At = 0.2 1x=1 = 0.25 4 using the formulas for B, C, D and E 2 1 [st 0.2 = Ax 0.25 C = 2-2 D= [+] F 12 = 0.64 2 =2-2×0.64 =2-120 -0.72 ΔΚ 2 = 0.64 E = 1 0.72 0.64 0 0.64 0.72 6.64 ○ [A]= 0 0.64 0.72 0.64 0 0 0 0.64 0.72 0.64 0.64 0.72 Part C: Assemble the Equations into a matrix form. Matrix Form {y} [A]{y}*+ {y} + {BC} у {y} = Y₂+ 43+ Удо 2 Yst DOO [A] = B C D O 0 D B C 0 10 080 D 0 B C D B C

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d) Initially, the string is at rest at the profile {y2  y3  y4}0 = {0.4  0.7  0.3}. Use the answer in 
part c) to find {y2  y3  y4}1. Show all the necessary calculations. 
 
e) Thus, further calculations to show one cycle (periodical process) are obtained by tabulating 
a table.  
 
f) Based on the problem statement given in part e), use MS Excel to plot a 
graph of displacement y against x. 

At = 0.2
1x=1 = 0.25
4
using the formulas for B, C, D and E
2
1
[st
0.2
=
Ax
0.25
C = 2-2
D= [+]
F
12
= 0.64
2
=2-2×0.64
=2-120 -0.72
ΔΚ
2
= 0.64
E = 1
0.72
0.64
0
0.64
0.72
6.64
○
[A]=
0
0.64
0.72
0.64
0
0
0
0.64
0.72
0.64
0.64
0.72
Transcribed Image Text:At = 0.2 1x=1 = 0.25 4 using the formulas for B, C, D and E 2 1 [st 0.2 = Ax 0.25 C = 2-2 D= [+] F 12 = 0.64 2 =2-2×0.64 =2-120 -0.72 ΔΚ 2 = 0.64 E = 1 0.72 0.64 0 0.64 0.72 6.64 ○ [A]= 0 0.64 0.72 0.64 0 0 0 0.64 0.72 0.64 0.64 0.72
Part C: Assemble the Equations
into
a matrix form.
Matrix Form {y} [A]{y}*+ {y} + {BC}
у
{y} =
Y₂+
43+
Удо
2
Yst
DOO
[A] =
B C
D O
0
D
B C
0
10
080
D
0
B
C
D
B
C
Transcribed Image Text:Part C: Assemble the Equations into a matrix form. Matrix Form {y} [A]{y}*+ {y} + {BC} у {y} = Y₂+ 43+ Удо 2 Yst DOO [A] = B C D O 0 D B C 0 10 080 D 0 B C D B C
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