AudioCircle
Audio/Video Gear and Systems => The Lab => Topic started by: timbley on 3 Jan 2019, 06:13 am
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Is the pressure in front of the speaker cone highest when the cone is moving fastest, or when it is at its furthest excursion point? Or somewhere in-between?
I've been pondering this ever since reading this article about doppler distortion by Rod Elliot:
http://sound.whsites.net/doppler.htm
He asserts ( if I read correctly ) that the pressure is greatest when the cone is at its furthest excursion point. This seems odd to me because the cone is not moving at that point. Sound pressure is moving away at the speed of sound, and the cone never moves anywhere near fast enough to keep up.
I did some web searching but couldn't find a definitive answer. Any help would be greatly appreciated!
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Neither - SPL is proportional to cone acceleration. Without knowing specifically which part of the article you are referring to, I suspect you have misinterpreted what Eliot has said.
:)
Interesting topic though, I must have another look.
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Acceleration would be maximum at the peaks of excursion. Is that right? It's doing a full 180 course change at that point.
I'm looking at this animation of a pendulum showing acceleration force maxing out at the max excursion points:
https://physics.stackexchange.com/questions/233193/direction-of-velocity-and-acceleration-for-a-pendulum
If it's true that max SPL happens when the cone is at max acceleration then that would imply the pressure is greatest when the cone velocity is zero and it is at its maximum acceleration back in the other direction. How can that be? It seems intuitive to me that pressure will be greatest in front of a diaphragm when it's moving the fastest.
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Here's part of Rod's article that's giving me pause, specifically the part I colored red:
"If the shift is a genuine Doppler shift, then we will see the signal compressed or stretched at the zero crossing points of the LF waveform, and at the peaks of the LF waveform the HF signal should be in phase, since LF cone movement is effectively zero (for a moment in time, at least). "
Interestingly, his measurements seem to suggest that he is correct in this assertion. I just can't understand how it could be true. Could it be that there's a phase shift between those frequencies from his speaker and it's throwing off the measurement? Or maybe the recording room volume is small enough and the bass frequency low enough that the pressure really is highest at peak excursion? At 50Hz the wavelength is 22.5 feet, so if he did this in a really small lab it might be that the whole room was evenly pressurizing from the cone movement, which would make his assertion correct.
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The cone doesn't have to be moving at the speed of sound in order to induce a doppler effect. Remember, the doppler effect was first noticed with the sound of sirens on ambulances. Ambulances certainly don't move at the speed of sound.
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I agree!
What I was trying to say is that the air pressure moves away at the speed of sound unless it is contained, so to get max pressure at max excursion you'd have to get the cone there impossibly fast.
The question is: When is the air pressure in front of the cone the highest? When it's moving fastest, or when it's at it's peak of excursion? It seems to me that it's when the cone is moving fastest, unless the air is contained in a small enough volume. In that case pressure will be greatest at peak excursion. I think I've heard this called the oil can effect.
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The question is: When is the air pressure in front of the cone the highest? When it's moving fastest, or when it's at it's peak of excursion?
Neither - it's when the acceleration is greatest. If the signal is a sine wave then the peak excursion does also happen to be the maximum acceleration. (With the sign reversed.)
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Thanks John R for answering my question. How did you learn this about maximum acceleration of the cone being the point of maximum pressure? I can't seem to find that information anywhere. Anything you could recommend that I read?
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Thanks John R for answering my question. How did you learn this about maximum acceleration of the cone being the point of maximum pressure? I can't seem to find that information anywhere. Anything you could recommend that I read?
Possibly explained here:
https://www.klippel.de/fileadmin/_migrated/content_uploads/Klippel__Schlechter__Distributed_parameter.pdf
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Thanks John R for answering my question. How did you learn this about maximum acceleration of the cone being the point of maximum pressure? I can't seem to find that information anywhere. Anything you could recommend that I read?
Hi, this is just the physics: velocity is the derivative of distance (displacement), acceleration is the derivative of velocity. If displacement is a sine function, velocity is cosine, and acceleration is -sine.
Actually the pendulum animation you posted before illustrates it.
[Update] Here is a graph:
(https://www.spaceagecontrol.com/calcsinm01.gif)
From https://www.spaceagecontrol.com/calcsinm.htm
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Thanks Doublej and JohnR. That Klippel paper is very interesting. I think I'll be looking at that for a while!
