Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I know, very easy for all of you, but I'm a beginner in physics ... ;)

I have to work with the mathematical pendulum. After some experiments (changing mass, chaning pendulum's length etc.), I could see that the changes in mass have no effects on the cycle duration ($T$) but changing the length of the pendulum ($\ell$) has. So I have to find out how exactly they correlate, right?

What I thought about: We know (from Huygen's law) that the formula is:

$$T = 2\pi\sqrt{\frac{\ell}{g}}$$

So I have to find out that $T$ is proportional to $\sqrt \ell$. How do I do that mathematically? Just guessing and seeing that it's correct is probably not the right solution, is it?

And how do I then find the constant of proportionality? Just dividing $T$ by $\sqrt\ell$?

Thanks a lot in advance!

share|improve this question
4  
Welcome to Physics Stack Exchange, Marco! I think this may actually be a more insightful question than you realize :-) –  David Z Mar 5 '12 at 20:13
    
Thanks for the warm welcome, David! :) And sorry for not using LaTeX! The problem is that I need an easy explanation/deduction and not something with differential equations as I've found it somewhere else. But I doubt there is an easy one ... –  Marco W. Mar 5 '12 at 20:16
    
Well I think this is a good place to find a conceptual explanation ;-) By the way, generally there are plenty of friendly editors (anyone on the site can propose edits) who are willing to fix equations to use MathJax, so it's not a big deal if you don't. But it's probably worthwhile to learn how it works: $...$ for inline equations, $$...$$ for display style, and most of the common mathematical macros are supported in the middle. –  David Z Mar 5 '12 at 20:21
2  
This is related to Finding coefficient of proportionality and the advice given in Qmechanic's answer therein is worth reading. –  dmckee Mar 5 '12 at 21:38

1 Answer 1

up vote 5 down vote accepted

The basic thing to realize is that straight lines are easy to identify and quantify, so you want to form a linear relationship somehow or another.


Because you think that you know what you're going to get you can go ahead with...

$$ T = 2 \pi \sqrt{\frac{g}{\ell}} $$

square it to get

$$ T^2 = 4 \pi^2 \frac{g}{\ell} $$

Substitute $U = T^2$ and $s = 1/\ell$ (letters picked from thin air, BTW) to get

$$ U = 4 \pi g s $$

which is linear, So try plotting $T^2$ against $1/\ell$ and read the slope off the graph which you can equate to $4 \pi g$.


In the olden days{*}, of course, we would have plotted them on log-log paper if we suspected a power relationship or semi-log if we expected a exponential relationship, seen which one gave a straight line, read the slope to get the power and the intercept to get the coefficient and then if needed jumped through the equation manipulation hoops above to find any constant that might need to be added (and get a more accurate determination of the coefficient).

These days you could also feed the data into a math package of some kind and try fitting various functional forms until you got a good reduced Chi-squared (and don't fret if you haven't heard of that...it just means "a good fit" in carefully quantified language that a statistician would recognize).


{*} I got my formal training just as this style of analysis was going out of fashion--rendered unnecessary by increasingly powerful computers and analysis packages--but my Dad had taken me through the basics for my secondary school science projects. I think it is worth playing with just for the insight it provides.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.