Sound question

[studiot]

 

Studiot, can you measure the intensity on both sides when the apparatus is in water, A medium of higher bulk modulous?

 

I'm sorry, I no longer have the sensors capable of submersion.

I last studied and used underwater acoustic profiling and positioning systems in the late 1970s/ early 1980s.

 

Here is the meter I was using for this thread. It is a very versatile multimeter that I keep in the toolbox.

 

Edited March 17, 2014 by studiot

[rktpro]

Studiot-Sir, can you predict if the difference in intensity between two sides, or say the difference in amplidue (mean); in air and in water, would have any relation with the bulk modulous of media.

[studiot]

 

Studiot-Sir, can you predict if the difference in intensity between two sides, or say the difference in amplidue (mean); in air and in water, would have any relation with the bulk modulous of media.

 

 

I doubt that anyone would award me a knighthood, but thanks.

 

The short answer is yes but I will have to give it some thought, before a fuller reply.

Obviously the modulus controls the strength of the restoring force that generates the oscillation and thus affects the amplitude.

But there is also a big difference in density between water and air and it is the density variation that distorts voices in other gasses such as helium by affecting the speed of sound.

Edited March 17, 2014 by studiot

[Function]

 

So what have I said that was controversial?

 

*****************************************************************************************************************************************************

 

 

Function, I don't know if you are still following this thread but here is a simple experimental analogy to what is happening.

 

 

I passively read and follow everything that is said as much as my level of English permits me to

 

Your example makes sense.

[Enthalpy]

[...] This is because the air inside the pump is confined, whereas the free air in the atmosphere is not.

Just as I said in post#4 about the difference between the air inside and outside the bell.

This analogy is not exact but follows the same principles. [...]

 

That's the wrong direction. Quite the opposite: less pressure builds up inside the bell than outside.

 

The first thing to understand is that the bell is smaller than a wavelength in air. The second thing is that its vibration has opposite antinodes: at least 2 in phase and 2 in antiphase, for the lowest mode - and these antinodes are close together and very well matched.

 

The result is that radiation by a bell is inefficient; not only as a small radiator, but also because radiation by each antinode is widely compensated by its neighbours, whose phase it opposite. This is desired, to a controlled degree, so a bell sounds for long.

 

Now, the antinodes with opposite phase interact inside the bell more easily than outside. That's because the path between them is shorter (straight line) and more open, while outside, the path goes around the bell curvature. As a result, the compensation is more efficient inside the bell than outside.

 

----------

 

The other effect is more observational. Near the rim, the inner and outside areas are close to an other, so the air can pass easily, and the pressure is locally small. To get more pressure, you have to hear at a position farther from the rim - but these positions are accessible only outside, not inside.

I last studied and used underwater acoustic profiling and positioning systems in the late 1970s/ early 1980s.

 

I designed a long-range military sonar. But that was easy, compared with music instruments, which are the difficult part of acoustics.

[studiot]

Enthalpy, we seem to have difficulty communicating to an extent that you have attributed several statements to me that I did not make and then proceeded to argue against them.

 

So let us try discussing one statement at a time.

 

In your post#15 you seem to be suggesting that the fact that the speed of sound in steel is some 15 to 20 times that in air is irrelevant.

 

In truth this fact is absolutely vital to the nature and distribution of the sound waves generated by the bell, both in the air inside and outside the bell.

 

It should further be noted that the the bell itself is a single oscillator and not undergoing wave motion.

Edited March 23, 2014 by studiot

[davidivad]

this may be a bit off topic but still bathroom related so i feel it must not be missed.

did you know that a toilet bowl has it's own ring when you flush it?

the catch is to be at the right distance from the bowl.

[Enthalpy]

In your post#15 you seem to be suggesting that the fact that the speed of sound in steel is some 15 to 20 times that in air is irrelevant.

 

In truth this fact is absolutely vital to the nature and distribution of the sound waves generated by the bell, both in the air inside and outside the bell.

 

You have to understand first what a flexural wave is.

 

Because the wave is flexion in the metal but can't be it in air, comparing the speeds of compression waves is irrelevant.

 

Finding resonant frequencies like 8KHz (or worse, 40Hz) for a bell should have alarmed you. Applying a 6000m/s speed from a compression wave is obviously wrong. Why insist?

 

I not only seem to be suggesting. Flexion is the basic and widespread standard model of bells. I recommend that you read about bells and flexion waves, before posting two pages of obviously flawed personal considerations.

