Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains
Chapter 3-11
Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains
Chapter 3-11
,Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11
Chapter 3
Modeling of Physical Processes
Problem 3.1: Ground Level
Determine the solutions to the following problems.
a 𝑑𝑑𝑑 1
= − 𝑑𝑑
𝑑𝑡 𝜏
𝑑𝑑 = 0 , 𝑡 = 0
b 𝑑2𝑥 𝑑𝑥
𝑚 +𝑐 +𝑘∙𝑥 =0
𝑑𝑡2 𝑑𝑡
𝑑𝑥
𝑥 =0, =0, 𝑡 =0
𝑑𝑡
c 𝜕2𝑇 𝜕2𝑇
+ =0
𝜕𝑥2 𝜕𝑦2
𝑇 = 0 on all boundaries
d 𝜕𝑇 𝜕2𝑇
=𝛼
𝜕𝑡 𝜕𝑥2
𝜕𝑇
=0 , 𝑥=0
𝜕𝑥
𝑇=0 , 𝑥=𝐿
𝑇 =0, 𝑡 =0
Solution
All these cases have zero forcing functions, thus they all remain in equilibrium.
a) 𝑑𝑑 = 0
b) 𝑥 = 0
c) 𝑇 = 0
d) 𝑇 = 0
1
,Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11
Problem 3.2: Mathematical Models
Consider the following systems of equations. In each case identify if the mathematical model is
formulated properly in order to allow for the possibility of a unique solution. If the system is not
properly formulated, what additions or changes are necessary?
a 𝑑𝑑𝑑 1
= − 𝑑𝑑 + 𝑆
𝑑𝑡 𝜏
b 𝑑𝑑𝑑 1
= − 𝑑𝑑 + 𝑆
𝑑𝑡 𝜏
𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0
𝑑𝑑 = 𝑑𝑑1 , 𝑡 = 𝑡1
c 𝑑𝑥
2 𝑑𝑥
𝑚 +𝑐 + 𝑘 ∙ 𝑥 = 𝑓(𝑡)
𝑑𝑡2 𝑑𝑡
𝑥 = 𝑥0 , 𝑡 = 0
d 𝑑𝑥
2 𝑑𝑥
𝑚 +𝑐 + 𝑘 ∙ 𝑥 = 𝑓(𝑡)
𝑑𝑡2 𝑑𝑡
𝑥 = 𝑥0 , 𝑡 = 0
𝑥 = 𝑥1 , 𝑡 = 𝑡1
e 𝜕𝑇 𝜕 𝑇 𝑔(𝑥, 𝑡)
2
=𝛼 + , 0<𝑥<𝐿
𝜕𝑡 𝜕𝑥2 𝜌𝑐
𝜕𝑇
−𝑘 = 𝑞" (𝑡) , 𝑥 = 0
𝜕𝑥 0
f 𝜕𝑇 𝜕2𝑇 𝑔(𝑥, 𝑡)
=𝛼 + , 0<𝑥<𝐿
𝜕𝑡 𝜕𝑥2 𝜌𝑐
𝜕𝑇
−𝑘 = 𝑞" (𝑡) , 𝑥 = 0
𝜕𝑥 0
𝑇 = 𝑇0(𝑡) , 𝑥=0
𝑇 = 𝑇0 , 𝑡 = 0
Solution
a Requires an initial condition 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0
b Too many initial conditions. Remove 𝑑𝑑 = 𝑑𝑑1 , 𝑡 = 𝑡1
c Requires a second initial condition 𝑑𝑥 = 𝑣 0 , 𝑡 = 0
𝑑𝑡
d Remove 𝑥 = 𝑥1 , 𝑡 = 𝑡1
Requires a second initial condition 𝑑𝑥 = 𝑣 0 , 𝑡 = 0
𝑑𝑡
e Requires a boundary condition at x=L.
Requires an initial condition.
f Requires a boundary condition at x=L.
Over specified at x+L. Remove one of the conditions.
2
, Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11
Problem 3.3: Coupled Mass-Spring-Dampers
Consider the following system with two sets of mass-spring-dampers. A force f1 is applied directly
to mass-1 while a force f2 is applied to mass-2. The spring constants are k1 and k2, and the damping
coefficients are b1 and b2. The equilibrium positions with f1=f2=0 correspond to x1=x2=0.
x1 x2
b2
f1 f2
k2
Derive the complete mathematical model of this system.
Solution
Draw free body diagrams of the two masses and apply Newton’s 2nd law.
Equation of Motion for Mass-1
𝑚1𝑥1 = � 𝑓𝑜𝑟𝑐𝑒𝑠
𝑚1𝑥1 = −𝑏1𝑥1 − 𝑏2(𝑥1 − 𝑥2) − 𝑘1𝑥1 − 𝑘2(𝑥1 − 𝑥2) + 𝑓1
𝑚1𝑥1 + 𝑏1𝑥1 + 𝑏2(𝑥1 − 𝑥2) + 𝑘1𝑥1 + 𝑘2(𝑥1 − 𝑥2) = 𝑓1
Equation of Motion for Mass-2
𝑚2𝑥2 = � 𝑓𝑜𝑟𝑐𝑒𝑠
𝑚2𝑥2 = −𝑏2(𝑥2 − 𝑥1) − 𝑘2(𝑥2 − 𝑥1) + 𝑓2
𝑚2𝑥2 + 𝑏2(𝑥2 − 𝑥1) + 𝑘2(𝑥2 − 𝑥1) = 𝑓2
Initial conditions at t=0
𝑥1 = 𝑥1,0
𝑥2 = 𝑥2,0
𝑥1 = 𝑣1,0
𝑥2 = 𝑣2,0
3
Chapter 3-11
Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains
Chapter 3-11
,Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11
Chapter 3
Modeling of Physical Processes
Problem 3.1: Ground Level
Determine the solutions to the following problems.
