How to change the system in 4x4 system of first-order equations? [duplicate]
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How to change the system in four first order equations?
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Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$y_1''=-frac{(k_1+k_2)}{m_1}y_1+frac{k_2}{m_1}y_2$
$y_2''=frac{k_2}{m_2}y_1-frac{(k_1+k_2)}{m_2}y_2$
Assume that the masses are $m_1 = 2$, $m_2 = 9/4$, and the spring constants are $k_1=1,k_2=3,k_3=15/4$.
a)use 4x4 system of first order equations to model this system of two second order equations. (hint: $x_1=y_1,x_2=y_2,x_3=y_1',x_4=y_2'$)
ordinary-differential-equations
asked Dec 5 '18 at 11:33
LTYLTY
285
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Dec 5 '18 at 12:35
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This question already has an answer here:
How to change the system in four first order equations?
2 answers
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$y_1''=-frac{(k_1+k_2)}{m_1}y_1+frac{k_2}{m_1}y_2$
$y_2''=frac{k_2}{m_2}y_1-frac{(k_1+k_2)}{m_2}y_2$
Assume that the masses are $m_1 = 2$, $m_2 = 9/4$, and the spring constants are $k_1=1,k_2=3,k_3=15/4$.
a)use 4x4 system of first order equations to model this system of two second order equations. (hint: $x_1=y_1,x_2=y_2,x_3=y_1',x_4=y_2'$)
ordinary-differential-equations
asked Dec 5 '18 at 11:33
LTYLTY
285
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marked as duplicate by LutzL differential-equations
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Dec 5 '18 at 12:35
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$begingroup$
This question already has an answer here:
How to change the system in four first order equations?
2 answers
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$y_1''=-frac{(k_1+k_2)}{m_1}y_1+frac{k_2}{m_1}y_2$
$y_2''=frac{k_2}{m_2}y_1-frac{(k_1+k_2)}{m_2}y_2$
Assume that the masses are $m_1 = 2$, $m_2 = 9/4$, and the spring constants are $k_1=1,k_2=3,k_3=15/4$.
a)use 4x4 system of first order equations to model this system of two second order equations. (hint: $x_1=y_1,x_2=y_2,x_3=y_1',x_4=y_2'$)
ordinary-differential-equations
asked Dec 5 '18 at 11:33
LTYLTY
285
$endgroup$
This question already has an answer here:
How to change the system in four first order equations?
2 answers
Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:
$y_1''=-frac{(k_1+k_2)}{m_1}y_1+frac{k_2}{m_1}y_2$
$y_2''=frac{k_2}{m_2}y_1-frac{(k_1+k_2)}{m_2}y_2$
Assume that the masses are $m_1 = 2$, $m_2 = 9/4$, and the spring constants are $k_1=1,k_2=3,k_3=15/4$.
a)use 4x4 system of first order equations to model this system of two second order equations. (hint: $x_1=y_1,x_2=y_2,x_3=y_1',x_4=y_2'$)
This question already has an answer here:
How to change the system in four first order equations?
2 answers
ordinary-differential-equations
ordinary-differential-equations
asked Dec 5 '18 at 11:33
LTYLTY
285
asked Dec 5 '18 at 11:33
LTYLTY
285
asked Dec 5 '18 at 11:33
LTYLTY
285
asked Dec 5 '18 at 11:33
LTYLTY
285
asked Dec 5 '18 at 11:33
LTYLTY
285
285
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Dec 5 '18 at 12:35
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1 Answer
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Introducing a whole new set of $x_i$ variables is somewhat confusing. Simpler to just introduce two new variables $y_3$ and $y_4$ where $y_3=y_1'$ and $y_4=y_2'$. So you now have:
$y_1' = y_3$
$y_2' = y_4$
$y_3' = y_1'' = dots$
$y_4'=y_2''= dots$
answered Dec 5 '18 at 11:57
gandalf61gandalf61
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Introducing a whole new set of $x_i$ variables is somewhat confusing. Simpler to just introduce two new variables $y_3$ and $y_4$ where $y_3=y_1'$ and $y_4=y_2'$. So you now have:
$y_1' = y_3$
$y_2' = y_4$
$y_3' = y_1'' = dots$
$y_4'=y_2''= dots$
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
$endgroup$
add a comment |
$begingroup$
Introducing a whole new set of $x_i$ variables is somewhat confusing. Simpler to just introduce two new variables $y_3$ and $y_4$ where $y_3=y_1'$ and $y_4=y_2'$. So you now have:
$y_1' = y_3$
$y_2' = y_4$
$y_3' = y_1'' = dots$
$y_4'=y_2''= dots$
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
$endgroup$
add a comment |
$begingroup$
Introducing a whole new set of $x_i$ variables is somewhat confusing. Simpler to just introduce two new variables $y_3$ and $y_4$ where $y_3=y_1'$ and $y_4=y_2'$. So you now have:
$y_1' = y_3$
$y_2' = y_4$
$y_3' = y_1'' = dots$
$y_4'=y_2''= dots$
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
$endgroup$
Introducing a whole new set of $x_i$ variables is somewhat confusing. Simpler to just introduce two new variables $y_3$ and $y_4$ where $y_3=y_1'$ and $y_4=y_2'$. So you now have:
$y_1' = y_3$
$y_2' = y_4$
$y_3' = y_1'' = dots$
$y_4'=y_2''= dots$
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
answered Dec 5 '18 at 11:57
gandalf61gandalf61
8,841725
8,841725
add a comment |
add a comment |
