A plank of length $l$ and mass $m$ is placed on two parallel springs, each with spring constant $k$ and equidistant from the plank's horizontal center of gravity. When the plank is displaced from it's equilibrium position, it's expected to oscillate at this frequency:

$$ f = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} $$

Now, if the plank isn't of uniform density, such that at equilibrium, the force acting on spring 1 is $amg$ and that on spring 2 is $bmg$ where $a + b = 1$. At what frequency will the plank oscillate at when it's displaced from its equilibrium position?

Is it possible to obtain an estimate of its horizontal center of gravity at equilibrium (or rather, constants $a$, $b$) by observing the oscillations, say by taking a fourier transform of the plank's oscillation waveform?

  • $\begingroup$ What drives the response is the location of the center of mass, and the mass moment of inertia of the plank. The response will be a displacement and a rotation and each will be at different frequencies. Are you going to be ignoring rotations and measuring only the displacement at the center? $\endgroup$ Commented Mar 11, 2016 at 13:40
  • $\begingroup$ No, I actually don't want to ignore rotations. How do I compute the frequencies that each spring of the two springs will be oscillating at? $\endgroup$ Commented Mar 11, 2016 at 13:53
  • $\begingroup$ I'm able to measure the frequencies at each spring (say using an accelerometer module with an embedded gyroscope). I'd like to use that to estimate position of horizontal center of mass at equlibrium. $\endgroup$ Commented Mar 11, 2016 at 14:13
  • $\begingroup$ If the springs are identical, then won't the spring bearing more of the weight (i.e., the one closer to the center of mass) be compressed more? I'd think that this would be an easier way to actually determine $a$ and $b$. $\endgroup$ Commented Mar 11, 2016 at 21:21
  • $\begingroup$ You're right, and I'm actually measuring the compressive displacement of the springs too, but in the particular case of my project, that unfortunately turns out to be much more cumbersome and expensive than say, observing the oscillation. $\endgroup$ Commented Mar 12, 2016 at 7:27

1 Answer 1


First you decided on the independent degrees of freedom. I choose the center of the plank and I track the translation $x$ and rotation $\theta$ from the equilibrium conditions.

The displacements of each spring are:

$$\begin{align} x_1 & = x - \frac{\ell}{2} \theta \\ x_2 & = x + \frac{\ell}{2} \theta \end{align} $$

Each spring force is:

$$\left. \begin{align} F_1 & = a m g - k x_1 \\ F_2 & = b m g - k x_2 \end{align} \right\} \begin{aligned} F_1 & = a m g + \frac{\ell}{2} k \theta - k x \\ F_2 & = b m g - \frac{\ell}{2} k \theta - k x \end{aligned}$$

The displacement of the center of mass is $x_C = x + \frac{\ell}{2} (b-a) \theta$ and hence the center of mass acceleration (needed for the equations of motion) is $ \ddot{x}_C = \ddot{x} + \frac{\ell}{2} (b-a) \ddot{ \theta}$.

The EOM are:

$$\begin{align} m \ddot{x}_C & = F_1 + F_2 - m g \\ I_C \ddot{\theta} & = a \ell F_1 - b \ell F_2 \end{align} $$

where $I_C$ is the mass moment of inertia about the center of mass. The above is solved by $x(t) = X \sin \omega t$ and $\theta(t) = \Theta \sin \omega t$. This produces the system of equations of

$$ \begin{align} 2 k X & = m \omega^2 \left( X + \Theta \frac{\ell}{2} (b-a) \right) \\ k X \ell (a-b) + k \frac{\ell^2}{2} \Theta (a+b) & = I_C \Theta \omega^2 \end{align} $$

This has two solutions for frequency $\omega_T$ and $\omega_R$ for translational and rotational modes of vibration.

The two degrees of freedom are coupled with

$$ -\frac{X}{\Theta} = \frac{\frac{\ell}{2} m \omega^2 (a-b)}{ (2k-m \omega^2)}$$

The left hand side of this equation is the center of rotation position (distance) from the center of the plank. Pure translation occurs when $\omega^2 = 2 \frac{k}{m}$ and pure rotation when $a=b$.

$$\begin{align} \omega^2_T & = \frac{k}{m} \left(1+ \frac{m \ell^2 (1-2 a b)}{2 I_C} + \sqrt{ 1 + \left( \frac{m \ell^2 (1-2 a b)}{2 I_C} \right)^2 - \frac{2 a b m \ell^2}{I_C} } \right) \\ \omega^2_R & = \frac{k}{m} \left(1+ \frac{m \ell^2 (1-2 a b)}{2 I_C} - \sqrt{ 1 + \left( \frac{m \ell^2 (1-2 a b)}{2 I_C} \right)^2 - \frac{2 a b m \ell^2}{I_C} } \right) \end{align} $$

Edit 1 To estimate $a$ $b$ from the resulting motion, maybe you can solve the equations of motion using the normalized frequency $n^2 = \frac{\omega^2}{2 k/m}$ and center of rotation location $r=-\frac{X}{\Theta}$.

$$ \begin{align} a &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r+\ell)}{n^2 \ell^2} \\ b &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r-\ell)}{n^2 \ell^2} \end{align} $$

  • $\begingroup$ There seems to be an ongoing discussion here, which I've moved to chat. $\endgroup$
    – David Z
    Commented Mar 15, 2016 at 15:27

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