Yes, the pendulum diagram I showed indeed shows what you've been saying, that acceleration of the cone is greatest at the peaks of excursion.
What I'm thinking is that there's a confusion here. We need to consider the acceleration of the air molecules in front of the cone instead of the cone's own acceleration. When do the air molecules reach their peak acceleration? That's when we get peak pressure. I think it's at the zero crossing point of the driver, when it's acceleration is at a minimum and its speed at max, ramming into the air and squishing it.
I'm trying to think up an experiment to make sure I'm wrong about this.
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Neither - it's when the acceleration is greatest. If the signal is a sine wave then the peak excursion does also happen to be the maximum acceleration. (With the sign reversed.)
Peak acceleration in the sine wave of the cone movement is at the x zero. At peak excursion acceleration is the lowest.
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Peak acceleration in the sine wave of the cone movement is at the x zero. At peak excursion acceleration is the lowest.
You are writing "acceleration" when you should be writing "velocity". Please refer to the graph I posted above.
I'm trying to think up an experiment to make sure I'm wrong about this.
Not sure what kind of experiment you had in mind, but suppose you played a tone at some frequency and SPL and measure the excursion. Then play a tone at half the frequency and same SPL and measure the excursion. If SPL is proportional to velocity then the excursion will be double. If SPL is proportional to acceleration then excursion will be 4x.
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THe peak air pressure would be somewhere before maximum excursion forward, of the cone.
I am certain there is some complex math which could show the air pressure in front of the cone!
As a thought experiment alone, The air cannot move out of the way as fast as the cone is moving forward from the start at the lowest cone position. As the cone accelerates forward the air pressure is building up from a negative to mean air pressure.. past the average in the room probably a little before the cone reaches the midpoint (at least some..) and still building as the cone is moving forward..
As the cone reaches nears the end of the excursion the wave in front of it reaches a maximum, and is already decreasing in pressure at the piston surface as the piston reaches the max forward point (since the velocity of the piston is dropping rapidly).
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Not sure what kind of experiment you had in mind, but suppose you played a tone at some frequency and SPL and measure the excursion. Then play a tone at half the frequency and same SPL and measure the excursion. If SPL is proportional to velocity then the excursion will be double. If SPL is proportional to acceleration then excursion will be 4x.
Measured total SPL is directly proportional to acceleration of the cone. I totally agree with that, but that's not what I'm asking.
I'm asking when during the cone's excursion is pressure greatest in front of the cone. I would need to do something like put a pressure meter on the speaker cone and film it in slow motion playing a 20hz tone.
One simple example might be to put a bubble wand on the end of a long heavy pendulum and get it swinging. At what point in the pendulum's swing do you think the most bubbles will get made? It takes pressure across the wand to blow bubbles. I'm pretty sure it's going to happen when the pendulum crosses through its zero crossing point, where it's moving fastest, not at the peaks of it's excursion. It's experiencing max acceleration out there, but it's not moving through the air so it can't be transferring energy to accelerate the air and create pressure.
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THe peak air pressure would be somewhere before maximum excursion forward, of the cone.
I am certain there is some complex math which could show the air pressure in front of the cone!
As a thought experiment alone, The air cannot move out of the way as fast as the cone is moving forward from the start at the lowest cone position. As the cone accelerates forward the air pressure is building up from a negative to mean air pressure.. past the average in the room probably a little before the cone reaches the midpoint (at least some..) and still building as the cone is moving forward..
As the cone reaches nears the end of the excursion the wave in front of it reaches a maximum, and is already decreasing in pressure at the piston surface as the piston reaches the max forward point (since the velocity of the piston is dropping rapidly).
This all makes sense to me. The exact point of max pressure may change with the relationship of the max speed of the cone compared to the speed of sound, and maybe some other factors, such as the presence of a horn and compression chamber. But it can't be at the end of the piston's max forward point, when it has momentarily ceased all motion.
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It's experiencing max acceleration out there, but it's not moving through the air so it can't be transferring energy to accelerate the air and create pressure.
You're saying that despite the fact that the cone is at its maximum acceleration, the air is not accelerating at all.
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It might be easiest to look at it in terms of the cone has finished doing all the work on the air once it's at it's max excursion. At that instant it's not doing any more work on the air, yet at the same time the air has been compressed or refracted to it's maximum.
