What determines how much of a sound wave reflects vs. transmits through a rock wall?

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Let's say a sound wave is propagating through air. At some point it hits a rock wall, some of the wave continues on through the wall and some reflects back.
What factors determine how much reflects back and how much is absorbed?
 
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Originally posted by Rockazella
Let's say a sound wave is propagating through air. At some point it hits a rock wall, some of the wave continues on through the wall and some reflects back.
What factors determine how much reflects back and how much is absorbed?

i'd say the density of the rock.

is this wrong?
 
Two factors

Density and compression modulus, and how that compares with the density and compressibility of air.
 
Density and compression modulus, and how that compares with the density and compressibility of air.

Could you please explain?

Honestly I don't even get why a wave reflects at all.
Maybe a good explanation of that could help me understand.
 
Reflections are best understood by thinking about what is essentially the resistance the wave experiences in traveling through a material. For example, attach a jump rope to brick wall.

Hold the free end of the rope, and wave it up and down for one cycle. The wave will propagate down the rope, reflect off the wall, and come back to you.

Why did it reflect? Because there was a large discontinuity in the 'flexibility' of the medium where the rope met the wall. The wall is not very flexible at all, while the rope is very flexible. The flexibility is sometimes known by other names, such as impedance.

The energy in the wave effectively just takes the path of least resistance when it hits the discontinuity. Some of the energy actually does go into making the wall move, but most of it follows the path of least resistance back to your hand.

Reflection phenomena happen all over the place -- with light, sound waves, even voltage signals on wires. When a signal on a wire with a low impedance hits a device (like a resistor) of very high impedance, the signal is reflected.

The best transmission of power from one medium to another occurs when their impedances are identical.

- Warren
 
The energy in the wave effectively just takes the path of least resistance when it hits the discontinuity. Some of the energy actually does go into making the wall move, but most of it follows the path of least resistance back to your hand.

Ok so waves follow the path of least resistance...makes sense I guess. If that true though, why would any of the engery go into making the wall move? I would imagine that doing anything to the wall would be much harder than just reflecting all the energy back to your hand.


When a signal on a wire with a low impedance hits a device (like a resistor) of very high impedance, the signal is reflected.

Well this is a new one to me. Are you saying current is reflected or just the AC wave?
 
What do you mean by "current" and "AC wave"?
Any electrical signal is reflected when there's an impedance mismatch along the way and this process is frequency dependent. So yes, the AC wave (part of it) gets reflected.
 
Well let's say you have a flat 5V, are you saying that's a signal? If there were a resistor in the circuit are you saying some current bounces back twords the negative?

I don't know much about electrical physics. When I hear electrical signal or wave I just think up a voltage fluctuation. When you say that can reflect like any other wave I don't really understand what's do the reflecting..?
 
Well, a flat 5V is not a wave -- so it won't show any sort of wavey phenomena like reflection.

Any time-varying signal is, however, a wave. It doesn't have to be alternating current, however -- it could be a wave like v(t) = sin(wt) + 1, which is always greater than or equal to zero. However, your choice of zero potential is arbitrary, so really, it suffices to just consider "any time-varying signal." It also doesn't have to be simple or periodic, because, as Fourier showed us, any arbitrary signal can be decomposed into a sum of pure tones.

The reflection problem often occurs in radio systems. You have a transmitter which has some characteristic output impendance. Depending upon the topology, type of transistors, and so on, your transmitter's amplifier might have a very high or very low output impedance.

You connect your transmitter to an antenna (let's ignore the connecting cables). The antenna, depending upon its design, may have an input impedance very different from the output impedance of the transmitter.

(By the way, the word 'impedance' is used to describe a load which is not simply resistive, but also includes some inductance or capacitance, which are together called 'reactance.' Most engineers choose to represent impedances on the complex plane and use complex arithmetic to work with them.)

If your transmitter and antenna are not 'matched,' some (or even most!) of your transmitter's power will reflect from the antenna, rather than being radiated away into the air. The reflected wave does nothing but heat up your cables and stress your amplifier. As a result, it's important to match your transmitter and antenna. Ham radio operators will often refer to devices called 'matching networks,' which are frequently automatic boxes with variable capacitors or inductors used to match transmitters and antennas. They'll also refer to a number called the Standing Wave Ratio (SWR), which is a measure of how much power is being reflected by antenna.

- Warren
 
Originally posted by Rockazella
Ok so waves follow the path of least resistance...makes sense I guess. If that true though, why would any of the engery go into making the wall move? I would imagine that doing anything to the wall would be much harder than just reflecting all the energy back to your hand.
The phrase 'follows the path of least resistance' is a little misleading, and I apologize for having used it without qualification. Let's see if I can be a little more precise.

To give another example, consider two resistors in parallel, each with a different resistance. Or for a third example, think about two pipes carrying water from the same source, one very large, and one very small.

Most people get the pipe problem correct by intuition. When you turn on the water source, what happens? Most of the water comes out the big pipe, but a small amount dribbles from the small one.

Most people get the resistor problem wrong -- even though it's the same problem. Most of the current flows through the small resistor, but some still flows through the large one.

In the case of a rope tied to a wall, the amount of energy invested in wagging the wall versus the amount reflected is proportional to the ratio of the impedance between the two materials. If the wall is 1000 times harder to wag than the rope, only 1/1000th of your wave energy will go into the wall.

- Warren
 
chroot,

how do you deal with impeadince without the complex plane?
 
On Rad:

Any 2-tuple would do. For some reason people like to the use complex plane... perhaps because people already know what it is.

- Warren
 

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