# Step by Step rearranging with square roots

1. Apr 23, 2004

### SLiM6y

Hello,

I have a simple equation (maybe simple for some) that I can't understand how to get from one point to the next when re-arranging. If you can help by giving an answer in english I would be thankful...

The equation T = 2 pi square root L / g (sorry couldn't find the square root etc...)

I want to solve for L

So the answer will be L = g / T 4 pi^2

Why is that? How do I get from the first equation to solving for L. Can you please explain in simplistic terms that way I can see my errors.

Thanks.

Maybe you can tell me how to do the square root and pi symbols...

2. Apr 23, 2004

### NSX

i don't know how to use LaTeX, so i can't make the equations loko nice:

:P

Anyways (this the eq'n for the period of a simple pendulum?):

T = 2pi * sqrt(L/g)

Divide by 2pi on both sides:

T / 2pi = sqrt(L/g) = [L/g]^(1/2)

Square both sides (i.e. raise to the exponent 2):

[T/2pi]^2 = [L/g]^{(1/2)(2)}

Then multiply by g:

L = g * [T/2pi]^2

I didn't get your L = g / T 4 pi^2

just checking over my work for any errors right now

Last edited: Apr 24, 2004
3. Apr 23, 2004

### SLiM6y

Thanks for the reply... I got the same as you the first and the second time, but the book got something different....

4. Apr 23, 2004

### SLiM6y

OK - I have looked through it and I can't get the same answer as the book...

The question is Calculate the length of cable required to give a clock (grandfather clock) a frequency of 1.0 Hz. So T = 2 pi * sq rt L / g

They work the answer through as L = g / T 4 pi^2 = 9.8 / 4 pi^2 = 0.25 m (a reasonable answer, but how the hell do we get to it?)

5. Apr 23, 2004

### SLiM6y

DOH - I get it!!!

6. Apr 23, 2004

### SLiM6y

But then in that case... that would mean EVERY grandfather clock has a pendulum with a cable length of 0.25 m - does that sound correct? Because the mass doesn't change the period, and the height it is dropped doesn't change the period so then only the lenght of the cable will change the period... Is this true?

7. Apr 24, 2004

### krab

Yes. That's why pendulum clocks keep good time, while mechanical watches (which use springs instead) eventually have to be adjusted. You are basically asking a physics question. It is related to the fact that different weights fall at the same speed. A more massive object is harder to get moving, but OTOH since its weight is basically the force that drives it, a more massive object falls with greater force. The two effects exactly cancel. This is sometimes called Einstein's equivalence principle.

8. Apr 24, 2004

### krab

You are almost there! Just put backslashes in front of pi and sqrt, and use tex in square brackets where you have b in square brackets. That's it!