acousticsresonancestringwavelength
We all heard that acoustic guitar body acts as the amplifier of the sound created by wire plucking and strumming. This is because an acoustic guitar body is some kind of resonator.
Every resonator amplifies just certain frequencies while it inhibits all others.
So why then when an acoustic guitar is played we hear all notes amplified and not just certain ones? Shouldn't there be some notes louder and some other notes quieter?
First, I believe you have a mistake in your equations. The transmission should be $$T = \frac{2Z_0}{Z_1 + Z_0}.$$
The transmission in Scenario 1 is the transmission through the one boundary:
$$T_1 = \frac{2Z_W}{Z_G + Z_W},$$
while the total transmission in Scenario 2 is the product of two boundaries:
$$T_2 = \frac{2Z_W}{Z_P + Z_W} \frac{2Z_P}{Z_G + Z_P}.$$
Using the corrected equations, the transmissions are 11.8% in Scenario 1 and 14.6% in Scenario 2.
I was curious if Scenario 2 could ever give a lower transmission, so I worked it out. If we assume $T_1$ is bigger, we can figure out under what circumstances that is satisfied as follows:
$$T_1 \stackrel{?}{>} T_2,$$
$$\frac{2Z_W}{Z_G + Z_W} \stackrel{?}{>} \frac{2Z_W}{Z_P + Z_W} \frac{2Z_P}{Z_G + Z_P},$$
$$Z_W(Z_P+Z_W)(Z_G+Z_P) \stackrel{?}{>} 2 Z_W Z_P(Z_G+Z_W),$$
$$(Z_P+Z_W)(Z_G+Z_P) \stackrel{?}{>} 2 Z_P (Z_G+Z_W),$$
$$Z_P Z_G + Z_W Z_G + Z_W Z_P + Z_P^2 \stackrel{?}{>} 2 Z_P Z_G + 2 Z_P Z_W,$$
$$ Z_W Z_G + Z_P^2 \stackrel{?}{>} Z_P Z_G + Z_PZ_W,$$
$$ Z_P^2 - Z_P (Z_G + Z_W) + Z_W Z_G \stackrel{?}{>} 0,$$
$$ (Z_P - Z_G)(Z_P - Z_W) \stackrel{?}{>} 0.$$
By inspection we see that $Z_P = Z_W$ and $Z_P = Z_G$ are roots. This parabola is upward facing (positive coefficient of $Z_P^2$), so at $Z_P = \{Z_W,Z_P\}$ it will cross zero and go positive outside of there and be negative between those two points.
Thus the transmission of Scenario 2 will be lower only if $Z_P < Z_W$ or $Z_P > Z_G$. In other words, the transmission would be worse if you put the glass between the water and polymer or sandwich water between polymer and glass.
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 $v = \sqrt{T/\mu}$ 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.
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.
Best Answer
This is true only for very simple resonators. The shape of the guitar body is such that it has a different size at different angles. This corresponds to different resonant frequencies. In addition, the top has a supporting bracing that is very different on different models and is critical for the sound.
Furthermore, a guitar is nor necessarily a resonator, but a converter of the mechanical energy of the strings to the acoustic energy of the air. You are absolutely correct that a strong resonance at a certain frequency would help this frequency at the expense of other frequencies and thus would be detrimental for the sound of a broad spectrum instrument. Therefore the objective of the guitar design (counter intuitively at first) is not to create, but to avoid strong resonances.
There are two parts in a guitar relevant here. One is the top made either of softer ceder for a warmer sound or harder spruce for a brighter sound (sometimes a two-layer top has both). The purpose of the top is to move air with its large area thus converting the mechanical energy of the strings to the acoustical energy of the sound. However, the top radiates both forward and back, plus the sound radiated back is out of phase with the sound radiated forward.
The second important part is the body. Historically the most popular material for the body had been the Brazilian rosewood before Brazil prohibited exporting it. Similar results are achieved with the rosewood from India, Madagascar, and other areas. Mahogany is also popular along with other wood species.
The main purpose of the body is to reflect forward (through the sound hole) the sound emitted back by the top and do it while inverting the phase, so that the sound emitted by the sound hole would be in phase with the sound emitted forward by the desk. Thus the guitar body is not a resonator, but a phase inverter very similar to common speakers with a PORT. Phase inversion depends on the internal dimensions of the body and is a compromise in order to optimize all frequencies equally. Although this is more important for lower frequencies (again, similarly to speakers).
The materials, shape, and bracing have been perfected by generations of guitar makers resulting in specific sound signatures. One example is the 1A model with a ceder top by Jose Romirez. Its sound is mesmerizing.
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