Why is this equation dimensionally correct?

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In summary, the equation y = (2m)cos(kx), where k = 2m^-1, is dimensionally correct because the variables x and y are lengths and the argument of the cosine function, kx, is dimensionless. In general, mathematical functions take dimensionless variables and return dimensionless values.
  • #1
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Homework Statement


Show that this equation is dimensionally correct.
y = (2m)cos(kx), where k = 2m^-1


Homework Equations


None that I know of.


The Attempt at a Solution



I don't know why this is correct. I have looked through the chapter and all of the dimensional analysis examples(there are only three) list what the variables represent. The only information given is what I have typed, so I am just assuming that m represents meters.

Are trig. functions dimensionless?
 
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  • #2
Yes, trigonometric functions take arguments that are dimensionless.
 
  • #3
Pengwuino said:
Yes, trigonometric functions take arguments that are dimensionless.

Thank you for answering my question.

So, does it just boil down to the equation being correct because y and 2m are both lengths?
 
  • #4
Were you asking if kx is dimensionless or if cos kx is dimensionless?
 
  • #5
vela said:
Were you asking if kx is dimensionless or if cos kx is dimensionless?

cos(kx)
This is my first physics class and our book doesn't mention trig. functions in the section about dimensional analysis. I am just a little confused.
 
  • #6
OK, Pengwuino answered the other question: kx needs to be dimensionless. The value of the cosine is dimensionless as well, which is the answer to your question.

In general, the standard mathematical functions map a unitless quantity to a unitless result. The main exception is the logarithm because you can do things like log (a/b) = log a - log b, but you should be able to combine logarithms in a way so that you have a unitless argument.
 
  • #7
Actually, whether this is "dimensionally correct" depends upon what dimensions x has!

Assuming that x is a distance, measured in meters, then, since k has dimensions of "m^{-1}", kx is dimensionless. In general, mathematical functions take dimensionless variables and return dimensionless values. When you are told that [itex]f(x)= x^2[/itex] and asked "what is f(3)", you don't have to ask if it is 3 meters or 3 feet, the answer is "9"!
 

FAQ: Why is this equation dimensionally correct?

1. Why is it important for equations to be dimensionally correct?

Dimensional correctness ensures that the units on both sides of an equation are consistent and accurate, which is crucial for obtaining reliable results and making meaningful comparisons.

2. What does it mean for an equation to be dimensionally correct?

An equation is dimensionally correct if the units on both sides of the equation are the same. This means that the equation obeys the fundamental principles of dimensional analysis.

3. How can I check if an equation is dimensionally correct?

To check if an equation is dimensionally correct, you need to look at the units on both sides of the equation and make sure they are the same. If there are any discrepancies, then the equation is not dimensionally correct.

4. What happens if an equation is not dimensionally correct?

If an equation is not dimensionally correct, it means that there is an error in the units used or the equation itself. This can lead to incorrect results and may also indicate a fundamental flaw in the underlying theory or assumptions.

5. Can an equation be dimensionally correct but still be wrong?

Yes, an equation can be dimensionally correct but still be wrong. Dimensional correctness is necessary but not sufficient for an equation to be considered valid. Other factors, such as the accuracy of the constants and assumptions used, also play a crucial role in determining the accuracy of an equation.

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