Basic theory on wave propogation

[Vaughan Odendaa]

Hi there,

I would like to pose a few questions on this as there are some things that I'm not 100% sure of. I've read the Master Handbook of Acoustics and I think I've been able to grasp many of the concepts well. However, some of the simpler, less technical theories are not explained enough.

I know that when a sound wave propogates, it has areas of low and high pressure. Sound in front is compressed, molecules are pulled together, and at 1/2 wavelength, there is a rarefaction low pressure region where molecules are pulled apart.

Simple. But can someone explain the why behind the low pressure region ? I know it probably is simple, but I don't think I got an answer as to the why behind the molecules pulling apart, hence the low pressure region.

I suspect this has something to do with the second law of thermodynamics, but if someone could explain this to me, I would appreciate this very much.

Second thing, wavelength, or cycles. I know that a cycle is a 360 degree round trip and it's the number of compression peaks made per second.

But the specifics of this are not clear, at least to me. Does a full cycle mean that the wave will travel from the speaker, to the rear wall, and then back to the front wall ? Or is it from the speaker, to the rear wall, and back to the speaker ?

Sorry if this seems very basic, but I just would like to have a more solid grasp on this. Sometimes the simple questions can be more difficult to explain properly. The Master Handbook of Acoustics is an excellent book that explains the how but not the why, in my opinion.

Your help on these two questions would be much appreciated.

Thanks.

--Sincerely,

 

[Vaughan Odendaa]

Sorry, some more questions. I recently attended a seminar and there was an engineer there talking about sound pressure and decibals. He said that a doubling of power would yield a 6 dB difference in output.

This troubled me somewhat. I thought that one would get a 3 dB gain, not 6 dB. And how is this number reached ? How does one calculate this ? If I have 100 dB plus another 100 dB, will this give me 103 dB ?

I'm not sure. In the Handbook of Acoustics, it tells me that a 3 dB difference in output would be gained from doubling power.

Why ?

If someone could explain why you get a 3 dB difference and an easy way to calculate it, I would appreciate it.

Thanks.

(btw, although these questions seem simple, I'm sure this thread will help those who don't have a solid grasp on the subject and would like a more informative, more detailed explanation).

--Sincerely,

 

[Savant]

Simple. But can someone explain the why behind the low pressure region ? I know it probably is simple, but I don't think I got an answer as to the why behind the molecules pulling apart, hence the low pressure region.

Any vibrating (sound-producing) body in a medium (air) produces sound pressure waves that fluctuate above and below the ambient (atmospheric) pressure. In the simplest case, a body will produce a sinusoidal wave, where the fluctations correspond to the compression and rarefaction of sound particles. The wave "squeezes" the air molecules together to a pressure slightly* above atmospheric and alternately "pulls apart" the air molecules to a pressure slightly below atmospheric. The distance between compressions (or rarefactions) is the wavelength. The speed of sound in the medium divided by the wavelegth will give you the frequency. But this you already know from Everest.

I suspect this has something to do with the second law of thermodynamics, but if someone could explain this to me, I would appreciate this very much.

More like the First Law of Thermodynamics. A full derivation showing this relationship shouldn't be necessary, I hope! If you feel so inclined, however, you might check out some of the more theoretical texts in acoustics like Theoretical Acoustics by Morse & Ingard, or Fundamentals of Acoustics by Kinsler, Frey, et al. (These are grad- to post-doc-level texts, btw.)

Second thing, wavelength, or cycles. I know that a cycle is a 360 degree round trip and it's the number of compression peaks made per second.

But the specifics of this are not clear, at least to me. Does a full cycle mean that the wave will travel from the speaker, to the rear wall, and then back to the front wall ? Or is it from the speaker, to the rear wall, and back to the speaker ?

