I did a bit of discussion on this subject in this thread on Music.SE.
The fundamental doesn't necessarily have the strongest amplitude.
As said by Alfred Centauri, it depends on the initial configuration: ideally, the string returns to exactly that configuration after each $\tfrac1{\nu_1}$, and the amplitude of each harmonic in frequency space is proportional to its amplitude in the initial configuration in wavenumber space. Now the initial configuration for plucked instruments happens to be roughly what I depicted in the first figure in the Music.SE answer: something between a triangle and a sawtooth, and as we know both have a monotonically descreasing1 sequence of Fourier coefficients ($\mathcal{O}(\tfrac1{n^2})$ for triangle, $\mathcal{O}(\tfrac1{n})$ for sawtooth), so the fundamental does tend to have the strongest amplitude on the string in the beginning. But this may not hold true for an actual instrument sound for a variety of reasons: on the guitar, if you strike the string quickly with a pick, the initial configuration may be more like a smoothened Dirac peak, for which all of the lower frequencies have the same amplitude. Similarly for a loud piano note. Then, resonances in the instruments' bodies may lead to the fundamental decaying faster than some of the other harmonics, so after a while they are stronger, though instrument builders classically try to prevent this. Similarly, as said by John Rennie, the bodies may be more efficient at conducting some of the harmonics to the air than the fundamental and likewise the microphone at measuring the sound.
Finally, you should assure yourself of what this Fourier spectrum of yours actually shows: in audio production, spectrum analysers do not display something proportional to the amplitude $A(\omega)$ at frequency $\omega$, but in fact to $A(\omega)\cdot\sqrt{\omega}$ so that pink noise appears to have constant amplitude. (For both historic / technical and convenience reasons.) So if in such a spectrum the fundamental turns up slightly weaker than e.g. the second harmonic, it may in actuality still be stronger.
1Actually, the sequence of Fourier coefficients of a triangular wave itself is not monotonic, but alternates between a monotonic sequence (for odd $n$) and zero. Since the fundamental has odd $n=1$, that amounts to the same result.
The classical string equation that you are referring to, is formulated by making a number of assumptions, which include that the vibration of the string does not affect its tension. This makes Young's modulus irrelevant for results calculated from the idealized equation.
In the real world, materials with low moduli of elasticity will follow the ideal equation more closely, since the tensions will change less during vibration. For materials with a higher modulus of elasticity, I would expect that:
- vibration frequency will not be independent of the amplitude, and
- when comparing two materials with the same rest tension, stiffer materials will vibrate at a slightly higher frequency, since the restoring force will be incrementally larger.
Best Answer
Yes. You can imagine that each particle on the string is oscillating at 440 Hz. As each particle oscillates, it drags neighboring air particles along with its motion, so each air particle around the string will also vibrate at 440 Hz. Since the sound created comes from the air particles' oscillations, you would hear a 440 Hz pitch sound.
You would indeed hear a 440 Hz sound that correponds to the string's fundamental frequency. However, take note that you will likely hear higher frequencies of sounds corresponding to the higher harmonics - unless the string is oscillating perfectly at its fundamental mode with no higher harmonics.