Cone movement vs. frequency

[sreten]

Originally posted by AndrewT

yes, all three of you forgot to read the question.

Hmmm.......

No, I read it and answered the question in terms of the conditions given.

Seems to me a complete fluke that your total misunderstanding of the
situation gives you a get out clause for a sealed box below resonance
rolling off at 12dB/octave. Nothing to do with the original question.


/sreten.

[AndrewT]

Hi Sreten,
Sorry, you are due an apology for my misunderstanding and using that to misinterpret your answer.

But I still read your answer as constant SPL and not constant voltage.

I wonder if I am confusing power with energy?

Any thoughts?

[sreten]

Originally posted by AndrewT


But I still read your answer as constant SPL and not constant voltage.

I wonder if I am confusing power with energy?

Any thoughts?

Hi,

that's because in terms of the question I didn't really care whether
its an "idealised" driver with flat response or a real driver EQed to
have a flat response in the frequency range of interest.

I took amplitude constant to mean either of the above, an area of flat response.

Energy = power x time, so I don't follow your question.

/sreten.

[Pano]

This is an interesting thread. Sort of a "basics of drivers".

To me the original question isn't clear - maybe that's why the answers haven't been.

Originally posted by Nichol1997
Let’s say if a subwoofer was held in free-air and the amplitude was constant and the frequency was swept from 100 Hertz down, does the cone move more as the frequency is lowered?

Amplitude of what? Sound pressure? That is what we might assume. But could it be the "amplitude" of the voltage, the current, the power? Nichol obviously isn't implying movement amplitude, because he asks if that changes.

Since I'm an amp guy, I tend to think of amplitude in terms of voltage. If I have an amplifier with a gain of 10, I want the output voltage to be exactly 10X the input voltage no matter what the frequency (within the BW).

So when I read Nichol’s question - I think voltage. If a constant voltage is swept across the voice coil, will cone excursion change with frequency? That seems to be what Andrew is getting at with his DC questions.

If the cone excursion at a given voltage does change with frequency - why does it change? I suspect the answer may be hidden in the nest of formulas posted above, but I can't see it.

(My apologies to Nichol if that is not what he was asking - it's just what I read when I see the question.)

[sreten]

Hi,

there is a simple model you can apply to a 2nd order alignment,
sealed box, open baffle, tweeters, or a reflex box below port frequency.
(note a reflex box uses the driver Vas not the effective box stiffness)

The equivalent circuit of 2nd order is a LC (+some R) circuit.
L=mass, C= stiffness, R=damping 2nd order high pass.


Below the resonant frequency the river is in the stiffness region
where C dominates. As described by AT the excursion remains
constant, consequently output falls at 12/dB octave. Above the
resonant frequency is the mass region where L dominates, the
output remains constant, the cone velocity is constant, so this
means the excursion much reduce with increased frequency,
or increase with reduced frequency. If it stayed the same
output would rise at 12dB/octave. the other way of looking at
it is Sv's Fsquared or 1/Fsquared relationships for excursion.

To ask the right question you almost need to know the answer
already. So to answer a simplistic question you have to make
assumptions as to what is meant. My assumption is the basic
behaviour above box resonance was being enquired about.

/sreten.

[AndrewT]

But he said

held in free-air

[sreten]

Hmmm......, this is getting tedious, and your point is ? /sreten.

[Pano]

So as frequency increases, so must cone speed. The cone has to “get there and back” faster.
That increase in speed means increased acceleration.
Higher acceleration needs more power.
But as the voltage is constant, there isn't more power applied.
Thus excursion must decrease.

Is that it?

Originally posted by sreten
and your point is?

Andrew's point is that the question was about a driver "held in free air" and you keep talking about boxes. If it's the same thing, say so.

[sreten]

Originally posted by sreten
Hi,

there is a simple model you can apply to a 2nd order alignment,
sealed box, open baffle, tweeters, or a reflex box below port frequency.
(note a reflex box uses the driver Vas not the effective box stiffness)

/sreten.

open baffle = free air

[Pano]

Newton's second law, then?