When the cone is at it's highest velocity it's right in the middle of doing work on the air, and still has 50% left to go.
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Hm...
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This might help inform or confuse the topic:
https://www.har-bal.com/wp-content/uploads/Speaker_theory.pdf
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This might help inform or confuse the topic:
https://www.har-bal.com/wp-content/uploads/Speaker_theory.pdf
I'll have to take a long look at that.
Thanks everybody for the responses. I think we're settled on the fact that the peak of excursion does not correspond to the sound wave pressure peak, assuming the driver is emitting into free space. Or are we?
I don't know what this means for the article on phase and doppler modulation. My take is that his figure 1.3 in the article is the correct expected display of both phase and doppler modulation, which are exactly the same thing. Figure 1.4 should not be expected normally, but that's the reading he is showing that he got. He is using an open baffle so that rules out the small space and long wavelength theory I had before.
The problem might be that the measured low frequency waveform is significantly misshapen compared to the electrical input signal, so lining them up in any meaningful way to show the subtle phase shifting of the higher frequencies is futile. I can't see for the life of me how figure 2.3 is a close up of figure 2.1. The tops of the waveforms don't line up at all.
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Well, I think you will continue to struggle with the article as long as you insist on starting from an incorrect premise. I pointed out the obvious contradiction in your reasoning... do you not see that contradiction?
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There has been several mentions of acceleration. Remember that acceleration is greatest when the net force applied is the greatest. F=MA
It is independent of velocity of the cone or the wave. But it will be affected by the reverse push of the air particles as they pile up. Particles will pile up in front of the moving cone fastest whenever the cone is moving the fastest, and the wavefront will move off at the speed of sound.
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Hi Scott, so what would you think of the explanation here that says "pressure is proportional to the cone acceleration":
https://physics.stackexchange.com/questions/208882/relationship-between-sound-pressure-level-and-amplitude-of-signal/367877#367877
and this followup explanation where it says that "presence of the i shows that the velocity is out of phase with the pressure" and rewrites in terms of acceleration:
https://physics.stackexchange.com/questions/405191/why-is-this-equation-true-for-the-sound-pressure-a-loudspeaker-creates
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After posting this I realize I made an error :lol:
Well, I think you will continue to struggle with the article as long as you insist on starting from an incorrect premise. I pointed out the obvious contradiction in your reasoning... do you not see that contradiction?
JohnR, I think you have finally clarified this problem for me and I can't thank you enough!!!
I think I see where the apparent contradiction is coming from. This first equation isn't appropriate to answer my question because it doesn't address a specific point in time:
sound pressure level = ( (air density x surface area of the cone) / (2 x pi x distance from the cone) ) x acceleration of the cone
But the equation in the link you provided below does address air pressure at a specific location and specific point in time!
This is what I've been looking for!!!
and this followup explanation where it says that "presence of the i shows that the velocity is out of phase with the pressure" and rewrites in terms of acceleration:
https://physics.stackexchange.com/questions/405191/why-is-this-equation-true-for-the-sound-pressure-a-loudspeaker-creates
If you take the equation he's showing there and make the distance from the speaker cone zero, ϕ comes out to zero too.
edit: I'm WRONG!! ϕ comes out to the diameter of the cone.
So the whole equation boils down to:
pressure(at time t) = density of air x speed of sound x velocity of cone (at time t)
EUREKA!!!
Let me try again.
PressureAtTimeT = DenisityOfAir x SpeedOfSound x (1 - e^(-i x K x diameter of cone)) x VelocityOfConeAtTimeT
That K thing: waves per unit distance? Something corresponding to wavelength and frequency.
This makes more sense because obviously the frequency and diameter of the cone need to be taken in to account.
I think this still shows on all counts that the Pressure at the cone at time T must be zero if the velocity of the cone is also zero at time T.
It really boils down to: Pressure at time T = (a bunch of complicated stuff) x Velocity of cone at time T.
p(t)=iρckϕv(t)
This is the far field reduction of the equation. The guy says the presence of i puts the pressure out of phase with the velocity.
i x 0 = 0. So even with this equation, when velocity goes to zero at time T, so does pressure at time T. This means it's not 90 degrees out of phase, it has to be 180 degrees out of phase. The zero crossing points still line up.
EUREKA!!! ???
Please correct me as needed.
