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Why following are true in both cases?.

1- Ultrasound passing through a metal block(solid medium) will not pass through a cavity (air medium) inside the block.

I got some explanation as speed of sound is less in air so air resist the moving wave therefore wave bounces back or reflects back.

2- Sound wave passing in air does not pass through walls of a room or pass through walls in less amount and reflects back in greater amount.

I got some explanation as vibrating air molecules are less capable of vibrating the solid matter.

Both logics seem to contradict each other. What will be the correct way to explain both?

I have some idea of behaviour of sound waves- like sound waves reflect if medium changes similar to light waves, but I am not very clear how a medium permits or reflects the wave for e.g. why wall does not let pass the sound wave?

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  • $\begingroup$ If I take anology of sound as a car and medium as traffic then only 1st explanation seems to be correct. Can anyone explain above question using similar analogy? $\endgroup$
    – Level1
    Commented Sep 14, 2022 at 15:16

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There is an impedence mismatch. At the interface the pressure on both sides of the interface must match and also the displacement of the air and the metal must match. But in the air the ratio of the pressure to the displacement --- the impedence -- is much smaller than in the metal. Because of this mismatch it is not possible satisfy both the pressure and the displacement boundary conditions without having a reflected wave.

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  • $\begingroup$ Do the explanations in the question tell the same thing in simpler terms or do they contain mistakes? $\endgroup$
    – Level1
    Commented Sep 15, 2022 at 2:18
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    $\begingroup$ Since the "explanations" you quote are so vague, it is impossible to know whether they are valid or not. Imepdence matching is a very general idea. You probably know it from the need to match the impdence of the cables, loudspeakers, and amplifier when you connect up your hi-fi components. If you get it wrong the amplifier output is reflected back and can damage the amplifier. $\endgroup$
    – mike stone
    Commented Sep 15, 2022 at 11:38

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