Applying Dimensional Analysis to correct the equation

[Byeongok]

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.

 

[Samy_A]

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.

 

[haruspex]

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.

 

[Byeongok]

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!

 

[haruspex]

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

 

[Byeongok]

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?

 

[haruspex]

[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?

 

[Byeongok]

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?

 

[haruspex]

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.