1
AA-312 FINAL EXAMINATION (Solutions)
Problem 1 (100 points)
x1(t)
k
c
x2(t)
k
c
m
F(t)
m
Smooth floor
Figure 1: A 2-DOF System.
(1)
(10 points) Show that equations of motion for the 2-DOF system (Figure 1) are
m
0
0
2c
x (t) +
m
c
c
2k
x (t) +
c
k
k
x

1.2
x
Derive the solution of m + kx = 0 and plot the result for at least two periods for the case
with n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s.
Solution:
Given:
(1)
m! + kx = 0
x
rt
2 rt
!
Assume: x (t) = ae . Then: x = are and ! = ar e . Substitute into

1.88
Repeat Problem 1.87 using plastic (E = 1.40 109 N/m2) and rubber (E = 7 106
N/m2). Are any of these feasible?
Solution:
From problem 1.53, k = 10 3 N/m =
EA
l
For plastic, E = 1.40 ! 109 N/m 2
So, l = 140 m
For rubber, E = 7 ! 10 6 N/m 2
So, l = 0.7

2- 29
2.40
Consider the base excitation problem for the configuration shown in Figure P2.40. In this
case the base motion is a displacement transmitted through a dashpot or pure damping
element. Derive an expression for the force transmitted to the suppor

AA-312
1
AA-312 Homework # 4 (Solutions)
Problem A
Using partial fraction expansion and inverse Laplace transform table, nd the response x(t)
for the following problems,
1. X (s) =
120
s(s + 1) (100s2 + 15s + 2500)
Partial fraction expansion yields
X (s)

AA-312 Homework # 6 (Solutions)
Additional Problem:
Given a two-mass-spring system described by the following equations of motion
20
05
x1 (t)
200 100
+
x2 (t)
100 100
x1 (t)
0
=
x2 (t)
F (t)
where the applied force is a harmonic excitation given by F (t)

Problem 6.17
We divide the problem into two parts:
(1) For 0 x l1 , the bar response is w(x,t) = a Cos
E1
1
where c1
(2) For l1 x l1
E2
2
where c2
w
c1
x
b Sin
w
c1
x
.
l2 , the bar response is w(x,t) = c Cos
w
c2
x
w
c2
d Sin
x
.
There are two boundary c

1
AA-312 Structural Vibrations MIDTERM (Solutions)
Problem 1 (10 points)
Express the response x(t) = 2 sin 2t 5 cos 2t in the form x(t) = A sin(t + ).
x( t ) =
2
5
29 sin 2t cos 2t
29
29
We have
A=
29 = 5.3852
= 2 (rad/sec)
5
= tan1
2
= 1.9513
x(t) = 5.