[studiot]

You have to understand first what a flexural wave is.

You have alluded several times to my educational background.

It seems that not only did we have different professors of mechanics at university; we also had different professors of etiquette.

 

I have not once disputed that the bell has flexural modes of vibration, or that the one we are both concentrating on is flexural.

In fact I described such vibrations very early on in posts# 4, 8 and 10 and posted links to many pictures of the same.

I did not explicitly state the mode of vibration, just described it as complex, which it is, although this information was available in the links. It was a good addition to the thread for you to state this explicitly as flexural or bending in your post# 15.

 

Flexural vibrations are not sound waves. Sound waves are pressure waves that travel at the speed of sound in the medium. Flexural waves have a different speed and equation.

 

However we are discussing sound in the air, not in the bell.

 

 

Because the wave is flexion in the metal but can't be it in air, comparing the speeds of compression waves is irrelevant.

Finding resonant frequencies like 8KHz (or worse, 40Hz) for a bell should have alarmed you. Applying a 6000m/s speed from a compression wave is obviously wrong. Why insist?

 

OK let us examine the mechanics of the situation.

 

I stated in post# 10 that the bell launches a spherical wave into the surrounding air.

 

This is in agreement with my measurements around my test bell. That is the readings possess spherical symmetry. There are no measurable nodes or antinodes and the wave is not, as you stated in post# 21 composed of standing waves. There is no measurable interference.

 

This implies that the bell should be considered as a single point source, emitting a single expanding spherical wave.

 

To see how this can arise and why the relative speeds of sound are important look at diagram (1)

 

Points A and B are on opposite sides of a bell, which is immersed in undisturbed air.

 

The constant that has the dimensions of velocity in the differential equation describing flexural vibrations of plates and bells has a value of a few thousand m/s at the frequencies observed.

The speed of sound has a value approaching six thousand m/s, as already noted.

 

The theory of elasticity requires that any elastic disturbance, imparted at A, is transmitted as rapidly as possible through the metal of the bell to B. It further requires that the maximum speed is the speed of sound.

 

So any disturbance imparted at A will create a disturbance locally in the adjacent air.

The disturbance through the metal of the bell will reach B some 10 to 20 times faster than the disturbance around the outside of the bell since the airborne disturbance travels at 340m/s, since both paths are essentially the same length.

 

When the disturbance in the metal of the bell reaches B, the elastic response at B generates a disturbance in its own right in the air local to B.

 

The point about this is that the response at B is effectively instantaneous so B can be considered to be launching its part of the wave at the same moment as A.

Another way of putting this is to say that the wave at B is in phase with the wave at A or that time zero is the same for both points or that the wave centres from A and B are coincident.

 

This is important because it is our justification for the statement that the bell is a single point source generating an expanding spherical wave centred somewhere in the middle of the bell.

 

It is also in complete accordance with conventional wisdom that sound sources smaller than the wavelength of the sound generate spherical waves based on a single point source.

 

So we have an expanding sound field generated by the bell acting as a single point vibrating source, that looks like diagram (2).

 

There is no interference since a single wave cannot directly interfere with itself.

 

Which brings us nicely to the issue of interference.

 

The principle of superposition states that the resultant motion is the sum of the individual motions acting separately.

 

So for any two waves Ytotal = y1 + y2

 

Now a spherical acoustic wave has both real and imaginary parts in its equation of motion. But I am going to take a plane wave since it has only real parts for simplicity in the next analysis.

 

 

[math]{y_1} = {A_1}{\omega _1}\left( {x - ct} \right)[/math]

 

 

[math]{y_2} = {A_1}{\omega _2}\left( {x - ct} \right)[/math]

 

In general since omega1 is not equal to omega 2 the sum of these two travelling waves yields a modulated travelling wave.

In general the sum may not have periodic zeros.

 

However by suitable choice of constants and making omega1 equal to omega2 we find after a bit of trigonometric reduction on the sum

 

 

[math]Y = {y_1} + {y_2} = B\sin \omega x\cos \omega ct[/math]

 

This is the equation of a standing wave since the first (sine) term has periodic zeros distanced in space between points in space specified by omega times x, regardless of the time t.

The point is that the wave (a cosine) is multiplied by a sine wave, which has periodic zeros.

 

Thus this equation is the equation of a standing wave and is how nodes arise.

 

Now you have stated categorically that standing waves interfere.

 

So please show mathematically how the superposition (sum) of two such waves can interfere to form zeros or nodes.

