About the propagation of sound

In summary: I'm sorry, I cannot provide a summary for this part of the conversation as it is not a question or statement for me to summarize.
  • #1
Lammey
7
0
I'm trying to understand how a vibrating body produces oscillations in sound pressure. I've been through derivations and solutions of the wave equation for a string, but I don't understand the transition from waves on a string to sound pressure waves. How are the waves on a string or drum membrane mathematically related to the sound waves they are supposed to create? Suppose ##y(x,t)## is the displacement of a string from equilibrium position. Does perhaps ##y=d## where ##d## is the average particle displacement from equilibrium position?

Also, I've been through Feymann's derivation of the wave equation where he assumes the wave fronts are planar, and only uses one spatial dimension. This seems extremely idealistic to me. As I understand it, waves radiate out spherically from a point source. Are the dimensions of the eardrum and typical distances from a sound source such that the segment of the spherical wave fronts that comes into contact with it are able to be considered planar?
 
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  • #2
Strings are rather poor at transferring energy to the air. That's why stringed instruments have sound boards.
The vibrating string of a violin causes the bridge to vibrate, which in turn vibrates the front surface of the violin body, and thus the body as a whole. This vibrating shell causes alternating compression and decompression in the surrounding air (and inside the shell).

Yes, the otic orifice is very small compared to to typical distances from the instrument, so the waves are effectively planar by then.
 
  • #3
haruspex said:
Strings are rather poor at transferring energy to the air. That's why stringed instruments have sound boards.
The vibrating string of a violin causes the bridge to vibrate, which in turn vibrates the front surface of the violin body, and thus the body as a whole. This vibrating shell causes alternating compression and decompression in the surrounding air (and inside the shell).
Thanks for the reply. I understand this, but I was looking for something more explicit. If I'm explicitly given a mathematical expression describing the vibration of a guitar string (or body), I want the mathematical expression describing the generated sound wave.

haruspex said:
Yes, the otic orifice is very small compared to to typical distances from the instrument, so the waves are effectively planar by then.
Ah ok, this simplfies things a bit!
 
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  • #4
I can't say I have an excellent answer, but I do have a hint as to where you might want to look next. In reality, materials are made up of periodic arrangements of atoms. These arrangements oscillate for many reasons, but mainly because of interactions with other atoms nearby. The oscillations of the structured lattice, or arrangement, of atoms produces what are known as phonons. Where the phonons represent an oscillating "state" of the system.

It might help to think of the ocean waves bombarding a sandy beach. As the waves approach the beach they move the sand. Similarly as oscillating groups of air molecules "collide" with the surface of a material they induce oscillations in the material that in turn propagate the sound. There are also other ways that oscillations in a material can be produced. Consider a piezoelectric speaker. In this case an oscillating electrical current produces vibrations within the crystal structure that in turn collides with the atoms in the air thereby producing a sound.

In summary, if you want to know more about phonons I recommend you get a good book on solid state physics or you could try a question in that category on this forum. Also, you might like the Wikipedia page about phonons given here.
 
  • #5
Lammey said:
Thanks for the reply. I understand this, but I was looking for something more explicit. If I'm explicitly given a mathematical expression describing the vibration of a guitar string (or body), I want the mathematical expression describing the generated sound wave.
You have generic expressions for each, based on the properties of the media. What parameter will they have in common?
 
  • #6
lofgran said:
I can't say I have an excellent answer, but I do have a hint as to where you might want to look next. In reality, materials are made up of periodic arrangements of atoms. These arrangements oscillate for many reasons, but mainly because of interactions with other atoms nearby. The oscillations of the structured lattice, or arrangement, of atoms produces what are known as phonons. Where the phonons represent an oscillating "state" of the system.

It might help to think of the ocean waves bombarding a sandy beach. As the waves approach the beach they move the sand. Similarly as oscillating groups of air molecules "collide" with the surface of a material they induce oscillations in the material that in turn propagate the sound. There are also other ways that oscillations in a material can be produced. Consider a piezoelectric speaker. In this case an oscillating electrical current produces vibrations within the crystal structure that in turn collides with the atoms in the air thereby producing a sound.

In summary, if you want to know more about phonons I recommend you get a good book on solid state physics or you could try a question in that category on this forum. Also, you might like the Wikipedia page about phonons given here.
Thanks for the recommendation - I'll look into it!

haruspex said:
You have generic expressions for each, based on the properties of the media. What parameter will they have in common?
I don't know, that's my question! As the wave on the string (and consequently the body) changes, the sound you hear changes. For example, what determines the amplitude of the pressure variations? I'm guessing it's related to the amplitude of the vibrations of the string. How so?
 
