Sound Power Challenge

[HARK!]

The amount of energy contained in two different sinusoids, of the same amplitude, of different frequencies, is directly proportional to their frequencies.

For those who aren't familiar with the scientific language; I'll provide this example:
we have two different continuous waves. One peaks and falls at twice the rate of the other; but both waves are the same height. The wave that peaks and falls faster contains twice the energy of the wave that is rising and falling at half the rate of the other.

Let's take two electronic sinusoids and pass them through a coil. Lower frequencies meet less impedance in a coil rather than do higher frequencies. This means that it would take a greater amplitude, at a higher frequency, to pass the same amount of energy as that of a lower frequency, through a coil of the same value; as at a lower frequency, more current would pass through the coil, all amplitudes equal.

Let's convert electronic power into sound power.

Audio drivers (speakers) Are driven by passing current through a coil which is in proximity to a fixed magnetic field. The efficiency of drivers at any particular frequency across their ranges gets very complicated.

Let's simplify it for the sake of the question. The difference that I will speak of is dramatic enough; that we won't need to analyze this to the Nth degree. Drivers are most efficient at their resonant frequency. Let's leave it there4.

We have two drivers with different resonant frequencies, but their efficiencies are equal at their respective resonant frequencies. Both are of the same impedance. Yes, it's a perfect theoretical world.

We run a sinusoid through each of the drivers which matches the resonant frequency of each driver.

So far, because of the higher impedance met by the higher frequency we have less work being done by the high frequency driver. More work is being done by the low frequency driver.

At this point one might expect that we would have more sound power from the low frequency driver. This is not the case in reality. Even at a lower impedance. it takes far more electrical power to deliver the same amount of sound power at a lower frequency.

Why is this true?

Back in the late '80s, early '90s I made some phone calls and visits to try to get an answer to this question. I spoke with numerous engineers and scientists who had no idea. After years of reading and asking others, I finally found a Physicist who worked for JHU, who answered my question. I'm satisfied with his answer; but I thought it would be fun to present this challenge; and if any answers would match the answer that I finally accepted. BTW, of all of the people I spoke with only, I received only two different answers. I accepted both of them; but the answer of "I don't know." didn't help me to move onto the next step toward my invention idea.

 

[SkyWriting]

The amount of energy contained in two different sinusoids, of the same amplitude, of different frequencies, is directly proportional to their frequencies.

For those who aren't familiar with the scientific language; I'll provide this example:
we have two different continuous waves. One peaks and falls at twice the rate of the other; but both waves are the same height. The wave that peaks and falls faster contains twice the energy of the wave that is rising and falling at half the rate of the other.

Let's take two electronic sinusoids and pass them through a coil. Lower frequencies meet less impedance in a coil rather than do higher frequencies. This means that it would take a greater amplitude, at a higher frequency, to pass the same amount of energy as that of a lower frequency, through a coil of the same value; as at a lower frequency, more current would pass through the coil, all amplitudes equal.

Let's convert electronic power into sound power.

Audio drivers (speakers) Are driven by passing current through a coil which is in proximity to a fixed magnetic field. The efficiency of drivers at any particular frequency across their ranges gets very complicated.

Let's simplify it for the sake of the question. The difference that I will speak of is dramatic enough; that we won't need to analyze this to the Nth degree. Drivers are most efficient at their resonant frequency. Let's leave it there4.

We have two drivers with different resonant frequencies, but their efficiencies are equal at their respective resonant frequencies. Both are of the same impedance. Yes, it's a perfect theoretical world.

We run a sinusoid through each of the drivers which matches the resonant frequency of each driver.

So far, because of the higher impedance met by the higher frequency we have less work being done by the high frequency driver. More work is being done by the low frequency driver.

At this point one might expect that we would have more sound power from the low frequency driver. This is not the case in reality. Even at a lower impedance. it takes far more electrical power to deliver the same amount of sound power at a lower frequency.

Why is this true?

Back in the late '80s, early '90s I made some phone calls and visits to try to get an answer to this question. I spoke with numerous engineers and scientists who had no idea. After years of reading and asking others, I finally found a Physicist who worked for JHU, who answered my question. I'm satisfied with his answer; but I thought it would be fun to present this challenge; and if any answers would match the answer that I finally accepted. BTW, of all of the people I spoke with only received two answers. I accepted both of them; but, "I don't know." didn't help me to move onto the next step toward my invention idea.

It takes less power to move the mass of the cone a short distance back and forth.
Far more power to drive it father and longer distance, quickly, and reverse the direction of the moving mass.

Higher frequencies require little power. Why ear buds (used to) sound tinny.

 

[HARK!]

It takes less power to move the mass of the cone a short distance back and forth.
Far more power to drive it father and longer distance, quickly, and reverse the direction of the moving mass.

