Solving Dimension Matching Problem with Brad - 65 Characters

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The discussion revolves around solving a dimension matching problem involving the relationship between the period τ, pendulum length l, and gravitational acceleration g. The user initially struggles with the dimensional analysis but is guided to express l and g in their dimensional forms. By setting up equations for the exponents based on the dimensional equation T1 = L^r * T^(-2s), the user finds the necessary relationships: 1 = -2s and 0 = r + s. Ultimately, the user successfully derives that τ is proportional to (l/g)^(1/2). The exchange highlights the importance of correctly handling dimensions in physics problems.
brad sue
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Hi,

I have a problem about dimension matchig.

the dimensional relation between the period τ , the pendul length l, and the acceleration of the gravity g takes the form:
[ τ ]=[l^r] [g^s]

Use the fact that the dimendion of τ is [T], that of l is [L], and that of g is [L/T^2] to show that

τ is proportional to (l/g)^(1/2)

I don't get the soltuion right...
Thank you

brad
 
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Show what you did. (Start by replacing l and g with their dimensional equivalents.)
 
Doc Al said:
Show what you did. (Start by replacing l and g with their dimensional equivalents.)

Lr*Ls/T2s equivalent to


Lr*Ls/T-2s . from here I get lost in my calculation I tried to separated the exponent of T into T-s*T-s but really it does not make sense to me.
please give me some hint so I can go further

Thank
Brad
 
You have T1 = Lr*Ls*T-2s. So now you can set up equations for the exponents:
1 = -2s
0 = r + s (note that having no L factor is equivalent to having L0)
 
Thank you SO much!

I got the solution now

Doc Al said:
You have T1 = Lr*Ls*T-2s. So now you can set up equations for the exponents:
1 = -2s
0 = r + s (note that having no L factor is equivalent to having L0)
 
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