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A vibration on a string can theoretically be examined by the back and forth moving sound waves. The two reflecting boundary conditions capture some part of the vibration energy in the region which causes some part of it to be stored. Those waves propagating in different directions interfere perfectly at some frequencies, they together form a standing wave where the physical positions of the peaks and the nulls are stationary. In this case, if the power source is nonzero for a certain amount of time, some portion of this power is continually stored every second in the region between the reflecting boundaries. This is why the stored energy increases greatly in time. This causes the mechanical vibrations such that we hear the musical tone of the instrument. Why would free single string and another equivalent string of a guitar would produce sound levels differ greatly.

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A string does not transfer much energy into the air directly, because its diameter is too small to move much air, and the two "sides" of the string produce equal and opposite sound waves in the air. If one "side" is moving in the direction which creates a positive pressure in the air, the other side is creating an equal negative pressure, and the two effects destructively interfere with each other.

In an acoustic guitar, the vibration of the string is transmitted through the bridge of the guitar and makes the whole of the front of the guitar's body vibrate. That large area is much more efficient at transferring energy into the air. The vibration of the "inside" of the guitar body only affects the air inside the instrument, and does not cancel out the vibrations caused by the "outside" of the body.

Actually, the last statement is above over-simplified, because at some frequencies the air inside the guitar will resonate, and then transfer its energy to the outside air by "pumping" air in and out through the sound hole in the guitar - but the two sound sources, from the whole of the front of the guitar and from the sound hole, don't cancel out as accurately as the sound from the two "sides" of a thin string.

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  • $\begingroup$ I completely agree with your second and third paragraphs which make it clear that the vibration energy transferred to the body can escape into free space much easier than the strings. I also agree that the vibration on a string is almost lossless so that it radiates much less. $\endgroup$ – GokhunTanyer May 13 '17 at 17:07
  • $\begingroup$ But the first statement, if I understood right, assumes that pressure change at a single point do not radiate energy if it is too quick. This conflicts Huygen's Law, Maxwell's Law for Electromagnetic Waves etc. Oscillation at a single point will radiate sound waves around its neighborhood which later will radiate again and again at a wavelength given by v/f. $\endgroup$ – GokhunTanyer May 13 '17 at 17:07

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