The experiment was relating the period of one "bounce" when you hang a weight on a spring and let it bounce. I have this data here, one being mass and one being time.

The time is an average of 5 trials, each one being and average of 20 bounces, to minimize human error.



The mass is the mass that was used in each trial (they aren't going up in exact differences because each weight has a slight difference, nothing is perfect in the real world)



My problem is that I need to find the relationship between them, I know $m = \frac{k}{4\pi^2}\times T^2$ so I can work out $k$ like that but we need to graph it. In this equation, $k$ is the spring constant, $m$ is the mass hanging on the spring and $T$ is the period.

I can assume that the relationship is a sqrt relation, not sure on that one. But it appears to be the reverse of a square. Should it be $\frac{1}{x^2}$ then?

Either way my problem is still present, I have tried $\frac{1}{x}$, $\frac{1}{x^2}$, $\sqrt{x}$, $x^2$, none of them produce a straight line.

The problem for SU is that when I go to graph the data on Excel I set the y axis data (which is the weights) and then when I go to set the x axis (which is the time) it just replaces the y axis with what I want to be the x axis, this is only happening when I have the sqrt of $m$ as the y axis and I try to set the x axis as the time.

The problem of math/physics is that, am I even using the right thing? To get a straight line it would need to be $x = y^{1/2}$ right? I thought I was doing the right thing, it is what we were told to do. I'm just not getting anything that looks right.

  • $\begingroup$ I can guess that $m$ is the mass, $T$ is the period you have measured, and $k$ is the spring constant; what are the other variables? What is $\Pi$? What are you plotting as your $x$ and $y$? $\endgroup$ Commented Oct 19, 2012 at 2:49
  • $\begingroup$ @ColinMcFaul I have edited the question to include explanation of the variables. $\Pi$ was supposed to be $\pi$: en.wikipedia.org/wiki/Pi $\endgroup$
    – Michiel
    Commented May 12, 2013 at 6:51

2 Answers 2


The equation you present, $m = \frac{k}{4\Pi^2}\times T^2$ is of the form $y = \mathrm{slope}\times x + \mathrm{intercept}$ for a straight line. To put it in that form, you can make the identifications $m \mapsto x$, $T^2 \mapsto y$, $\frac{4\Pi}{k} \mapsto \mathrm{slope}$. Your equation is also predicting $\mathrm{intercept} = 0$. I'm using mass as the x-variable because that is the independent variable you are actually controlling in your experiment. Time is then the dependent variable, so it gets plotted as the y-variable. As I mentioned in my comment, I'm guessing that $k$ is the spring constant, and I don't know what $\Pi$ is. Those are important for the physical interpretation of the slope, but aren't that important for getting the fit correct.

I'm plotting mass as my x-variable, and $T^2$ as my time variable. I didn't try to actually fit that to a straight line, but I did plot a straight line with no y-intercept on the same plot. Eye-balling those together, the fit looks nearly perfect. I don't see any evidence of any deviation from the law you're trying to verify.


Your balance plate weighed about 6-7 grams and you need to take this into account. In order to get a good relation, you need to add a constant to all the weights. This is not exactly right, because the center of mass also changes a little when you add weights, but it's the major effect. The only thing you need to do is drop the first point, where there is a 10-15% error due to the weight of the balance.

  • $\begingroup$ The base weight was the first one, 50.59g $\endgroup$
    – FabianCook
    Commented Oct 18, 2012 at 23:38
  • $\begingroup$ @SmartLemon: It isn't quite--- there is a shift either in the mass or in the length --- I found that I had to add an offset to make it match the square-root law, although it is approximately correct. $\endgroup$
    – Ron Maimon
    Commented Oct 19, 2012 at 0:37

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