Is there any reasonable atomic theory which can provide a rational reason for the existence of sonority in metals?

Almost all the non-metals do not exhibit sonority. Can it be correlated to the material's atomic structure? Are there any metals which do not show sonority? If not, then why not?

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    $\begingroup$ 'Sonority' is not a common term in physics. If you mean resonance (closely related in meaning), then many materials resonate including such non-metals as good wine glasses. Unclear what nuclear-physics or atomic-physics has to do with this question. Please clarify. $\endgroup$ – Jon Custer Jul 24 '19 at 13:26
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    $\begingroup$ Can you give which definition of "sonority" you're using? There are many possible definitions for this term, many of which make the claim "There is almost negligible evidence of sonority in non metals" incorrect. $\endgroup$ – probably_someone Jul 24 '19 at 13:26
  • $\begingroup$ 'Sonority' is defined here as the ability of a material to produce a ringing sound when struck hard. $\endgroup$ – Shishir Maharana Jul 24 '19 at 13:29
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    $\begingroup$ The physics property that underlies your "sonority" is elasticity. Highly elastic materials store potential energy when they are deformed, and they return the energy when allowed to relax back to their natural shape. Inelastic materials tend to turn the energy of deformation into heat. Materials that make good bells (e.g., glass, some ceramics, some kinds of stone, and most metals) are all highly elastic. (If that sounds counter intuitive, it's because the every-day meaning of "elastic" is based on misunderstanding the physics meaning.) $\endgroup$ – Solomon Slow Jul 24 '19 at 13:58
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    $\begingroup$ I agree with @SolomonSlow. For a high Q-factor, damping has to be small. Glass would also be good, but it is just too fragile for percussion instruments or tuning forks. $\endgroup$ – Pieter Jul 24 '19 at 15:24

The metals being sonorous has deep connections with their damping capacity which is lower as compared to non-metals. This implies that the impulse vibrations that create the sound to last longer. This would become clear through a careful analysis of stress-strain analysis of metals.

For metals under small loads, the stress and strain are in phase. The phase angle $\delta$ is given by $$\tan\delta=\frac{E''}{E'}$$ where $E''$ is the loss modulus and $E'$ is the storage modulus. So for metals, the phase angle is (nearly) zero. So the loss modulus is zero and the loading and unloading curves are superposed on each other. This means that the intervening area is zero and there is no hysteresis loss. Thus low loss modulus means that there is low damping which in turn means that metals are sonorous.

  • $\begingroup$ But a metal like lead would not be a good material to make bells of, or xylophones or triangles. Too soft. It needs a hard alloy, like brass or steel. And I have attended a concert with percussion instruments made out of ice! $\endgroup$ – Pieter Jul 24 '19 at 15:13
  • $\begingroup$ Why are some metals like lead not sonorous? $\endgroup$ – Shishir Maharana Jul 25 '19 at 6:35
  • $\begingroup$ The properties of lead are different from other metals at room temperature. Lead is extremely soft at room temperature so that it can easily deform when struck and it absorbs the energy of the strike resulting in a dull knocking sound. However if dipped in liquid nitrogen and cooled to $-196^\circ $C, lead becomes stiffer and as a result the sound changes, increases both in pitch and duration. $\endgroup$ – Richard Jul 25 '19 at 7:43
  • $\begingroup$ @ShishirMaharana, read about elasticity and Yield point. If you bend something made of spring steel, it will snap back to its original shape when you let go of it. That's what "elasticity" means. If you bend something made of lead, it will stay bent. The ability of a material to "snap back," if you haven't bent it past its yield point, is what makes it possible for a bell made of the stuff to ring. $\endgroup$ – Solomon Slow Jul 25 '19 at 14:18
  • $\begingroup$ Yes. Exactly. Thanks for elucidating the answer @SolomonSlow $\endgroup$ – Shishir Maharana Jul 25 '19 at 15:31

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