Mechanical Vibrations - Assignment

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Mechanical Vibrations
Student’s Name
Institutional Affiliation
Date

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Question 1
a)
b)
Force acting at A,
Fa=Ka x
Force acting at B,
Fb=Kb x
Force acting at D
Fd=C ̇x
a=m ̈x
Summing the forces,
Fd + Fb=Fa

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m ̈x +C ̇x + Kb x=Ka x
m ̈x +C ̇x + Kb x−Ka x=0
m ̈x +C ̇x + ( K b−Ka ) x=0
10 ̈x+ 200 ̇x + ( 150−53 ) x=0
10 ̈x+ 200 ̇x +97 x=0
c)
∑ T =I o ̈θ
−a Fa−b Fb−L Fd=I o ̈θ
−Ka a ( ya ) −Kb a ( yb ) b−CL ( ̇yd )= ( mL2 ) ̈θ
Where,
I o=mk2=mL2
ya=aθ
yb=bθ
yd =Lθ
−Ka a2 θ−Kb b2 θ−C L2 ̇θ=mL2 ̈θ
mL2 ̈θ+C L2 ̇θ+ ( Ka a2 + Kb b2 ) θ=0
Substituting the given values we have,
10(0.3)2 ̈θ+200(0.3)2 ̇θ+ (53( 0.2)2 +150(0.1)2 ) θ=0
0.9 ̈θ+18 ̇θ+3.62 θ=0
d)
Solving the equation of motion,
0.9 ̈θ+18 ̇θ+3.62 θ=0
θ=−b ± √b2 −4 ac
2 a , a=0.9, b=18 , c=3.62
θ=−18 ± √ 182−4 (0.9)(3.62)
1.8
θ=−18 ± √324−13.032
1.8

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θ=−18 ± 17.63
1.8
θ=−19.79∨−0.21
Solution,
θ ( t ) =c1 e−19.79t + c2 e−0.21 t
θ ( 0 ) =c1+ c2=15
̇θ ( t ) =−19.79 c1 e−19.79 t −0.21 c2 e−0.21 t
̇θ ( 0 ) =−19.79 c1−0.21 c2=0
19.79 c1 +0.21 c2=0
c2=94.24 c1
From,
c1 +c2=15
c1 +94.24 c1=15
95.24 c1 =15
c1=0.1575
c2=94.24 ( 0.1575 ) =14.8428
θ ( t )=0.1575 e−19.79 t +14.8428 e−0.21 t
Question 2
i.