Higher frequencies require little power. Why ear buds (used to) sound tinny.

What covers more distance? Walking a mile, or walking a half mile then walking back again?

In this theoretical model, the efficiencies of the drivers are equal. Try again.

 

[HARK!]

Why ear buds (used to) sound tinny.

Did you know that in free air, that a driver can only reproduce a frequency down to that which has a wave length that is equal to the diameter of its' cone?

My answer to this statement provides a telling clue to the answer that I accepted to my question.

 

[Mark Quayle]

The amount of energy contained in two different sinusoids, of the same amplitude, of different frequencies, is directly proportional to their frequencies.

For those who aren't familiar with the scientific language; I'll provide this example:
we have two different continuous waves. One peaks and falls at twice the rate of the other; but both waves are the same height. The wave that peaks and falls faster contains twice the energy of the wave that is rising and falling at half the rate of the other.

Well, no. The RMS value of one is the same as the RMS value of the other, if the amplitude is the same. Your formula gives no further information; you have voltage, but not current —no reference to load.

Let's take two electronic sinusoids and pass them through a coil. Lower frequencies meet less impedance in a coil rather than do higher frequencies. This means that it would take a greater amplitude, at a higher frequency, to pass the same amount of energy as that of a lower frequency, through a coil of the same value; as at a lower frequency, more current would pass through the coil, all amplitudes equal.

Let's convert electronic power into sound power.

Audio drivers (speakers) Are driven by passing current through a coil which is in proximity to a fixed magnetic field. The efficiency of drivers at any particular frequency across their ranges gets very complicated.

Let's simplify it for the sake of the question. The difference that I will speak of is dramatic enough; that we won't need to analyze this to the Nth degree. Drivers are most efficient at their resonant frequency. Let's leave it there4.

We have two drivers with different resonant frequencies, but their efficiencies are equal at their respective resonant frequencies. Both are of the same impedance. Yes, it's a perfect theoretical world.

We run a sinusoid through each of the drivers which matches the resonant frequency of each driver.

So far, because of the higher impedance met by the higher frequency we have less work being done by the high frequency driver. More work is being done by the low frequency driver.

At this point one might expect that we would have more sound power from the low frequency driver. This is not the case in reality. Even at a lower impedance. it takes far more electrical power to deliver the same amount of sound power at a lower frequency.

Why is this true?

Back in the late '80s, early '90s I made some phone calls and visits to try to get an answer to this question. I spoke with numerous engineers and scientists who had no idea. After years of reading and asking others, I finally found a Physicist who worked for JHU, who answered my question. I'm satisfied with his answer; but I thought it would be fun to present this challenge; and if any answers would match the answer that I finally accepted. BTW, of all of the people I spoke with only received two answers. I accepted both of them; but, "I don't know." didn't help me to move onto the next step toward my invention idea.

While you may have it clear in your mind what you are trying to say, and what you are trying to ask, you have clouded the issue. "Sound power" is not specific enough, and "electric power" even less specific. As your question as stated, it is bogus. I can only answer your question with, "But it is not true." or, "It is only true if you mean.....".

Take for example, "Even at a lower impedance. it takes far more electrical power to deliver the same amount of sound power at a lower frequency." The word "even" is misleading: Lower impedance always means higher current, if voltage is consistent. Frequency is irrelevant. The word "far" is hyperbolic. More is simply more. Less is less, and the voltage/current relationship is always proportionate to the impedance.

This is a bit like 'the answer to life, the universe, and everything' that was posed in Douglas Adam's book. (The answer is 42, (and it only took 7.5 million years for the supercomputer to figure that out), but we don't know the question!).

 

[HARK!]

Well, no. The RMS value of one is the same as the RMS value of the other, if the amplitude is the same. Your formula gives no further information; you have voltage, but not current —no reference to load.

I mentioned later, in the OP, that a higher frequency will meet greater resistance through a coil, but for this part of the statement we can break this down into terms that are very easy to understand.

If you apply a positive voltage to a speaker, the cone will move out. The greater the voltage, the more it will move out. If you apply a negative voltage, the cone will move in, the greater the voltage, the more it will move in. If we alternate the positive and negative voltages one time per second; we have one cycle per second. If we alternate the voltages twice per second; we have two cycles per second. At two cycles per second, twice as much work is done than at one cycle per second. As a theoretical ideal, it takes twice as much energy to do twice as much work.

 

[HARK!]

While you may have it clear in your mind what you are trying to say, and what you are trying to ask, you have clouded the issue. "Sound power" is not specific enough, and "electric power" even less specific. As your question as stated, it is bogus. I can only answer your question with, "But it is not true." or, "It is only true if you mean.....".

