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Dimensional analysis question

  • Thread starter mjolnir80
  • Start date
  • #1
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Homework Statement


determin the dimensions of [tex]\alpha[/tex] in the following
a)Sin([tex]\alpha[/tex]X[tex]^{}2[/tex]) (alpha* X squared) (X is a distance)
b)10[tex]\alpha[/tex]t3
c)cot([tex]\alpha[/tex]X2/R) (R is a radius)
d)e(hf/[tex]\alpha[/tex]T - 1 (h is plancks constant with units J*s) ( f is frequency


Homework Equations





The Attempt at a Solution


so are these all supposed to be dimensionless?

attempt at a: [L2 [tex]\alpha[/tex] ] = 1 therefore [tex]\alpha[/tex]= [1/L2 ] (where L is length)

id appreciate some help :)
 

Answers and Replies

  • #2
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Yes, you're right and the solution is correct. All of those example functions must have dimensionless arguments, otherwise they don't make sense, sort of "apples plus oranges = peaches" or something like that.
 
  • #3
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just to clarify for b & d, does it matter that the dimensions are in the exponent?
 
  • #5
tiny-tim
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Hi mjolnir80! :smile:

(have an alpha: α and a squared: ² and a cubed: ³ :smile:)
just to clarify for b & d, does it matter that the dimensions are in the exponent?
No, it's all the same … 10αt³ and sin(αt³) need the αt³ to be dimensionless for exactly the same reason. :smile:
 
  • #6
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one more quick thing about dimensional analysis :)
in an equation lets say X=Vit + 1/2 a t2

if we wanted to prove that this equation is dimensionally correct, how would the + between the 2 terms on the r.h.s effect the analysis would we have to ignore the + and just try to make it so that the overall dimensions canel each other out to give lenghth?
 
  • #7
tiny-tim
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Homework Helper
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in an equation lets say X=Vit + 1/2 a t2

if we wanted to prove that this equation is dimensionally correct, how would the + between the 2 terms on the r.h.s effect the analysis would we have to ignore the + and just try to make it so that the overall dimensions canel each other out to give lenghth?
Hi mjolnir80! :smile:

No … with one or more +s, each part must have the same dimensions …

in this case, X must have the same dimensions as Vit and as 1/2 a t2 :smile:
 

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