 

You have also mentioned that the harmonics of the vibration die away more quickly than the fundamental.

Yet you claim energy analysis is irrelevant at best and misleading at worst.

 

The energy in the harmonics of an oscillator increases with harmonic number.

The higher the harmonic more energetic it is.

Thus as energy is lost to the vibrating system through dispersion and/or dissipation, the energy to power the highest harmonic is lost first. And so on down the list.

This is commonly observed in vibrating systems.

 

OK so we have generated an interference free expanding spherical wave in the room, but we are here to explain the why the sound level is lower inside the bell.

 

What is different?

 

The only thing that is different is that the air inside the bell is significantly more confined that outside.

 

Here are a few calculations to show how this makes a difference.

The room volume is V₀ = 3x2x10 = 60 cubic metres

 

The bell volume is

[math]{V_2} = \pi \left( {{{0.2}^2}} \right)\left( {0.15} \right) = .02[/math]cubic metres

 

And the bell surface area is 0.2 square metres

 

Say the average movement is 0.01 mm = 2x10^-5 metres

 

Then the volume change on compression is –2x10^-6 cubic metres (note negative for compression)

 

Now the volume and pressure changes follow the adiabatic gamma law with gamma = 1.4

 

So

 

For the room

[math]{P_0}V_0^{1.4} = {P_1}V_1^{1.4}[/math]

 

and for the inside of the bell

[math]{P_0}V_2^{1.4} = {P_3}V_3^{1.4}[/math]

 

Since the original pressure P0 is the same in both equations solve each for P₀ and equate.

[math]\frac{{{P_1}}}{{{P_3}}} = \frac{{V_3^{1.4}V_0^{1.4}}}{{V_1^{1.4}V_2^{1.4}}}[/math]

Now if we observe that V₀ = 60 cubic metres and V₁ = (60 - .000002) cubic metres we can approximate V₀/V₁ = 1

 

Thus

 

[math]\frac{{{P_1}}}{{{P_3}}} = {\left( {\frac{{20000}}{{20002}}} \right)^{1.4}} = 1.00014[/math]

 

Which very clearly states that the pressure generated outside the bell (P₁) is greater than the pressure inside (P₃) and of the correct order since sound pressure levels are measured in parts per million relative to normal absolute (1bar).

 

But we should further note that these are absolute pressures in order to use the adiabtic gas laws.

 

The sound pressure level (spl) is actually a pressure difference between the quiescent pressure and the wave pressure.

 

That is spl outside = (P₁ - P₀)

and inside = (P₃- P₀)

 

We could calculate these by using a standard value for P₀ eg (1bar) and I will leave that as an exercise.

 

So is detailed discussion of the modes of vibration of the bell or standing wave interference really necessary to answer the question?

Edited March 25, 2014 by studiot

[Enthalpy]

I stated in post# 10 that the bell launches a spherical wave into the surrounding air.

 

Which was already horribly wrong. That's the very basics of a bell. Find an elementary book.

You have alluded several times to my educational background.

 

Obviously, you're weak on acoustics. It looks like you don't realize that. Your long contribution here is just repeatedly wrong.

I stated in post# 10 that the bell launches a spherical wave into the surrounding air.

 

This is in agreement with my measurements around my test bell. That is the readings possess spherical symmetry. There are no measurable nodes or antinodes and the wave is not, as you stated in post# 21 composed of standing waves. There is no measurable interference.

 

In case you really made the measurements, then do they again, and better.

The nodes and antinodes can be heard.

The constant that has the dimensions of velocity in the differential equation describing flexural vibrations of plates and bells has a value of a few thousand m/s at the frequencies observed.

The speed of sound has a value approaching six thousand m/s, as already noted.

 

As already noted, you should first learn what a flexural mode is. Its propagation speed is not constant, and is much slower than the compression wave you're mixing it up with.

 

Once you know this absolute minimum of acoustics, you can try to analyze bells.

The theory of elasticity requires that any elastic disturbance, imparted at A, is transmitted as rapidly as possible through the metal of the bell to B. It further requires that the maximum speed is the speed of sound.

 

You obviously ignore the basics of the theory of elasticity, too. That's nonsense.

[math]Y = {y_1} + {y_2} = B\sin \omega x\cos \omega ct[/math]

 

This is the equation of a standing wave

 

sin(omega*distance) is nonsense, because the units are not consistent. So is cos(omega*c*time).

That little bit of trigonometry is already too much for you.