Sound power or acoustic power is the rate at which sound energy is emitted, reflected, transmitted or received, per unit time.[1] It is defined[2] as "through a surface, the product of the sound pressure, and the component of the particle velocity, at a point on the surface in the direction normal to the surface, integrated over that surface." The SI unit of sound power is the watt (W).[1] It relates to the power of the sound force on a surface enclosing a sound source, in air.

I'm fairly sure that you know that W = I X E.

Or Watts are equal to voltage times current.

 

[HARK!]

Take for example, "Even at a lower impedance. it takes far more electrical power to deliver the same amount of sound power at a lower frequency." The word "even" is misleading: Lower impedance always means higher current, if voltage is consistent. Frequency is irrelevant. The word "far" is hyperbolic. More is simply more. Less is less, and the voltage/current relationship is always proportionate to the impedance.

It seems that you don't understand that frequency is a variable in the formula for inductive reactance.

 

[Hans Blaster]

Sound power or acoustic power is the rate at which sound energy is emitted, reflected, transmitted or received, per unit time.[1] It is defined[2] as "through a surface, the product of the sound pressure, and the component of the particle velocity, at a point on the surface in the direction normal to the surface, integrated over that surface." The SI unit of sound power is the watt (W).[1] It relates to the power of the sound force on a surface enclosing a sound source, in air.

I'm fairly sure that you know that W = I X E.

Or Watts are equal to voltage times current.

And if you read on:

In a medium, the sound power is given by

P = A p[sup]2[/sup] cos θ / (ρ c)

where

• A is the area of the surface;
• ρ is the mass density;
• c is the sound velocity;
• θ is the angle between the direction of propagation of the sound and the normal to the surface.
• p is the sound pressure.

Which tells us that for equal sound pressure (the amplitude of the sound wave) there is no difference in power with frequency.

[Too many edits. Real equations need \LaTeX.]

 

[FrumiousBandersnatch]

The amount of energy contained in two different sinusoids, of the same amplitude, of different frequencies, is directly proportional to their frequencies.

For those who aren't familiar with the scientific language; I'll provide this example:
we have two different continuous waves. One peaks and falls at twice the rate of the other; but both waves are the same height. The wave that peaks and falls faster contains twice the energy of the wave that is rising and falling at half the rate of the other.

Let's take two electronic sinusoids and pass them through a coil. Lower frequencies meet less impedance in a coil rather than do higher frequencies. This means that it would take a greater amplitude, at a higher frequency, to pass the same amount of energy as that of a lower frequency, through a coil of the same value; as at a lower frequency, more current would pass through the coil, all amplitudes equal.

Let's convert electronic power into sound power.

Audio drivers (speakers) Are driven by passing current through a coil which is in proximity to a fixed magnetic field. The efficiency of drivers at any particular frequency across their ranges gets very complicated.

Let's simplify it for the sake of the question. The difference that I will speak of is dramatic enough; that we won't need to analyze this to the Nth degree. Drivers are most efficient at their resonant frequency. Let's leave it there4.

We have two drivers with different resonant frequencies, but their efficiencies are equal at their respective resonant frequencies. Both are of the same impedance. Yes, it's a perfect theoretical world.

We run a sinusoid through each of the drivers which matches the resonant frequency of each driver.

So far, because of the higher impedance met by the higher frequency we have less work being done by the high frequency driver. More work is being done by the low frequency driver.

At this point one might expect that we would have more sound power from the low frequency driver. This is not the case in reality. Even at a lower impedance. it takes far more electrical power to deliver the same amount of sound power at a lower frequency.

Why is this true?

Back in the late '80s, early '90s I made some phone calls and visits to try to get an answer to this question. I spoke with numerous engineers and scientists who had no idea. After years of reading and asking others, I finally found a Physicist who worked for JHU, who answered my question. I'm satisfied with his answer; but I thought it would be fun to present this challenge; and if any answers would match the answer that I finally accepted. BTW, of all of the people I spoke with only received two answers. I accepted both of them; but, "I don't know." didn't help me to move onto the next step toward my invention idea.

If you want an informed discussion on this you could do worse than ask on the Audioholics Loudspeaker forum.

 

[HARK!]

And if you read on:

In a medium, the sound power is given by

P = A p[sup]2[/sup] cos θ / (ρ c)

where

• A is the area of the surface;
• ρ is the mass density;
• c is the sound velocity;
• θ is the angle between the direction of propagation of the sound and the normal to the surface.
• p is the sound pressure.

Which tells us that for equal sound pressure (the amplitude of the sound wave) there is no difference in power with frequency.

That applies at the point at were the power is measured.


[Edit: You seemed like you might catch on too quickly. I didn't want too give away to many clues too soon; but here goes: Let's assume that the point where the amplitude of the waves being measured, is not at the origins of the waves being measured. Let's also work in free air, to get an accurate representation of what is actually happening under the inverse square law.]

 

[HARK!]

