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some answers:
1a) 2 cycles per second, so
ω
= 4
π
1b)
m
= 100 000,
k
= 80 000
g/.
02 = 4(10)
6
g
.02
80 000g
ky
y
1c) Since
y
=
.
003 sin(
ωt
), we get
my
±±
+
ky
=
.
003(

ω
2
m
+
k
) sin(
ωt
), so
f
0
=
.
003(

ω
2
m
+
k
). (It was not required but this works out to
.
003(

16
π
2
10
5
+
4(10)
6
g
)
.
=
.
003(

16 + 40)(10)
6
.
= 70 000 Newtons approx.)
1d) The empty car has unforced equation 20 000
y
±±
+
ky
= 0 with circular
frequency
q
k
20 000
=
q
80 000
g
(
.
02)20 000
, while the pendulum has equation
θ
±±
=

g
l
θ
with circular frequency
p
g
l
. These agree when
1
l
=
4
.
02
, or
l
=
.
005 meter.
2) The characteristic polynomial is 2
r
2
+ 5
r
+ 3 = (2
r
+ 3)(
r
+ 1), so the
general solution to the homogeneous equation is
y
c
=
c
1
e

x
+
c
2
e

3
2
x
. Un
determined coeﬃcient method gives a particular solution
y
p
=

4
25
sin(2
x
)

8
25
cos(2
x
)

2. Adding these, then applying initial conditions we ﬁnd
y
(
x
) =
15
25
e

x
+
18
25
e

3
2
x
+
y
p
(
x
).
3) Integrate four times (deﬁnite integrals save space and time) to ﬁnd
EIy
(
x
) =
Lx
5
120

x
6
360

L
3
x
3
36
+
L
4
x
2
24
and at the end
EIy
(
L
) =
L
6
(
1
120

1
360

1
36
+
1
24
)
. Since 120
<
360 and 24
<
36, the number in parenthesis is positive.
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This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 TERRELL,R

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