So why is acceleration a better metric than speed of cone when determining how loud the sound will be? I think I understand that! For any given cone size acting like a point source, acceleration will be proportional to the sound level regardless of the frequency! Speed varies with frequency at the same sound level. More speed is required at lower frequencies, but the acceleration doesn't go up because the number of turn arounds required per unit time goes down.
But don't forget, at any given frequency, sound pressure level is directly proportional to the speed of the cone.
I found Kinsler 4th edition here if you feel like reading:
https://the-eye.eu/public/WorldTracker.org/Physics/Fundamentals%20of%20Acoustics%204th%20ed%20-%20L.%20Kinsler%2C%20et%20al.%2C%20%28Wiley%2C%202000%29%20WW.pdf
I realize that the question I'm asking is not often asked. Generally people don't care about this particular concern. Rod Elliott tackled the issue and I applaud him for it. It may become a practical concern if DSP is to be implemented on sophisticated speaker systems to effectively eliminate doppler distortion.
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I'm not sure pressure is a function of velocity at all as Pmax is at Vmin... right when the cone has stopped... and not at Vmax. From what you've wrote, you haven't changed your premise at all.
I do think V matters, but likely as a corrective factor rather than a main factor, at least within the range of human hearing.
While the air will move the instant the cone starts moving, it is compressible and we can look at the simplified ideal that the air is acted on evenly and pressure built evenly by the cone while it's in motion.
If this is the case, then when the cone is at Vmax it's still in the process of compressing the air and the pressure is still going up. It's not until the cone stops moving that it stops compressing the air, which is at Vmin=0 and Amax.
Also, another real life factor may be that the Pmax is right before Vmin because as V approaches zero, at some point the velocity will become slow with respect to the speed of sound, and V will be effectively zero even though it's not quite there yet.
As far as the math, I can do math... but, well... I'd much rather not do math unless I really have to. :green:
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I'm not sure pressure is a function of velocity at all as Pmax is at Vmin... right when the cone has stopped... and not at Vmax. From what you've wrote, you haven't changed your premise at all.
You are right. I haven't changed my premise. I've just had it clarified! The math is backing me up after all. The math that didn't seem to be backing me up was simply misapplied. Pmax is at Vmin? If you're talking about Vmin of a speaker cone I don't know that. Haven't seen any clear evidence of that. The textbooks don't seem to indicate that. Doesn't make any intuitive sense to me. If you are talking about the venturi effect, then I'm with you. But this is not a venturi effect. We're talking about a cone running into air that doesn't want to stay put when it gets squeezed.
I do think V matters, but likely as a corrective factor rather than a main factor, at least within the range of human hearing.
While the air will move the instant the cone starts moving, it is compressible and we can look at the simplified ideal that the air is acted on evenly and pressure built evenly by the cone while it's in motion.
If this is the case, then when the cone is at Vmax it's still in the process of compressing the air and the pressure is still going up. It's not until the cone stops moving that it stops compressing the air, which is at Vmin=0 and Amax.
Also, another real life factor may be that the Pmax is right before Vmin because as V approaches zero, at some point the velocity will become slow with respect to the speed of sound, and V will be effectively zero even though it's not quite there yet.
The cone never even gets remotely close to the speed of sound. Any pressure built up in front of the diaphragm escapes at the speed of sound in free space, leaving the cone far, far behind. It won't sit around and wait for the diaphragm to finish it's work. The only time peak excursion would correlate with peak pressure is in the case of a very small enclosure in front of the diaphragm. In that case the pressure can't escape so it just builds up as the diaphragm reduces the volume of the chamber.
As far as the math, I can do math... but, well... I'd much rather not do math unless I really have to. :green:
I'm like you, but in this case we have to do the math! :scratch:
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Great find on Kinsler :thumb: Worth further study (me three on the math) but note quickly page 181 "The quantity r1 has physical meaning only if the ratio a/lambda is large enough that r1 > 0. Indeed, if a = lambda/2, then r1 = 0 and there is no nearfield. At still lower frequencies the radiation from the piston approaches that of a simple source." r1 is the nearfield-farfield transition, a is cone radius. In other words, if the wavelength is long enough only the farfield / point source equation applies.
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It really boils down to: Pressure at time T = (a bunch of complicated stuff) x Velocity of cone at time T.
p(t)=iρckϕv(t)
This is the far field reduction of the equation. The guy says the presence of i puts the pressure out of phase with the velocity.
i x 0 = 0.