That depends on the frequency in relation to the speaker to rear wall and rear wall to ??? distances. Is this question related to modes? It seems to me that would be the "hidden question" behind your questions. If so, you might also benefit from Philip Newell's Recording Studio Design. While the book is written from the perspective of recording studio design (duh), it has a lot of useful sections and chapters on room & loudspeaker interaction - all of which are applicable to home theaters; the physics are the same. The only things that might not be applicable are the chapters on recording studio designs - but I'm sure they'd interest you if you're interested in acoustics as a general topic. IMO, Newell is a perfect supplement to Everest.

I hope this helps!

* "Slightly" is probably overstating it: Even very "loud" sounds only fluctuate atmospheric pressure to a very small degree. E.g., the sound pressure from a jet engine (when standing several feet away) may have a sound pressure on the order of 20 Pa. This is less than 0.02% of atmospheric pressure [p(atm) = ~101 kPa].​

 

[Savant]

Sorry, some more questions. I recently attended a seminar and there was an engineer there talking about sound pressure and decibals. He said that a doubling of power would yield a 6 dB difference in output.

This troubled me somewhat. I thought that one would get a 3 dB gain, not 6 dB. And how is this number reached ? How does one calculate this ? If I have 100 dB plus another 100 dB, will this give me 103 dB ?

Yes. A doubling of acoustic power yields a 3 dB increase in level.

I'm not sure. In the Handbook of Acoustics, it tells me that a 3 dB difference in output would be gained from doubling power.

Why ?

10*log(2) = ~3

 

[Ethan Winer]

Vaughan,

> He said that a doubling of power would yield a 6 dB difference in output. <

That's wrong, as Jeff explained. But this is a common mistake. In electrical terms, doubling the voltage gives a 6 dB increase in power. Half (3 dB) of that is due to the doubled voltage, and the other 3 dB is because the current also doubles along with the increased voltage.

By the way, electrical voltage is equivalent to acoustic velocity, and electrical current is equal to acoustic pressure. I tend to see things in mechanical terms rather than as math, so yet another equivalent is water flowing in a pipe. Water pressure is equal to volts and the flow in gallons per minute is the same as electrical current. Yet another parallel is an automobile transmission - RPM is the same as volts, and torque is the same as current. Cool eh?

--Ethan

 

[Vaughan Odendaa]

Thanks for posting guys !

Savant,

The wave "squeezes" the air molecules together to a pressure slightly* above atmospheric and alternately "pulls apart" the air molecules to a pressure slightly below atmospheric.

I know that. But why ? Although I am not a math guy like many, I don't like having to just accept answers from books and not understand the reasons behind them. I am one of those people who need, absolutely, need an answer to certain questions.

Not just restricted to acoustics, but other subjects too. I am consistently told things, things that I must just accept but there isn't underlying reasons as to why.

This is why I made the thread. Because I know that experts like yourself and Ethan would post and set me straight.

The distance between compressions (or rarefactions) is the wavelength.

Okay. So if I have a wave, lets say it's a frequency (40 hz), then would the rarefaction valley be 1/2 of it's wavelength ? And every 2/4 of it's wavelength and multiples of that ?

And compression peaks are where ? Sorry to ask you these things but I might just be a little confused. Might.

But this you already know from Everest.

Yep. Heh.

More like the First Law of Thermodynamics. A full derivation showing this relationship shouldn't be necessary, I hope!

Could you try to explain this to me in a practical, empirical manner ? Why do you suggest that this is related to the First Law and could you briefly explain what it entails and how it fits into the low pressure regions that sound waves have ? Thanks !

If you feel so inclined, however, you might check out some of the more theoretical texts in acoustics like Theoretical Acoustics by Morse & Ingard, or Fundamentals of Acoustics by Kinsler, Frey, et al. (These are grad- to post-doc-level texts, btw.)

But would these texts delve straight into equations, or would there be readable text that explains in a practical, or semi-practical way what actually happens, or rather, why this-or-that happens in this-or-that way ?

Sorry.

That depends on the frequency in relation to the speaker to rear wall and rear wall to ??? distances. Is this question related to modes?

