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As a kid I was bemused at why soundboards worked. A small sound could be demonstrably amplified simply by attaching the source to a surface that is rigid and not too thick. How could the volume increase so much given that there was no extra energy added?

As an adult I kind-of-think I know, but there are still many nagging questions. I assume it has to do with the waves propogating from a vibrating object actually being a compression on one side of the object just as they are a decompression on the other side, and something about that lack of coherence limits the volume. Exactly why remains a mystery to me. Is separating the pocket of compression and decompression so that the boundary along which they meet is quite small part of the issue?

My question is what are the physics that make a soundboard work?

Interesting specifics that would be nice-to-knows would be why does a hollow one (like a violin) work better than a solid one (imagine a filled in violin)? How important are the harmonics of the solid? But the real question is what are the physics that make a soundboard work?

P.S. I am a mathematician, so feel free to wax very mathematical if it is necessary to give a good explanation.

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  • $\begingroup$ I'm only guessing, and it's not a mathematical answer. A sound board vibrates at the same frequency your string vibrates, so there's constructive interference. I think it's not that the same energy is able to produce a louder sound with a sound board, but more that that energy would have gone to waste without a sound board. $\endgroup$
    – markovchain
    Commented Feb 22, 2014 at 7:18
  • $\begingroup$ Another possibility (not necessarily incompatible with yours): the soundboard serves to match impedances, like the bell of a brass instrument. $\endgroup$
    – user10851
    Commented Feb 22, 2014 at 9:17

3 Answers 3

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The comments above that say the sound is louder because the soundboard itself begins to vibrate are correct. This is called resonance. It sounds louder because the motion of the board is mechanically more efficient at converting the energy of the system into sound waves than the string alone. The board is an effective radiator of sound energy. A louder sound wave has a larger amplitude.

If you want a mathematical analysis that shows resonance, try the ODE for a one dimensional SHO being driven by sine function. You'll see the amplitude of your oscillator increases near the natural resonance period of the system. You could also do a 2D analysis on a thin plate such as a drum head.

Strings and thin plates are relatively good sound radiators because relatively low force is required to cause a larger displacement, and so a sound wave of larger amplitude.

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  • $\begingroup$ If this is the primary mechanism, then soundboards only work at frequencies for which they have harmonics. Is that true? $\endgroup$
    – John Robertson
    Commented Mar 2, 2014 at 1:26
  • $\begingroup$ Yes but a sound board would have multiple harmonics. $\endgroup$
    – Mark Rovetta
    Commented Mar 2, 2014 at 2:02
  • $\begingroup$ Oh, I know. Teaching a semester of PDE included working out the harmonic states of a hollow sphere. Cool stuff, for sure. $\endgroup$
    – John Robertson
    Commented Mar 2, 2014 at 2:17
  • $\begingroup$ Having had time to think about it I think this explains why type of sound made becomes that of the soundboard, i.e. the harmonics of the better soundmaker dominate. But I think Olin is onto something. If one were to use the voilin to drive the string instead one would get a greater response at the strings resonating frequencies, but it would not cause the violin to produce a much louder sound at those frequencies. $\endgroup$
    – John Robertson
    Commented Mar 3, 2014 at 7:34
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The soundboard resonates with the same frequencies as the source. It takes it energy form the vibrating source. As the soundboard distributes this energy over a larger volume of air, the sound is louder, but the energy is depleted quicker, limiting the time you hear the sound. Try this with a tuning fork. Hold it by your ear and time the duration 0of the sound. Repeat the measurement with the tuning fork held against a glass plate, a window will suffice. Even without a stopwatch you will find the duration of the amplified sound shorter.

Furthermore the shape and material of the (hollow) soundboard determines the relative amplification of the produced frequencies. That's why certain violins sound better than others (and are worth more).

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  • $\begingroup$ If this is the major mechanism, then the proportion decrease in the time the tuning fork vibrates should correspond pretty well to the amplitude of the soundwave. I am a little dubious about that, because human ear responses are (coarsely) logarithmic, so a very audible increase in volume should correspond to a quite substantial increase in wave amplitude. That should mean the tuning fork should go quiet in a small fraction of the time that it did without the soundboard. I agree your answer is a physical factor. I am not sure it is the dominant factor. Feel free to correct me if I am wrong. $\endgroup$
    – John Robertson
    Commented Mar 2, 2014 at 1:32
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You are right that a soundboard adds no energy to the system. However, it does allow the existing energy to be converted to sound better. The greater area of the soundboard causes more air to be pushed than the string alone can, even though the displacement amplitude of the soundboard is less than the string.

This exactly the same reason speakers have cones. Only a small ring at the center of the speaker is actually driven. That by itself makes little sound because it is so small and therefore doesn't move much air. By causing it to move the speaker cone, which in turn moves much more air, you get a much louder sound.

Another way to think of this is as a impedance match. The string vibrates at high amplitude, but the air presents a very high impedance to it, so little energy is transferred. The soundboard puts a little more drag on the string, but this lower impedance allows more of the energy from the vibrating string to be transferred to the air.

Yes, the thickness of the soundboard definitely matters. Ideally you want something with low stiffness and mass so that energy isn't dissipated in the soundboard itself, but rather more of it is tranferred to the air. This is why speaker cones are made of paper or other light and thin material.

The wood of a violin has other functions it must perform, and was also something that needed to be manufacturable by hand 100s of years ago. The shape, type of wood, thickness, and even the glue used in violins are all part of a black art with much accumulated lore about how to get the "best" sound. In this case "best" isn't the loudest or most efficient coupling to the air, but what has become expected of a violin over the centuries. Air cavities that resonate as sub-harmonics also have a lot to do with it.

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  • $\begingroup$ Your point about "the energy not being dissipated in the soundboard itself" is well made, as well as harmonics in air cavities. I think your point is the soundboard makes much physically smaller and lower energy vibrations (which is the drag and energy loss to the string), but those very low energy vibrations of the soundboard do a much better job of creating waves in the air, as most of the vibrating objects motion through the air goes to turbulent noise, very little contributing to the coherent sound wave. Correct? Certainly I think that is believable for a vibrating string. $\endgroup$
    – John Robertson
    Commented Mar 2, 2014 at 2:22
  • $\begingroup$ @John: Yes, that's basically what I am saying. The little area of the vibration string makes less sound than the large area of the soundboard, even though the displacement of the soundboard is less than that of the string. $\endgroup$
    – Olin Lathrop
    Commented Mar 2, 2014 at 14:37
  • $\begingroup$ Way less than that of the string. In fact, it needs to not only be physically smaller, but be much lower energy. Part of what would need to be the case is that in fact many soundmakers generate soundwaves incredibly inefficiently. Otherwise the transfer to the soundboard would not be so beneficial. $\endgroup$
    – John Robertson
    Commented Mar 3, 2014 at 7:26

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