# Microsoft Excel not graphing, $x = y^{1/2}$

Im not sure if this is a math question or a su question.

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.

t

0.3049s
0.3982s
0.4838s
0.5572s
0.6219s
0.6804s
0.7362s
0.7811s
0.8328s
0.869s


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)

m

50.59g
100.43g
150.25g
200.19g
250.89g
301.16g
351.28g
400.79g
450.43g
499.71g


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.

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 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.

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If you want to produce something that kind of looks like a straight line, the conventional thing to do is to say put $m$ on the $x$-axis, and $T^2$ on the $y$-axis. Or else $\sqrt{m}$ on the $x$-axis, and $T$ on the $y$-axis. Or else reverse roles of $x$ and $y$. Whatever you do, you will not get a straight line, but there will be a line of best fit, obtainable through an Excel command that I have forgotten. –  André Nicolas Oct 18 '12 at 22:54
So the relationship is still a sqrt/sqr relationship right? I just realised what I was thinking, I think I just need to identify the relationship, not produce a straight line. –  FabianCook Oct 18 '12 at 22:56
We have for your data $T^2\approx c m$ for some constant $c$. Or equivalently $T\approx a \sqrt{m}$, or equivalently $m \approx b T^2$. It is for your data only a rough approximation. Hooke's Law is only a rough model to begin with. The fit seems to be worst at the light end. –  André Nicolas Oct 18 '12 at 23:00
I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Oct 19 '12 at 6:51

You can find the proper dependence using dimensional analysis. The units of $k$, the spring constant, are force/displacement. In the metric system, this is dyne/cm=g/s^2. The units of mass are g, and period are s. To get the units right, $T \propto \sqrt {\frac mk}$ and you are looking for the constant. If you divide T by $\sqrt m$, your values cluster around $.39$ The 50g data is out of family, and being on the end, I would just exclude it. Then the average is $T \approx 0.039724\sqrt m$