No, this isn't related to modes, or at least I didn't intend for it. I was just thinking, because no where does it specifically explain what a cycle entails as far as waves propogating in a room is concerned.

I've heard time and again that a full wavelength is a 360 degree trip that the wave must travel. But does that entail from the speaker, to the rear wall and back to the speaker ? Or does that entail from the speaker, to the rear wall and back to the front wall ?

I'm not sure. I just want a general answer to this. I know that the distance the wave would travel correlates to frequency, but in general, for any given frequency, what would the cycle consist of in terms of travel from the speaker ?

Hope you can steer me in the right direction.

It seems to me that would be the "hidden question" behind your questions. If so, you might also benefit from Philip Newell's Recording Studio Design. While the book is written from the perspective of recording studio design (duh), it has a lot of useful sections and chapters on room & loudspeaker interaction - all of which are applicable to home theaters; the physics are the same.

Cool. I'll order that book. But until then D), ahem.

The only things that might not be applicable are the chapters on recording studio designs - but I'm sure they'd interest you if you're interested in acoustics as a general topic.

Oh definitely, I take this subject very seriously and constantly want to improve my understanding. Your help is appreciated, thanks !

--Sincerely,

 

[Vaughan Odendaa]

10*log(2) = ~3

How does one arrive at this ? Could you break this down any further ? At the moment I'm just looking at a few numbers. In fact, Everast's book has the same answer, but again, I have no idea how one reaches that conclusion.

Thanks.

--Sincerely,

 

[Vaughan Odendaa]

Hi Ethan !

That's wrong, as Jeff explained. But this is a common mistake. In electrical terms, doubling the voltage gives a 6 dB increase in power. Half (3 dB) of that is due to the doubled voltage, and the other 3 dB is because the current also doubles along with the increased voltage.

What books would you recommend that delves into this ?

By the way, electrical voltage is equivalent to acoustic velocity, and electrical current is equal to acoustic pressure.

How is electrical voltage equivalent to acoustic velocity ? Not disagreeing with you, but I would appreciate it if you could expound on this. The current to pressure correlation I think I understand.

I tend to see things in mechanical terms rather than as math, so yet another equivalent is water flowing in a pipe. Water pressure is equal to volts and the flow in gallons per minute is the same as electrical current.

Cool. Thanks for the explanation.

Yet another parallel is an automobile transmission - RPM is the same as volts, and torque is the same as current. Cool eh?

Definitely. Thanks again.

--Sincerely,

 

[Ethan Winer]

Vaughan,

> What books would you recommend that delves into this ? <

This is Basic Electricity 101. The formula for power is:

Power = Amps x Volts​

And the formula for Amps is:

Amps = Volts / Ohms​

So when the voltage is doubled the amps drawn is also doubled. And when both are doubled the power quadruples.

> How is electrical voltage equivalent to acoustic velocity ? <

I see this stuff in terms of water flow and gear ratios. The "volts / wave velocity / water pressure / RPM" part is the potential for work, but only when a sufficient load is provided. It takes little power to keep a spindle spinning or to put 120 volts at an AC power outlet. But once you open the water valve or load down the AC power source, then the "work" is done which requires power. That's my simplistic understanding. Jeff can probably explain what's really happening physically in more appropriate terms.

--Ethan

 

[Vaughan Odendaa]

I just did a search on google and I saw this bit of information :

http://www.csun.edu/~lg48405/virtual/trackC/CHERNUCHIN/ac.literature_review.html

"One cycle of a sound wave in air consists of one compression of the air together with the subsequent rarefaction that occurs. The air molecules are forced together (compression or compaction) and then subsequently (in accordance with the 2nd law of thermodynamics) they immediately begin returning to their equilibrium state. The equilibrium state of the air molecules is the state in which they were before the compression under observation occurred.[/b]

So it is the second law that causes molecules to revert back to their original position and hence less pressure than atmospheric ?

