Does the sign matter for proportionality?
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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
New contributor
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$begingroup$
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
New contributor
$endgroup$
$begingroup$
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 15 mins ago
Brian
1,529516
1,529516
New contributor
asked 3 hours ago
Sebi SSebi S
182
182
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New contributor
$begingroup$
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
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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.
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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.
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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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
12.3k11341
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$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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
9709
9709
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Sebi S is a new contributor. Be nice, and check out our Code of Conduct.
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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