Why can't sound pass through closed doors?
All those silly people knocking on doors, thinking that you can hear it on the other side. Laughable!
Hahaha, that made me chuckle quite a lot.
it's undoubtly laughable but it also makes quite sense.we know that sound travels faster through solids and many a time less in gases(air).so when a door is closed the sound waves striking the door move very fast within the door(solid medium) but then it is again subjected to travel through air(gas) before reaching our ears which involves quite time and ultimately the sound wave loses its intensity.
but don't u think of keeping your ears close to the door when someone is knocking at it. you will certainly hear the sound earlier than any body else in the room.
I'm not an acoustics expert, but there are reflection losses at the two interfaces between door and air, just as there are for light between two materials of significantly different refractive index. I would guess that's the main loss, but maybe someone else can correct me.
A.T.'s answer is perfectly adequate but I talk too much so will expound on it slightly. The problem that I think is confusion you is that you expect sound which originates OUTSIDE of the solid (the door) and in a very weak medium (air) be impinge on the solid sufficiently to induce sound in it. That's not gonna' happen much at all. If you START sound in the door, by for example knocking on it, the it travels quite nicely and quickly and was induced forcefully enough that it can regenerate itself in the air on the other side of the door.
If you can't hear sounds behind closed door in a sealed room:
- Your parents gave you a sound-proof room with a sound-proof door or;
- You have hearing problems (genetic, war veteran, music nut, etc)
I sure can hear people say things about me behind closed doors even if they're whispering!
I vote the answer given by johng23 as the best one so far. It's same reason that sonar works so well underwater but if you are swimming underwater, you can't hear well some standing out the water yelling at you: strong reflections at the air/water interface. That's why a helicopter with a sonar system (sends and receives sound waves to map out) must lower it on a cable until it dips in the water.
The answer is actually already given in the OPs question itself:
If the speed of sound would be the same in air and door, there would be no reflection.
it's not so much the speed of propagation being different that causes reflections. it's the difference in the characteristic impedance of propagation. reflections happen when that parameter changes suddenly.
Why do you think that faster speed means higher amplitude? Basically, speed of sound depends on young's modulus and stuff like that. When it is very stiff, sound could travel faster.
But I think when it is very stiff it requires larger amount of energy to make the same amplitude, same as the case of harmonic oscillator. Your throat is not likely to generate sufficient amount energy. Only when you put your ear on the door, you could hear a little bit, otherwise the weak sound would again become more dispersed that you can hardly hear any far from the door.
I was only musing a few days ago why sound is able to pass through not just one but a series of rigid barriers. And not just pass through but still maintain fine detail for example in the modulations of the human voice. So I also found the premise of this thread to be puzzling.
It is important to point out that virtually no doors are actually airtight. Sounds can get through very small gaps. An actual airtight door will have a dramatic effect on sound passing from outside to inside.
Can you have reflection, if the speed of propagation is the same in both media?
Yes, in principle, if they have different densities. The acoustic impedance of the medium is Z=cρ where c is the speed of sound and ρ the density. The reflection coefficient depends on the difference between the impedance of the two media.
Acoustic impedance mismatch...
At room temperature air has a specific acoustic impedance of about 420 Pa·s/m
That for wood (pine) is about 1.57 Pa·s/m 10^6
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