1
Lecture 21
1.1
Free undamped motion
mx
00
+
kx
= 0
:
Set
k
m
=
!
2
0
.
Then we have
x
00
+
!
2
0
x
= 0
:
Characteristic equation:
r
2
+
!
2
0
= 0
)
r
=
&
!
0
i
.
General solution:
x
(
t
) =
A
cos
!
0
t
+
B
sin
!
0
t:
We can rewrite the foregoing formula as follows:
x
(
t
) =
A
cos
!
0
t
+
B
sin
!
0
t
=
p
A
2
+
B
2
&
A
p
A
2
+
B
2
cos
!
0
t
+
B
p
A
2
+
B
2
sin
!
0
t
±
:
Because
&
A
p
A
2
+
B
2
±
2
+
&
B
p
A
2
+
B
2
±
2
= 1
;
there is
&
such that
cos
&
=
A
p
A
2
+
B
2
and
sin
&
=
B
p
A
2
+
B
2
:
So, we have
x
(
t
) =
p
A
2
+
B
2
(cos
&
cos
!
0
t
+ sin
&
sin
!
0
t
)
=
p
A
2
+
B
2
cos (
!
0
t
±
&
)
:
Set
C
=
p
A
2
+
B
2
:
x
(
t
) =
C
cos (
!
0
t
±
&
) =
C
cos
!
0
&
t
±
&
!
0
±
1
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View Full Documentis called simple
C
!
0
&
2
^
/
g
0
J
/
g
0
x
00
+
!
2
0
x
= 0
:
x
(
t
) =
C
cos
!
0
&
t
&
&
!
0
±
Let
T
denote the time (in seconds) needed to perform one cycle. Complete
cycle corresponds to the angle
2
±
radians. So,
T
=
2
±
!
0
is the period of vibration in seconds.
1
T
=
!
0
2
±
& frequency of vibration in Hertz=cycle/second.
Example 1
Determine the period and frequency of the simple harmonic mo
tion of a body of mass 0.75 kg on the end of a spring with spring constant
k
= 48
n=m
. Neglect friction.
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 Fall '08
 Staff
 Differential Equations, Equations, Constant of integration, 1 m, 0 m

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