Abstract
We derive and numerically demonstrate that perfect absorption of elastic waves can be achieved in two types of ultra-thin elastic meta-films: one requires a large value of almost pure imaginary effective mass density and a free space boundary, while the other requires a small value of almost pure imaginary effective modulus and a hard wall boundary. When the pure imaginary density or modulus exhibits certain frequency dispersions, the perfect absorption effect becomes broadband, even in the low frequency regime. Through a model analysis, we find that such almost pure imaginary effective mass density with required dispersion for perfect absorption can be achieved by elastic metamaterials with large damping. Our work provides a feasible approach to realize broadband perfect absorption of elastic waves in ultra-thin films.
Highlights
We derive and numerically demonstrate that perfect absorption of elastic waves can be achieved in two types of ultra-thin elastic meta-films: one requires a large value of almost pure imaginary effective mass density and a free space boundary, while the other requires a small value of almost pure imaginary effective modulus and a hard wall boundary
Acoustic metamaterials have been applied to enhance the absorption of acoustic wave energy
We further show that when the almost pure imaginary effective mass density is inversely proportional to frequency, or the almost pure imaginary effective modulus is proportional to frequency, the perfect absorption effect becomes broadband
Summary
We derive and numerically demonstrate that perfect absorption of elastic waves can be achieved in two types of ultra-thin elastic meta-films: one requires a large value of almost pure imaginary effective mass density and a free space boundary, while the other requires a small value of almost pure imaginary effective modulus and a hard wall boundary. On the other hand, when the hard wall boundary is applied, perfect absorption can be obtained when the ultra-thin elastic meta-film satisfies tu = tτ = 2/3 and ru = − rτ = 1/3.
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