But why ? Why does this happen ? I just would like to know in a easy to understand way why once sound compresses, there is also a rarefaction and what causes the rarefaction.

Thank you.

--Sincerely,

 

[Vaughan Odendaa]

Ethan, thank you for contributing. BTW, you wouldn't be able to explain to me how to calculate the 3 dB difference you get when you add two powers together, would you ?

Thanks.

--Sincerely,

 

[Vaughan Odendaa]

10 log 2 = 3.02 (from the Master Handbook of Acoustics). Now what if someone does not have a grasp on the math but would like to know how one actually gets that figure ?

The last time I did math was a long, long time ago and truth be told, my math wasn't great to begin with. I never was math inclined either. Which makes things worse, I know. But I wouldn't mind learning.

--Sincerely,

 

[jaxvon]

The scale of loudness that we used is based on the Bel. More commonly we work with 1/10 of this unit, the deciBel. This scale is a logarithmic scale based on powers of 10. This means that for every increase in 1 bel (or 10dB), the sound is 10 times louder. Due to the way we hear, this "sounds" about twice as loud.

 

[Vaughan Odendaa]

Jaxvon, I know full well what a decibel is. What I don't understand yet is how one reaches the "3.02" dB for every doubling of power. How does one calculate that ? No one has tried to answer that one for me.

Concerning wave cycles, is that from the speaker to the rear wall and back to the speaker, or is that from the speaker to the rear wall and back to the front wall ?

I just want a general picture. I know that frequency will determine the distance of each wave cycle but, in general, am I correct in what I posted above ? Please correct me if I'm wrong.

Thank you.

--Sincerely,

 

[FLZapped]

Oh geez...

10*log(2) = ~3

How does one arrive at this ? Could you break this down any further ? At the moment I'm just looking at a few numbers. In fact, Everast's book has the same answer, but again, I have no idea how one reaches that conclusion.

Thanks.

--Sincerely,

Another way to find squares and square roots is through base 10 logs. If you take the log of a number, divide it by 2, then take the anti-log (10^x) of the result, you get the square root. 10 because it is base 10 and x is the result of log(n)/2 = x.

So you do this:

log(4)/2 = 0.30103

Then 10^0.30103 = 2.000 (that's 10 raised to the power of 0.301)


Why is this important? We'll get there.....slowly

By definition E(volts) = I(amps)*R(resistance)

Power is P(watts) = I*E

Let's do some math.

Let:

I=2
E=2
R=1

E= I*R = 2*1 = 2

Okay, that worked as it should.

Then P=2*2 = 4 watts

Great.

Now lets double our current:

E=I*R = 4*1 = 4 volts

Our voltage ALSO doubled.

Therefore, power is now:

P=I*E = 4*4 = 16 watts, It quadroupled. this is known as the "Square Law Relationship" in math.

What if you only wanted to double your power.

Well, if you noticed we have I*E to get power. What if you only know one of those two values? Since the two values BOTH increase, by substitution you can write the forumla P= E*E or I*I for a normalized resistance of 1 ohm. So P would be equal to either E^2 or I^2(E square, or I squared). In the real world R isn't going to be equal to 1 and that is the only place this is valid. Without going through the gyrations, you end up with P= I^2/R or P= E^2 * R

There's that square law again......

Now if you know the power, but need to find the current, you have to manipulate the formula and you get:

I = (P/R) ^1/2 (square root of power divided by resistance) -is it becoming clear yet?

In order to get a doubling of the power x2 -you actually need to multiply the current and voltage by the square root of 2- remember that little trick?

log(2)/2 = 0.1505 and 10^0.1505 = 1.414

So if we remember that I and E increase equally for a fixed resistance:

P = (2*1.414) * (2*1.414) = 2.828 * 2.282 = 7.99 watts (essentially double)

I hope this shows you the link between base 10 logs and the other formulas.

