An essential part of a guitar is its hollow body. Without it, the strings wouldn't be very loud; as far as I know, the purpose of the body is to set up some sort of resonance and make the sound louder.

How does this work? How can an isolated system amplify sound? Where is the energy coming from?

  • 2
    $\begingroup$ It's not amplification, it's impedance matching from the string to the air. I'm working on a full answer explaining this but it might take me a while to work it all out in detail. $\endgroup$ – DanielSank Dec 5 '14 at 3:21

It is not amplification! The purpose of the guitar body is to impedance and mode match between the string and the surrounding air.


When a an object vibrates it pushes on the surrounding air creating pressure waves which we hear as sound. A string vibrating alone without the body of the instrument doesn't make a very loud sound because exchange of energy from the vibrating string to the air pressure waves is inefficient. Why is it inefficient?

The fundamental reason is that the string is a lot stiffer than the surrounding air and has a small cross sectional area. This means that as the string vibrates with a given amount of energy, it doesn't actually displace much air. With the same energy in the motion, a larger, more mechanically compliant object (e.g. the acoustic guitar body) would do a better job at transferring the energy into the air and thus into your ears.


The equation of motion of a vibrating string with vertical displacement $u$ and horizontal position $x$ is

$$\frac{\partial^2 u}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 u}{\partial t^2}$$

where $v = \mu/T$ is the velocity of sound in the string, $\mu$ is the linear mass density, and $T$ is the tension. If you consider, for example, the fundamental mode, then the solution is of the form

$$u(x,t) = \sin(k x) f(t)\,.$$

Plugging this into the equation of motion gives you

$$ \begin{align} -k^2 f &= \frac{1}{v^2} \ddot{f} \\ 0&= \ddot{f} + \omega_0^2 f \end{align} $$ where $\omega_0^2 = (vk)^2 = (2\pi)^2T/(\mu \lambda^2)$ and $\lambda$ is the wavelength of the mode ($k \equiv 2\pi / \lambda$). This is just the equation of a harmonic oscillator.

Now suppose we add air friction. We define a drag coefficient $\gamma$ by saying that the friction force on a piece of the moving string of length $\delta x$ is

$$F_{\text{friction}} = -\delta x \, \gamma \, \dot{u} \, .$$

Note that $\gamma$ has dimensions of force per velocity per length. Drag coefficients are usually force per velocity; the extra "per length" comes in because we defined $\gamma$ as the friction force per length of string.

Adding this drag term, re-deriving the equation of motion, and again specializing to a single mode, we wind up with

$$0 = \ddot{f} + \omega_0^2 f + \frac{\gamma}{\mu} \dot{f} \, .$$

Now we have a damped harmonic oscillator. The rate at which this damped oscillator decays tells you how fast (i.e. how efficiently) that oscillator transfers energy into the air. The energy loss rate $\kappa$ for a damped harmonic oscillator is just the coefficient of the $\dot{f}$ term, which for us is $\kappa = \gamma / \mu$.$^{[1]}$

The quality factor $Q$ of the resonator, which is the number of radians of oscillation that happen before the energy decays to $1/e$ of its initial value, is $$Q = \omega_0 / \kappa = \frac{2\pi}{\lambda} \frac{\sqrt{T \mu}}{\gamma} \, .$$

Lower $Q$ means less oscillations before the string's energy has dissipated away as sound. In other words, lower $Q$ means louder instrument. As we can see, $Q$ decreases if either $\mu$ or $T$ decreases. This is in perfect agreement with our intuitive discussion above: lower tension would allow the string to deflect more for a given amount of vibrational energy, thus pushing more air around and more quickly delivering its energy to the air.

Impedance and mode matching

Alright, so what's going on when we attach the string to a guitar body? We argued above that the lower tension string has more efficient sound production because it can move farther to push more air. However, you know this isn't the whole story with the guitar body because you plainly see with your eyes that the guitar body surface does not deflect even nearly as much as the string does. Note that the guitar surface has much more area than the string. This means that for a given velocity, the frictional force is much higher, i.e. $\gamma$ is larger than for the string.

So there you have it: the guitar body has lower $T$ and higher $\gamma$ than the string. These both contribute to making the $Q$ lower, which means that the vibrating guitar body more efficiently transfers energy to the surrounding air than does the bare string.

Lowering the $T$ to be more mechanically compliant like the air is "impedance matching". In general, two modes with similar response to external force (or voltage, or whatever), more efficiently transfer energy between themselves. This is precisely the same principle at work when you use index matching fluid in an immersion microscope to prevent diffraction, or an impedance matching network in a microwave circuit to prevent reflections.

Increasing the area to get larger $\gamma$ is "mode matching". It's called mode matching because you're taking a vibrational mode with a small cross section (the string) and transferring the energy to one with a larger cross section (the guitar body), which better matches the waves you're trying to get the energy into (the concert hall). This is the same reason horn instruments flare from a tiny, mouth sides aperture at one side, to a large, "concert hall" sized aperture at the other end.

[1] I may have messed up a factor of 2 here. It doesn't matter for the point of this calculation.

  • $\begingroup$ I just thought of something so I'm coming back to this answer. Does this mean that the hollow body is not actually very important? Would a string attached to a large flat wooden board make roughly as much sound as a guitar? $\endgroup$ – Javier May 31 '15 at 20:01
  • $\begingroup$ My guess is no, because if that were the case we wouldn't make guitars the way we do. I left a step out of my discussion. The guitar body forms a resonator. When the string drives the body, the resonance makes it so that it's easier for the string to transfer energy into the wobbly face of the body (this is basically fancy impedance matching). $\endgroup$ – DanielSank May 31 '15 at 22:21
  • $\begingroup$ Your answer only barely touches on one important idea: Namely, that the more efficiently the energy is coupled to the air, the more quickly the note decays. Solid-body, electric guitars are known for being able to sustain a note much longer than an acoustic guitar. That's because relatively little of the string's energy is lost, either to the air or, to the electric pickup. $\endgroup$ – Solomon Slow Jul 19 '16 at 22:53
  • $\begingroup$ @jameslarge Of course there must be a trade-off between sustain and volume. I don't know if in practice the coupling of string vibrational energy into sound is usually the limiting factor on sustain. Also, you mention electric guitars; they use amplifiers, so it's perfectly ok if the string's vibrational energy couples out into the air slowly. $\endgroup$ – DanielSank Jul 20 '16 at 1:07

Simple version: When something vibrates, it puts pressure on the air molecules around it, making them vibrate. So if just 1 or 2 strings that vibrate, it won't make very many molecules vibrate because it's not touching many of them. So you can imagine that when a large flat piece of wood is vibrating as well, it will make a louder sound because more molecules are moving.


protected by Qmechanic Jul 19 '16 at 20:58

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