Applying Dimensional Analysis to correct the equation

AI Thread Summary
The discussion revolves around checking the dimensional correctness of the equation for the frequency of a pendulum, f = L/2π √(g/L). Participants clarify that 'g' represents gravitational field strength with dimensions [L][T]^-2, while the frequency f has dimensions [T]^-1. The user attempts to manipulate the equation but struggles with the square root and dimensional analysis. After several iterations, it is concluded that the initial equation is not dimensionally correct, with suggestions to simplify the square root further for clarity. The conversation emphasizes the importance of careful dimensional analysis in verifying physical equations.
Byeongok
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Homework Statement


Check if the equation is dimensionally correct
The frequency f of vibration of a pendulum of length L is given by the equation:

f = L/2π √(g/L)

I'm assuming 'g' is the gravitation field strength (?), which is [L]1[T]-2

The Attempt at a Solution


[/B]
I did not know what the physical quantity was for f, so i looked up and it was [T]-1
so i did:

[T]-1 = [L] x √([L]1[T]-2/[L])
simply
[T]-1 = [L] x √([T]-2)From here I'm stuck on what to do with that square-root.
 
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Byeongok said:

Homework Statement


The frequency f of vibration of a pendulum of length L is given by the equation:

f = L/2π √(g/L)

I'm assuming 'g' is the gravitation field strength (?), which is [L]1[T]-2

The Attempt at a Solution


[/B]
I did not know what the physical quantity was for f, so i looked up and it was [T]-1
so i did:

[T]-1 = [L] x √[L]1[T]-2/[L]From here I'm stuck on what to do with that square-root.
I assume the equation to check is ##f=\frac{L}{2\pi}\sqrt{\frac{g}{L}}##.

EDIT: please write the complete question in part 1 of the template next time. For the rest, see haruspex post.
 
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You have not stated what you are tasked with. Is the idea to check the given equation is dimensionally correct? Was the equation provided as part of the problem statement? You might want to check you have copied it out correctly.
With regard to how to proceed from your last equation above, be careful about the scope of the square root sign. The way you have written it it only appears to apply to the leading [L].
There is some cancellation you can do inside the square root.
 
haruspex said:
You have not stated what you are tasked with. Is the idea to check the given equation is dimensionally correct? Was the equation provided as part of the problem statement? You might want to check you have copied it out correctly.
With regard to how to proceed from your last equation above, be careful about the scope of the square root sign. The way you have written it it only appears to apply to the leading [L].
There is some cancellation you can do inside the square root.

Thanks, I've edited the question in. It was to dimensionally see if the equation was correct
I've followed your advice and ended up at
[T]-1 = [L] x √([T]-2)
Could you please help me from here?

Thanks!
 
Byeongok said:
Thanks, I've edited the question in. It was to dimensionally see if the equation was correct
I've followed your advice and ended up at
[T]-1 = [L] x √([T]-2)
Could you please help me from here?

Thanks!
You might find it obvious if you write T-2 as 1/T2
 
haruspex said:
You might find it obvious if you write T-2 as 1/T2

Thanks for the reply!

So I've gone ahead where you left me from,
[T]-1 = [L] x √([T]-2)
[T]-1 = [L] x √(1/[T]2)
[T]-1 = [L] x 1/√T2
Rationalized it
[T]-1 = [L] x [T]2

Does that mean the equation at the start is not dimensionally correct?
 
Byeongok said:
[T]-1 = [L] x 1/√T2
Rationalized it
[T]-1 = [L] x [T]2
you were doing well until that last step. What happened to the square root?
 
haruspex said:
you were doing well until that last step. What happened to the square root?

Oh oops!

is it
[T]-1 = [L] x √(1/[T]2)
[T]-1 = [L]/√T2

Therefore it not dimensionally correct?
 
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Byeongok said:
Oh oops!

is it
[T]-1 = [L] x √(1/[T]2)
[T]-1 = [L]/√T2

Therefore it not dimensionally correct?
Correct, but you should simply the √T2.
 

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