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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.
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.
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