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Assume the following are constant:

  • Surface area of both spheres
  • The spheres are both made of titanium (Ti).
  • Mechanism that strikes a given sphere delivers the same energy per strike.
  • Spheres are suspended in a vacuum by a flexible string possessing negligible mass and great strength.
  • Both Spheres are massive enough in comparison to the force of the strike, such that, the energy is primarily translated into vibration as opposed to movement in 3D space.
  • Hollow sphere contains a vacuum.

Of primary interest is whether the duration of the vibration would be significantly longer for the hollow sphere than the solid sphere.

Hypothesis: The solid sphere would attenuate the vibration more rapidly as it contains more atoms to conduct and absorb the mechanical energy. The hollow sphere being bell-like contains fewer atoms and paths to atoms that could distribute the vibratory load.

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    $\begingroup$ Certainly. IMHO, the question would be more interesting if the striking energy were proportional to the mass. I think the hollow sphere would still win. I don't know how to calculate this, but I think the number of available vibration modes is very important. $\endgroup$
    – PM 2Ring
    Commented Jun 21, 2018 at 3:50
  • $\begingroup$ It might depend on the specific nature of the strike (i.e. force vs. time), compared to the resonances of each sphere. Forces attempting to drive either sphere significantly off-resonance will be quickly damped. $\endgroup$ Commented Jun 21, 2018 at 4:24
  • $\begingroup$ @probably_someone I think it's reasonable to treat the strike as a Delta function so that the spheres are excited by a broad range of frequencies. $\endgroup$
    – PM 2Ring
    Commented Jun 21, 2018 at 6:35
  • $\begingroup$ @PM2Ring Answers that consider a "striking energy proportional to the mass" are welcome. $\endgroup$
    – nu everest
    Commented Jun 21, 2018 at 14:14

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A precise answer to this question is surprisingly developed mathematically but fairly simple to find computationally (pick your favorite FEA software suite and chug away). I'll try to attack the problem the first way with the caveat that this explanation could be all wrong and I am being deceived by mathematical mistakes.

The first key aspect to note is that mechanical waves through solids are damped more as the wave frequency increases. As a result, you only usually hear the "lowest" characteristic frequencies of an object that is mechanically struck. In short, lower frequencies = less attenuation. Frequency-proportional damping in structural mechanics is usually called "Rayleigh damping" and is a fairly straightforward extension of the typical velocity-proportional damping model you see in mass-dashpot-spring models.

This means our task is to now find the characteristic frequencies of each sphere! Assuming these spheres aren't getting plastically bent when dinged, the displacements of both of the spheres follow the Navier-Lamé equations:

$$\rho\frac{\partial^2 \bf u}{\partial t^2} = (\lambda + \mu)\nabla(\nabla \cdot \bf{u} ) + \mu\nabla^2\bf{u} +\bf F$$

where $\bf u$ are the displacements, $\rho$ is the density, $\lambda$ and $\mu$ are material coefficients and $\bf F$ is the external force. Because both sphere's materials are the same, the difference is actually coming from the boundary conditions on each sphere. Because the hollow sphere has both an inside and outside surface with no-traction boundary conditions, the oscillatory parts of the solutions for $\bf u$ are going to have smaller frequencies and therefore damp slower.

In short, free space to vibrate $\rightarrow$ vibrations are slower $\rightarrow$ vibrations last more.

Hope this helps; feel free to verify this/prove me wrong by whacking a couple of titanium spheres!

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  • $\begingroup$ What does no-traction boundary conditions mean? $\endgroup$
    – nu everest
    Commented Jun 22, 2018 at 19:33
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    $\begingroup$ @nueverest It means that there are no imposed forces per surface area on the external boundaries or “edges” of the spheres. (This is of course not true when the spheres get dinged, but it is true for every moment immediately afterwards). This may not be exactly true in many physical cases (like if they’re tied to a string like a pendulum), but such a change is a comparatively tiny perturbation to this assumption. $\endgroup$ Commented Jun 22, 2018 at 19:39

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