What is the role of dimensional analysis in solving physics equations?

[mjolnir80]

Homework Statement

determin the dimensions of \alpha in the following
a)Sin(\alphaX^{}2) (alpha* X squared) (X is a distance)
b)10^\alphat3
c)cot(\alphaX²/R) (R is a radius)
d)e^(hf/\alphaT - 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 \alpha ] = 1 therefore \alpha= [1/L² ] (where L is length)

id appreciate some help :)

 

[Irid]

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.

 

[mjolnir80]

just to clarify for b & d, does it matter that the dimensions are in the exponent?

 

[mjolnir80]

anyone?

 

[tiny-tim]

Hi mjolnir80!

(have an alpha: α and a squared: ² and a cubed: ³ )

just to clarify for b & d, does it matter that the dimensions are in the exponent?

No, it's all the same … 10^αt³ and sin(αt³) need the αt³ to be dimensionless for exactly the same reason.

 

[mjolnir80]

one more quick thing about dimensional analysis :)
in an equation let's say X=V_it + 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?

 

[tiny-tim]

in an equation let's say X=V_it + 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!

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

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