# Relationship between frequency and power for sound?

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• ngn
ngn
TL;DR Summary
If I have two sine waves (pure tones) that differ in frequency but each have the same amplitude, then they would be equal in power. So, the two sounds would be equivalent in their intensities and sound pressures? But what about displacement and particle velocity?
Hello,
I have a problem wrapping my head around the relationship between frequency, power/pressure, and displacement. Let's say I have two sine waves that I generated in my computer: A 50 Hz tone and a 100 Hz tone. Let's say they both have an amplitude of 1. Therefore, they will both have the same RMS value of 0.707. That means that when I play these tones out of a loudspeaker (assuming a perfect loudspeaker with perfect performance and frequency response), both sounds will have the same intensity. Both sounds will also have the same pressure, since Intensity = pressure^2 / impedance.

Now here is where I'm having trouble understanding this. I am assuming that both tones when played out of the loudspeaker will produce the same particle displacement, because their amplitudes are the same. That is, if we could plot a waveform with particle displacement on the y axis, then we would find that both tones have equal amplitude.

But here is the issue. If the particle displacement is the same, then wouldn't the particle velocity be twice as high for the high frequency tone, because the particles need to cover the same distance in half the time. If that is true, then shouldn't power and pressure be higher for the high frequency tone?

If that is not true, then that would mean that displacement for the higher tone is half the lower tone. And if that is true, then wouldn't pressure be half and the sound should be less intense?

I believe I am missing some connection between what is meant by amplitude of a sine wave and how it maps onto particle displacement/velocity and how that translates into pressure.

I've been thinking about a pendulum and how the distance it swings translates into amplitude and the frequency it swings into frequency. It seems that for the same amount of power, a 100 Hz pendulum would have to swing half as far as a 50 Hz pendulum. But that would mean that the two sine waves would have different amplitudes, and hence different powers. So, I can't figure it out. If there's a way to explain it with a simple model like a pendulum, that would be helpful I think.

Thank you!

The amplitude of a pendulum swing is not related to real power. Energy is circulating between potential and kinetic, but there is no energy flow to or from the pendulum system. The circulating energy in the pendulum system, is described as reactive power, it is not real power.

Where you compare two amplitudes or frequencies, you need to do it in the same impedance, that is, the ratio of voltage to current needs to be the same. A resonant circuit has a frequency dependent impedance.

You will need to study the energy converted to heat in a resistor, or the wave travelling in one direction along an impedance matched transmission line. When comparing powers, that is the flow of energy, you must avoid resonance, since they are energy storage systems that do no real work.

DeBangis21
Baluncore said:
The amplitude of a pendulum swing is not related to real power. Energy is circulating between potential and kinetic, but there is no energy flow to or from the pendulum system. The circulating energy in the pendulum system, is described as reactive power, it is not real power.

Where you compare two amplitudes or frequencies, you need to do it in the same impedance, that is, the ratio of voltage to current needs to be the same. A resonant circuit has a frequency dependent impedance.

You will need to study the energy converted to heat in a resistor, or the wave travelling in one direction along an impedance matched transmission line. When comparing powers, that is the flow of energy, you must avoid resonance, since they are energy storage systems that do no real work.
Thank you for the reply! Is there a way to explain this in terms of sound? If I play from a loudspeaker two sine waves that are 100 Hz and 50 Hz, and they both have the same amplitude (in the computer which I am using to play the sounds), am I correct in assuming they will be equally intense coming out of the perfect loudspeaker? If so, the sound pressure will be equal for a listener at some distance (assuming the same impedance of air for both sounds). If that is the case, then what is going on with regard to the particle displacement and velocities and how do those relate to intensity and pressure? I am confused because when I think about the particles, it seems that the higher-frequency sound's particle velocity should be equal to the lower-frequency sound and thus its displacement should be lower.

ngn said:
Let's say they both have an amplitude of 1.
I think you need to be more precise what you refer to as amplitude -- it can't be a dimensionless number. You could consider the displacement of an average molecule, the rms speed of a parcel of air, or the pressure fluctuation. In section 4.4 of volume 3 of the Berkeley physics course I found some useful formulas for the pressure fluctuation $$p_\text{rms} = \sqrt{ Z \cdot I}$$ and the displacement $$A = \sqrt{ 2 I \over Z \omega^2}$$ in a sound wave of intensity ## I ##. ## Z = \sqrt{(c_p/c_v) p \rho} ## is the impedance of the air. For an intensity of ## \rm 1~\mu W~cm^{-2} ## I get ## p_\text{rms} \approx 4.3~\rm Pa ##, ## v_\text{rms} \approx 4.8~\rm mm~s^{-1} ##, and (at a frequency of ## 100~\rm Hz ##) ## A \approx 10.9~\rm \mu m ##.