Checking the units, too - you can't make physics that way.

Edited March 26, 2014 by Enthalpy

[Function]

 

Which was already horribly wrong. That's the very basics of a bell. Find an elementary book.

 

Obviously, you're weak on acoustics. It looks like you don't realize that. Your long contribution here is just repeatedly wrong.

 

In case you really made the measurements, then do they again, and better.

The nodes and antinodes can be heard.

 

As already noted, you should first learn what a flexural mode is. Its propagation speed is not constant, and is much slower than the compression wave you're mixing it up with.

 

Once you know this absolute minimum of acoustics, you can try to analyze bells.

 

You obviously ignore the basics of the theory of elasticity, too. That's nonsense.

 

[some other things...]

 

Your sentences show disrespect against studiot. Be polite and respectful and try to correct one another in a most respectful way.

Let's remember that each of us is only human and that we all make mistakes, not to say that any of you made any mistakes (I couldn't hardly follow any longer), but stay polite.

Edited March 26, 2014 by Function

[Enthalpy]

Comparing the volume of the bell with the volume of the room is just nonsense as well. We're talking about acoustics, not bicycle pumps.

 

The whole contribution brings only confusion to the subject.

[studiot]

Setting aside all the personal remarks I will take one statement and ask you for detailed backup.

 

 

 

Enthalpy

As already noted, you should first learn what a flexural mode is. Its propagation speed is not constant, and is much slower than the compression wave you're mixing it up with.

 

The figures I have quoted were published by the Applied Research Laboratory of Pennsylvania State University in Acoustics Today.

 

Studiot

The constant that has the dimensions of velocity in the differential equation describing flexural vibrations of plates and bells has a value of a few thousand m/s at the frequencies observed.

 

If you disagree with these then I look forward to your detailed alternative figures along with your substantiation of them.

 

I am still looking forward to your detailed mathematical analysis requested in my post#16.

Edited March 26, 2014 by studiot

[Enthalpy]

6000m/s is not a flexural wave. In a publication neither.

 

I have no time to make detailed answers to false random assertions.

http://en.wikipedia.org/wiki/Structural_acoustics#Bending_Waves_in_beams_and_plates

Edited March 26, 2014 by Enthalpy

[Function]

6000m/s is not a flexural wave. In a publication neither.

 

I have no time to make detailed answers to false random assertions.

http://en.wikipedia.org/wiki/Structural_acoustics#Bending_Waves_in_beams_and_plates

 

Well... Wikipedia...

[studiot]

 

Enthalpy

6000m/s is not a flexural wave. In a publication neither.

 

I have no time to make detailed answers to false random assertions.

http://en.wikipedia....eams_and_plates

 

 

 

What exactly am I supposed to make of this?

 

What is it a reply to, did you not read my post?

 

studiot

flexural vibrations of plates and bells has a value of a few thousand m/s at the frequencies observed.

 

Do you think 6000 is a few thousand?

 

I would say a few thousand is more than 1000 but not more than 3000. This is the figure you will find published as I noted.

 

I further note your Wikipedia link is self admittedly half finished and does not publish any figures.

Edited March 26, 2014 by studiot

[imatfaal]

!

Moderator Note

 

Enthalpy - the personal comments stop now. Any further posts including remarks addressed to the person rather than to the argument risk being deleted and sanctions being taken.

 

Do not respond to this modnote within the thread

 

[studiot]

 

I couldn't hardly follow any longer

 

 

Good morning function.

 

All sound sources much smaller than the wavelength can be considered as an equivalent sphere. There is indeed much theory using equivalent spheres as it is important to the theory of both binaural hearing and stereophonic sound reproduction.

 

As frequency rises the wavelength gets shorter and the sound emitted gets more directional.

The higher frequencies contain all the directional information.

 

A spherical wave contains no directional information.

 

This is why it does not matter where you place a bass loudpeaker in a room and indeed you only really need one bass speaker.

It is also why animals, such as bats and dolphins, use high frequency for location.

 

For interest here is a short table of frequency and wavelength

 

50 Hz 6.90 metres

500 Hz 0.69 metres

1000 Hz 0.34 metres

5000 Hz 0.07 metres

15000 Hz 0.02 metres

 

You can see that at 500Hz it need a pretty sizeable source to be bigger than the wave.

Edited March 27, 2014 by studiot

[rktpro]

 

Studiot-Sir, I am having a difficult understanding the point that a disturbance must instantaneously travel to point B from point A. If it takes time to reach from one point to another, only if it strictly does, then it might be possible to measure beats.