Now, the basic unit is the Bel which is a base 10 log of a ratio. To get decibels, we multiple by 10 or 10* log (ratio) = decibels

Therefore, 10*log(2/1) = 3.01 dB

If we only knew voltage (or current) is doing, because of the square law relationship, we get:

(2*10) * log(2/1) = 6.02dB Because it is a log, multiplying by 2 is the equivalent of squaring in linear numbers from our earlier examples.

And that's how those relationships exist.

Clear as mud, right?

-Bruce
(it probably is, but it's the best I can do for the moment......)

 

[Savant]

Probably reiterating, but 10*log(2) = ~3 because 10^(~0.3)=2. That's just the definition of a base-10 logarithm. Consult your favorite math text for more details on logarithms.

As for "why" compression and rarefaction occur the way they do, I don't know quite what you're looking for. When you compress something, in general, it will "want" to return to its equilibrium (uncompressed) state. In doing so, an elastic medium (like air) will inherently "overshoot" the mark and rarefy. From there, it will "overshoot" again and go back to compression, and so on. Depending on other factors like resonant frequency and damping present, this cycle could carry on indefinitely, or for barely one cycle. It just depends.

And, AFAIK, the wave equation is related more to the 1st law. I.e., you cannot get more energy out than you put in. The second law basically states that entropy never decreases. Since the cycling is more about changes between potential and kinetic energies, and their conservation, I am always more inclined to think of the wave equation as a specific solution to the first law. But one could argue that sound, in general, creates a zero or net positive change in entropy. I've just never really considered it that way. Again, AFAIK, I believe most acoustics texts that delve into the matter tend to treat the wave equation as an extention of the first law. I could dig a bit more, I guess, if that is not satisfactory...

The texts I referred to above (Morse & Ingard, and Kinsler, Frey, et al) are quite mathematical. This tends to be the case when "getting to the bottom" of anything related to physics. From a practical standpoint, the Newell book, as well as Sound System Engineering by Don & Carolyn Davis are your best bets. IMO, all your questions will be answered. (And, of course, you will wind up with 10x as many questions as when you started!!! )

 

[Vaughan Odendaa]

First, I just want to thank Bruce for giving me a nice explanation on logarithms. So thank you Bruce.

Savant, thank you very much for your contributions. You know, I know that rarefactions and compressions are natural things that occur but I just never understood the reasons for them occuring.

BTW, concerning wave cycles, was I correct in my assessments (above) ? Because that hasn't been addressed. I just want to know, for interests sake, if the sound wave travels from the speaker to the rear wall and back to the front wall (or is it back to the speaker again ? )

Please remember that I know that this is determined by frequencies, in terms of wavelength, but I want to get a general idea of what contitutes a cycle. A cycle is the distance from one compression peak to the next, which is great, but in real world terms, I just need to know how the wave travels in the room itself.

Final question, peaks and rarefactions; a peak occurs at full wavelength and rarefaction occurs at 1/2 wavelength ? Am I correct in my assessments ?

Thank you very much.

PS Thanks for recommending the books to me.

--Sincerely,

 

[Tomorrow]

So it is the second law that causes molecules to revert back to their original position and hence less pressure than atmospheric ?

But why ? Why does this happen ? I just would like to know in a easy to understand way why once sound compresses, there is also a rarefaction and what causes the rarefaction.

Thank you.

--Sincerely,

Vaughan,

The realm of science is always the effort to explain how things happen. The "Why?" question is always reserved for the metaphysical study of the universe. Churches will tell you 'why'. Engineers/scientists will tell you 'how'. If you are asking how the 2nd Law of Thermodynamics explains the rarefaction...you probably already know the answer. If, on the other hand, what you really want to know is why the universe operates in this fashion (action/reaction), you'll have to visit someone in a different set of cloth, lol.

 

[Vaughan Odendaa]

Well, thanks, I guess.

--Sincerely,

 

[Vaughan Odendaa]

Hopefully my question on wave cycles will be answered. I've only asked it several times now.

--Sincerely,