Consider the double mass / triple spring system shown beloun WWWWWWWWWW springs have spring constant k=7 and all masses have mass m=1kg. spring is subject to a damping force of Ffriction = -3X, Using the standard conversion to a first-order system, he would obtain the system Each 0100 -14 -3 7 0 0001 27 0 14-35 All W³³ (+) = W = Aw / where (t) = The eigenvalues of A are approximately. 712 -1.5±4.33; and 713,4 -1.5 ± 2.187 The eigenvector for 21-1.5 +4.331 is approximately eigenvector for 3-1.5 +2.18 is approximately ) Xilt) xỉ (t X2 ( t) X₂ (+)) 0.07 +0.2/ -007-021; 1 _0.21 -0.31 -0.210.31 1 +27 +211 √i - 1.5+4.33%- / (a). Write general solution wit) of system of the differential equation above (b). Find an initial condition that will result in the masses displaying simple oscillations at relatively high frequency I the fart mode) provided the relaity is zero at to. Jutify the your answer. and the (01 Suppose that there is no friction, i.e., Friction -0. Which of the following could be an eigenvalue of the now corresponding matrix A?. √21

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Consider the double mass / triple spring system shown belown
WWWWWWWW
All
springs
have spring constant k=7 and all masses have mass m=1kg.
Each spring is subject to a damping force of Ffriction = -3X, Using the
standard conversion to a first-order system, he would obtain the
system
0100
-14 -3 7 0
0001
0 14-3
W²³³ (+) =
7
W = Aw, where w(t) =
The eigenvalues of A are approximately.
71,2 -1.5±4.33; and 713,4 -1.5 ± 2.18%
The eigenvector for 2₁-1.5 +4.331 is approximately
eigenvector for 73 -1.5 +2.18; is approximately
Xilt)
xỉ (t
X2 (t)
X₂(+))
0.07 +0.21t
-1
-007-0.21₂
0.21 -0.31ż
1
-0.21 -0.31
and the
(a). Write general solution wilt) of system of the differential equation above
(b). Find an initial condition that will result in the masses displaying simple
oscillations at relatively high frequency I the fart mode) provided the relacity
is zero at t-o. Jutify the your answer.
[c]. Suppose that there is no friction, i.e., Friction -o. Which of the following
could be an eigenvalue of the new corresponding matrix A?.
तरा, नश + नये, √, - 1.5+4337.
Transcribed Image Text:Consider the double mass / triple spring system shown belown WWWWWWWW All springs have spring constant k=7 and all masses have mass m=1kg. Each spring is subject to a damping force of Ffriction = -3X, Using the standard conversion to a first-order system, he would obtain the system 0100 -14 -3 7 0 0001 0 14-3 W²³³ (+) = 7 W = Aw, where w(t) = The eigenvalues of A are approximately. 71,2 -1.5±4.33; and 713,4 -1.5 ± 2.18% The eigenvector for 2₁-1.5 +4.331 is approximately eigenvector for 73 -1.5 +2.18; is approximately Xilt) xỉ (t X2 (t) X₂(+)) 0.07 +0.21t -1 -007-0.21₂ 0.21 -0.31ż 1 -0.21 -0.31 and the (a). Write general solution wilt) of system of the differential equation above (b). Find an initial condition that will result in the masses displaying simple oscillations at relatively high frequency I the fart mode) provided the relacity is zero at t-o. Jutify the your answer. [c]. Suppose that there is no friction, i.e., Friction -o. Which of the following could be an eigenvalue of the new corresponding matrix A?. तरा, नश + नये, √, - 1.5+4337.
Expert Solution
Step 1: Writing down the given information

Given the system of DE as

w with rightwards arrow on top apostrophe open parentheses t close parentheses equals A w with rightwards arrow on top, where A equals open square brackets table row 0 1 0 0 row cell negative 14 end cell cell negative 3 end cell 7 0 row 0 0 0 1 row 7 0 cell negative 14 end cell cell negative 3 end cell end table close square brackets.

The eigenvalues and its corresponding eigenvectors are

lambda subscript 1 comma 2 end subscript equals negative 1.5 plus-or-minus 4.33 i comma space v subscript 1 equals open square brackets table row cell 0.07 plus 0.21 i end cell row cell negative 1 end cell row cell negative 0.07 minus 0.21 i end cell row 1 end table close square brackets comma space v subscript 2 equals open square brackets table row cell 0.07 minus 0.21 i end cell row cell negative 1 end cell row cell negative 0.07 plus 0.21 i end cell row 1 end table close square brackets and

lambda subscript 3 comma 4 end subscript equals negative 1.5 plus-or-minus 2.18 i comma space v subscript 3 equals open square brackets table row cell negative 0.21 minus 0.31 i end cell row 1 row cell negative 0.21 minus 0.31 i end cell row 1 end table close square brackets comma space v subscript 4 equals open square brackets table row cell negative 0.21 plus 0.31 i end cell row 1 row cell negative 0.21 plus 0.31 i end cell row 1 end table close square brackets.


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