Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I saw an exercise where you had to calculate the units of $C_i, i=1,2$ from an equation like this:

  • $v^2=2\cdot C_1x$ and
  • $x=C_1\cdot \cos(C_2\cdot t)$

where

  • $x$ means meters,
  • $t$ means seconds and
  • $v$ means velocity.

For $C_1$ I got $C_1=m/s^2$. But coming to $C_2$ the cosinus irritates me somehow:

$$x=C_1 \cdot \cos(C_2 t)\Rightarrow m=m/s^2 \cdot \cos(C_2 s)\Rightarrow s^2 = \cos(C_2 s)$$

Does this mean, that $C_2$ must have the unit $s$?

Thanks a lot!

share|improve this question
1  
Not exactly a duplicate, but answered by Fundamental question about dimensional analysis. In short the argument to a trigonometric function must be dimensionless (or a pure angle if you're one of those strange people who like to work in degrees), so either a conversion has been elided or there is a mistake. –  dmckee Mar 1 '12 at 23:06
2  
@dmckee: I see know where I made my mistake. Since the result of cos must be dimensionsless, $C_2$ must be $1/s$ and therefore $C_1$ must be $m$. The two equations do not correlate, as I assumed by mistake. Thanks for the link, helped a lot! –  Aufwind Mar 1 '12 at 23:22
    
An easy way to understand why functions such as $\sin$, $\cos$ or $\exp$ must have dimensionless arguments is to note that they have non-trivial power series, e.g. $\exp(x) = 1 + x + x^2/2 + \dots$. Since it doesn't make sense to add two quantities with different units, e.g. $1m + 3m^2$, we know that $x$ must be dimensionless. –  Lagerbaer Mar 2 '12 at 4:12

1 Answer 1

up vote 7 down vote accepted

Trigonometric functions don't "preserve" units. The expression under a trigonometric function must be dimensionless and so is the value of a trigonometric function.

Thus, C2 in your equations is in units of frequency: Hz or 1/s.

There is an error in one of the equations, perhaps a missing constant.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.