Step by Step rearranging with square roots

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Discussion Overview

The discussion revolves around rearranging the equation for the period of a simple pendulum, specifically T = 2π√(L/g), to solve for the length L. Participants explore the steps involved in this rearrangement and address discrepancies between their calculations and a book's answer.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on rearranging the equation to solve for L and expresses confusion over the steps involved.
  • Another participant provides a step-by-step breakdown of the rearrangement, arriving at L = g * (T/2π)², but questions the book's answer of L = g / (T * 4π²).
  • A later reply indicates that the book's answer is derived from the same equation but suggests a different interpretation of the parameters involved.
  • One participant expresses concern about the implication that all grandfather clocks would have the same pendulum length based on the derived formula, questioning the role of mass and height in determining the period.
  • Another participant responds by discussing the relationship between mass and gravitational force, referencing Einstein's equivalence principle as a relevant concept.
  • There is a technical note about using LaTeX for formatting equations, indicating a desire for clearer mathematical representation.

Areas of Agreement / Disagreement

Participants generally agree on the steps to rearranging the equation but disagree on the interpretation of the results and the implications regarding pendulum lengths and their dependence on mass and height. The discussion remains unresolved regarding the correctness of the book's answer and its implications.

Contextual Notes

Some participants express uncertainty about the mathematical steps and how they relate to the physical interpretation of the pendulum's behavior. There is also a lack of consensus on the implications of the derived formula for different pendulum clocks.

SLiM6y
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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...
 
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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:
Thanks for the reply... I got the same as you the first and the second time, but the book got something different...
 
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?)
 
DOH - I get it!
 
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 length of the cable will change the period... Is this true?
 
SLiM6y said:
Because the mass doesn't change the period, and the height it is dropped doesn't change the period so then only the length of the cable will change the period... Is this true?
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.
 
NSX said:
i don't know how to use LaTeX, so i can't make the equations loko nice:...
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!
 

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