Problem 6
edit
Problem Statement
edit
Realize spring-dashpot-mass systems in series as shown in Fig. p.1-4 with the similar characteristic as in (3) p.5-5, but with double real root
λ
=
−
3
{\displaystyle \lambda =-3}
Solution
edit
Recall the equation of motion for the spring dashpot mass system:
m
(
y
k
″
+
k
c
y
k
′
)
+
k
y
k
=
f
(
t
)
{\displaystyle m(y''_{k}+{\frac {k}{c}}y'_{k})+ky_{k}=f(t)\!}
y
k
″
+
k
c
m
y
k
′
+
k
m
y
k
=
f
(
t
)
{\displaystyle y''_{k}+{\frac {k}{cm}}y'_{k}+{\frac {k}{m}}y_{k}=f(t)\!}
λ
=
−
3
{\displaystyle \lambda =-3\!}
(
λ
+
3
)
2
=
λ
2
+
6
λ
+
9
=
0
{\displaystyle (\lambda +3)^{2}=\lambda ^{2}+6\lambda +9=0\!}
y
″
+
6
y
′
+
9
=
0
{\displaystyle y''+6y'+9=0\!}
y
″
⇒
1
=
1
{\displaystyle y''\Rightarrow 1=1\!}
y
′
⇒
k
c
m
=
6
{\displaystyle y'\Rightarrow {\frac {k}{cm}}=6\!}
y
⇒
k
m
=
9
{\displaystyle y\Rightarrow {\frac {k}{m}}=9\!}
k
=
18
{\displaystyle k=18\!}
c
=
3
2
{\displaystyle c={\frac {3}{2}}\!}
m
=
2
{\displaystyle m=2\!}
Solved and typed by - Egm4313.s12.team4.Lorenzo 20:04, 6 February 2012 (UTC)
Problem 7
edit
Problem Statement
edit
Develop the MacLaurin series (Taylor series at t=0) for:
e
t
{\displaystyle e^{t}\!}
cos
t
{\displaystyle \cos t\!}
sin
t
{\displaystyle \sin t\!}
Solution
edit
Recalling Euler's Formula:
e
i
ω
x
=
cos
ω
x
+
i
sin
ω
x
{\displaystyle e^{i\omega x}=\cos \omega x+i\sin \omega x\!}
e
x
{\displaystyle e^{x}}
x
=
0
{\displaystyle x=0}
e
x
=
∑
n
=
0
∞
x
n
n
!
{\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\!}
e
t
{\displaystyle e^{t}}
e
t
=
∑
n
=
0
∞
x
n
n
!
{\displaystyle e^{t}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}\!}
i
2
k
=
(
i
2
)
k
=
(
−
1
)
k
{\displaystyle i^{2k}=(i^{2})^{k}=(-1)^{k}\!}
i
2
k
+
1
=
(
i
2
)
k
i
=
(
−
1
)
k
i
{\displaystyle i^{2k+1}=(i^{2})^{k}i=(-1)^{k}i\!}
x
=
i
t
{\displaystyle x=it}
e
i
t
=
∑
n
=
0
∞
i
n
t
n
n
!
=
∑
k
=
0
∞
i
2
k
t
2
k
(
2
k
)
!
+
∑
k
=
0
∞
i
2
k
+
1
t
2
k
+
1
(
2
k
+
1
)
!
{\displaystyle e^{it}=\sum _{n=0}^{\infty }{\frac {i^{n}t^{n}}{n!}}=\sum _{k=0}^{\infty }{\frac {i^{2k}t^{2k}}{(2k)!}}+\sum _{k=0}^{\infty }{\frac {i^{2k+1}t^{2k+1}}{(2k+1)!}}\!}
e
i
t
=
∑
k
=
0
∞
(
−
1
)
k
t
2
k
(
2
k
)
!
+
∑
k
=
0
∞
(
−
1
)
k
t
2
k
+
1
(
2
k
+
1
)
!
{\displaystyle e^{it}=\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k}}{(2k)!}}+\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}\!}
⇒
e
i
t
=
cos
t
+
i
sin
t
{\displaystyle \Rightarrow e^{it}=\cos t+i\sin t\!}
cos
t
=
∑
k
=
0
∞
(
−
1
)
k
t
2
k
(
2
k
)
!
{\displaystyle \cos t=\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k}}{(2k)!}}\!}
sin
t
=
∑
k
=
0
∞
(
−
1
)
k
t
2
k
+
1
(
2
k
+
1
)
!
{\displaystyle \sin t=\sum _{k=0}^{\infty }{\frac {(-1)^{k}t^{2k+1}}{(2k+1)!}}\!}
Solved and typed by - Egm4313.s12.team4.Lorenzo 20:05, 6 February 2012 (UTC)
References
edit
