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R1.1: Spring-dash pot system in parallel with a mass and applied force f(t) [ edit ] Initial Information [ edit ] From lecture slide 1-4
Variables:
k
,
{\displaystyle k,}
c
,
{\displaystyle c,}
m
,
{\displaystyle m,}
f
(
t
)
,
{\displaystyle f(t),}
y
(
t
)
,
{\displaystyle y(t),}
y
k
,
{\displaystyle y_{k},}
y
c
,
{\displaystyle y_{c},}
f
i
,
{\displaystyle f_{i},}
f
k
,
{\displaystyle f_{k},}
f
c
{\displaystyle f_{c}}
Methods [ edit ] Kinematics :
y
=
y
k
=
y
c
{\displaystyle \displaystyle y=y_{k}=y_{c}}
Derived from (Eq.1)
Kinetics :
f
(
t
)
=
m
y
k
″
+
c
y
k
′
+
k
y
k
{\displaystyle \displaystyle f(t)=my''_{k}+cy'_{k}+ky_{k}}
Solution [ edit ] Final Equation :
f
(
t
)
=
m
y
k
″
+
c
y
k
′
+
k
y
k
{\displaystyle \displaystyle f(t)=my''_{k}+cy'_{k}+ky_{k}}
R1.2: Spring-mass-dashpot with applied force r(t) on the ball(Fig. 53, p.85, K2011) [ edit ] Initial Information [ edit ] Variables:
k
,
{\displaystyle k,}
c
,
{\displaystyle c,}
m
,
{\displaystyle m,}
f
(
t
)
,
{\displaystyle f(t),}
y
k
,
{\displaystyle y_{k},}
f
I
,
{\displaystyle f_{I},}
f
k
,
{\displaystyle f_{k},}
f
c
{\displaystyle f_{c}}
Methods [ edit ] Kinematics:
y
=
y
k
{\displaystyle \displaystyle y=y_{k}}
Kinetics:
C
d
2
v
c
d
t
2
=
d
i
d
t
{\displaystyle \displaystyle C{\frac {d^{2}v_{c}}{dt^{2}}}={\frac {di}{dt}}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \displaystyle (Eq.{1}') </p> |} ==Solution== It is blank here. =R1.3 Spring-dashpot-mass system FBD and Equation of Motion= <br /> Problem found on slide 1-6 ==Initial Information== From lecture slide 1-4, the spring-dashpot-mass system:<br /> [image, or link to image on earlier] ==Methods== ==Solution== =R1.4: RLC Circuit Modeling= ==Initial Information== From lecture slide 2-2, a general RLC circuit Kirchhoff's Voltage Law (KVL) equation, and two alternative formulations, are given: :{| style="width:100%" border="0" align="left" |- |<math>\displaystyle V = LC \frac{d^{2}v_{c}}{dt^{2}} + RC \frac{dv_{c}}{dt}+ v_{c}}
(
E
q
.2
)
{\displaystyle \displaystyle (Eq.2)}
V
′
=
L
I
″
+
R
I
′
+
1
C
I
{\displaystyle \displaystyle {V}'=L{I}''+R{I}'+{\frac {1}{C}}I}
(
E
q
.3
)
{\displaystyle \displaystyle (Eq.3)}
V
=
L
Q
″
+
R
Q
′
+
1
C
Q
{\displaystyle \displaystyle V=L{Q}''+R{Q}'+{\frac {1}{C}}Q}
(
E
q
.4
)
{\displaystyle \displaystyle (Eq.4)}
We are being asked to derive (3) and (4) from (2).
Methods [ edit ] From lecture slide 2-2, capacitance is defined as,
Q
=
C
v
c
⇒
∫
i
d
t
=
C
v
c
⇒
i
=
C
d
v
c
d
t
{\displaystyle \displaystyle Q=Cv_{c}\Rightarrow \int idt=Cv_{c}\Rightarrow i=C{\frac {dv_{c}}{dt}}}
(
E
q
.1
)
{\displaystyle \displaystyle (Eq.1)}
Solution [ edit ] Deriving (1), we get:
C
d
2
v
c
d
t
2
=
d
i
d
t
{\displaystyle \displaystyle C{\frac {d^{2}v_{c}}{dt^{2}}}={\frac {di}{dt}}}
(
E
q
.