If you want an informed discussion on this you could do worse than ask on the Audioholics Loudspeaker forum.

I'm fairly sure that the JHU Physicist that I spoke to, gave me the correct answer. (At least it seemed to make perfect sense.)

I just posted this as a fun brain teaser.

 

[AV1611VET]

The amount of energy contained in two different sinusoids, of the same amplitude, of different frequencies, is directly proportional to their frequencies.

Is this something like putting plastic wrap over a comb and playing the comb kazoo?

 

[Mark Quayle]

I mentioned later, in the OP, that a higher frequency will meet greater resistance through a coil, but for this part of the statement we can break this down into terms that are very easy to understand.

If you apply a positive voltage to a speaker, the cone will move out. The greater the voltage, the more it will move out. If you apply a negative voltage, the cone will move in, the greater the voltage, the more it will move in. If we alternate the positive and negative voltages one time per second; we have one cycle per second. If we alternate the voltages twice per second; we have two cycles per second. At two cycles per second, twice as much work is done than at one cycle per second. As a theoretical ideal, it takes twice as much energy to do twice as much work.

This is the kind of thing I am trying to say: here, in, "At two cycles per second, twice as much work is done than at one cycle per second" you fail to say, "in one second's passage". But even that is misleading. A square wave (as you describe) is not a sinusoid. You also do not mention how long each positive or negative excursion is, or does it immediately return to zero. You've introduced all sorts of questions with your square wave; as you have said, the coil may or may not be built to handle the sudden slope change. In your attempt at clarity, you confuse issues.

 

[Mark Quayle]

Sound power or acoustic power is the rate at which sound energy is emitted, reflected, transmitted or received, per unit time.[1] It is defined[2] as "through a surface, the product of the sound pressure, and the component of the particle velocity, at a point on the surface in the direction normal to the surface, integrated over that surface." The SI unit of sound power is the watt (W).[1] It relates to the power of the sound force on a surface enclosing a sound source, in air.

I'm fairly sure that you know that W = I X E.

Or Watts are equal to voltage times current.

Is the "RATE"?

Let's describe what happens first, on a purely resistive load, a resistor, not a coil. A voltage of a low frequency across that resistor, produces the same power as a voltage of a high frequency across that resistor. The coil does not, as it is not purely resistive. No doubt you will want to say, that is what you have been trying to say all along. Good. But that isn't what you have been saying. Watts is only watts. If you run equal sinusoidal voltages of low vs high frequency, the current departs more from sinusoidal the higher the frequency.

 

[Mark Quayle]

It seems that you don't understand that frequency is a variable in the formula for inductive reactance.

I could have said the same thing! Read your first post again. You are not specific enough when presenting your thesis. You speak in confusing generalities.

 

[HARK!]

This is the kind of thing I am trying to say: here, in, "At two cycles per second, twice as much work is done than at one cycle per second" you fail to say, "in one second's passage". But even that is misleading. A square wave (as you describe) is not a sinusoid. You also do not mention how long each positive or negative excursion is, or does it immediately return to zero. You've introduced all sorts of questions with your square wave; as you have said, the coil may or may not be built to handle the sudden slope change. In your attempt at clarity, you confuse issues.

The Inverse-Square Law is not about square waves. Paragraph three of the OP makes it clear that I'm talking about sine waves.

 

[HARK!]

Let's describe what happens first, on a purely resistive load, a resistor, not a coil. A voltage of a low frequency across that resistor, produces the same power as a voltage of a high frequency across that resistor. The coil does not, as it is not purely resistive. No doubt you will want to say, that is what you have been trying to say all along.

I didn't mention a resistor. I mentioned a coil; and I used the word impedance in the the OP to describe the load. Later I used the word resistance to describe the load, because it seemed that you didn't understand inductive reactance. BTW, you didn't ask me the minimum impedance of the coil, nor what frequencies were passing through the coil. That could make a huge difference in your calculations; but It's completely irrelevant to the question. The fact that the higher frequency might meet greater impedance through the coil was just a side not to accentuate the fact that it will take more more wattage to produce the same amount of sound power at a lower frequency. The answer is simple when you understand it; so there's no need to create a complicated model.

If you run equal sinusoidal voltages of low vs high frequency, the current departs more from sinusoidal the higher the frequency.

Huh?

There is a 90 degree phase shift through an inductor. See ELI the ICE man.

 

[HARK!]

I could have said the same thing! Read your first post again. You are not specific enough when presenting your thesis. You speak in confusing generalities.

What more information do you need to answer the question?

 

[HARK!]

Is this something like putting plastic wrap over a comb and playing the comb kazoo?

I don't want you to feel like I'm ignoring you.

I've been thinking since yesterday how I could walk through this with you using that model. I started come up with one way to explain it; but as I worked through it, the explanation seemed to be more complicated than the model in the OP.


I'll keep trying.