  • #7
Lammey said:
For example, what determines the amplitude of the pressure variations? I'm guessing it's related to the amplitude of the vibrations of the string. How so?

and largely the amplitude of the vibrations of the body of the instrument ...
remember what haruspex told you in post #2

haruspex said:
Strings are rather poor at transferring energy to the air. That's why stringed instruments have sound boards.
The vibrating string of a violin causes the bridge to vibrate, which in turn vibrates the front surface of the violin body, and thus the body as a whole. This vibrating shell causes alternating compression and decompression in the surrounding air (and inside the shell).
Dave
 
  • #8
Lammey said:
I don't know, that's my question! As the wave on the string (and consequently the body) changes, the sound you hear changes. For example, what determines the amplitude of the pressure variations? I'm guessing it's related to the amplitude of the vibrations of the string. How so?
Yes, a larger amplitude of string vibration will likely lead to a larger amplitude in the air, but there will be losses, and there's no simple way to predict these.
But there are other parameters: phase, frequency, wavelength, speed. Can you connect any of those across the media?
 
  • #9
haruspex said:
Yes, a larger amplitude of string vibration will likely lead to a larger amplitude in the air, but there will be losses, and there's no simple way to predict these.
But there are other parameters: phase, frequency, wavelength, speed. Can you connect any of those across the media?
Well I don't know, that's what I'm asking.
 
  • #10
Lammey said:
Well I don't know, that's what I'm asking.
I think you should be able to figure out that some of those will change, and that one certainly won't.
 
  • #11
haruspex said:
I think you should be able to figure out that some of those will change, and that one certainly won't.
I obviously competely misunderstood the format of these forums. I can't deal with this style of answering where I'm being addressed like a stupid schoolboy, even if that is how I appear to you! I'm not able to figure this out. Obviously the wave speed is different, and wave speed depends on wave length for sound waves. Frequency I'd guess would be the same. But none of this is "figuring out", and if you just tell me I am correct I will have learned nothing.
 
  • #12
Lammey said:
I obviously competely misunderstood the format of these forums. I can't deal with this style of answering where I'm being addressed like a stupid schoolboy, even if that is how I appear to you! I'm not able to figure this out. Obviously the wave speed is different, and wave speed depends on wave length for sound waves. Frequency I'd guess would be the same. But none of this is "figuring out", and if you just tell me I am correct I will have learned nothing.
I mostly dwell on the homework help forums, so I responded as is appropriate there. I suspect you know more than you realize you know, so I hoped to get you to discover that. That would be learning something.
But see, you have figured out that frequency must be conserved (as long as no Doppler).
I would have thought you'd know that the wave speed depends on the medium, and the equation relating speed, wavelength and frequency, so it follows that the wavelength will change too.
Phase is rather irrelevant since that changes with time and distance anyway.
 
  • #13
haruspex said:
I mostly dwell on the homework help forums, so I responded as is appropriate there. I suspect you know more than you realize you know, so I hoped to get you to discover that. That would be learning something.
But see, you have figured out that frequency must be conserved (as long as no Doppler).
I would have thought you'd know that the wave speed depends on the medium, and the equation relating speed, wavelength and frequency, so it follows that the wavelength will change too.
Phase is rather irrelevant since that changes with time and distance anyway.
Apologies for the large gaps between my replies.
The problem is I haven't figured out that frequency will be conserved, I just guessed it. I have no reason why that's the case! I want to understand the mechanism by which the waves on the string cause the waves in the air.
 
  • #14
Lammey said:
Apologies for the large gaps between my replies.
The problem is I haven't figured out that frequency will be conserved, I just guessed it. I have no reason why that's the case! I want to understand the mechanism by which the waves on the string cause the waves in the air.
You just have to think through the propagation process. The vibrating string exerts a varying force on the bridge. Each time the string near the bridge is towards the base of the bridge (i.e. close to the surface of the violin), the tension in the string is higher than average. This pushes down on the sounding board, causing it to flex in slightly. That in turn pushes on the air molecules near the board, raising the air pressure slightly. Conversely, when the string is further from the sounding board there is less force on the bridge, allowing the sounding board to flex back again, reducing the air pressure near it.
Note the 'each time' aspect. Over a period of time, the number of times the string goes through this cycle is the same as the number of times the air goes through it, so the frequencies are the same.
 