Ahhmm.... I don't think it works that way. i is the imaginary unit, v(t) is a complex sinusoid i.e. cos(wt) + isin(wt). When you multiply the i in front you then get i cos(wt) - sin(wt) ...
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Update: had to think about that. The reason your substitution doesn't work is because you are considering only the real part of v(t). However, when the real part is zero then the imaginary part is at its peak (+/- i). So your substitution should be something like
p(t) = i blah v(t)
= i blah (0 + i) | at some t
= blah (i * i) = -blah
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Ahhmm.... I don't think it works that way. i is the imaginary unit, v(t) is a complex sinusoid i.e. cos(wt) + isin(wt). When you multiply the i in front you then get i cos(wt) - sin(wt) ...
Well, how about that. It's never as simple as I want it to be. One thing is for sure, I'm going to be significantly more knowledgeable before I get this straightened out. I'm weak on imaginary numbers and Euler's identity is mind boggling to me. Had to check that multiplying i by zero was really zero. I try not to make assumptions about imaginary numbers.
I appreciate everyone's engagement with this!
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The cone never even gets remotely close to the speed of sound. Any pressure built up in front of the diaphragm escapes at the speed of sound in free space...
I'm not sure about that. The air is being compressed by a cone going under the speed of sound, and the air is moving less than the speed of the cone unless it's right next to the cone. This results in a pressure wave that propagates at the speed of sound, but the compressed air is not moving anywhere near that fast.... how can a cone moving much slower than the speed of sound propel the air faster than it's own velocity? There's a difference between a pressure wave propagating through the air vs the air actually moving as a result of cone motion. Both are obviously happening to some degree, as pressure requires the air molecules to move closer together.
I misspoke when I said cone motion relative to speed of sound, it's cone motion relative to how fast the cone needs to move to cause the air to compress significantly.
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I'm not sure about that. The air is being compressed by a cone going under the speed of sound, and the air is moving less than the speed of the cone unless it's right next to the cone. This results in a pressure wave that propagates at the speed of sound, but the compressed air is not moving anywhere near that fast.... how can a cone moving much slower than the speed of sound propel the air faster than it's own velocity? There's a difference between a pressure wave propagating through the air vs the air actually moving as a result of cone motion. Both are obviously happening to some degree, as pressure requires the air molecules to move closer together.
I misspoke when I said cone motion relative to speed of sound, it's cone motion relative to how fast the cone needs to move to cause the air to compress significantly.
Thanks for clearing that up about the cone speed relative to speed of sound. As for sound waves propagating, I'm having trouble visualizing it. It's strange to me that a cone moving so slowly can create any pressure at all when the pressure escapes so quickly. And why does that sound wave move so fast?
Most of the visual examples we have of wave propagation involve circumstances where the initial motion is faster than the propagation rate. Tap on the surface of water quickly, the wave propagates slowly. Flick one end of a garden hose quickly and watch the wave propagate slowly.
A Newton's cradle, now THAT starts to work for me. A slow swing of the ball on one end transmits the power to the other end very quickly. The power is transmitted much faster than the motion I input. If the balls are spaced further apart, the transmission speed slows. Well anyways, something to think about.
I Found this page that describes the speed of sound relative to different conditions.
http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/souspe3.html#c1
It is interesting to compare this speed [of sound] with the speed of molecules as a result of their thermal energy.
So I just learned that hot, humid air transmits sound faster than cold dry air at the same atmospheric pressure. I learned years ago while taking flying lessons that hot humid days provide less drag and lift than cold dry days at the same atmospheric pressure, making for long runway rolls and higher groundspeed on take off and landing.
Is there a relationship here? Does a speaker cone have to move farther and faster to create sound pressure on a hot, humid day, even though the atmospheric pressure is the same?
Sounds like an interesting experiment!!
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(https://www.audiocircle.com/image.php?id=188717)
ρ (rho) is the density of air. (From here (https://physics.stackexchange.com/questions/405191/why-is-this-equation-true-for-the-sound-pressure-a-loudspeaker-creates?answertab=votes#tab-top).)
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(https://www.audiocircle.com/image.php?id=188717)
ρ (rho) is the density of air. (From here (https://physics.stackexchange.com/questions/405191/why-is-this-equation-true-for-the-sound-pressure-a-loudspeaker-creates?answertab=votes#tab-top).)
