Pendulum: Deduce proportionality from experiment 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!
 A: 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.
