This preview shows pages 1–2. Sign up to view the full content.
Dr. Saxena
ME130_F’08
1
ME130 – APPLIED ENGINEERING ANALYSIS
Key to Exam #1: Fall 2008 – Section 02:
#1
A 5kg mass is hanging from a spring with a spring constant K = 30 N/m, and there is a dash pot
attached to the mass with a damping coefficient C = 25 Kg/sec.
The initial conditions are: y(0) = 1 m, and
y
′
(0) = 0.
(a) Derive the governing differential equation for the displacement y(t) of this mass
(b) Solve the equation to determine the displacement y(t).
(c) Calculate the percent reduction in the amplitude in 1 sec.
Solution:
(a) From the free body diagram
∑
F =
−
F
I
−
R
S
−
R
D
=
0, where
R
D
=
Spring force
=
K y(t),
K
=
spring constant
R
D
=
Viscous (damping) force
=
C dy/dt,
C
=
damping coefficient
F
I
=
Inertia force
=
m d
2
y/dt
2
So, the governing differential equation becomes
2
2
dy
M+
C
+
K
y
=
0
dt
dt
(Note: the weight is offset by initial stretch of the spring)
(b) This is a second order homogeneous differential equation, with constant coefficients, with the initial
conditions given as:
y(0)
=
1 m,
and y’(0)
=
0
The corresponding characteristic equation is:
m
2
+ am
+
b
=
0, where a = C/M
=
25/5
=
5, and b = K/M
=
30/5
=
6.
The two roots of this equation are:
()
( )
3
6
4
5
2
1
2
5
b
4
a
2
1
2
5
m
2
6
4
5
2
1
2
5
b
4
a
2
1
2
a
m
2
2
2
2
2
1
−
=
×
−
−
−
=
−
−
−
=
−
=
×
−
+
−
=
−
+
−
=
Therefore, the general solution of this equation is expressed as
t
3
t
2
t
m
t
m
e
B
e
A
e
B
e
A
)
t
(
y
2
1
−
−
+
=
+
=
Using the given initial conditions, the two constants are
A
=
3, and B
=
−
2.
Therefore, the specific
solution of the differential equation is
t
3
t
2
e
2
e
3
)
t
(
y
−
−
−
=
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 09/08/2010 for the course ME 130 at San Jose State University .
 '08
 Saxena,Dr.

Click to edit the document details