Thanks JohnR
We know this doesn't work for my question because it only works in the far field. As we get very close to the cone the pressure unrealistically approaches infinity. As we move away from the cone, at any given point in time we should see the pressure going positive, through zero, and then negative, up and down in a sine wave. This equation doesn't do that. The pressure just drops steadily as we move away. It is just a useful generalization of the sound level at a given distance and relating it to the cone acceleration required to get that sound level.
Now I could be missing something because P(t) in this equation is still a complex pressure, and I'll confess I'm fuzzy on what that means. By breaking that down maybe the pressure undulations with distance can be recovered?
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Hi, I think you missed the post I made above, which quoted Kinsler. According to that, and assuming I am interpreting your post correctly, the cyclic pressure variations with distance don't exist if the wavelength is greater than the cone diameter. Yes, the
question equation (darned auto-correct) is an approximation only, actual pressure at r=0 will be finite.
However, I was addressing the interesting heat/humidity question you asked, "Does a speaker cone have to move farther and faster to create sound pressure on a hot, humid day, even though the atmospheric pressure is the same?" Since a. density decreases with temperature b. density decreases with humidity (I didn't know that, my intuition told me otherwise, another example of why you should never use intuition to solve a physics problem ;) ) c. sound pressure is proportional to air density (per the equation) d. velocity is the integral of acceleration e. excursion is the integral of velocity, the answer to your question is YES!
Or, if you want your audio system to play louder, move to the North Pole ;)
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Thanks John for the responses. The thing about the speaker losing efficiency on a hot humid day is fascinating to me. Maybe we should make the insides of our speakers hot and humid to reduce the sound inside the box. Or, fill the box with hot helium!
Back to the question about when during the cone's excursion is the pressure in front of the cone the greatest, I got to thinking about how to test it, and came up with an idea. Play a 100hz tone and bring the microphone up to the cone until it just barely starts getting tapped by the cone. Record that and see where the tap happens. That should tell you when the cone is at maximum excursion relative to the pressure.
Here is the result I got.
(https://www.audiocircle.com/image.php?id=188930)
My interpretation is that the descending slope zero crossing point correlates to the cone's maximum excursion. The microphone vibrates on contact at a higher frequency than the cone and nicks the cone again a few times as it's moving away. Or it may just be that the cone oscillates a couple times from the tap.
Earlier I tried to measure doppler distortion by playing a 1000 hz tone on top a 50hz tone. I could not detect the difference in wavelength of the 1000hz tone at different points on the 50hz tone, nor could I see any phase shift. It seemed to be remain perfectly aligned to within a 44.1khz sample. It's a very subtle effect.
I should also note that when I played a 20hz tone, the tap did occur at the peak of the wave form, as I predicted. My room doesn't even have space for a 1/4 length of a 20hz wave, so there is no wave possible in my room. Just pressure going up and down uniformly throughout the room as the cone moves in and out.
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It takes a lot of excursion to be able to hear a doppler effect, but it's possible, just crank it up louder! :)
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Dave, I could easily hear the high frequency being modulated with it turned up loud. The bass waveform got pretty distorted. Frequency analysis showed the sidebands around the 1000khz. I just couldn't easily see the phase or frequency shifts when looking at the waveform.
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Back to the question about when during the cone's excursion is the pressure in front of the cone the greatest, I got to thinking about how to test it, and came up with an idea. Play a 100hz tone and bring the microphone up to the cone until it just barely starts getting tapped by the cone. Record that and see where the tap happens. That should tell you when the cone is at maximum excursion relative to the pressure.
Here is the result I got.
(https://www.audiocircle.com/image.php?id=188930)
Very clever! I wonder if it's possible to use an "obstruction" other than the mic that will make enough sound so the mic can be moved to different distances. Will try to do something this week.
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Very clever! I wonder if it's possible to use an "obstruction" other than the mic that will make enough sound so the mic can be moved to different distances. Will try to do something this week.
That's a good idea. It'll take a little more setup work than just holding the mic. But I guess you could just hold the object with the mic. on a stand. I noticed that I could get the click to happen down at the base of the waveform if I moved the mic further out on the cone. Not sure what that means but I'm guessing it has something to do with the cone flexing.