The setup:

4 spheres of diameter $d$ are suspended on a horizontal plane (magically). They are arranged in a square with length $l$ (the vertices are correspondent to the sphere's centers). The velocity of a wave in the medium they are suspended in is equal to $c$.


The spheres can be made of any material. There is no loss of energy in this system.

The task:

To find a configuration where if s0 is struck (and starts vibrating at a natural frequency), the vibration of s3 is maximized and the vibration of s2 is minimized.

The only free variable is the material (rather, the wave velocity of that material) from which the spheres are made.

Additional comments:

  • If the question does not make sense, please comment and I will clear up any confusion.
  • If the analysis of such a problem is very complicated, just state that.
  • As a side question, is there any physics simulators in which I could simulate such setup?
  • Are these sort of calculations even feasible in larger configurations?

EDIT The medium in which the spheres are, spans $0<x<(l+d); 0<y<(l+d); 0<z<d$. If a wave hits the boundary, it just bounces. No loss of energy. The starting coordinate for the square in which the spheres are arranged is $(d/2, d/2, d/2)$ and it lies in the $x-y$ plane.

  • $\begingroup$ I don't understand the "magical suspension" condition, so let me ask some specific questions. Does the medium extend to infinity in all directions? Why do you say all spheres touch except s1 and s3? Shouldn't s0 not touch s4? This is obvious but: the spheres and the surrounding medium are a different material, yes? What can we change about the material? Density? What else? $\endgroup$
    – DanielSank
    May 26, 2015 at 4:51
  • $\begingroup$ The medium does not extend infinitely in all directions. But that should not matter since waves can only travel between spheres (and inside spheres) $\endgroup$
    – Anton
    May 26, 2015 at 15:54
  • $\begingroup$ Indeed, s0 and s4 would not be touching, but I changed the question to allow for the length $l$ of the square to be different from the diameter of the spheres. $\endgroup$
    – Anton
    May 26, 2015 at 15:55
  • $\begingroup$ The spheres and the surrounding material can be different materials (or even the same). To my understanding (which could be wrong) the only thing that matters about changing the material is changing the velocity of waves inside that material. As to what you can change - change any property. Density, color, atomic composition, ... $\endgroup$
    – Anton
    May 26, 2015 at 15:58
  • $\begingroup$ Now something you're saying doesn't make sense. The material encapsulating the spheres: what is it's extent? Can you write it out like "material exists for $-l<x<l$...etc." in the question to make it unambiguous? $\endgroup$
    – DanielSank
    May 26, 2015 at 16:04

1 Answer 1


In very general terms, we can analyze this problem via a sort of normal mode analysis. We're going to gloss over a whole mess of details here, such as the precise vibrational modes in which the spheres are oscillating and how efficiently these vibrations are transmitted to & from the surrounding medium. Rather, we can assume that the potential energy of the entire system can be split up into a kinetic and potential term; that the potential term is quadratic in the amplitudes of the oscillations (call them $A_0$ through $A_3$); and the kinetic term is quadratic in the "velocities" of these amplitudes ($\dot{A}_0$ through $\dot{A}_3$.) In other words, the Lagrangian of the system will be something like $$ \mathcal{L} = \frac{1}{2} \sum_{i,j = 0}^3 T_{ij} \dot{A}_i \dot{A}_j - \frac{1}{2} \sum_{i,j = 0}^3 U_{ij} A_i A_j. $$

We now attempt to simplify this via symmetry. In the absence of any coupling between the spheres (i.e., no fluid), we would have $T_{ij}$ and $U_{ij}$ proportional to the identity matrix, since all the spheres are identical. In fact, we can pick our units for mass and time such that these matrices are exactly the identity matrix in the absence of coupling. In the presence of coupling, there will be a small perturbation to each of these matrices: $$ T_{ij} = \delta_{ij} + \kappa_{ij}, U_{ij} = \delta_{ij} + \upsilon_{ij}, $$ where the components of the matrices $\kappa$ and $\upsilon$ are supposedly "small". (This will turn out to be unnecessary, actually, but I found it easier to think about it this way.) Note that by construction above, these matrices must be symmetric. Now, from the symmetry of the situation, we make the following assertions:

  • $\kappa_{00} = \kappa_{11} =\kappa_{22} =\kappa_{33} \equiv s_\kappa$, and similarly for $\upsilon$. This says that the presence of the fluid shifts the KE and the PE for a given sphere's vibration by a fixed amount for all of the spheres.

  • $\kappa_{01} = \kappa_{12} = \kappa_{23} = \kappa_{30} = e_\kappa$, and similarly for $\upsilon$. This says that the coupling between all pairs of spheres that share an edge is the same.

  • $\kappa_{02} = \kappa_{13} = d_\kappa$, and similarly for $\upsilon$. This says that the coupling between the two diagonal pairs of spheres is the same.

Thus, we have $$ T_{ij} = \begin{bmatrix} 1 + s_\kappa & e_\kappa & d_\kappa & e_\kappa \\ e_\kappa & 1 + s_\kappa & e_\kappa & d_\kappa \\ d_\kappa & e_\kappa & 1 + s_\kappa & e_\kappa \\ e_\kappa & d_\kappa & e_\kappa & 1 + s_\kappa\end{bmatrix} $$ and similarly for $U_{ij}$. This is now a standard normal mode problem; the normal modes work out to be:

  • $A_0 = - A_2$, $A_1 = A_3 = 0$ (i.e., 0 & 2 out of phase, 1 & 3 stationary.)
  • $A_1 = - A_3$, $A_0 = A_2 = 0$ (i.e., 1 & 3 out of phase, 0 & 2 stationary.)
  • $A_0 = - A_1 = A_2 = - A_3$ (i.e., 0 & 2 in phase, 1 & 3 in phase, and the two pairs out of phase with each other.)
  • $A_0 = A_1 = A_2 = A_3$ (i.e., all four spheres in phase.)

(The frequencies of the first two modes are equal; the other two are distinct unless there's some accidental degeneracy in the coupling.)

What this means is that if we "strike" sphere 0 and set it vibrating, this will excite a combination of the normal modes of the coupled system. Since the amplitudes of $A_0$ and $A_2$ are the same in each of the normal modes, this implies (via Parseval's Theorem) that the mean-square amplitude of the overall vibration of sphere 2 will always be exactly the same as that of sphere 0. So under the above assumptions, what you're asking for can't be done.

In practical terms, what will happen is that the oscillation energy in sphere 0 will start to "slosh around" in the system: from sphere 0, into spheres 1 & 3, and then into sphere 2. Then the process will reverse itself. You can see this process with two coupled oscillators in this video.

Finally, as an aside, I should note that this sort of analysis of the properties of vibrations relied only on the symmetry properties of the system. This sort of analysis can be extended and formalized: by looking at the symmetries of a system from the perspective of group theory, we can extract a surprising amount of information about its vibrations without knowing the fine details. The interested reader is invited to search for "group theory molecular vibrations" for more information on this subject; Hamermesh's classic volume on the subject is a good one as well.


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