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I'm trying to non-dimensionalise the equation for the mass-spring system $$m\ddot{y}+ky=0$$ I've made a list of the constants and their dimensions, including the one I've added which is $g$. $$\begin{array}{ll} \text{Constant} & \text{Dimension} \\ \hline m & \small[ M \small] \\ k & \small[ M T^{-2} \small] \\ g & \small[ L T^{-2} \small]\end{array}$$

As well as the list of the variables and their dimensions

$$\begin{array}{ll} \text{Variable} & \text{Dimension} \\ \hline t & \small[ T \small] \\ y & \small[ L \small] \end{array}$$

Now, I would like to introduce the dimensionless time, $\tau$. The problem is that I do not know how to combine the constants that I have to achieve that. I always have a dimension left, such as $L$, for example. Please advise.

PS This is not a homework assignment, just in case.

The Gruffalo
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  • I am probably loosing my mind, but isn't $g$ express in $LT^{-2}$ and $k$ express in $L^2T^{-2}$? – C. Falcon Nov 23 '16 at 22:01
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    In $m y''+ky=F$ where $F$ has units of force and $y$ has the dimension of length, $k$ has the dimension of $M T^{-2}$ (so no $L$). Usually $g$ is just the acceleration due to gravity, not the force due to gravity, so it would be $L T^{-2}$. – Ian Nov 23 '16 at 22:06
  • @Ian Thank you, I wanted to write $k$ has dimension $MT^{-2}$, but I did not. What alerted me is that $\displaystyle\omega:=\sqrt{\frac{k}{m}}$ is usually defined to be the pulsation. – C. Falcon Nov 23 '16 at 22:09
  • Thanks guys, that was a momentary lapse of reason on my part. I've updated the post now. Still don't know how to get rid of L though :( – The Gruffalo Nov 24 '16 at 17:42
  • Oh no, I've got it, it's $\sqrt{\frac{k}{m}}$ – The Gruffalo Nov 24 '16 at 17:50

2 Answers2

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You set up a system of linear equations. Assuming your dimension labels are correct (which I think they are actually not, but I'll run with it), $m^{a_1} k^{a_2} g^{a_3} t^{a_4} y^{a_5}$ has dimensions $M^{a_1+a_2+a_3} T^{-2a_2-2a_3+a_4} L^{2a_2+a_3+a_5}$. The possible dimensionless quantities are those with the values of $a_i$ such that the exponents are all zero. So they satisfy:

$$a_1+a_2+a_3 = 0 \\ -2a_2-2a_3+a_4 = 0 \\ 2a_2 + a_3 + a_5 = 0.$$

There is a two dimensional subspace of such values of $a_i$ which you can find using Gaussian elimination.

For a dimensionless time, you would choose a particular solution to the above with $a_4=1$ (so that $\tau$ is proportional to $t$) and probably $a_5=0$ (so that $\tau$ is independent of $y$). With both these constraints, there is just one dimensionless time, up to a multiplicative constant.

Ian
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After reading the helpful suggestions above, and correcting my dimensions I was able to figure out that $t$ is scaled by $\sqrt{\frac{k}{m}}$ to get $\tau$. And for $y$ it's $\frac{k}{gm}$. Thanks!

The Gruffalo
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