1
′
)
{\displaystyle \displaystyle (Eq.{1}')}
Also, by solving (1) for
v
c
{\displaystyle v_{c}}
v
c
=
1
C
∫
i
d
t
{\displaystyle \displaystyle v_{c}={\frac {1}{C}}\int idt}
(
E
q
.
1
″
)
{\displaystyle \displaystyle (Eq.{1}'')}
Substituting equations (1), (1'), and (1") into (2)
V
=
L
I
′
+
R
I
+
1
C
∫
I
{\displaystyle \displaystyle V=L{I}'+RI+{\frac {1}{C}}\int I}
(
E
q
.
2
′
)
{\displaystyle \displaystyle (Eq.{2}')}
Which is an "integro-differential equation." Therefore, to eliminate the integral we differentiate (2') with respect to t , to get:
V
′
=
L
I
″
+
R
I
′
+
1
C
I
{\displaystyle \displaystyle {V}'=L{I}''+R{I}'+{\frac {1}{C}}I}
(
E
q
.3
)
{\displaystyle \displaystyle (Eq.3)}
Since
Q
=
∫
i
d
t
{\displaystyle Q=\int idt}
V
=
L
Q
″
+
R
Q
+
1
C
Q
{\displaystyle \displaystyle V=L{Q}''+RQ+{\frac {1}{C}}Q}
(
E
q
.4
)
{\displaystyle \displaystyle (Eq.4)}
R1.5: General Solution of ODE [ edit ] Initial Information [ edit ] From[1]
y
″
+
4
y
′
+
(
π
2
+
4
)
y
=
0
{\displaystyle \displaystyle {y}''+4{y}'+({{\pi }^{2}}+4)y=0}
(
E
q
.4
)
{\displaystyle \displaystyle (Eq.4)}
[1]
y
″
+
2
π
y
′
+
π
2
y
=
0
{\displaystyle \displaystyle {y}''+2\pi {y}'+{{\pi }^{2}}y=0}
(
E
q
.5
)
{\displaystyle \displaystyle (Eq.5)}
Find a general solution for Equations (4) and (5) and check the answer by substitution.
Methods [ edit ] Solution [ edit ] Initial Information [ edit ] We are asked to determine the order, linearity and whether the principle of superposition can be applied to the following examples.
order of a differential equation is found by looking at the highest occurring derivative of the dependent variable.
linear if the dependent variable and all of its derivatives occur linearly throughout the equation.
Falling Stone
Governing Equation:
y
″
=
g
=
c
o
n
s
t
a
n
t
{\displaystyle \displaystyle {y}''=g=constant}
Order: 2
Linearity: Yes
Parachutist
Governing Equation:
m
v
′
=
m
g
−
b
v
2
{\displaystyle \displaystyle {mv'}=mg-bv^{2}}
Order: 1
Linearity: No
Outflowing water from a tank
Govering Equation:
h
′
=
−
k
h
{\displaystyle \displaystyle {h'}=-k{\sqrt {h}}}
Order: 1
Linearity: No
Vibrating mass on a spring
Governing Equation:
m
y
″
+
k
y
=
0
{\displaystyle \displaystyle {my''+ky}=0}
Order: 2
Linearity: Yes
Beats of a vibrating system
Governing Equation:
y
″
+
ω
2
y
=
c
o
s
ω
t
{\displaystyle \displaystyle {y''+\omega ^{2}y}=cos\omega {t}}
Order: 2
Linearity: Yes
Current I in an RLC Circuit
Governing Equation:
V
′
=
L
I
″
+
R
I
′
+
1
C
I
{\displaystyle \displaystyle {V}'=L{I}''+R{I}'+{\frac {1}{C}}I}
Order: 2
Linearity: Yes
Beam Deformation
Governing Equation:
E
I
y
i
v
=
f
(
x
)
{\displaystyle \displaystyle {EIy^{iv}}=f(x)}
Order: 0
Linearity: No
Pendulum
Governing Equation:
L
θ
″
+
g
∗
s
i
n
(
θ
)
=
0
{\displaystyle \displaystyle {L\theta ''+g*sin(\theta )}=0}
Order: 2
Linearity: Yes
Solution [ edit ] References [ edit ]
^ a b Kreyszig, "Advanced Engineering Mathematics," John Wiley & Sons, 2011.
a
Solution [ edit ] a
a