a 𝑑𝑑𝑑 1
= − 𝑑𝑑
𝑑𝑡 𝜏
𝑑𝑑 = 0 , 𝑡 = 0
b 𝑑2𝑥 𝑑𝑥
𝑚 +𝑐 +𝑘∙𝑥 =0
𝑑𝑡2 𝑑𝑡
𝑑𝑥
𝑥 =0, =0, 𝑡 =0
𝑑𝑡
c 𝜕2𝑇 𝜕2𝑇
+ =0
𝜕𝑥2 𝜕𝑦2
𝑇 = 0 on all boundaries
d 𝜕𝑇 𝜕2𝑇
=𝛼
𝜕𝑡 𝜕𝑥2
𝜕𝑇
=0 , 𝑥=0
𝜕𝑥
𝑇=0 , 𝑥=𝐿
𝑇 =0, 𝑡 =0
Solution
All these cases have zero forcing functions, thus they all remain in equilibrium.
a) 𝑑𝑑 = 0
b) 𝑥 = 0
c) 𝑇 = 0
d) 𝑇 = 0
1
,Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11
Problem 3.2: Mathematical Models
Consider the following systems of equations. In each case identify if the mathematical model is
formulated properly in order to allow for the possibility of a unique solution. If the system is not
properly formulated, what additions or changes are necessary?
a 𝑑𝑑𝑑 1
= − 𝑑𝑑 + 𝑆
𝑑𝑡 𝜏
b 𝑑𝑑𝑑 1
= − 𝑑𝑑 + 𝑆
𝑑𝑡 𝜏
𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0
𝑑𝑑 = 𝑑𝑑1 , 𝑡 = 𝑡1
c 𝑑𝑥
2 𝑑𝑥
𝑚 +𝑐 + 𝑘 ∙ 𝑥 = 𝑓(𝑡)
𝑑𝑡2 𝑑𝑡
𝑥 = 𝑥0 , 𝑡 = 0
d 𝑑𝑥
2 𝑑𝑥
𝑚 +𝑐 + 𝑘 ∙ 𝑥 = 𝑓(𝑡)
𝑑𝑡2 𝑑𝑡
𝑥 = 𝑥0 , 𝑡 = 0
𝑥 = 𝑥1 , 𝑡 = 𝑡1
e 𝜕𝑇 𝜕 𝑇 𝑔(𝑥, 𝑡)
2
=𝛼 + , 0<𝑥<𝐿
𝜕𝑡 𝜕𝑥2 𝜌𝑐
𝜕𝑇
−𝑘 = 𝑞" (𝑡) , 𝑥 = 0
𝜕𝑥 0
f 𝜕𝑇 𝜕2𝑇 𝑔(𝑥, 𝑡)
=𝛼 + , 0<𝑥<𝐿
𝜕𝑡 𝜕𝑥2 𝜌𝑐
𝜕𝑇
−𝑘 = 𝑞" (𝑡) , 𝑥 = 0
𝜕𝑥 0
𝑇 = 𝑇0(𝑡) , 𝑥=0
𝑇 = 𝑇0 , 𝑡 = 0
Solution
a Requires an initial condition 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0
b Too many initial conditions. Remove 𝑑𝑑 = 𝑑𝑑1 , 𝑡 = 𝑡1
c Requires a second initial condition 𝑑𝑥 = 𝑣 0 , 𝑡 = 0
𝑑𝑡
d Remove 𝑥 = 𝑥1 , 𝑡 = 𝑡1
Requires a second initial condition 𝑑𝑥 = 𝑣 0 , 𝑡 = 0
𝑑𝑡
e Requires a boundary condition at x=L.
Requires an initial condition.
f Requires a boundary condition at x=L.
Over specified at x+L. Remove one of the conditions.
2
, Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11
Problem 3.3: Coupled Mass-Spring-Dampers
Consider the following system with two sets of mass-spring-dampers. A force f1 is applied directly
to mass-1 while a force f2 is applied to mass-2. The spring constants are k1 and k2, and the damping
coefficients are b1 and b2. The equilibrium positions with f1=f2=0 correspond to x1=x2=0.
x1 x2
b2
f1 f2
k2
Derive the complete mathematical model of this system.
Solution
Draw free body diagrams of the two masses and apply Newton’s 2nd law.
Equation of Motion for Mass-1
𝑚1𝑥1 = � 𝑓𝑜𝑟𝑐𝑒𝑠
𝑚1𝑥1 = −𝑏1𝑥1 − 𝑏2(𝑥1 − 𝑥2) − 𝑘1𝑥1 − 𝑘2(𝑥1 − 𝑥2) + 𝑓1
𝑚1𝑥1 + 𝑏1𝑥1 + 𝑏2(𝑥1 − 𝑥2) + 𝑘1𝑥1 + 𝑘2(𝑥1 − 𝑥2) = 𝑓1
Equation of Motion for Mass-2
𝑚2𝑥2 = � 𝑓𝑜𝑟𝑐𝑒𝑠
𝑚2𝑥2 = −𝑏2(𝑥2 − 𝑥1) − 𝑘2(𝑥2 − 𝑥1) + 𝑓2
𝑚2𝑥2 + 𝑏2(𝑥2 − 𝑥1) + 𝑘2(𝑥2 − 𝑥1) = 𝑓2
Initial conditions at t=0
𝑥1 = 𝑥1,0
𝑥2 = 𝑥2,0
𝑥1 = 𝑣1,0
𝑥2 = 𝑣2,0
3
