Dimensional alalysis to show functional dependence

  • Thread starter Thread starter General_Sax
  • Start date Start date
  • Tags Tags
    Functional
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
General_Sax
Messages
445
Reaction score
0

Homework Statement


Use the method of dimensional analysis to show that the functional dependence in equation (1) can be derived from an observational expression: [tex]lambda = k*mu*f^m*T^n[/tex].


Homework Equations


[tex]lambda=k\sqrt {{\frac {T}{\mu}}}{f}^{-1}[/tex] (1)
[tex]lambda = k*mu*f^m*T^n[/tex]
[tex]\mu={\frac {{\it kg}}{m}}[/tex]
[tex]T={\frac {{\it kg}\,m}{{s}^{2}}}[/tex]


The Attempt at a Solution



First I solve for n.

[tex]m={\frac {{m}^{n}}{m}}[/tex]
n = 2

Now I solve for m.

[tex]0=0={1/s}^{m}{s}^{-2\,n}[/tex]
m = -4

so now I have:

[tex]\lambda={\frac {ku{T}^{2}}{{f}^{-4}}}[/tex]

I don't understand what the question means by "show that the functional dependence in equation (1). . ."

In equation (1) we were told that [tex]\lambda[/tex] and T were variables. Well in equation I've derived [tex]\lambda[/tex] and T could very well be variables, but I don't think I understand the question. But, if the question is asking me to equate the two expressions and prove an "identity", then I can't do that. Any help welcomed and appreciated.

When I try to "work" the units out they don't work out at all.
 
Physics news on Phys.org
I tried to make it all pretty with the LaTeX, but it didn't work :(
 
Misread question. I thought the u had an exponent of 1, but it had an exponent of L.