ngn said:
when I think about the particles, it seems that the higher-frequency sound's particle velocity should be equal to the lower-frequency sound and thus its displacement should be lower.
Correct. For a frequency of ## 50~\rm Hz## the displacement ## A ## is twice as big.

hutchphd
This is very helpful. I see in the denominator for the displacement amplitude "ω" for frequency. This would mean that as frequency increases, displacement decreases. Okay, that confirms my mental model of the situation. But then I have a few questions to confirm:
1. Even though particle displacement for the higher frequency wave goes down (compared to the lower frequency wave), particle velocity stays the same. So therefore, pressure for both the high and low frequency waves are the same because their velocities are the same? If so, then pressure is not a measure of displacement independent of frequency? This is where I was getting tripped up. I thought there was a relationship between particle displacement and intensity/pressure regardless of frequency, meaning that the same displacement would mean the same pressure and intensity across all sounds. From your equation, it looks like there is a relationship between intensity/pressure and particle displacement but it depends on frequency.

2. Therefore, across all sounds, intensity/pressure is a measure of particle velocity? In other words, if two sounds of differing frequencies have the same particle velocity, they will have the same pressure/intensity -- BUT, they won't have the same particle displacement?

3. I think for amplitude, I was referring to values you start with in a computer program before you play them out of the loudspeakers. When I create sine waves in matlab for example to present from loudspeakers, the amplitude is a normalized value between -1 and 1. When I create two sounds that differ in frequency, but with the same amplitude in matlab, I expect them to have the same intensity when I play them from the loudspeaker. I was thinking that the amplitude in the computer program was related to displacement. But it seems that it is related to velocity? What is that amplitude related to? Power? Should I think of it as the current applied to the loudspeaker? Does it mean the velocity of the loudspeaker cone, and not its displacement?

Thank you!

ngn said:
I thought there was a relationship between particle displacement and intensity/pressure regardless of frequency, meaning that the same displacement would mean the same pressure and intensity across all sounds.
No. This is not correct.
ngn said:
Therefore, across all sounds, intensity/pressure is a measure of particle velocity?
Pressure, velocity, and intensity are all different things. When you multiply pressure (N/m^2) and velocity (m/s) you get a power flux (W/m^2). The formulas are valid only for pure sine waves. Double frequency means that a parcel of air can travel only half the distance in the shorter time.

ngn said:
When I create sine waves in matlab for example to present from loudspeakers, the amplitude is a normalized value between -1 and 1. When I create two sounds that differ in frequency, but with the same amplitude in matlab, I expect them to have the same intensity when I play them from the loudspeaker.
Your ears are not a measuring device. The loudness of sound depends very much on frequency, and typical sounds have a broad spectrum (contain very many frequencies). It may be instructive to listen to sounds created with Matlab, containing higher harmonics of a fundamental frequency, and for learning about Fourier series and transforms. But it is meant to produce just an approximation to the pressure fluctuations described by those sine waves and their superpositions.

ngn said:
If I play from a loudspeaker two sine waves that are 100 Hz and 50 Hz, and they both have the same amplitude (in the computer which I am using to play the sounds), am I correct in assuming they will be equally intense coming out of the perfect loudspeaker?
Forget the perfect loudspeaker - just consider the wave travelling through the air. For the same medium, carrying the sound, the actual sound power will be the same for two frequencies if the sound pressures are equal.
Achieving equal pressure from any loudspeaker is very hard so you have to believe the basic theory.

Trying to relate the energy situation in a pendulum and the energy in a travelling sound wave is chalk and cheese. The air molecules do not have a 'natural frequency' of vibration; they are just moved back and forth by the pressure on either side of them and they just pass the energy along, rather than storing it like the pendulum bob. (Stop calling me Bob!)

hutchphd
WernerQH said:
No. This is not correct.

