✔ 最佳答案
Clearly, materials with high densities insulate sound much better than those with low densities.
The energy per unit volume E of sound wave passing through a material is found to be proportional to the p(fA)^2, where p is the density of the material, f is the frequency of sound and A is the amplitude of sound wave.
Assuming the sound wave from air is totally transmitted into a material (i.e. no reflection), then we have,
E = p(air).[f.A(air)]^2 = p(material).[f.A(material)]^2
where p(air) and p(material) are the densities of air and the material respectively
A(air) and A(material) are the amplitude of sound wave in air and material respectively
Thus, A(material) = A(air).square-root[p(air)/p(material)]
since the term p(air)/p(material) < 1
the sound amplitude, hence its intensity, in the material is much smaller than that in air. The sound wave is thus diminshed after passing through the material, which serves the purpose of shielding off sound waves.
For example, if concrete of density 2400 kg/m^3 is used to shield off external noice, the amplitude of sound could be reduced by a factor of square-root[1.2/2400]. Because intensity of sound is proportional to sqaure of amplitude, the sound intensity thus could be reduced by a factor of [1.2/2400] or 0.0005