Engineering Vibration 5th Edition Inman Solution Manual 2025 I Complete Chapters 1-8 I 758 Pages I Verified Step-by-Step Solutions I A+ Guide

All 8 Chapters Covered
k k k




SOLUTION MANUAL
k

,ProblemskandkSolutionskSectionk1.1k(1.1kthroughk1.19)

1.1 ThekspringkofkFigurek1.2kisksuccessivelykloadedkwithkmasskandkthekcorrespondingk(stat
ic)kdisplacementkiskrecordedkbelow.k Plotkthekdatakandkcalculatekthekspring'skstiffness.k
Notekthatkthekdatakcontainksomekerror.k Alsokcalculatekthekstandardkdeviation.

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-bodykdiagram: Fromkthekfree-
bodykdiagramkandkstatickequilibrium:
kx
kxk =k mgk k (gk=k 9.81mk/ksk2)
k kk =k mgk/kx

ki
m k=k =k 86.164
n
mg

20
Theksamplekstandardkdeviationkinkcom
putedkstiffnesskis:
n
2
m 15 i

k k i=1
=k 0.164
=


10
0 1 2
x
Plotkofkmasskinkkgkversuskdisplacementkinkm
Computationkofkslopekfromkmg/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 Derivektheksolutionkofk m˙x˙k+kkxk=k0k andkplotkthekresultkforkatkleastktwokperiodskforkthekcase
withknk=k2krad/s,kx0k=k1kmm,kandkv0 5 mm/s.
k=


Solution:

Given:
mxk+kkxk =k0 (1)
Assume:k x(t)k=kaertk.k Then xk =kare and xk=kark e k.k Substitutekintokequationk(1)kto
rt 2 rt

:kget:
mar2ertk +kkaertk =k0
mr2k +kkk =k0
k
rk =k  i
m
Thusktherekarektwoksolutions:
k k
k
k k
x1k =kc1e ,k andk x2k =kc2e k


k
wherek(n k =k2k rad/s
= m

Theksumkofkx1kandkx2kiskalsokaksolutionksokthatkthektotalksolutionkis:

xk=k xk +kx =kck e2itk +kck e2it
1 2 1 2



Substitutekinitialkconditions:kx0k=k1kmm,kv0k= 5 mm/s

xk(0)k=kc1k+kc2k =kx0k =k1kkc2k =k1kkc1,k andk v ( 0 ) k=kxk(0)k=k2ic1k k2ic2k 5 mm/s
=kv0k =
k2c1k+k2c2k= 5ki.k Combiningkthektwokunderlinedkexpressionsk(2keqskink2kunkowns):
1k 1k 5
2c1 +k2kk2c = =k i, k andk c =k + i
5
5kikk 2 4

1 2
4
c1 2

Thereforektheksolutionkis:

k1 k 2it
5  2it k1 5
xk=k k ik e +k+
kk i k e

k 4 
4 
k2 k2
UsingkthekEulerkformulaktokevaluatekthekexponentialktermskyields:
k1 5  k1
xk=k k 5 
   2
k k
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, ik (cosk2tk+kiksink2tk)k+k k k+ ik (cosk2tkkiksink2tk)
4  k2 4 
3k
(kx(tk)k=kcosk2tk+ 5 sink2tk= sin(2tk+k0.7297)
2 2




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