Interesting. I’m not sure I follow though. How would stricter boundary conditions (having guitar walls) make the air more receptive to energy? Why doesn’t this theory predict that strings with just a fretboard would be the loudest?
If you had no boundary conditions at all (string with fretboard) then the air is receptive to sounds of all frequencies. It is also equally receptive to sound sources at every location. (X and X+1 are identical). The vibrating string couples to the normal modes, which in this case (no boundaries) are plane waves. All directions and locations are equal. Frequencies are unequal (high frequencies more strongly emit because their are a higher density of plane waves with a higher frequency).
Now, lets introduce a single infinite flat wall. (Infinitely high and wide. Made of perfectly reflective material for simplicity). The normal modes are now different, they all have a node at the location of the wall (x=0). The normal modes are no longer plane waves, but more like sine waves. If we imagine a sine wave coming out from x=0 then it has peaks and troughs. A string at one of the anti-nodes will radiate no sound at all into that mode. (So for sound with wavelength 10cm a string either 5cm, 10cm, 15cm, 20cm… from the wall would not radiate into that mode.) Meanwhile a string moved 2.5cm from any of those will be hitting the sine wave at its maximum/minimum (peak/trough), and radiate additional energy into that mode.
In the guitar example the guitar chamber has a normal mode with a very high intensity at the string location. This means that that specific normal mode will be strongly activated by the string. (Hitting the mode at its anti-node). The total sound output is determined by how well the string couples not just to that specific normal mode, but to all of them. But, there is no rule that they have to average out the same as they would without boundaries.
I think the biggest mental block in terms of understanding this is to think of the sound as “too small”. You are imagining that the string radiates little ripples of sound that “don’t know about” the walls until they crash into them. This is not a useful mental image. The wave is a spatially extended object, (otherwise it could not have a wavelength). The “it can’t fit” situation (a sound with wavelength 10cm in a pipe only 3cm across) is quite intuitive. The “it fits super well” is, I agree, harder to wrap the head around.
Something I got hung up on when learning this was trying to work out what made the air molecules next to the string different in one case from another. The answer is the air molecules right next to the string are not taking the energy by themselves, any more than the cashier at a shop is taking the money by themselves. All the money passes through the cashier’s hands, but the amount of money handed over is not dependent entirely on the cashier themselves, but also depends on the wider context of the shop, its location and the goods available. Those air molecules right next to the string are the immediate place the energy goes (like the coins in the cashier’s hand) but trying to zoom right in on them and look at the causality without the bigger picture will not be useful. They are ordinary air molecules, but the will absorb more energy if their context is different. Just like the same person at a cash register might get handed more money if they worked in a different shop. (This last paragraph is not an explanation of why more sound goes out, but instead a wooly sketch of why a particular mindset is not useful).
I am afraid I have not given enough detail for you to “get it” based just on my comments. Often the phrases “resonance”, “density of states” and “impedance matching” are used as catch all terms for describing the situation, although these phrases obviously don’t themselves bring any understanding.
All three terms are describing the same thing in different words. “Impedance matching” is gesturing to the fact that when a wave moves into a material where it travels with a different speed their is a reflection at the interface, so shaping a cavity that effectively slows the sound waves down by making them ricochet reduces the amount of sound that the air bounces back into the string at that interface. Resonance is getting the same thing in the frequency picture. Density of states is saying the same thing again, this time using a Fourier transform to talk about the density of different possible waves in position instead of the speed of waves. (Essentially if the waves are walking on a monopoly board a slower speed implies a greater density of squares → more states. So something that likes putting sound counters into nearby squares will deposit more in the slow areas because their are more squares to put them in.)
Interesting. I’m not sure I follow though. How would stricter boundary conditions (having guitar walls) make the air more receptive to energy? Why doesn’t this theory predict that strings with just a fretboard would be the loudest?
If you had no boundary conditions at all (string with fretboard) then the air is receptive to sounds of all frequencies. It is also equally receptive to sound sources at every location. (X and X+1 are identical). The vibrating string couples to the normal modes, which in this case (no boundaries) are plane waves. All directions and locations are equal. Frequencies are unequal (high frequencies more strongly emit because their are a higher density of plane waves with a higher frequency).
Now, lets introduce a single infinite flat wall. (Infinitely high and wide. Made of perfectly reflective material for simplicity). The normal modes are now different, they all have a node at the location of the wall (x=0). The normal modes are no longer plane waves, but more like sine waves. If we imagine a sine wave coming out from x=0 then it has peaks and troughs. A string at one of the anti-nodes will radiate no sound at all into that mode. (So for sound with wavelength 10cm a string either 5cm, 10cm, 15cm, 20cm… from the wall would not radiate into that mode.) Meanwhile a string moved 2.5cm from any of those will be hitting the sine wave at its maximum/minimum (peak/trough), and radiate additional energy into that mode.
In the guitar example the guitar chamber has a normal mode with a very high intensity at the string location. This means that that specific normal mode will be strongly activated by the string. (Hitting the mode at its anti-node). The total sound output is determined by how well the string couples not just to that specific normal mode, but to all of them. But, there is no rule that they have to average out the same as they would without boundaries.
I think the biggest mental block in terms of understanding this is to think of the sound as “too small”. You are imagining that the string radiates little ripples of sound that “don’t know about” the walls until they crash into them. This is not a useful mental image. The wave is a spatially extended object, (otherwise it could not have a wavelength). The “it can’t fit” situation (a sound with wavelength 10cm in a pipe only 3cm across) is quite intuitive. The “it fits super well” is, I agree, harder to wrap the head around.
Something I got hung up on when learning this was trying to work out what made the air molecules next to the string different in one case from another. The answer is the air molecules right next to the string are not taking the energy by themselves, any more than the cashier at a shop is taking the money by themselves. All the money passes through the cashier’s hands, but the amount of money handed over is not dependent entirely on the cashier themselves, but also depends on the wider context of the shop, its location and the goods available. Those air molecules right next to the string are the immediate place the energy goes (like the coins in the cashier’s hand) but trying to zoom right in on them and look at the causality without the bigger picture will not be useful. They are ordinary air molecules, but the will absorb more energy if their context is different. Just like the same person at a cash register might get handed more money if they worked in a different shop. (This last paragraph is not an explanation of why more sound goes out, but instead a wooly sketch of why a particular mindset is not useful).
I don’t get it yet. Thanks though! Might come back. (I’m probably not devoting enough attention to get it.)
I am afraid I have not given enough detail for you to “get it” based just on my comments. Often the phrases “resonance”, “density of states” and “impedance matching” are used as catch all terms for describing the situation, although these phrases obviously don’t themselves bring any understanding.
All three terms are describing the same thing in different words. “Impedance matching” is gesturing to the fact that when a wave moves into a material where it travels with a different speed their is a reflection at the interface, so shaping a cavity that effectively slows the sound waves down by making them ricochet reduces the amount of sound that the air bounces back into the string at that interface. Resonance is getting the same thing in the frequency picture. Density of states is saying the same thing again, this time using a Fourier transform to talk about the density of different possible waves in position instead of the speed of waves. (Essentially if the waves are walking on a monopoly board a slower speed implies a greater density of squares → more states. So something that likes putting sound counters into nearby squares will deposit more in the slow areas because their are more squares to put them in.)