Isn't the wave equation used dimensionally incorrevt? That, as Enthalpy points, might have created an error in the superimposed wave.

Edited March 27, 2014 by rktpro

[studiot]

Oh dear, rktpro, yes the equation is wrong. That blooming velocity, c keeps turning up in the wrong place.

 

Unfortunately I don't speak LaTex so I write all my equations out in MathType and then paste them into the text.

This time I wrote it in Word and pasted them in there and finally pasted it again into SF.

 

Mea culpa.

 

The correct equation is

 

[math]Y = B\sin \omega \left( {\frac{x}{c}} \right)\cos \omega t[/math]

 

The original two equations should have x/c as well.

 

This does not alter the points made, however since the point is that the original waves contain additions (or subtractions) within the angle, of a time and distance function, reduced to common units.

The composite of the two waves into a single standing wave results in the separation of time and distance into the product of a time function and a distance function.

Edited March 27, 2014 by studiot

[arc]

this may be a bit off topic but still bathroom related so i feel it must not be missed.

did you know that a toilet bowl has it's own ring when you flush it?

"the catch is" to be at the right distance from the bowl.

 

And to not get your toes wet.

[davidivad]

yep

[rktpro]

Oh dear, rktpro, yes the equation is wrong. That blooming velocity, c keeps turning up in the wrong place.

 

Unfortunately I don't speak LaTex so I write all my equations out in MathType and then paste them into the text.

This time I wrote it in Word and pasted them in there and finally pasted it again into SF.

 

Mea culpa.

 

The correct equation is

 

[math]Y = B\sin \omega \left( {\frac{x}{c}} \right)\cos \omega t[/math]

 

 

The original two equations should have x/c as well.

 

This does not alter the points made, however since the point is that the original waves contain additions (or subtractions) within the angle, of a time and distance function, reduced to common units.

The composite of the two waves into a single standing wave results in the separation of time and distance into the product of a time function and a distance function.

I mentioned a question regarding disturbance travelling instantaneously. Help needed.

I feel that beats can be detected initially.

[studiot]

 

Studiot-Sir, I am having a difficult understanding the point that a disturbance must instantaneously travel to point B from point A. If it takes time to reach from one point to another, only if it strictly does, then it might be possible to measure beats.

Isn't the wave equation used dimensionally incorrevt? That, as Enthalpy points, might have created an error in the superimposed wave.

 

 

I am not sure what you think is modulating what to create beats.

 

Nothing actually travels instantaneously, I believe I said it was sensibly so.

 

Consider the distance from A to B is about 0.3m

 

An elastic signal travelling in the steel travels at 6000 m/s

 

An elastic signal travelling in the air travels at 340 m/s

 

The bell in struck with a hammer (at A), not continuously excited.

 

So at the moment of imapct at A the part of the sidewall at A deflects.

 

This causes a local pressuse disturbance around A.

 

At this instant B is undeflected as is the local air around it.

 

From the diagram you can see that the disturbance in the steel has to travel about the same distance as the disturbance in the air (=0.3m)

 

So in the time for the disturbance in the steel to reach B and cause an elastic deflection there the distrubance in the air around A travels

 

0.3 x (340/6000) = 0.017m so it has not reached B, by a large margin

 

This is why I am suggesting that the elastic disturbances at A and B in the air can be considered contemporaneous.

 

This is an additional step to the usual analysis you will find for a piston in air because the piston is being driven at all points simultaneously.

 

What the above shows is that the impact at one point on my bell has the same effect as if the whole bell was excited at once.

 

The usual analyis a la a piston can then be applied and it become a question of geometry. That is the size of the radiator relative to the size of the wave.

 

When the radiator size is comparable to (or much bigger than) the wave size, an off axis listener can receive waves from different parts of the radiator at different times because she is at different distances from different parts of the radiator. These waves can interfere and cause directionality as I previously noted.

This effect is most marked at right angles to the main axis.

 

But remember that the size of my bell is smaller than the wavelength of the signal I am observing.

 

I can draw some helpful diagrams if you are interested.

[Enthalpy]

An elastic signal travelling in the steel travels at 6000 m/s

 

Very wrong for flexural waves, which is the case here. Flexural waves are the absolute pre-requisite to understand bells. I've provided a link to Wiki.

 

By the way, 6000m/s would give nonsense resonant frequencies for a bell. Checking that figure tells that it's the wrong way.