Pressure, velocity, and intensity are all different things. When you multiply pressure (N/m^2) and velocity (m/s) you get a power flux (W/m^2). The formulas are valid only for pure sine waves. Double frequency means that a parcel of air can travel only half the distance in the shorter time.Your ears are not a measuring device. The loudness of sound depends very much on frequency, and typical sounds have a broad spectrum (contain very many frequencies). It may be instructive to listen to sounds created with Matlab, containing higher harmonics of a fundamental frequency, and for learning about Fourier series and transforms. But it is meant to produce just an approximation to the pressure fluctuations described by those sine waves and their superpositions.
Although pressure and velocity are different things, I was working under the assumption that pressure = velocity/impedance (p = v x z), where z = specific acoustic impedance, and thus this equation may be different under conditions of flux. As such, pressure and velocity would be equivalent, holding specific acoustic impedance constant. I will have to look into how this relationship might change under conditions of acoustic flow impedance (e.g., sound traveling through little pores in a surface).

I also wanted to mention that I was not referring to using my ears for loudness. I was referring to the intensity of the sound itself (Power/Area) and thus, holding area constant, the power of the sound. When I create two sine waves of equal amplitude in Matlab and play them out of a perfect loudspeaker, I assume their intensities are equal. Thus my confusion. If their intensities are equal, then their particle displacements must be unequal. Thus, the amplitude in matlab does not refer to particle displacement and must refer to something else. The amplitude equates to voltage, I believe? Since I would use the 20*log(amplitude) to convert units in matlab to decibels, I am assuming that the amplitude is not power. If it is voltage, I would be interested in how voltage equates to sound pressure since I would also use 20*log(pressure) to convert to decibels. Voltage and pressure are similar concepts. Is there something similar to particle displacement in electricity?

In any case, if the units in matlab equate to pressure, then that means that the amplitude is equivalent to pressure and thus the displacement for a 100 Hz sine wave would be half the 50 Hz sine wave when their amplitudes are equivalent in matlab and the sound is played out of a perfect loudspeaker.

I also found this online:
One other important thing to note here is that the velocity is also related to frequency (which is discussed below). If we maintain the same peak pressure, the higher the frequency, the faster the particles have to move back and forth, therefore the higher the peak velocity. So, remember that particle velocity is proportional both to pressure (and therefore displacement) and frequency.

Now I'm confused again. Would two sine waves that differ in frequency but have the same peak pressure differ in power? If particle velocity increases, then that would mean greater distance traveled (displacement) per unit time, which would mean greater power for the higher frequency wave. Shouldn't greater power = greater pressure? I keep going in circles.

ngn said:
As such, pressure and velocity would be equivalent, holding specific acoustic impedance constant.
That's correct. You can express the intensity either way $$I = p_\text{rms}^2 / Z \quad\text{ or }\quad I = Z \times v_\text{rms}^2 ~~.$$ For sound waves, the impedance ## Z = p_\text{rms} / v_\text{rms} ## happens to be independent of frequency. But the intensity always grows with the square of the amplitude.
ngn said:
I am assuming that the amplitude is not power.
Correct. The amplitude is proportional to the square root of the energy flux.
ngn said:
Voltage and pressure are similar concepts. Is there something similar to particle displacement in electricity?
Yes, for transmission lines there are analogous quantities obeying similar relations: $$P = U \cdot I \quad\text{ and }\quad Z = U / I$$ and for electromagnetic waves (see Poynting vector) $$\mathbf{S} = \mathbf{E} × \mathbf{H} \quad\text{ and }\quad Z = \rm E / H ~~.$$
ngn said:
In any case, if the units in matlab equate to pressure, then that means that the amplitude is equivalent to pressure and thus the displacement for a 100 Hz sine wave would be half the 50 Hz sine wave when their amplitudes are equivalent in matlab and the sound is played out of a perfect loudspeaker.
It's a bad idea to think of "displacement" in this context. A static (permanent) displacement of the loudspeaker of 1 mm doesn't mean anything. It's the movement (the rapid speed variations) of the loudspeaker that matters. And it is important to keep in mind that there are always two quantities involved which are proportional to each other: voltage and current, or pressure and velocity fluctuations.

ngn said:
Might I suggest that sound is rather a complicated system, as is light This is why most textbooks initiate the study of wave behavior using transverse waves on a tensioned string. I would recommend a similar approach for your edification. Otherwise it rapidly becomes the study of everything. This is why it is studied in every introductory text, and there is "no royal road".

sophiecentaur

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