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Does the sign matter for proportionality?
What does this formula do?what is the constant of proportionality?I did not understand the approach needed to solve this problem on inverse proportionalityCalculating Volume for the Perfect CompostWhy does proportionality yield multiplicitiveness and not addition?Why does 1:2 = 1/2?Why does 100% represent the whole of somethingSimple Ratios Shortcut - Why does it work?How do I figure out the proportionality between two aspect ratiosHow is the joint proportionality (joint variation) equation (i.e. $y = kxz$) logically correct?
$begingroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
edited 1 hour ago
Brian
1,529516
asked 4 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
edited 1 hour ago
Brian
1,529516
asked 4 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago
add a comment |
$begingroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
edited 1 hour ago
Brian
1,529516
asked 4 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).
Therefore, it can be said that:
$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium
Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.
Both are equivalent right?
ratio
ratio
edited 1 hour ago
Brian
1,529516
asked 4 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Brian
1,529516
asked 4 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Brian
1,529516
edited 1 hour ago
Brian
1,529516
edited 1 hour ago
Brian
1,529516
1,529516
asked 4 hours ago
Sebi SSebi S
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Sebi SSebi S
182
asked 4 hours ago
Sebi SSebi S
182
182
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago
add a comment |
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago
$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 3 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 3 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 3 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
add a comment |
$begingroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 3 hours ago
John DoeJohn Doe
12.3k11341
$endgroup$
Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.
answered 3 hours ago
John DoeJohn Doe
12.3k11341
answered 3 hours ago
John DoeJohn Doe
12.3k11341
answered 3 hours ago
John DoeJohn Doe
12.3k11341
answered 3 hours ago
John DoeJohn Doe
12.3k11341
12.3k11341
add a comment |
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
add a comment |
$begingroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
$endgroup$
Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
answered 3 hours ago
Adam LatosińskiAdam Latosiński
9809
9809
add a comment |
add a comment |
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$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago