Please help to explain: speaker cone mechanical energy

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I wouldn't say clear yet but it makes much, much more sense to me now. Geoffroy you really have the gift for educating people, i.e. for grasping what they struggle with and making it straight on their current level of understanding.

ErnieM: Do you mean I'm not the only one who's not a physics expert here? Thank goodness! 🙂
 
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Geoffroy you really have the gift for educating people, i.e. for grasping what they struggle with and making it straight on their current level of understanding.

ErnieM: Do you mean I'm not the only one who's not a physics expert here? Thank goodness! 🙂

Yes he does! :up: 😀

Hardly! I understand it in a mostly common sense sort off way, but couldn't come up with any sort of response like Geoffroy [sic?] did and John K and others with the math just cause my eyes to glaze over. 🙁

GM
 
A simple experiment to try.

Wave your open hand in the air side to side.
If your hand is horizontal, you hardly notice the resistance of the air.
Put it vertically, a bit more air resistance.
Use a magazine, the difference is even greater.

This should give people a more intuitive feel for the mechanical to acoustical inefficiency.
 
A simple experiment to try.

Wave your open hand in the air side to side.
If your hand is horizontal, you hardly notice the resistance of the air.
Put it vertically, a bit more air resistance.
Use a magazine, the difference is even greater.

This should give people a more intuitive feel for the mechanical to acoustical inefficiency.
This is misleading because the work being done is overwhelmingly moving the air around rather than creating sound. In order to create a significant amount of sound you need to move your hand fast enough for the inertia of the disturbed air to be large enough for enough of it to stay where it is and compress creating sound rather than quietly spinning out the way without compressing.

Mathematically if you oscillate your hand in air to create waves around an average state you will get 5 of them: 2 vorticity waves, 2 acoustic waves and an entropy wave. Interesting stuff if you are that way inclined but few are these days compared to when I was young.
 
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If I'm getting this right, this mechanical system would keep in motion once some mechanical energy was put into it? (no losses - vacuum, no friction, etc.)

I must admit this was the non-intuitive part for me (I almost feel the hard work needed to move the piston all the time) but fers' sentence "the energy has nowhere to go in the mechanical system" has made it. Now I simply see no other option.

What is misleading by intuition (at least for me) is that if you want to stop something moving by your hands, you "feel" you have to do some work, i.e. to push it again the motion. But in the physical terms, you are really only dissipating the energy that the moving thing is getting rid of (and maybe some more for your muscles to actually work). Right? 😱
 

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This is misleading because the work being done is overwhelmingly moving the air around rather than creating sound. In order to create a significant amount of sound you need to move your hand fast enough for the inertia of the disturbed air to be large enough for enough of it to stay where it is and compress creating sound rather than quietly spinning out the way without compressing

Moving air around, compressing air and making sound are the same thing. It's just the frequency range that is different.
 
Moving air around, compressing air and making sound are the same thing. It's just the frequency range that is different.
This seems to be the next step beyond not being interested (in fluid mechanics in this case) in making assertions about matters that one presumably knows are not understood. Nobody likes to be wrong but when I was younger responses like this were rare because of the discomfort felt when seen to be wrong and drawing attention to it. This level seems to be much lower today to the extent the UK and the US are both lead by people that are quite openly serial liars in a way which would have terminated their careers a few years ago. Interesting if somewhat disconcerting stuff.
 
What is misleading by intuition (at least for me) is that if you want to stop something moving by your hands, you "feel" you have to do some work, i.e. to push it again the motion. But in the physical terms, you are really only dissipating the energy that the moving thing is getting rid of (and maybe some more for your muscles to actually work). Right? 😱

No. I think the problem here are that people don't have much of a grip on thermodynamics and how work and energy are related. You are doing work when you apply a force to stop a moving object. Two ways to figure the work are W = int(F dx), and W = delta (KE). So if the kinetic energy of the moving mass is m x v^/2 and you bring it to rest, v = 0, the work you have done is doing so is equal to the initial KE.

If I'm getting this right, this mechanical system would keep in motion once some mechanical energy was put into it? (no losses - vacuum, no friction, etc.)

Yes.
 
No. I think the problem here are that people don't have much of a grip on thermodynamics and how work and energy are related. You are doing work when you apply a force to stop a moving object. Two ways to figure the work are W = int(F dx), and W = delta (KE). So if the kinetic energy of the moving mass is m x v^/2 and you bring it to rest, v = 0, the work you have done is doing so is equal to the initial KE.
As the force is in the opposite direction of the velocity vector, the work gets negative, doesn't it (?), i.e. the energy is in fact received, not spend. So in the end after all the kinetic energy is released and everything is still, it must all get converted to heat in the muscles (i.e. more energy in the body -> rise of temperature). If it isn't so, then I just give up.

IOW, as I see it, me stopping a ball is the same as a bullet hitting a wall. Kinetic energy is released in making the deformation/breaking of/heating up (or whatever, including the sound). Are you saying that the wall is loosing its inner energy by stopping the bullet?
 
It's a matter of accounting. As Earl said, no e energy is lost or created. Does the moving mass do work on you or do you do work on the moving mass? Does the bullet do work on the wall or the wall do work on the bullet? The wok done is the change in KE. Since the final state is zero KE and the initial stat is some finite KE amount the work done on the moving mass is negative, thus energy is given up to the wall. Positive work is done on the wall. Does the work or energy end up as heat, sound, chunks of wall flying off, what ever? But ultimately all energy is conserved, but it may change form, like to heat. What is heat, or temperature. It just a measure of the energy of the atoms vibrational state. Look at your example of the fly wheel and piston. Starting from rest work is done on the system to get it moving. What happens then. Depending on the position of the flywheel and piston, energy is being transferred between the two. When the piston is at an extreme where it velocity is zero and starts moving the flywheel is doing work on the piston to increase it's KE and the flywheel slows down. When the piston reaches it's maximum velocity and starts to slow down it is because it is doing work on the flywheel, transferring it's kinetic energy back to the fly wheel and speeding it up.

When you stop a mass the, yes, you are absorbing the energy and converting it to some other form. Replace yourself with a spring and you have a different result. Replace the spring with friction and yet a different result.

We talk about friction as an energy dissipation mechanism. That has to be qualified. Friction does not dissipate energy. It is an energy conversion device. It converts mechanical energy to heat, but the energy is conserved.
 
Suppose we have a woofer with a moving mass m and force factor Bl. Current through its voice coil will cause force on a cone F = Bl * I, which will give it an acceleration a = F/m = (Bl * I) / m.
The prime statement in the original post is too simplistic. The acceleration of the cone is not simply BLI/m. Here's my take! 🙂

The magnetic driving force is not constant but varies sinusoidally with time so it will be labelled F(t).

The periodic driving force F(t) has to do work against the restoring force, Fr, of the suspension and the damping force, Fd, of the air.

The resultant force, F, on the cone is therefore F(t) - Fr - Fd

Hence, a = F/m = [F(t) - Fr - Fd]/m

The physics here is that of 'forced oscillation' where the frequency of the damped simple harmonic motion of the cone is forced to be of the same frequency as the sinusoidally varying driving force.

The driving force F(t) = BLI cos ωt (where ω = 2πf and f is the frequency of the driving signal)

Hence a = [BLI cos ωt -Fr - Fd]/m and not simply BLI/m
 
... When the piston is at an extreme where it velocity is zero and starts moving the flywheel is doing work on the piston to increase it's KE and the flywheel slows down. When the piston reaches it's maximum velocity and starts to slow down it is because it is doing work on the flywheel, transferring it's kinetic energy back to the fly wheel and speeding it up.
Exactly, and that is the reason why I don't do any work in stopping the ball. The energy flow is unambiguous, otherwise the wheel wouldn't keep spinning.
 
It's a matter of accounting. As Earl said, no e energy is lost or created. Does the moving mass do work on you or do you do work on the moving mass? Does the bullet do work on the wall or the wall do work on the bullet? The wok done is the change in KE.
OK, I think I finally see your point. I can choose whatever coordinate system I want - it can be the one moving with the ball. Then the situation will look different from that perspective, but the sum will be equal. OK.
 
This seems to be the next step beyond not being interested (in fluid mechanics in this case) in making assertions about matters that one presumably knows are not understood. Nobody likes to be wrong but when I was younger responses like this were rare because of the discomfort felt when seen to be wrong and drawing attention to it. This level seems to be much lower today to the extent the UK and the US are both lead by people that are quite openly serial liars in a way which would have terminated their careers a few years ago. Interesting if somewhat disconcerting stuff.

Put a woofer in free air and make it move.
Moving a hand in free air.
It's the exact same thing.
You simply move a surface in free air. That's it, nothing more nothing less. The exact same physics must applie.
 
I seem to remember someone here saying that the electro-mechanical system of a loudspeaker is 100% efficient. That cannot possibly be true since the vast majority of the energy (95-99%) is dissipated in the electromechanical system. Or did I read/remember this wrong?

In the equivalent circuit of a loudspeaker the air resistance is very small and putting the driver in a vacuum simply removes this resistance (and a small mass). But this makes virtually no difference to the circuit. It still dissipates virtually all of the energy in the mechanical system.
 
Exactly, and that is the reason why I don't do any work in stopping the ball. The energy flow is unambiguous, otherwise the wheel wouldn't keep spinning.

Again, simply put W = change in KE. If the piston is accelerating , its velocity is increasing and it KE is increasing. Thus positive work is being done on the piston. Conservation of energy then says the KE of the wheel must be decreasing, thus negative work is being done on the wheel. The KE of the system is constant, thus no work is being extracted from the system. Work done by or done to is just an accounting of the direction of energy flow.

I seem to remember someone here saying that the electro-mechanical system of a loudspeaker is 100% efficient. That cannot possibly be true since the vast majority of the energy (95-99%) is dissipated in the electromechanical system. Or did I read/remember this wrong?

In the equivalent circuit of a loudspeaker the air resistance is very small and putting the driver in a vacuum simply removes this resistance (and a small mass). But this makes virtually no difference to the circuit. It still dissipates virtually all of the energy in the mechanical system.

Earl, this thread is like talking to a bullet about to hit a wall. 🙂
 
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