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Dimensioning Problem?

  • #1
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Homework Statement


If x and t represent position and time, respectively, and A and B are constants in x = At2{1-exp(-t2/B)}, what are the dimensions of A and B?

2. The attempt at a solution

I just slept through my first class and found that I have many homework problems related to dimensioning; this was supposed to be the problem my professor went over in class this morning. The book is no help and I'm pretty lost, conceptually. I apologize if my syntax for the equation is a bit erroneous.
 

Answers and Replies

  • #2
gneill
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Can you make some attempt at a solution? What's the objective of choosing "correct" units for the constants? What variables in the equation do you already know the units for? What are they?
 
  • #3
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Can you make some attempt at a solution? What's the objective of choosing "correct" units for the constants? What variables in the equation do you already know the units for? What are they?
Well, arbitrarily speaking, we can say x has a unit of meters and t has a unit of seconds, so the equation gives meters as a function of time squared.
 
  • #4
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Are A and B meant to resolve the variable of time into a unit of distance?
 
  • #5
haruspex
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Are A and B meant to resolve the variable of time into a unit of distance?
Pace gneill, but I wouldn't introduce units into the discussion. The question, as was asked, is about dimension.
The argument to exponentiation needs to be dimensionless. So if you have the argument t2/B, what dimension must B have?
 
  • #6
gneill
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Well, arbitrarily speaking, we can say x has a unit of meters and t has a unit of seconds, so the equation gives meters as a function of time squared.
Almost. It gives distance in a function of time. The fact that the time variable is squared in the formula is something that has to be dealt with by using suitable constants. Dimensions are usually spoken of in terms of their fundamental quality rather than specific units. So for example a distance has the dimension of length [L]. Time is, well, time: [T]. A speed would have dimensions of length over time: [L]/[T], or expressed another way: [L][T]-1. An acceleration would have dimensions [L][T]-2.

Are A and B meant to resolve the variable of time into a unit of distance?
In the big picture that's true. At the detail level though, the individual constants resolve unit issues within the formula. For example, functions like sin() and cos() don't take any arguments other than angular ones (radians, degrees, and so forth), and they return a dimensionless number as a result. The sine of 10 seconds is meaningless, as would be the cosine of six meters. It's true for the exponential function ##e^x## too, where x needs to be dimensionless or a complex angle in radians, and the function returns a pure (dimensionless) number.

Note that the lone "1" (one) in your equation has no dimensions just as the result of the exponential function ##e^x## has no dimensions. In order to add things together they must have the same dimensions (no adding apples to oranges) or no dimensions at all (pure number). Thus the sub-expression in parentheses returns no dimensions and it's up to the preliminary ##At^2## to resolve the dimensions of the overall expression to the required result, namely a distance (length). So what dimensions must the constant A have in order to make the final result [L] ?
 
  • #7
gneill
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Pace gneill, but I wouldn't introduce units into the discussion. The question, as was asked, is about dimension.
Yes, thanks H. I realized that as soon after I hit <post> on my first response, then endeavored to write up something a bit more substantial. :)
 
  • #8
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Pace gneill, but I wouldn't introduce units into the discussion. The question, as was asked, is about dimension.
The argument to exponentiation needs to be dimensionless. So if you have the argument t2/B, what dimension must B have?
If my thinking is correct, in order to get a dimensionless exp^(), B must have units of [T]^2.

As well as, in order to resolve At^2 to a dimension of [L], A must have a dimension of [L]/[T]^2
 
  • #9
gneill
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Yes, that looks good.
 
  • #10
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I appreciate the help, thanks everyone!
 

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