University Vibration Analysis Homework: Problems and Solutions

This document presents a comprehensive solution set for a mechanical engineering homework assignment focused on vibration analysis. The assignment addresses several key concepts, including the equation of motion for a vibrating system, and the calculation of system response under various initial conditions. The solutions also cover the determination of initial conditions to achieve oscillation at a single frequency, and the analysis of damped and undamped natural frequencies in a vibration system. Furthermore, the assignment explores a base excitation problem, examining the effects of base motion transmitted through a damper. The solutions provide detailed steps for each problem, offering a thorough understanding of the underlying principles of vibration analysis.

Frequency response
Question 2
a. ,
m ¨x + kx = 10 cos 10 t ……………… …………………………….Equation 1
1000 ¨x + 2000x = 10 cos 10 t
The complimentary solution is shown below for displacement
u ( t ) =c1 sos ( ω0 t ) +c2 sin ( ω0 t ) …………………… ……………………….Equation 2
ω0 is the natural frequency. it shall be assumed that ω ≠ ω0
The particular solution will be given as shown below
up ( t )= A cos ( ωt ) + B sin ( ωt ) ……………… …………………………Equation 3
Differentiating equation 3 we have
f 0 cos ( ωt ) =¿ ( cos ( ωt ) )∗( −mω2 A+ Ak ) + ( sin ( ωt ) ) ∗( −m ω2 B+ Bk ) … . ¿..Equation 4
Grouping the like terms and equating the coefficients
( cos ( ωt ) )∗(−mω2 A+ Ak ) ¿ f 0 cos ( ωt )
−m ω2 A + Ak= f 0
A= f 0
−mω2 +k
(−mω2 B+Bk ) + ( sin ( ωt ) )= 0
B=0
Particular equation now becomes
up ( t )= f 0
k−mω2 cos ( wt )
Now the whole solution can be written as shown below

u(t )=Rcos( ω0 t−δ)+ F 0
m ( ω0
2 −ω2 ) cos( ωt)
ω0= √ k
m
= √ ( 2000
100 )
= 4.4721
u(t )=c 1cos (10 t)+c 2sin (5 t)+ 10
2 ( 10 ) ( 10 ) tsin(10t )
After feeding the initial conditions
u(t )=0.1 cos( 10t )+ 0.01sin (5 t)+ 10
2 ( 10 ) ( 10 ) tsin(10 t )
Now evaluating the value of R
R= √ ( ( 0.1 ) 2 + ( 0.01 ) 2 )
= 0.0101
δ=tan−1
( 0.01
0.1 )
= 5.71
The displacement now becomes
u(t )=0.0101 cos(10 t +5.71)+ ( 1
20 )tsin(10 t)
b. ,
c. The displacement is
d. u(t )=0.0101 cos(10 t +5.71)+ ( 1
20 ) tsin(10 t) + 0.05

Question 3
f 0 cos ( ωt ) =¿ ( cos ( ωt ) )∗(−mω2 A+ Ak ) + ( sin ( ωt ) )∗( −m ω2 B+ Bk ) ¿
Can be written as
The first two terms in the right hand side can be dropped
To give equation 5 below
2 mω0 ( Bcos ( ωt ) − Asin ( ωt ) )=F0 cos (ωt ) ………………….Equation 5
now
A=0
B= f 0
2ω0 m
The particular equation now is
uparticular = f 0
2ω0 m tsin ( t ω0 ) …………………..…..Equation 6
The displacement now is given as
u(t )=Rcos(t ω0 −δ)+ F 0
2 ω0 m tsin(t ω0 )
Question 4
¨x +α ˙x +64 x=3 cos 6 t
x ( 0 )=x0 , ˙x ( 0 )= ˙x0
complimentary solution
uc (t)=c 1 e (−20+ 20 √ 2 ) t + c 2 e ( −20−20 √ 2 ) t
uc ( t )=c 1 e−8.2t +c 2 e−48.28t
the particular solutionis as follows
uparticular ( t ) = A cos 6t + B sin 6 t

Inserting into the differential equation it becomes
250A cos (5t) – 250 B sin (6t) = 3cos 6t
Grouping the like terms and sorting the coefficient
uparticular =( 3
250 )cos 6 t
The general solution now becomes
uc ( t )=c 1 e−8.2 t +c 2 e−48.28t + ( 3
250 ) cos 6 t
now applying the initial conditions
u ( t )=0.1455 e−8.2 t +¿0.431323e−48.28 t + ( 3
250 )cos 6 t
Question 5
The general equation is given as shown below
m d2 x
d t2 +c dx
dt +kx=0 since there is no external force
treating the above as an auxillary differential equation
Mr2 +cr +k=0 and the roots become
R1= −c + √ c2−4 mk
2 m or r2 = −c− √ c2−4 mk
2 m
Where r = x
Therefore the force will be given as
Force = k *(−c + √c2−4 mk
2 m )
Or
Force = k* (−c− √c2−4 mk
2 m )