  • #15
Lammey said:
I want to understand the mechanism by which the waves on the string cause the waves in the air.

it's the same process as with a loudspeaker ...

sound_waves.jpg
_154068.jpg
as the speaker cone ( guitar etc string) moves back and forwards, it compresses and rarefies the air that is in contact with it
the frequency produced is determined by how many times it does this per second. A wavelength ( one cycle) can be measured by the distance
between 2 consecutive compressions

here is a video that may help you as well ...

Dave
 
  • #16
Thanks for the help guys.

haruspex said:
You just have to think through the propagation process. The vibrating string exerts a varying force on the bridge. Each time the string near the bridge is towards the base of the bridge (i.e. close to the surface of the violin), the tension in the string is higher than average. This pushes down on the sounding board, causing it to flex in slightly. That in turn pushes on the air molecules near the board, raising the air pressure slightly. Conversely, when the string is further from the sounding board there is less force on the bridge, allowing the sounding board to flex back again, reducing the air pressure near it.
Note the 'each time' aspect. Over a period of time, the number of times the string goes through this cycle is the same as the number of times the air goes through it, so the frequencies are the same.
Ok this make sense. But frequency isn't the only variable of a sound wave, how can we determine the amplitude at a specific time?
davenn said:
it's the same process as with a loudspeaker ...

View attachment 94473 View attachment 94474as the speaker cone ( guitar etc string) moves back and forwards, it compresses and rarefies the air that is in contact with it
the frequency produced is determined by how many times it does this per second. A wavelength ( one cycle) can be measured by the distance
between 2 consecutive compressions

here is a video that may help you as well ...

Dave

Thanks for this, will take a look at the video
 
  • #18
Lammey said:
Thanks for the help guys.
Ok this make sense. But frequency isn't the only variable of a sound wave, how can we determine the amplitude at a specific time?
As I wrote, there are losses at each transfer. But it's not just losses; attenuation also occurs because of weak linkage. With no sounding board, the transfer of energy from a vibrating string to the air would be feeble. You wouldn't get much amplitude in the air, but the string would vibrate for longer. At the other extreme, make the linkage too strong and you get plenty of amplitude but the note dies away too quickly as it saps energy from the string too fast.
Predicting the transfer rate and efficiency would be quite challenging.
 
  • #19
haruspex said:
As I wrote, there are losses at each transfer. But it's not just losses; attenuation also occurs because of weak linkage. With no sounding board, the transfer of energy from a vibrating string to the air would be feeble. You wouldn't get much amplitude in the air, but the string would vibrate for longer. At the other extreme, make the linkage too strong and you get plenty of amplitude but the note dies away too quickly as it saps energy from the string too fast.
Predicting the transfer rate and efficiency would be quite challenging.

that doesn't really answer the Q

He asked how to measure the amplitude, not what affects it

so as I said, use a sound level meter :wink:Dave
 
  • #20
davenn said:
that doesn't really answer the Q
He asked how to measure the amplitude, not what affects it
Dave
You may be right, but if you follow the thread from the first post you will see that is not at all clear.
 
  • #21
haruspex said:
You may be right, but if you follow the thread from the first post you will see that is not at all clear.

yeah I have been
you may notice I have done a number of posts :wink:

his last Q about "how to measure amplitude" was one of his clearer questions
and not directly related to earlier comments/questions on how sound is transferred from source to air :smile:

Dave
 

What is sound propagation?

Sound propagation is the movement of sound waves through a medium, such as air, water, or solids. It is the way in which sound travels from its source to the listener.

How does sound propagate?

Sound propagates through a series of compressions and rarefactions in the medium. When an object vibrates, it creates a disturbance in the surrounding air molecules, causing them to move back and forth. This creates sound waves that travel through the air and are detected by our ears.

What factors affect sound propagation?

The speed of sound is affected by the properties of the medium it travels through, such as temperature, density, and humidity. The distance between the source and the listener also affects sound propagation, as well as any obstacles or barriers in between.

How does sound travel in different mediums?

Sound travels fastest through solids, followed by liquids, and then gases. This is because the molecules in solids are closer together, allowing sound waves to travel more quickly. In liquids and gases, the molecules are more spread out, causing sound waves to travel at a slower speed.

What is the application of understanding sound propagation?

Understanding sound propagation is important in various fields, such as acoustics, communication, and engineering. It helps us design better sound systems, study the effects of noise pollution, and develop methods for efficient sound transmission in different environments.

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