meqsminm2munkowns):
1m 1m 5
2c1 m2mm2  m i, m andm c m  i
5
c1 5mimm  2 2 4
4
c1 2

Thereforemthemsolutionmis:

m1 m 2it
5  2i m1 5
xmm m  e
it m m
mm i m e

m
4 
m2 4  m2
UsingmthemEulermformulamtomevaluatemthemexponentialmtermsmyields:
m1 5  m1
xmm m 5 
    2
m m

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, im cosm2tmmimsinm2tmmm m 
m im cosm2tmmimsinm2tm
4  m2 4 
3m
mx(tm)mmcosm2tm5 sinm2tm sin2tmm0.7297
2 2




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