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?

  • 4
    $\begingroup$ 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 $\endgroup$
    – Mathusalem
    Commented May 17, 2013 at 16:05
  • $\begingroup$ 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. $\endgroup$
    – pppqqq
    Commented May 17, 2013 at 19:55
  • 2
    $\begingroup$ $m' \ddot{x} = \ldots$ is wrong. You cannot take a massive spring and a linear force law. You have to solve a differential equation. $\endgroup$ Commented May 18, 2013 at 5:04
  • $\begingroup$ 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. $\endgroup$ Commented May 18, 2013 at 5:47
  • 1
    $\begingroup$ I have written this but it yields a ratio of $\frac{4}{\pi^2}=0.405$. $\endgroup$ Commented May 18, 2013 at 6:03

1 Answer 1


You can understand in a simple way the factor $1/3$ which gives you the approximate solution in the low frequency regime (more on this later) in the following way. Start by writing the kinetic energy of your system as:

$$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \int_0^\ell \frac{1}{2} \lambda \dot{\delta}(u)^2 du$$

where $\delta(u)$ is the displacement of the point of the spring which is in the $x=u$ position in the equilibrium configuration and $\lambda=m_0/\ell$ the linear mass density of the spring. The spring has length $\ell$ when unstretched.

If you suppose an harmonic motion for the mass at a very low frequency, the stretching of the spring will be approximately uniform, which means

$$\delta(u) = \frac{u}{\ell} \delta(\ell)$$

Accepting this approximation by substituting in the expression for kinetic energy one gets

$$K = \frac{1}{2} m \dot{\delta}(\ell)^2 + \frac{1}{2} \frac{m_0}{\ell} \dot{\delta}(\ell)^2 \int_0^\ell \frac{u^2}{\ell^2} du$$

and after an integration

$$K = \frac{1}{2} \left( m +\frac{1}{3} m_0 \right) \dot{\delta}(\ell)^2$$

which is the expected result.

The system has an infinite number of degrees of freedom, which means that it will have an infinite number of oscillation modes. If $m\gg m_0$ the lowest frequency mode will be approximately described as an oscillation of the mass with an uniform stretch of the spring. In the higher frequency modes the mass will be nearly fixed, and there will be a nearly stationary elastic wave on the spring.

The flaw in your reasoning consist in supposing that the external force applied to the system mass+spring is $-k(x-x_0)$. The applied force is really the tension of the spring at his fixed point, which is not $-k(x-x_0)$ for a spring with mass when there are accelerations.


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