Sound Waves -- propagating, dispersing, volume effects?

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FScheuer
Generally sound waves are depicted as simple sine waves, where volume is related to amplitude, and there is periodic motion. Realistically sound waves aren’t as simple. I attached a picture of a dissipating sound wave. I would appreciate if you guys could answer a couple questions I have about it.

1.) Aren’t waves defined by periodic motion? Can real sound waves really be described as waves if they change in amplitude over time?

2.) How loud would the sound wave be at each of the three points? Would they each have the same volume as perfect non-dissipating sine waves would if they were drawn out to go through those points?

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1.) Aren’t waves defined by periodic motion? Can real sound waves really be described as waves if they change in amplitude over time?
Yes, a decaying wave is still a wave.
How loud would the sound wave be at each of the three points? Would they each have the same volume as perfect non-dissipating sine waves would if they were drawn out to go through those points?
You would need to measure the amplitude at these points and use the formula for converting sound pressure into decibels.
$$L_{p}=20log_{10}\left(\frac{p_{rms}}{p_{ref}}\right)$$
where ##p_{ref}## is the standard reference sound pressure of 20 micropascals in air.

FScheuer
i am not expert in sound wave field, but how is the wave amplitude being measured? connecting the peaks with a line will show you the type of decay function.

is a wave periodic? in general no. continuous waves that form patterns typically have a mathematical function to define it, but how would you describe a single ripple that travels?

model it like pebble in infinite pond. thus take your XY graph and now rotate it 360 degree to get your spatial representation.

and following NFuller post, db is that log scale, 10db is 10x, 20db is 100x, so if the point is -10db from origin the intensity will be -10x, if its -20db the intensity will be -100x.

FScheuer
FScheuer
Yes, a decaying wave is still a wave.

You would need to measure the amplitude at these points and use the formula for converting sound pressure into decibels.
$$L_{p}=20log_{10}\left(\frac{p_{rms}}{p_{ref}}\right)$$
where ##p_{ref}## is the standard reference sound pressure of 20 micropascals in air.
But what is the amplitude at point B? The amplitude is generally taught to be the highest point of a wave, but if the wave is decaying this wouldn’t be as easy.

i mentioned decay function. to find this function the sound pressure(x) has to be recorded at multiple spots along wave path. it would become a simple factor in the equation.

$$L_{p}=20log_{10}\left(\frac{p_{rms}}{p_{ref}}\right)*{decay(x)}$$

But what is the amplitude at point B? The amplitude is generally taught to be the highest point of a wave, but if the wave is decaying this wouldn’t be as easy.
Notice the value ##p_{rms}## is a root-mean-square value meaning it is a type of average of the sound pressure at a given point in time. So ##p_{rms}## is not zero at point B.
i mentioned decay function. to find this function the sound pressure(x) has to be recorded at multiple spots along wave path. it would become a simple factor in the equation.

$$L_{p}=20log_{10}\left(\frac{p_{rms}}{p_{ref}}\right)*{decay(x)}$$
You should not multiply a decay function with this equation. Use the equation as is.

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FScheuer
FScheuer
Notice the value ##p_{rms}## is a root-mean-square value meaning it is a type of average of the sound pressure at a given point in time. So ##p_{rms}## is not zero at at point B.

You should not multiply a decay function with this equation. Use the equation as is.
Wouldn’t the root-mean-square value be constant, and therefore indicate a constant volume at any point on the wave?

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The rms value is defined over an interval from ##t_{1}## to ##t_{2}##.
$$p_{rms}=\sqrt{\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}\left[p(t)\right]^{2}dt}$$
Generally speaking, the time interval should be short enough to respond to changes in amplitude but not shorter than several periods of the signal. So if the signal is slowly attenuated over many rms measurement intervals, the rms value will steadily decrease. If the signal very rapidly decays such that the decay is captured in one rms measurement interval, which is roughly the case in your drawing, there will be a sudden drop in the rms value.

FScheuer
JBA
Gold Member
Air has mass and the propagation of sound waves depends upon moving that mass which requires energy drawn from the sound wave and therefore the reduction of the amplitude of the wave form with its distance from the source point.

FScheuer
The rms value is defined over an interval from ##t_{1}## to ##t_{2}##.
$$p_{rms}=\sqrt{\frac{1}{t_{2}-t_{1}}\int_{t_{1}}^{t_{2}}\left[p(t)\right]^{2}dt}$$
Generally speaking, the time interval should be short enough to respond to changes in amplitude but not shorter than several periods of the signal. So if the signal is slowly attenuated over many rms measurement intervals, the rms value will steadily decrease. If the signal very rapidly decays such that the decay is captured in one rms measurement interval, which is roughly the case in your drawing, there will be a sudden drop in the rms value.
Would this just approximate the volume of the wave? I don’t have the knowledge to fully grasp what you’re saying, but the times chosen for the time interval seem somewhat arbitrary.

Would this just approximate the volume of the wave?
The whole idea of volume in sound is sort of an approximation. The decibel system isn't perfect; it is meant to help quantify loudness perceived by the human ear. Our ears will not pick up a rapidly decaying waveform as a decrease in loudness but rather as a unique sound. The waveform in your drawings is similar to the sound of a drum being hit: a load sound which rapidly falls off in amplitude.
the times chosen for the time interval seem somewhat arbitrary.
They are, I think a lot of rms devices like multimeters use an interval from 0.1-0.5 seconds.