3
$\begingroup$

We just finished our sound waves class at my university. I was taught that reflection of sound waves from an elastic boundary is accompanied by a phase change of $\pi$ while that from a rigid boundary is not.

Let us assume that the ear membrane is an elastic boundary ( I really am unable to figure out what type of boundary it would be so please correct me if I am wrong). In that case the sound reflected would have a phase change of $\pi$.

Let us additionally assume that this wave had come from a continuous sound source $A$. This source would still be producing sound when our wave comes back after reflection.

Shouldn't our reflected wave interfere destructively with incoming sound waves from A and stop us from listening all further sounds ?

$\endgroup$
  • $\begingroup$ I have come up with this : when the reflected wave comes out of the ear it encounters a rarer boundary (air) and thus the wave transmitted through this rarer boundary is such that it does not cause any destructive interference. Am I right ? $\endgroup$ – Raghav Jan 28 '17 at 6:37
1
$\begingroup$

The eardrum is elastic, andon the inward movement, it transmits the pressure wave into your brain, ultimately. But it is not perfectly elastic, so even if it did reflect sound, it would not be a exact reflection of the incoming wave. So destructive effects would not occur.

In particular, the reflected wavelength would be lengthened and the conical shape of the ear canal would alter the characteristics of the sound wave that emerges from it.

A phase change of $\pi $ is not likely unless you assume the eardrum is completely rigid, which of course it is not.

$\endgroup$
0
$\begingroup$

The ear membrane is very light weight. That is why it moves with the air (waves). In order to produce "distructiv" waves the membrane would have to move in the exact oposite direction of the incoming sound. That would require separate source of energy, which the membrane does not have.

Any imperfection in the otherwise synchronous movement of the membrane would produce a wave of much less amplitude than the incoming sound, that it would be practically negligible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.