# Effective mass in Spring-with-mass/mass system

Suppose you have a particle of mass $m$ fixed to a spring of mass $m_0$ that, in turn, is fixed to some wall. I'm trying to calculate the effective mass $m'$ that appears in the law of motion of the particle (suppose the system is isolated):$$m'\ddot x=-k(x-x_0).$$ I've read somewhere that this should be $m'=m+m_0/3$, but I'm getting a different result.

My reasoning is as follows. Suppose the particle is at position $x$. The lenght of the spring is $x$ and we can suppose that its center of mass is at $x/2$. So the spring/particle center of mass is at: $$X= \dfrac {\frac{m_0}{2} + m}{m+m_0}x$$ Differentiating two times we get $$\ddot X = \dfrac {\frac{m_0}{2} + m}{m+m_0}\ddot x$$ Now, the only external force causing acceleration to the center of mass is the ceiling reaction to elastic force, that is exactly $-k(x-x_0)$. Thus:$$-k(x-x_0)=(m+m_0)\ddot X=(\frac{m_0}{2} + m)\ddot x$$and so I'm getting:$$m'=\frac{m_0}{2} + m.$$

Could you please point out where am I wrong (if I am) and possibly how is the result demonstrated?

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You are wrong when supposing you can view the problem as an enhanced mass with a shifted center of mass. The spring behaves much differently. The solution to this is rather lengthy : look at this website for a extensive solution : mathrec.org/old/2001dec/solutions.html The answer you were provided with comes from taylor expanding the solution to the eqs. of motion when m >> m0 –  Mathusalem May 17 '13 at 16:05
Thank you for the link, it helped a lot. But I don't understand yet where my reasoning is flawed. Is it just a coincidence that it works for the static case, but it doesn't for the dynamic? I mean, in the static case there seem to be no error. I also suggest that you post this as an answer so I can accept it. –  pppqqq May 17 '13 at 19:55
$m' \ddot{x} = \ldots$ is wrong. You cannot take a massive spring and a linear force law. You have to solve a differential equation. –  ja72 May 18 '13 at 5:04
I have wondered about the same, and I remember solving this problem a few years back. But how, I do not remember. I also remember that the 1/3 fraction is only valid for low frequencies (< nat. freq.). As the frequencies increase the ratio increases also. –  ja72 May 18 '13 at 5:47
I have written this but it yields a ratio of $\frac{4}{\pi^2}=0.405$. –  ja72 May 18 '13 at 6:03